EDEXCEL HIGHERS ENGINEERING THERMODYNAMICS H2 NQF LEVEL 4. TUTORIAL No. 1 PRE-REQUISITE STUDIES FLUID PROPERTIES

EDEXCEL HIGHERS ENGINEERING HERMODYNAMICS H NQF LEEL 4 UORIAL No. PRE-REQUISIE SUDIES FLUID PROPERIES INRODUCION Before you study the four outcomes that make u the module, you should be cometent in finding the roerties of liquids, gases and vaours. If you are already cometent in this, you should ski this tutorial. his tutorial is designed to teach you the basic concets of thermodynamics and the roerties of fluids. On comletion of this tutorial you should be able to the following. Use the correct thermodynamic symbols. Determine the roerties of a gas. Determine the roerties of vaours. Determine the roerties of liquids. We will start by examining the symbols used. D.J.Dunn freestudy.co.uk

. SYMBOLS he symbols used throughout in these tutorials are S.I. (International System). hese are not always the ones used by the examiner. hese are the main S.I. Symbols. he S.I. also recommends the use of / to mean divide rather than the use of negative indices favoured by examiners. For examle ms- becomes m/s. Quantity units Derived Unit S.I. symbol Length m various Mass kg m ime s t olume m or Q Secific olume m/kg v olume Flow Rate m/s Density kg/m ρ Force kg m/s N F Weight kg m/s N W Pressure Head m h Altitude m z Area m A Seed of Sound m/s a Secific Heat Ca. N m/kg K Joule/kg K c Energy N m Joule Enthaly N m Joule H Internal Energy N m Joule U Secific Enthaly N m/kg J/kg h Secific Int. Energy N m/kg J/kg u Mass flow rate kg/s Polytroic Index n Adiabatic Index γ Pressure N/m Pascal Heat ransfer N m Joule Q Work N m Joule W Heat ransfer Rate N m/s Watt Φ Work Rate (ower) N m/s Watt P Char. Gas Const N m/kg K J/kg K R Universal Gas Constant J/kmol K Ro Entroy J/K S Secific Entroy J/kg K s Absolute emerature K Celsius emerature oc θ elocity m/s v or u Dynamic iscosity N s/m Pa s η or µ Kinematic iscosity m/s ν Now we will examine the basic concets required to do this tutorial successfully. D.J.Dunn

. BASIC CONCEPS hroughout these tutorials you will use roerties which are either EXENSIE or INENSIE. An extensive roerty is one which is divisible. For examle olume when divided by a number becomes smaller. Other examles are mass and energy. An intensive roerty is a roerty of a mass which remains the same value when the mass is divided into smaller arts. For examle the temerature and density of a substance is unchanged if it is divided into smaller masses. hroughout the tutorials you will use OAL and SPECIFIC quantities which relate only to extensive roerties. A total quantity is always denoted by a higher case letter such as for volume (m) and H for enthaly (J). A secific quantity reresents the quantity er kg and is obtained by dividing the roerty by the mass. Such roerties are always designated by lower case letters such as v for secific volume (m/kg) and h for secific enthaly (J/kg). Secific volume is mainly used for gas and vaours. he inverse of secific volume is density (ρ) (kg/m) and this is mainly used for liquids and solids but also for gases. Note ρ/v. Because the same letters are used to designate more than one roerty, often alternative letters are used. For examle v for secific volume may occur in the same work as v for velocity so often u or c is used for velocity. h is used for height, head and secific enthaly so z is often used for height instead. he unit of Force and Weight is the kg m/s. his comes from Newton's Second Law of Motion (Force mass x acceleration). he derived name for the unit is the Newton. In the case of Weight, the acceleration is that of gravity and in order to convert mass in kg into weight in Newtons, you must use W mg where g is normally 9.8 m/s. Now we will examine forms of energy a fluid may have. D.J.Dunn

. ENERGY FORMS A fluid may ossess several forms of energy. All fluids ossess energy due to their temerature and this is called INERNAL ENERGY. All ossess GRAIAIONAL or POENIAL ENERGY due to their elevation relative to some datum level. If the fluid is moving it will ossess KINEIC ENERGY due to its velocity. If it has ressure then it will ossess FLOW ENERGY. Often ressure and temerature are the main two governing factors and we add internal energy to flow energy in order to roduce a single entity called ENHALPY. Let us look at each in more detail... GRAIAIONAL or POENIAL ENERGY In order to raise a mass m kg a height z metres, a lifting force is required which must be at least equal to the weight mg. he work done raising the mass is as always, force x distance moved so Work mgz Since energy has been used to do this work and energy cannot be destroyed, it follows that the energy must be stored in the mass and we call this gravitational energy or otential energy P.E. here are many examles showing how this energy may be got back, e.g. a hydro-electric ower station. P.E. mgz. KINEIC ENERGY When a mass m kg is accelerated from rest to a velocity of v m/s, a force is needed to accelerate it. his is given by Newton's nd Law of Motion F ma. After time t seconds the mass travels x metres and reaches a velocity v m/s. he laws relating these quantities are a v/t and x vt/ he work done is W Fx max mv/ Energy has been used to do this work and this must be stored in the mass and carried along with it. his is KINEIC ENERGY. K.E. mv/ D.J.Dunn 4

. FLOW ENERGY When fluid is umed along a ie, energy is used to do the uming. his energy is carried along in the fluid and may be recovered (as for examle with an air tool or a hydraulic motor). Consider a iston ushing fluid into a cylinder. he fluid ressure is N/m. he force needed on the iston is F A he iston moves a distance x metres. he work done is W Fx Ax Since Ax and is the volume umed into the cylinder the work done is W Since energy has been used doing this work, it must now be stored in the fluid and carried along with it as FLOW ENERGY. F.E..4 INERNAL ENERGY his is covered in more detail later. he molecules of a fluid ossess kinetic energy and otential energy relative to some internal datum. Usually this is regarded simly as the energy due to the temerature and very often the change in internal energy in a fluid which undergoes a change in temerature is given by U mc he symbol for internal energy is U kj or u kj/kg. Note that a change in temerature is the same in degrees Celsius or Kelvin. he law which states internal energy is a function of temerature only is known as JOULE'S LAW..5 ENHALPY When a fluid has ressure and temerature, it must ossess both flow and internal energy. It is often convenient to add them together and the result is ENHALPY. he symbol is H kj or h kj/kg. H F.E. + U Next you need to study the roerties of fluids and the laws relating them. D.J.Dunn 5

4 GAS LAWS In this section you will do the following. 4. HEORY Derive basic gas laws. Examine the characteristic gas law. Examine the universal gas law. Define the mol. Solve gas law roblems. A gas is made of molecules which move around with random motion. In a erfect gas, the molecules may collide but they have no tendency at all to stick together or reel each other. In other words, a erfect gas in comletely inviscid. In reality there is a slight force of attraction between gas molecules but this is so small that gas laws formulated for an ideal gas work quite well for real gas. Each molecule in the gas has an instantaneous velocity and hence has kinetic energy. he sum of this energy is the internal energy U. he velocity of the molecules deends uon the temerature. When the temerature changes, so does the internal energy. he internal energy is for all intents and uroses zero at - 7o C. his is the absolute zero of temerature. Remember that to convert from Celsius to absolute, add on 7. For examle 40oC is 40 + 7 Kelvins. 4. PRESSURE If a gas is comressed it obtains ressure. his is best exlained by considering a gas inside a vessel as shown. he molecules bombard the inside of the container. Each roduces a momentum force when it bounces. he force er unit area is the ressure of the gas. Remember that ressure Force/area F/A N/m or Pascals Note that 0 Pa kpa 06 Pa MPa 05 Pa bar D.J.Dunn 6

4. CONSAN OLUME LAW If the gas is heated the velocity of the molecules increases. If the container is rigid, then the molecules will hit the surface more often and with greater force so we exect the ressure to rise roortional to temerature. c when is constant. WORKED EXAMPLE No. A mass of gas has a ressure of 500 kpa and temerature of 50oC. he ressure is changed to 900 kpa but the volume is unchanged. Determine the new temerature. SOLUION Using constant volume law find / c / where 50 + 7 4 K 500 000 900 000 / 900 000 x 4/500 000 76.4 K 4.4 CHARLE'S LAW If we ket the ressure constant and increased the temerature, then we would have to make the volume bigger in order to sto the ressure rising. his gives us Charle's Law: c when is constant WORKED EXAMPLE No. A mass of gas has a temerature of 50oC and volume of 0. m. he temerature is changed to 50oC but the ressure is unchanged. Determine the new volume. SOLUION Using Charles s law we find /c / where 50 + 7 4 K 0. 50 + 7 K / x 0./5 0. m D.J.Dunn 7

4.5 BOYLE'S LAW If we kee the temerature constant and increase the volume, then the molecules will hit the surface less often so the ressure goes down. his gives Boyle's Law: c/ when is constant. WORKED EXAMPLE No. A mass of gas has a ressure of 800 kpa and volume of 0. m. he ressure is changed to 00 kpa but the temerature is unchanged. Determine the new volume. SOLUION Using Boyle's law we find c where 800 x0 0. 00 x0 / 800 x0 x 0./00 x0.4m. 4.6 GENERAL GAS LAW Consider a gas which undergoes a change in and from oint () to oint () as shown. It could have gone from () to (A) and then from (A)to () as shown. Process () to (A) is constant volume A/A / Process (A) to () is constant temerature A Hence A/ / and A /...() For the rocess (A) to () Boyle's Law alies so AA Since A then we can write A So A /...() Equating () and () we get / / constant D.J.Dunn 8

his is the GENERAL GAS LAW to be used to calculate one unknown when a gas changes from one condition to another. WORKED EXAMPLE No.4 A mass of gas has a ressure of. MPa, volume of 0.0 m and temerature of 00oC. he ressure is changed to 400 kpa and the volume is changed to 0.06 m. Determine the new temerature. SOLUION Using the general gas law we find / /. x06 0.0 400 x0 where 00 + 7 7 K / 400 x0 x 0.06 x 7/(. x06 x 0.0) 48.7 K 4. 7 CHARACERISIC GAS LAW he general gas law tells us that when a gas changes from one ressure, volume and temerature to another, then / constant hinking of the gas in the rigid vessel again, if the number of molecules was doubled, keeing the volume and temerature the same, then there would be twice as many imacts with the surface and hence twice the ressure. o kee the ressure the same, the volume would have to be doubled or the temerature halved. It follows that the constant must contain the mass of the gas in order to reflect the number of molecules. he gas law can then be written as / mr where m is the mass in kg and R is the remaining constant which must be unique for each gas and is called the CHARACERISIC GAS CONSAN. If we examine the units of R they are J/kg K. he equation is usually written as Since m/ is the density ρ, it follows that Since /m is the secific volume,v, then mr ρ/r vr/ D.J.Dunn 9

WORKED EXAMPLE No.5 A mass of gas has a ressure of. MPa, volume of 0.0 m and temerature of 00oC. Given the characteristic gas constant is 00 J/kg K find the mass. SOLUION From the characteristic gas law we have mr where. x 06 N/m 0.0 m 00 + 7 7 K m /R. x 06 x 0.0/00 x 7 0. kg SELF ASSESSMEN EXERCISE No. All ressures are absolute.. Calculate the density of air at.0 bar and 5 oc if R 87 J/kg K. (.6 kg/m). Air in a vessel has a ressure of 5 bar, volume 0. m and temerature 0oC. It is connected to another emty vessel so that the volume increases to 0.5 m but the temerature stays the same. aking R 87 J/kg K. Calculate i. the final ressure. (0 bar) ii. the final density. (.89 kg/m). dm of air at 0oC is heated at constant ressure of 00 kpa until the volume is doubled. Calculate i. the final temerature. (586 K) ii. the mass. (.56 g) 4. Air is heated from 0oC and 400 kpa in a fixed volume of m. he final ressure is 900 kpa. Calculate i. the final temerature.(659 K) ii. the mass. (4.747 kg) 5.. dm of gas is comressed from bar and 0oC to 7 bar and 90oC. aking R 87 J/kg K calculate i. the new volume. ( cm) ii. the mass. (.47 g) D.J.Dunn 0

4.8 HE UNIERSAL GAS LAW he Characteristic Gas Law states mr where R is the characteristic constant for the gas. his law can be made universal for any gas because R Ro/Mm. where Mm is the mean molecular mass of the gas (numerically equal to the relative molecular mass). he formula becomes mro/mm. Ro is a universal constant with value 84. J/kmol K. It is worth noting that in the exam, this value along with other useful data may be found in the back of your fluids tables. he kmol is defined as the number of kg of substance numerically equal to the mean molecular mass. yical values are GAS Symbol Mm. Hydrogen H Oxygen O Carbon Dioxide CO 44 Methane CH4 6 Nitrogen N 8 Dry Air 8.96 Hence kmol of hydrogen (H) is kg kmol of oxygen (O) is kg kmol of Nitrogen is 8 kg and so on. For examle if you had kmol of nitrogen (N) you would have x 8 84 kg It follows that the Mm must have units of kg/kmol In order to calculate the characteristic gas constant we use R Ro/Mm For examle the characteristic gas constant for air is Examine the units R 84./8.96 87 R Ro/Mm (J/kmol K)/(kg/kmol)J/kg K D.J.Dunn

WORKED EXAMPLE No.6 A vessel contains 0. m of methane at 60oC and 500 kpa ressure. Calculate the mass of Methane. SOLUION mro/mm. Mm 6 500 000 x 0. m x 84. x (7 + 60)/6 m 0.578 kg SELF ASSESSMEN EXERCISE No.. A gas comressor draws in 0.5 m/min of Nitrogen at 0oC and 00 kpa ressure. Calculate the mass flow rate. (0.595 kg/min). A vessel contains 0.5 m of Oxygen at 40oC and 0 bar ressure. Calculate the mass. (6.48 kg) Next we will examine the meaning of secific heat caacities. D.J.Dunn

5. SPECIFIC HEA CAPACIIES In this section you will do the following. Learn how to calculate the change in internal energy of gases and liquids. Learn how to calculate the change in enthaly of gases and liquids. Define the secific heats of fluids. Relate secific heat caacities to the characteristic gas constant. 5. SPECIFIC HEA CAPACIIES he secific heat caacity of a fluid is defined in two rincial ways as follows:. Constant olume he secific heat which relates change in secific internal energy 'u' and change in temerature '' is defined as : cv du/d If the value of the secific heat caacity cv is constant over a temerature range then we may go from the differential form to the finite form cv u/ J/kg Hence ucv J/kg For a mass m kg the change is Umcv Joules his law indicates that the internal energy of a gas is deendant only on its temerature. his was first stated by Joule and is called JOULE'S LAW.. Constant Pressure. he secific heat which relates change in secific enthaly 'h' and change in temerature '' is defined as : c dh/d If the value of the secific heat caacity c is constant over a temerature range then we may go from the differential form to the finite form c h/ J/kg Hence hc J/kg For a mass m kg the change is Hmc Joules he reasons why the two secific heats are given the symbols cv and c will be exlained next. hey are called the PRINCIPAL SPECIFIC HEAS. D.J.Dunn

5. CONSAN OLUME HEAING When a fluid is heated at constant volume, the heat transfer 'Q' must be the same as the increase in internal energy of the fluid U since no other energy is involved. It follows that : Q U mcv Joules he change in internal energy is the same as the heat transfer at constant volume so the symbol cv should be remembered as alying to constant volume rocesses as well as internal energy. 5. CONSAN PRESSURE HEAING When a fluid is heated at constant ressure, the volume must increase against a surrounding ressure equal and oosite to the fluid ressure. he force exerted on the surroundings must be F A Newtons he work done is force x distance moved hence : Work Done F x Ax where is the volume change. he heat transfer Q must be equal to the increase in internal energy lus the work done against the external ressure. he work done has the same formula as flow energy Enthaly was defined as H U + he heat transfer at constant ressure is also: Q U + Since secific heats are used to calculate heat transfers, then in this case the heat transfer is by definition : Q mc It follows that H Q mc For the same temerature change it follows that the heat transfer at constant ressure must be larger than that at constant volume. he secific heat caacity c is remembered as linked to constant ressure. D.J.Dunn 4

5.4 LINK BEWEEN c v, c AND R. From the above work it is aarent that H mc U + We have already defined U mcv Furthermore for a gas only, mr Hence mc mcv + mr Hence c cv + R 5.5 LIQUIDS Since the volume of a liquid does not change much when heated or cooled, very little work is done against the surrounding ressure so it follows that cv and c are for all intents and uroses the same and usually the heat transfer to a liquid is given as : Q mc Where c is the secific heat caacity. 5.6 APOURS aour is defined as a gaseous substance close to the temerature at which it will condense back into a liquid. In this state it cannot be considered as a erfect gas and great care should be taken alying secific heats to them. We should use tables and charts to determine the roerties of vaours and this is covered in the next section. WORKED EXAMPLE No.7 Calculate the change in enthaly and internal energy when kg of gas is heated from 0oC to 00oC. he secific heat at constant ressure is. kj/kg K and at constant volume is 0.8 kj/kg K. Also determine the change in flow energy. SOLUION i. Change in enthaly. Hmc x. x 80 648 kj ii. Change in internal energy. Hmcv x 0.8 x 80 4 kj iii. Change in flow energy FE H - 6 kj D.J.Dunn 5

WORKED EXAMPLE No.8 A vertical cylinder contains dm of air at 50oC. One end of the cylinder is closed and the other end has a frictionless iston which may move under the action of weights laced on it. he weight of the iston and load is 00 N. he cylinder has a cross sectional area of 0.05 m. he outside is at atmosheric conditions. Determine i. the gas ressure. ii. the gas mass. iii. the distance moved by the iston when the gas is heated to 50oC. For air take c 005 J/kg K and cv 78 J/kg K. Atmosheric ressure 00 kpa SOLUION he ressure of the gas is constant and always just sufficient to suort the iston so Weight/Area + atmosheric ressure 00/0.05 + 00 kpa 0 kpa + 00 kpa 0 kpa 50 + 7 K 0.00 m R c - cv 005-78 J/kg K m /R 0 000 x 0.00/( x ) 0.0048 kg 50 + 7 4 K / but / 0.0 x 4/ 0.06 m Distance moved olume change/area (0.06-0.0)/0.05 0.4 m D.J.Dunn 6

WORKED EXAMPLE No.9 Convert the rincial secific heats and characteristic gas constant for dry air into molar form. SOLUION he normal values for dry air are found in the back of your fluids tables and are: c.005 kj/kg K cv 0.78 kj/kg K R 0.87 kj/kg K In order to convert these into molar form we must multily them by the molar mass. he molar mass for dry air is a mean value for a gas mixture and is found on the back age of your fluids tables and is 8.96 kg/kmol. In molar form c.005 [kj/kg K] x 8.96 [kg/kmol] 9. kj/kmol K cv 0.78 [kj/kg K] x 8.96 [kg/kmol] 0.8 kj/kmol K R 0.87 [kj/kg K] x 8.96 [kg/kmol] 8. kj/kmol K Note that the last value is the universal gas constant Ro. SELF ASSESSMEN EXERCISE No. For air take c 005 J/kg K and cv 78 J/kg K unless otherwise stated.. 0. kg of air is heated at constant volume from 40oC to 0oC. Calculate the heat transfer and change in internal energy. (.49 kj for both). 0.5 kg of air is cooled from 00oC to 80oC at a constant ressure of 5 bar. Calculate the change in internal energy, the change in enthaly, the heat transfer and change in flow energy. (-4 kj), (-60. kj), (-7. kj). kg/s of water is heated from 5oC to 80oC. Calculate the heat transfer given c 486J/kg K. (8.7 MW) 4. Air is heated from 0oC to 50oC at constant ressure. Using your fluid tables (ages 6 and 7) determine the average value of c and calculate the heat transfer er kg of air. (0.5 kj) D.J.Dunn 7

5. he diagram shows a cylinder fitted with a frictionless iston. he air inside is heated to 00oC at constant ressure causing the iston to rise. Atmosheric ressure outside is 00 kpa. Determine : i. the mass of air. (.9 g) ii. the change in internal energy. (.57 kj) iii. the change in enthaly. (.55 kj) iv. the ressure throughout. (500 kpa) v. the change in volume. (. dm) 6. Define the meaning of a mole as a means of measuring the amount of a substance. Calculate the volume occuied at a temerature of 5oC and a ressure of bar, by 60 kg of (i) Oxygen gas, O, (ii)atomic oxygen gas,o, and (iii) Helium gas, He. he resective molar masses,m, and the molar heats at constant volume, Cv, of the three gases, and the molar (universal) gas constant, RM, are as follows: M (kg/kmol) cv(kj/kmol K) RM(kJ/kmol K) O 0.786 8.44 O 6.476 8.44 He 4.476 8.44 Go on to calculate the values of the secific heats C and Cv. Using these values, calculate the secific gas constant R for all three gases. Show not only numerical work, but also the maniulation of units in arriving at your results. You should now be able to determine the roerties of gases. Next we will examine the roerties of liquids and vaours. D.J.Dunn 8

6. PROPERIES OF LIQUIDS AND APOURS In this section you will do the following. Learn about the roerties and definitions concerning vaours. Learn how to find the roerties of vaours and liquids from your tables and charts. You should ensure that you have a coy of 'hermodynamic and ransort Proerties of Fluids' by Mayhew and Rogers. 6. GENERAL HEORY When a liquid changes into a vaour by the rocess of evaoration, it undergoes a change of state or hase. he reverse rocess is called liquefaction or condensing. he following work should lead you to an understanding of this rocess and by the end of it you should be able to find the same quantities and do the same tye of roblems as you have already done for gas. When a liquid is heated, the temerature rises directly roortional to the heat transferred, 'Q'. Q is given by Q mc he secific heat c is reasonably constant but changes significantly if the ressure or temerature change is very large. When discussing heat transfer and energy of a fluid, we may wish to consider the internal energy U or the enthaly H. In the following, the energy of the fluid may be construed as either. In tables this is tabulated as secific internal energy or enthaly u and h. When the liquid receives enough heat to bring it to the boiling oint, the energy it contains is called SENSIBLE ENERGY. In tables this is denoted as uf or hf. A liquid starts to evaorate because it becomes saturated with heat and can absorb no more without changing state (into a vaour and hence gas). For this reason the boiling oint is more correctly described as the SAURAION EMPERAURE and the liquid in this state is called SAURAED LIQUID. he saturation temerature is denoted as ts in tables. If a boiling liquid is sulied with more heat, it will evaorate and vaour is driven off. he vaour, still at the saturation temerature is called DRY SAURAED APOUR. A vaour is a gas near to the temerature at which it will condense. In order to convert liquid into vaour, extra heat must be transferred into it. he amount of enthaly and internal energy required to evaorate kg is denoted hfg and ufg in tables and this is called the LAEN ENHALPY and LAEN INERNAL ENERGY resectively. D.J.Dunn 9

he energy contained in kg of dry saturated vaour must be the sum of the sensible and latent energy and this is denoted hg and ug. It follows that : hg ug hf + hfg uf + ufg ABLE FOR HE ENHALPY OF HIGH PRESSURE WAER em o C Pressure in bar 0 5 50 75 00 5 50 75 00 50 0 0.5 5 7.5 0.6 5 7.5 0 5 0 84 86 87 9 9 96 98 00 0 05 07 40 68 70 7 74 76 79 8 8 85 87 90 60 5 5 55 57 59 6 64 66 68 70 7 80 5 7 9 4 4 45 47 49 5 5 55 00 49 4 4 45 47 48 40 4 44 46 49 0 504 505 507 509 5 5 54 56 58 59 5 40 589 59 59 594 595 597 599 600 60 60 605 60 675 677 678 680 68 68 684 686 687 688 690 80 76 764 765 767 767 769 790 77 77 774 776 00 85 85 854 855 856 857 858 859 86 86 86 All enthaly values are given in kj/kg D.J.Dunn 0

he temerature at which evaoration occurs 'ts' deends uon the ressure at which it takes lace. For examle we all know that water boils at 00oC at atmosheric ressure (.0 bar). At ressure below this, the boiling oint is less. At higher ressures the boiling oint is higher. If we look u the values of ts and for water in the tables and lot them we get the grah below. It should also be noted that if the temerature of a liquid is ket constant, it may be made to boil by changing the ressure. he ressure at which it boils is called the SAURAION PRESSURE and is denoted as s in the tables. he grah below also shows the freezing oint of water lotted against ressure (ressure has little effect on it). he two grahs cross at 0.0oC and 0.006 bar. his oint is called the RIPLE POIN. he grah shows the three hases of ice, water and steam. At the trile oint, all three can occur together. Below the trile oint, ice can change into steam without a liquid stage (and vice versa). All substances have a trile oint. If you did the exercise of lotting the grah of ts against for water/steam you would find that the tables sto at. bar and 74.5oC. Above this ressure and temerature, the henomenon of evaoration does not occur and no latent energy stage exists. his oint is called the CRIICAL POIN and every substance has one. If vaour is heated, it becomes hotter than the boiling oint and the more it is heated, the more it becomes a gas. Such vaour is referred to as SUPERHEAED APOUR, excet when it is a substance at ressures and temeratures above the critical oint when it is called SUPERCRIICAL APOUR. D.J.Dunn

6. CONINUOUS EAPORAION A simle boiler or evaorator as shown is needed to continuously roduce vaour from liquid. he liquid is umed in at the same rate at which the vaour is driven off. he heat transfer rate needed to do this must suly the internal energy to the rocess and the flow energy. In other words, the heat transfer is equal to the increase in the enthaly from liquid to vaour. his is why enthaly is such an imortant roerty. 6. WE APOUR Wet vaour is a mixture of dry saturated vaour and liquid drolets. It may also be thought of as a artially evaorated substance. In order to understand its roerties, consider the evaoration of kg of water illustrated with a temerature - enthaly grah. Starting with water at atmosheric ressure and 0.0oC, the enthaly is arbitrarily taken as zero. Keeing the ressure constant and raising the temerature, the enthaly of the water rises to 49. kj/kg at 00oC. At this oint it is saturated water and the sensible enthaly is hf49. kj/kg. he addition of further heat will cause the water to evaorate. During evaoration, the temerature remains at 00oC. When the latent enthaly hfg (56.7 kj/kg) has been added, the substance is dry saturated vaour and its secific enthaly hg is 675.8 kj/kg. Further addition of heat will cause the temerature to rise and the substance becomes suerheated vaour. D.J.Dunn

his grah may be drawn for any ressure and the same basic shae will be obtained but of course with different values. At the critical ressure it will be found that hfg is zero. he oint of interest is the enthaly value at some oint along the evaoration line. Any oint on this line reresents wet steam. Suose only fraction x kg has been evaorated. he latent enthaly added will only be xhfg and not hfg. he enthaly of the water/steam mixture is then h hf + xhfg he fraction x is called the DRYNESS FRACION but it is rarely given as a fraction but rather as a decimal. If no evaoration has started, then x 0. If all the liquid is evaorated then x. x cannot be larger than as this would mean the vaour is suerheated. he same logic alies to internal energy and it follows that u uf + xufg 6.4 OLUMES he secific volume of saturated water is denoted vf. he secific volume of dry saturated steam is denoted vg. he change in volume from water to steam is vfg. It follows that the secific volume of wet steam is v vf + xvfg Since the value of vf is very small and the secific volume of dry steam is very large (in all but the extreme cases), then vfg is ractically the same as vg and vf is negligible. he secific volume of steam is then usually calculated from the formula v xvg 6.5 OAL ALUES All the formula above reresent the values for kg (secific values). When the mass is m kg, the values are simly multilied by m. For examle the volume of m kg of wet steam becomes mxvg 6.6 SAURAION CURE If we lot a grah of hf and hg against either temerature or ressure, we get a roerty chart. he grah itself is the SAURAION CURE. aking the -h grah as an examle, temeratures and dryness fractions may be drawn on it and with the resulting grah, the enthaly of water, wet, dry or suerheated steam may be found. he ressure - enthaly chart is oular for refrigerants but not for steam. A -h chart is enclosed for arcton. D.J.Dunn

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6.7 USE OF ABLES It is vitally imortant for you to be able to use the fluid tables in order to find the roerties of steam. he tables are sulied in the exam but you must have a coy and become comletely roficient in their use. Regarding water/steam, the tables contain a section on saturated water/steam and a section on suerheated/suercritical steam. he saturated water/steam tables are laid out as follows with an examle of values. Check this out for yourself on age 4. ts vg uf ug hf hfg hg sf sfg sg 0 79.9 0.944 76 584 76 05 778 Don't worry about the columns headed s at this stage. his is the roerty called entroy which is dealt with later. he latent internal energy ufg is not listed because of lack of room. However you do need to remember that it is the difference between uf and ug. Note that in all cases the value of hfg is the difference between the values on either side of it. he suerheat tables are laid out differently. In this case the roerty value deends uon the ressure and temerature since the steam can exist at any ressure and temerature above the saturation values. his by necessity makes the tables very concise. More detailed tables are ublished. Interolation is required to find values between those tabulated. In the suerheat tables (e.g. age 6), you must locate the temerature along the to and the ressure down the side. his results in set of values at these co-ordinates giving v, u, h and s. D.J.Dunn 5

WORKED EXAMPLE No.0 Find the secific enthaly, internal energy and volume of steam at bar and 00oC. SOLUION On age 6 of your tables locate the column with 00oC at the to and come down the age to the row with bar at the left side. At this oint you have a block of 4 figures. he enthaly value is the third figure down and is 866 kj/kg. he second figure down is the internal energy and is 65 kj/kg. he first figure is the volume and is 0.766 m/kg. You don't need the fourth figure at this stage. /bar t 50 00 50 00 50 0.766...volume (.5) 65...int. energy 866...enthaly 7...entroy D.J.Dunn 6

WORKED EXAMPLE No. Find the enthaly, internal energy and volume of kg of steam at bar and dryness 0.75. SOLUION From age 4 of the steam tables determine the row corresonding to bar and look u the following values. hf 78 kj/kg hfg000 kj/kg hg78 kj/kg uf780 kj/kg ug586 kj/kg vg0.774 m/kg Next deduce ufg 586-780 806 kj/kg Now find the enthaly. H m(hf + xhfg) (78 + 0.75 x 000) 684 kj Next find the internal energy in the same way. U m(uf + xufg) (780 + 0.75 x 806) 640.5 kj Finally the volume. mxvg x 0.75 x 0.774 0.99 m D.J.Dunn 7

SELF ASSESSMEN EXERCISE NO. 4. Using your steam tables, lot a grah of hf and hg against ressure horizontally and mark on the grah the following: i. the suerheat region ii. the wet steam region. iii. the liquid region. iv. the critical oint. Also label the saturation curve with dry saturated steam and saturated water.. Using your steam tables, lot a grah of vg horizontally against ressure vertically. Also lot vf Show on the grah: i. the suerheated steam region. ii. the wet vaour region. iii. the liquid region. iv. the critical oint. Also label the saturation curve with dry saturated steam and saturated water. D.J.Dunn 8

SELF ASSESSMEN EXERCISE No. 5 Use tables and charts to do the following.. What is the saturation temerature at bars?. What is the secific enthaly and internal energy of saturated water at 6 bars?. What is the secific enthaly and internal energy of dry saturated steam at 6 bars? 4. Subtract the enthaly in from that in and check that it is the latent enthaly hfg at 6 bars in the tables. 5. What is the secific enthaly and internal energy of suerheated steam at 0 bar and 400oC? 6. What is the secific volume of dry saturated steam at 0 bars? 7. What is the volume of kg of wet steam at 0 bars with dryness fraction x0.7? 8. What is the secific enthaly and internal energy of wet steam at 0 bars with a dryness fraction of 0.7? 9. What is the secific volume of suerheated steam at 5 bars and 500oC. 0. What is the volume and enthaly of kg of wet steam at 5 bar with dryness fraction 0.9.. Using the -h chart for arcton (freon ) determine a. the secific enthaly at bar and 70% dry. (x 0.7). b. the secific enthaly at 5 bar and 0 K c. the secific enthaly of the liquid at 8 bars and 00 K.. What is the enthaly of.5 kg of suerheated steam at 8 bar and 50oC?. What is the internal energy of. kg of dry saturated steam at bars? 4. What is the volume of 0.5 kg of suerheated steam at 5 bar and 400oC? D.J.Dunn 9

Answers to Assignment 5. Comare your answers with those below. If you find your answers are different, go back and try again referring to the aroriate section of your notes.. 7.4oC.. 859 and 857 kj/kg.. 794 and 596kJ/kg.. 95 kj/kg and 79 kj/kg. 5. 64 and 957 kj/kg. 6. 0.09957 m/kg. 7. 0.0697 m/kg. 8. and 09. kj/kg. 9. 0.5 m 0..0 m, 7.6 MJ, 7. MJ. a. 90 kj/kg. b.86 kj/kg. c. 0 kj/kg. 4.74 MJ.. 5.69 MJ. 4. 0.0m. D.J.Dunn 0

EDEXCEL HIGHERS ENGINEERING HERMODYNAMICS H NQF LEEL 4 OUCOME UORIAL No. HERMODYNAMIC SYSEMS hermodynamic systems Polytroic rocesses: general equation v n c, relationshis between index `n' and heat transfer during a rocess; constant ressure and reversible isothermal and adiabatic rocesses; exressions for work flow hermodynamic systems and their roerties: closed systems; oen systems; alication of first law to derive system energy equations; roerties; intensive; extensive; two-roerty rule Relationshis: R c c v. and γ c /c v When you have comleted this tutorial you should be able to do the following. Exlain and use the First Law of hermodynamics. Solve roblems involving various kinds of thermodynamic systems. Exlain and use olytroic exansion and comression rocesses. D.J.Dunn www.freestudy.co.uk

. ENERGY RANSFER here are two ways to transfer energy in and out of a system, by means of work and by means of heat. Both may be regarded as a quantity of energy transferred in Joules or energy transfer er second in Watts. When you comlete section one you should be able to exlain and calculate the following. Heat transfer. Heat transfer rate. Work transfer Work transfer rate (Power).. HEA RANSFER Heat transfer occurs because one lace is hotter than another. Under normal circumstances, heat will only flow from a hot body to a cold body by virtue of the temerature difference. here are mechanisms for this, Conduction, convection and radiation. You do not need to study the laws governing conduction, convection and radiation in this module. Fig. A quantity of energy transferred as heat is given the symbol Q and it's basic unit is the Joule. he quantity transferred in one second is the heat transfer rate and this has the symbol Φ and the unit is the Watt. An examle of this is when heat asses from the furnace of a steam boiler through the walls searating the combustion chamber from the water and steam. In this case, conduction, convection and radiation all occur together. Fig. D.J.Dunn www.freestudy.co.uk

SELF ASSESSMEN EXERCISE No.. kg/s of steam flows in a ie 40 mm bore at 00 bar ressure and 400oC. i. Look u the secific volume of the steam and determine the mean velocity in the ie. (7.9 m/s) ii. Determine the kinetic energy being transorted er second. (. W) iii. Determine the enthaly being transorted er second. (89 W) D.J.Dunn www.freestudy.co.uk

.. WORK RANSFER Energy may be transorted from one lace to another mechanically. An examle of this is when the outut shaft of a car engine transfers energy to the wheels. A quantity of energy transferred as work is 'W' Joules but the work transferred in one second is the Power 'P' Watts. An examle of ower transfer is the shaft of a steam turbine used to transfer energy from the steam to the generator in an electric ower station. Fig. It is useful to remember that the ower transmitted by a shaft deends uon the torque and angular velocity. he formulae used are P ω or P πn ω is the angular velocity in radian er second and N is the angular velocity in revolutions er second. D.J.Dunn www.freestudy.co.uk 4

WORKED EXAMPLE No. A duct has a cross section of 0. m x 0.4 m. Steam flows through it at a rate of kg/s with a ressure of bar. he steam has a dryness fraction of 0.98. Calculate all the individual forms of energy being transorted. SOLUION Cross sectional area 0. x 0.4 0.08 m. olume flow rate mxvg at bar olume flow rate x 0.98 x 0.8856.6 m/s. velocity c olume/area.6/0.08.5 m/s. Kinetic Energy being transorted mc/ x.5 / 584 Watts. Enthaly being transorted m(hf + x hfg) H (505 + 0.98 x 0) 7988.9kW Flow energy being transorted ressure x volume Flow Energy x05 x.6 50 000 Watts Internal energy being transorted m(uf + x ufg) U (505 + 0.98 x 05)7468.5 kw Check flow energy H - U 7988.9-7468.5 50 kw D.J.Dunn www.freestudy.co.uk 5

SELF ASSESSMEN EXERCISE No.. he shaft of a steam turbine roduces 600 Nm torque at 50 rev/s. Calculate the work transfer rate from the steam. (88.5 W). A car engine roduces 0 kw of ower at 000 rev/min. Calculate the torque roduced. (95.5 Nm) D.J.Dunn www.freestudy.co.uk 6

. HE FIRS LAW OF HERMODYNAMICS When you have comleted section two, you should be able to exlain and use the following terms. he First Law of hermodynamics. Closed systems. he Non-Flow Energy Equation. Oen systems. he Steady Flow Energy Equation.. HERMODYNAMIC SYSEMS In order to do energy calculations, we identify our system and draw a boundary around it to searate it from the surroundings. We can then kee account of all the energy crossing the boundary. he first law simly states that he nett energy transfer nett energy change of the system. Energy transfer into the system E(in) Energy transfer out of system E (out) Fig. 4 Nett change of energy inside system E(in) - E (out) E his is the fundamental form of the first law. hermodynamic systems might contain only static fluid in which case they are called NON-FLOW or CLOSED SYSEMS. Alternatively, there may be a steady flow of fluid through the system in which case it is known as a SEADY FLOW or OPEN SYSEM. he energy equation is fundamentally different for each because most energy forms only aly to a fluid in motion. We will look at non-flow systems first. D.J.Dunn www.freestudy.co.uk 7

. NON-FLOW SYSEMS he rules governing non-flow systems are as follows. he volume of the boundary may change. No fluid crosses the boundary. Energy may be transferred across the boundary. When the volume enlarges, work (-W) is transferred from the system to the surroundings. When the volume shrinks, work (+W) is transferred from the surroundings into the system. Energy may also be transferred into the system as heat (+Q) or out of the system (-Q). his is best shown with the examle of a iston sliding inside a cylinder filled with a fluid such as gas. Fig.5 he only energy ossessed by the fluid is internal energy (U) so the net change is U. he energy equation becomes Q + W U his is known as the NON-FLOW ENERGY EQUAION (N.F.E.E.) D.J.Dunn www.freestudy.co.uk 8

. SEADY FLOW SYSEMS he laws governing this tye of system are as follows. Fluid enters and leaves through the boundary at a steady rate. Energy may be transferred into or out of the system. A good examle of this system is a steam turbine. Energy may be transferred out as a rate of heat transfer Φ or as a rate of work transfer P. Fig.6. he fluid entering and leaving has otential energy (PE), kinetic energy (KE) and enthaly (H). he first law becomes Φ + P Nett change in energy of the fluid. Φ + P (PE)/s + (KE)/s + (H)/s his is called the SEADY FLOW ENERGY EQUAION (S.F.E.E.) Again, we will use the convention of ositive for energy transferred into the system. Note that the term means change of and if the inlet is denoted oint () and the outlet oint (). he change is the difference between the values at () and (). For examle H means (H-H). D.J.Dunn www.freestudy.co.uk 9

WORKED EXAMPLE No. A steam turbine is sulied with 0 kg/s of suerheated steam at 80 bar and 400oC with negligible velocity. he turbine shaft roduces 00 knm of torque at 000 rev/min. here is a heat loss of. MW from the casing. Determine the thermal ower remaining in the exhaust steam. SOLUION Shaft Power πn π(000/60) x 00 000 6.8 x 06 W 6.8 MW hermal ower sulied H at 80 bar and 400oC H 0(9) 9470 kw 94.7 MW otal energy flow into turbine 94.7 MW Energy flow out of turbine 94.7 MW SP + Loss + Exhaust. hermal Power in exhaust 94.7 -. - 6.8 0.4 MW SELF ASSESSMEN EXERCISE No.. A non-flow system receives 80 kj of heat transfer and loses 0 kj as work transfer. What is the change in the internal energy of the fluid? (60 kj). A non-flow system receives 00 kj of heat transfer and also 40 kj of work is transferred to it. What is the change in the internal energy of the fluid? (40 kj). A steady flow system receives 500 kw of heat and loses 00 kw of work. What is the net change in the energy of the fluid flowing through it? (00 kw) 4. A steady flow system loses kw of heat also loses 4 kw of work. What is the net change in the energy of the fluid flowing through it? (-6 kw) 5. A steady flow system loses kw of heat also loses 0 kw of work. he fluid flows through the system at a steady rate of 70 kg/s. he velocity at inlet is 0 m/s and at outlet it is 0 m/s. he inlet is 0 m above the outlet. Calculate the following. i. he change in K.E./s (-0.5 kw) ii. he change in P.E/s (-.7 kw) iii. he change in enthaly/s (. kw) D.J.Dunn www.freestudy.co.uk 0

D.J.Dunn www.freestudy.co.uk

. MORE EXAMPLES OF HERMODYNAMIC SYSEMS When we examine a thermodynamic system, we must first decide whether it is a nonflow or a steady flow system. First, we will look at examles of non-flow systems.. PISON IN A CYLINDER Fig. 7 here may be heat and work transfer. he N.F.E.E. is, Q + W U Sometimes there is no heat transfer (e.g. when the cylinder is insulated). Q 0 so W U If the iston does not move, the volume is fixed and no work transfer occurs. In this case Q U For a GAS ONLY the change in internal energy is U mcv... SEALED EAPORAOR OR CONDENSER. Fig. 8 Since no change in volume occurs, there is no work transfer so Q U D.J.Dunn www.freestudy.co.uk

WORKED EXAMPLE No.4 0 g of gas inside a cylinder fitted with a iston has a temerature of 5oC. he iston is moved with a mean force of 00 N so that that it moves 60 mm and comresses the gas. he temerature rises to oc as a result. Calculate the heat transfer given cv 78 J/kg K. SOLUION his is a non flow system so the law alying is Q + W U he change in internal energy is U mcv 0.0 x 78 x ( - 5) U 9.4 J he work is transferred into the system because the volume shrinks. W force x distance moved 00 x 0.06 J Q U - W 7.4 J Now we will look at examles of steady flow systems. D.J.Dunn www.freestudy.co.uk

.. PUMPS AND FLUID MOORS he diagram shows grahical symbols for hydraulic ums and motors. he S.F.E.E. states, Fig.9 Φ + P KE/s + PE/s + H/s In this case, esecially if the fluid is a liquid, the velocity is the same at inlet and outlet and the kinetic energy is ignored. If the inlet and outlet are at the same height, the PE is also neglected. Heat transfer does not usually occur in ums and motors so Φ is zero. he S.F.E.E. simlifies to P H/s Remember that enthaly is the sum of internal energy and flow energy. he enthaly of gases, vaours and liquids may be found. In the case of liquids, the change of internal energy is small and so the change in enthaly is equal to the change in flow energy only. he equation simlifies further to P FE/s Since FE and is constant for a liquid, this becomes P WORKED EXAMPLE No.5 A um delivers 0 kg/s of oil of density 780 kg/m from atmosheric ressure at inlet to 800 kpa gauge ressure at outlet. he inlet and outlet ies are the same size and at the same level. Calculate the theoretical ower inut. SOLUION Since the ies are the same size, the velocities are equal and the change in kinetic energy is zero. Since they are at the same level, the change in otential energy is also zero. Neglect heat transfer and internal energy. P m/ρ 0/780 0.056 m/s 800-0 800 kpa P 0.056 x 800000 0 480 W or 0.48 kw D.J.Dunn www.freestudy.co.uk 4

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WORKED EXAMPLE No.6 A feed um on a ower station ums 0 kg/s of water. At inlet the water is at bar and 0oC. At outlet it is at 00 bar and 40oC. Assuming that there is no heat transfer and that PE and KE are negligible, calculate the theoretical ower inut. In this case the internal energy has increased due to frictional heating. he SFEE reduces to P H/s m(h - h) he h values may be found from tables. h 504 kj/kg his is near enough the value of hf at 0 o C bar in steam tables. h 60 kj/kg P 0 (60-504) 969 kw or.969 MW If water tables are not to hand the roblem may be solved as follows. h u + f.e. u c where c 4.8 kj/kg K for water u 4.8 (40-0) 8.6 kj/kg f.e. he volume of water is normally around 0.00 m/kg. f.e. 0.00 x (00 - ) x 05 9 900 J/kg or 9.9 kj/kg hence h u + fe 8.6 + 9.9 0.5 kj/kg P m h 0 x 0.5 070 kw or.07 MW he discreancies between the answers are slight and due to the fact the value of the secific heat and of the secific volume are not accurate at 00 bar. D.J.Dunn www.freestudy.co.uk 6

.4. GAS COMPRESSORS AND URBINES. Figure 0 shows the basic construction of an axial flow comressor and turbine. hese have rows of aerofoil blades on the rotor and in the casing. he turbine asses high ressure hot gas or steam from left to right making the rotor rotate. he comressor draws in gas and comresses it in stages. Fig. 0 Comressing a gas normally makes it hotter but exanding it makes it colder. his is because gas is comressible and unlike the cases for liquids already covered, the volumes change dramatically with ressure. his might cause a change in velocity and hence kinetic energy. Often both kinetic and otential energy are negligible. he internal energy change is not negligible. Figure shows grahical symbols for turbines and comressors. Note the narrow end is always the high ressure end. Fig. D.J.Dunn www.freestudy.co.uk 7

WORKED EXAMPLE No.7 A gas turbine uses 5 kg/s of hot air. It takes it in at 6 bar and 900oC and exhausts it at 450oC. he turbine loses 0 kw of heat from the casing. Calculate the theoretical ower outut given that c 005 J/kg K. First identify this as a steady flow system for which the equation is Φ + P K.E./s + P.E./s + H/s For lack of further information we assume K.E. and PE to be negligible. he heat transfer rate is -0 kw. he enthaly change for a gas is H mc H 5 x 005 x (450-900) -6000 W or -.6 MW P H - Φ -6 - (-0) -4 kw he minus sign indicates that the ower is leaving the turbine. Note that if this was a steam turbine, you would look u the h values in the steam tables. D.J.Dunn www.freestudy.co.uk 8

.5 SEADY FLOW EAPORAORS AND CONDENSERS A refrigerator is a good examle of a thermodynamic system. In articular, it has a heat exchanger inside that absorbs heat at a cold temerature and evaorates the liquid into a gas. he gas is comressed and becomes hot. he gas is then cooled and condensed on the outside in another heat exchanger. Fig.. For both the evaorator and condenser, there is no work transferred in or out. K.E. and P.E. are not normally a feature of such systems so the S.F.E.E. reduces to Φ H/s On steam ower lant, boilers are used to raise steam and these are examles of large evaorators working at high ressures and temeratures. Steam condensers are also found on ower stations. he energy equation is the same, whatever the alication. D.J.Dunn www.freestudy.co.uk 9

WORKED EXAMPLE No.8 A steam condenser takes in wet steam at 8 kg/s and dryness fraction 0.8. his is condensed into saturated water at outlet. he working ressure is 0.05 bar. Calculate the heat transfer rate. SOLUION Φ H/s m(h - h) h hf + x hfg at 0.05 bar from the steam tables we find that h 8 + 0.8(4) 5 kj/kg h hf at 0.05 bar 8 kj/kg hence Φ 8(8-5) -5896 kw he negative sign indicates heat transferred from the system to the surroundings. D.J.Dunn www.freestudy.co.uk 0

SELF ASSESSMEN EXERCISE No.4. Gas is contained inside a cylinder fitted with a iston. he gas is at 0oC and has a mass of 0 g. he gas is comressed with a mean force of 80 N which moves the iston 50 mm. At the same time 5 Joules of heat transfer occurs out of the gas. Calculate the following. i. he work done.(4 J) ii. he change in internal energy. (- J) iii. he final temerature. (9.9 o C) ake cv as 78 J/kg K. A steady flow air comressor draws in air at 0oC and comresses it to 0oC at outlet. he mass flow rate is 0.7 kg/s. At the same time, 5 kw of heat is transferred into the system. Calculate the following. i. he change in enthaly er second. (70.5 kw) ii. he work transfer rate. (65.5 kw) ake c as 005 J/kg K.. A steady flow boiler is sulied with water at 5 kg/s, 00 bar ressure and 00oC. he water is heated and turned into steam. his leaves at 5 kg/s, 00 bar and 500oC. Using your steam tables, find the following. i. he secific enthaly of the water entering. (856 kj/kg) ii. he secific enthaly of the steam leaving. (7 kj/kg) iii. he heat transfer rate. (7.75 kw) 4. A um delivers 50 dm/min of water from an inlet ressure of 00 kpa to an outlet ressure of MPa. here is no measurable rise in temerature. Ignoring K.E. and P.E, calculate the work transfer rate. (.4 kw) 5. A water um delivers 0 dm/minute (0. m/min) drawing it in at 00 kpa and delivering it at 500 kpa. Assuming that only flow energy changes occur, calculate the ower sulied to the um. (860 W) D.J.Dunn www.freestudy.co.uk

6. A steam condenser is sulied with kg/s of steam at 0.07 bar and dryness fraction 0.9. he steam is condensed into saturated water at outlet. Determine the following. i. he secific enthalies at inlet and outlet. ( kj/kg and 6 kj/kg) ii. he heat transfer rate. (46 kw) 7. 0. kg/s of gas is heated at constant ressure in a steady flow system from 0oC to 80oC. Calculate the heat transfer rate Φ. (7.4 kw) C. kj/kg K 8. 0. kg of gas is cooled from 0oC to 50oC at constant volume in a closed system. Calculate the heat transfer. (-6.8 kj) Cv 0.8 kj/kg. D.J.Dunn www.freestudy.co.uk

4. POLYROPIC PROCESSES. When you comlete section four you should be able to do the following. Use the laws governing the exansion and comression of a fluid. State the names of standard rocesses. Derive and use the work laws for closed system exansions and comressions. Solve roblems involving gas and vaour rocesses in closed systems. We will start by examining exansion and comression rocesses. 4. COMPRESSION AND EXPANSION PROCESSES. A comressible fluid (gas or vaour) may be comressed by reducing its volume or exanded by increasing its volume. his may be done inside a cylinder by moving a iston or by allowing the ressure to change as it flows through a system such as a turbine. For ease of understanding, let us consider the change as occurring inside a cylinder. he rocess is best exlained with a ressure - volume grah. When the volume changes, the ressure and temerature may also change. he resulting ressure deends uon the final temerature. he final temerature deends on whether the fluid is cooled or heated during the rocess. It is normal to show these changes on a grah of ressure lotted against volume. (- grahs). A tyical grah for a comression and an exansion rocess is shown in fig.. Fig. It has been discovered that the resulting curves follows the mathematical law n constant. D.J.Dunn www.freestudy.co.uk

Deending on whether the fluid is heated or cooled, a family of such curves is obtained as shown (fig.4). Each grah has a different value of n and n is called the index of exansion or comression. Fig.4 he most common rocesses are as follows. CONSAN OLUME also known as ISOCHORIC A vertical grah is a constant volume rocess and so it is not a comression nor exansion. Since no movement of the iston occurs no work transfer has taken lace. Nevertheless, it still fits the law with n having a value of infinity. CONSAN PRESSURE also known as ISOBARIC A horizontal grah reresents a change in volume with no ressure change (constant ressure rocess). he value of n is zero in this case. CONSAN EMPERAURE also known as ISOHERMAL All the grahs in between constant volume and constant ressure, reresent rocesses with a value of n between infinity and zero. One of these reresents the case when the temerature is maintained constant by cooling or heating by just the right amount. When the fluid is a gas, the law coincides with Boyle's Law constant so it follows that n is. When the fluid is a vaour, the gas law is not accurate and the value of n is close to but not equal to. D.J.Dunn www.freestudy.co.uk 4

ADIABAIC PROCESS When the ressure and volume change in such a way that no heat is added nor lost from the fluid (e.g. by using an insulated cylinder), the rocess is called adiabatic. his is an imortant rocess and is the one that occurs when the change takes lace so raidly that there is no time for heat transfer to occur. his rocess reresents a demarcation between those in which heat flows into the fluid and those in which heat flows out of the fluid. In order to show it is secial, the symbol γ is used instead of n and the law is γ C It will be found that each gas has a secial value for γ (e.g..4 for dry air). POLYROPIC PROCESS All the other curves reresent changes with some degree of heat transfer either into or out of the fluid. hese are generally known as olytroic rocesses. HYPERBOLIC PROCESS he rocess with n is a hyerbola so it is called a hyerbolic rocess. his is also isothermal for gas but not for vaour. It is usually used in the context of a steam exansion. WORKED EXAMPLE No.9 A gas is comressed from bar and 00 cm to 0 cm by the law.constant. Calculate the final ressure. SOLUION. If. C then. C. hence x 00. x 0. x (00/0). 8. bar D.J.Dunn www.freestudy.co.uk 5

WORKED EXAMPLE No.0 aour at 0 bar and 0 cm is exanded to bar by the law. C. Find the final volume. SOLUION.. C. 0 x 0. x. (59.) /. 04.4 cm WORKED EXAMPLE No. A gas is comressed from 00 kpa and 0 cm to 0 cm and the resulting ressure is MPa. Calculate the index of comression n. SOLUION. 00 x 0 n 000 x 0 n (0/0) n 000/00 5 4 n 5 nlog4 log 5 n log5/log4.6094/.86.6 Note this may be solved with natural or base 0 logs or directly on suitable calculators. D.J.Dunn www.freestudy.co.uk 6

SELF ASSESSMEN EXERCISE No. 5. A vaour is exanded from bar and 50 cm to 50 cm and the resulting ressure is 6 bar. Calculate the index of comression n. (0.6).a. A gas is comressed from 00 kpa and 00 cm to 800 kpa by the law.4c. Calculate the new volume. (.4 cm ).b. he gas was at 50oC before comression. Calculate the new temerature using the gas law / C. (07 o C).a. A gas is exanded from MPa and 50 cm to 50 cm by the law.5 C. Calculate the new ressure. (506 kpa).b. he temerature was 500oC before exansion. Calculate the final temerature. (4 o C) D.J.Dunn www.freestudy.co.uk 7

4.. COMBINING HE GAS LAW WIH HE POLYROPIC LAW. For gases only, the general law may be combined with the law of exansion as follows. and so Since for an exansion or comression n n n Substituting into the gas law we get n and further since n substituting into the gas law gives n o summarise we have found that n n In the case of an adiabatic rocess this is written as γ γ For an isothermal rocess n and the temeratures are the same. WORKED EXAMPLE No. A gas is comressed adiabatically with a volume comression ratio of 0. he initial temerature is 5oC. Calculate the final temerature given γ.4 SOLUION C K o 475.5 or 748.5 98(0).4 γ γ D.J.Dunn www.freestudy.co.uk 8

WORKED EXAMPLE No. A gas is comressed olytroically by the law. C from bar and 0oC to bar. calculate the final temerature. SOLUION 9 n n 0.67 ( ) 9(.5) 44.K ( ) 9. WORKED EXAMPLE No.4 A gas is exanded from 900 kpa and 00oC to 00 kpa by the law. C. Calculate the final temerature. SOLUION n 00 7 900. 7 n 0.08 ( 0.) 7( 0.60) 86.9K D.J.Dunn www.freestudy.co.uk 9

SELF ASSESSMEN EXERCISE No. 6. A gas is exanded from MPa and 000oC to 00 kpa. Calculate the final temerature when the rocess is i. Isothermal (n) (000 o C) ii Polytroic (n.) (594 o C) iii. Adiabatic (γ.4) (86 o C) iv. Polytroic (n.6) (64 o C). A gas is comressed from 0 kpa and 5oC to 800 kpa. Calculate the final temerature when the rocess is i. Isothermal (n) (5 o C) ii. Polytroic (n.) (7 o C) iii Adiabatic (γ.4) ( o C) iv. Polytroic (n.5) (69 o C). A gas is comressed from 00 kpa and 0oC to. MPa by the law.c. he mass is 0.0 kg. c005 J/kg K. c v 78 J/kg K. Calculate the following. i. he final temerature. (44 K) ii. he change in internal energy (.0 kj) iii. he change in enthaly (.84 kj) 4. A gas is exanded from 900 kpa and 00oC to 0 kpa by the law.4 C. he mass is 0.05 kg. c00 J/kg K cv 750 J/kg K Calculate the following. i. he final temerature. (88 K) ii. he change in internal energy (-7.5 kj) iii. he change in enthaly (-0.7 kj) D.J.Dunn www.freestudy.co.uk 0

4.. EXAMPLES INOLING APOUR Problems involving vaour make use of the formulae n C in the same way as those involving gas. You cannot aly gas laws, however, unless it is suerheated into the gas region. You must make use of vaour tables so a good understanding of this is essential. his is best exlained with worked examles. WORKED EXAMPLE No.5 A steam turbine exands steam from 0 bar and 00oC to bar by the law. C. Determine for each kg flowing: a. the initial and final volume. b. the dryness fraction after exansion. c. the initial and final enthalies. d. the change in enthaly. SOLUION he system is a steady flow system in which exansion takes lace as the fluid flows. he law of exansion alies in just the same way as in a closed system. he initial volume is found from steam tables. At 0 bar and 00oC it is suerheated and from the tables we find v 0.55 m/kg Next aly the law. C.. 0 x 0.55. x. Hence.5m/kg Next, find the dryness fraction as follows. Final volume.5 m/kg xvg at bar. From the tables we find vg is.694 m/kg hence.5.694x x 0.899 We may now find the enthalies in the usual way. h at 0 bar and 00oC is 05 kj/kg h hf + xhfg at bar (wet steam) h 47 + (0.899)(58) 447 kj/kg D.J.Dunn www.freestudy.co.uk

he change in enthaly is h - h -578 kj/kg SELF ASSESSMEN EXERCISE No.7. kg/s of steam is exanded in a turbine from 0 bar and 00oC to.5 bar by the law.c. Determine the following. i. he initial and final volumes. (0.68 m and m ) ii. he dryness fraction after exansion. (0.86) iii. he initial and final enthalies. (89 kj/kg and 88 kj/kg) iv. he change in enthaly. -4 kw)..5 kg/s of steam is exanded from 70 bar and 450oC to 0.05 bar by the law. C. Determine the following. i. he initial and final volumes. (0.066 m /kg and 7.4 m /kg) ii. he dryness fraction after exansion. (0.4) iii. he initial and final enthalies. (87 kj/kg and 5 kj/kg) iv. he change in enthaly. (-8 kw). A horizontal cylindrical vessel is divided into two sections each m volume, by a non-conducting iston. One section contains steam of dryness fraction 0. at a ressure of bar, while the other contains air at the same ressure and temerature as the steam. Heat is transferred to the steam very slowly until its ressure reaches bar. Assume that the comression of the air is adiabatic (γ.4) and neglect the effect of friction between the iston and cylinder. Calculate the following. i. he final volume of the steam. (.9 m ) ii. iii. he mass of the steam. (.97 kg) he initial internal energy of the steam. (05 kj) iv he final dryness fraction of the steam. (0.798) v. he final internal energy of the steam. (47 kj) D.J.Dunn www.freestudy.co.uk

vi. he heat added to the steam. (9 kj) D.J.Dunn www.freestudy.co.uk

4.4. CLOSED SYSEM WORK LAWS 4.4.. EXPANSION OF PRESSURE WIH OLUME We will start by studying the exansion of a fluid inside a cylinder against a iston which may do work against the surroundings. A fluid may exand in two ways. a) It may exand raidly and uncontrollably doing no useful work. In such a case the ressure could not be lotted against volume during the rocess. his is called an UNRESISED EXPANSION b) It may exand moving the iston. he movement is resisted by external forces so the gas ressure changes in order to overcome the external force and move the iston. In this case the movement is controlled and the variation of ressure with volume may be recorded and lotted on a - grah. Work is done against the surroundings. his rocess is called a RESISED EXPANSION. Consider the arrangement shown in fig. 5. Assume that there is no ressure outside. If the string holding the weight was cut, the gas ressure would slam the iston back and the energy would be dissiated first by acceleration of the moving arts and eventually as friction. he exansion would be unresisted. Fig. 5 If the weights were gradually reduced, the gas would ush the iston and raise the remaining weights. In this way, work would be done against the surroundings (it ends u as otential energy in the weights). he rocess may be reeated in many small stes, with the change in volume each time being d. he ressure although changing, is at any time. D.J.Dunn www.freestudy.co.uk 4

his rocess is characterised by two imortant factors.. he rocess may be reversed at any time by adding weights and the otential energy is transferred back from the surroundings as work is done on the system. he fluid may be returned to its original ressure, volume, temerature and energy.. he fluid force on one side of the iston is always exactly balanced by the external force (in this case due to the weights) on the other side of the iston. he exansion or comression done in this manner is said to be REERSIBLE and CARRIED OU IN EQUILIBRIUM. 4.4.. WORK AS AREA UNDER HE - DIAGRAM If the exansion is carried out in equilibrium, the force of the fluid must be equal to the external force F. It follows that F A. When the iston moves a small distance dx, the work done is dw dw - F dx - Adx - d. he minus sign is because the work is leaving the system. For an exansion from oints to it follows that the total work done is given by W We must remember at this stage that our sign convention was that work leaving the system is negative. It should be noted that some of the work is used to overcome any external ressure such as atmosheric and the useful work is reduced. Consider the system shown in fig.5 again but this time suose there is atmosheric ressure on the outside a. In this case It follows that F + a A A. F A. - a A d When the iston moves a small distance dx, the the useful work done is F dx - F dx - (Adx - a Adx) - ( - a )d. For an exansion from oints to it follows that the useful work done is given by D.J.Dunn www.freestudy.co.uk 5

W ( a ) d 4.4.. WORK LAWS FOR CLOSED SYSEMS If we solve the exression W d we obtain the work laws for a closed system. he solution deends uon the relationshi between and. he formulae now derived aly equally well to a comression rocess and an exansion rocess. Let us now solve these cases. CONSAN PRESSURE W W d d W - ( - ) CONSAN OLUME If is constant then d 0 W 0. HYPERBOLIC his is an exansion which follows the law C and is ISOHERMAL when it is a gas. Substituting C - the exression becomes W d C d C ln Since C then W ln since W ln In the case of gas we can substitute mr and so D.J.Dunn www.freestudy.co.uk 6

W mr ln mr ln D.J.Dunn www.freestudy.co.uk 7

POLYROPIC In this case the exansion follows the law n C. he solution is as follows. W W W C n+ n+ [ ] [ ] d n n but C d W C n + Since C or For gas only we may substitute mr and so ADIABAIC -n W mr [ ] n Since an adiabatic case is the secial case of a olytroic exansion with no heat transfer, the derivation is identical but the symbol γ is used instead of n. W [ ] γ For gas only we may substitute mr and sow mr [ ] γ his is the secial case of the olytroic rocess in which Q0. Q 0 W mr γ Substituting for Q and U in the NFEE we find Q + W U Since R C Cv mr 0 + mcv γ C Cv Cv( γ ) C Cv R Cv γ γ his shows that the ratio of the rincial secific heat caacities is the adiabatic index. It was shown earlier that the difference is the gas constant R. hese imortant relationshis should be remembered. C Cv R γ C/Cv D.J.Dunn www.freestudy.co.uk 8

WORKED EXAMPLE No.5 Air at a ressure of 500 kpa and volume 50 cm is exanded reversibly in a closed system to 800 cm by the law. C. Calculate the following. SOLUION a. he final ressure. b. he work done. 500 kpa 50 x 0-6 m 800 x 0-6 m. W.6x0 ( ) n W 47Joules. 500x0 or.6kpa.6x0 (50x0 6 ). (800x0 6 x800x0 500x0. 6 ). x50x0 6 WORKED EXAMPLE No.6 Steam at 6 bar ressure and volume 00 cm is exanded reversibly in a closed system to dm by the law. C. Calculate the work done. SOLUION 6 bar 00 x 0-6 m x 0 - m.. 00x0 6x x0 5 5 6 ( ) ( 0.648x0 xx0 6x0 x00x0 ) W n W 5.Joules 6. 0.648bar. D.J.Dunn www.freestudy.co.uk 9

SELF ASSESSMEN EXERCISE No.8. 0 g of steam at 0 bar and 50oC exands reversibly in a closed system to bar by the law.c. Calculate the following. i. he initial volume. (0.008 m ) ii. he final volume. (0.00974 m ) iii. he work done. (-.9 kj). 0 g of gas at 0oC and bar ressure is comressed to 9 bar by the law.4 C. aking the gas constant R 87 J/kg K calculate the work done. (Note that for a comression rocess the work will turn out to be ositive if you correctly identify the initial and final conditions). (.67 kj). Gas at 600 kpa and 0.05 dm is exanded reversibly to 00 kpa by the law.5 C. Calculate the work done. (-.8 kj) 4. 5 g of gas is comressed isothermally from 00 kpa and 0oC to MPa ressure. he gas constant is 87 J/kg K. Calculate the work done. (.9 kj) 5. Steam at 0 bar with a volume of 80 cm is exanded reversibly to bar by the law C. Calculate the work done. (-84. kj) 6. Gas fills a cylinder fitted with a frictionless iston. he initial ressure and volume are 40 MPa and 0.05 dm resectively. he gas exands reversibly and olytroically to 0.5 MPa and dm resectively. Calculate the index of exansion and the work done. (.46 and -.4 kj) 7. An air comressor commences comression when the cylinder contains g at a ressure is.0 bar and the temerature is 0 o C. he comression is comleted when the ressure is 7 bar and the temerature 90 o C. (.4 and 944 J) he characteristic gas constant R is 87 J/kg K. Assuming the rocess is reversible and olytroic, calculate the index of comression and the work done. D.J.Dunn www.freestudy.co.uk 40

WORKED EXAMPLE No.7 0. kg of gas at 00 oc is exanded isothermally and reversibly from MPa ressure to 00 kpa. ake Cv 78 J/kg K and R 87 J/kg K. Calculate i. he work transfer. ii. he change in internal energy. iii. he heat transfer. SOLUION W ln mr ln mr ln 6 x0 W 0.x87x7ln 4900 J or - 49. kj 5 0 x he work is leaving the system so it is a negative work transfer. Since is constant U 0 Q - 49. 0 Q 49. kj Note that 49. kj of heat is transferred into the gas and 49. kj of work is transferred out of the gas leaving the internal energy unchanged. WORKED EXAMPLE No.8 Reeat worked examle 7 but for an adiabatic rocess with γ.4 Calculate SOLUION 00 x 0 7 x 6 x 0 W mr W 580 J Q + W U Check ( ) For an adiabatic rocess Q 0 U mc 9 K 0. x 87 x hence U -580 J γ ( 9-7) 0.4 0. x 78 x (9-7) - 5848 J D.J.Dunn www.freestudy.co.uk 4

WORKED EXAMPLE No.9 Reeat worked examle 7 but for a olytroic rocess with n.5 Calculate SOLUION 00 x 0 7 x 6 x 0 W mr W 605 J U mc Q U - W Q ( ) -977.7 - (-60) n 5. K 0. x 87 x 8. J ( 5. - 7) 0.4 0. x 78 x (5. - 7) -977.7 J D.J.Dunn www.freestudy.co.uk 4

SELF ASSESSMEN EXERCISE No.9 ake Cv 78 J/kg K and R 87 J/kg K throughout.. dm of gas at 00 kpa and 0oC is comressed to. MPa reversibly by the law. C. Calculate the following. i. he final volume. (0.6 dm ) ii. he work transfer. (57 J) iii. he final temerature. (70oC) iv. he mass. (.89 g) v. he change in internal energy. (8 J) vi. he heat transfer. (-8 J). 0.05 kg of gas at 0 bar and 00oC is exanded reversibly to bar by the law. C in a closed system. Calculate the following. i. he initial volume. (9.85 dm ) ii. he final volume. (58 dm ) iii. he work transfer. (-7 kj) iv. he change in internal energy. (-0. kj) v. he heat transfer. (6.7 kj). 0.08 kg of air at 700 kpa and 800oC is exanded adiabatically to 00 kpa in a closed system. aking γ.4 calculate the following. i. he final temerature. (65.4 K) ii. he work transfer. (6. kj) iii. he change in internal energy. (-6. J) D.J.Dunn www.freestudy.co.uk 4

4. A horizontal cylinder is fitted with a frictionless iston and its movement is restrained by a sring as shown (Figure 6.) Figure 6 a. he sring force is directly roortional to movement such that F/ x k Show that the change in ressure is directly roortional to the change in volume such that / k/a b. he air is initially at a ressure and temerature of 00 kpa and 00 K resectively. Calculate the initial volume such that when the air is heated, the ressure volume grah is a straight line that extends to the origin. (0.5 dm ) c. he air is heated making the volume three times the original value. Calculate the following. i. he mass. (0.58 g) ii. he final ressure. (00 kpa) iii. he final temerature. (700 K) iv. he work done. (-00 kj) v. he change in internal energy. (97 J) vi. he heat transfer. (. kj) D.J.Dunn www.freestudy.co.uk 44

4.5. CLOSED SYSEM PROBLEMS INOLING APOUR he solution of roblems involving steam and other vaours is done in the same way as for gases with the imortant roviso that gas laws must not be used. olumes and internal energy values should be obtained from tables and roerty charts. his is best illustrated with a worked examle. WORKED EXAMPLE No.0 kg of steam occuies a volume of 0. m at 9 bar in a closed system. he steam is heated at constant ressure until the volume is 0.44 m. Calculate the following. i. he initial dryness fraction. ii. he final condition. iii. he work transfer. iv. he change in internal energy. v. he heat transfer. SOLUION First find the initial dryness fraction. 0. mx v g at 9 bar x 0./( x 0.49) x 0.9 (initial dryness fraction). Now determine the secific volume after exansion. 9 bar (constant ressure) 0.44 m. mv v 0.44/ 0.44m /kg First, look in the suerheat tables to see if this value exists for suerheat steam. We find that at 9 bar and 50oC, the secific volume is indeed 0.44 m/kg. he final condition is suerheated to 50oC. Note that if v was less than v g at 9 bar the steam would be wet and x would have to be found. Next find the work. W -( - ) -9x 05 (0.44-0. ) -0950 J W -0.95 kj (Energy leaving the system) D.J.Dunn www.freestudy.co.uk 45

Next determine the internal energy from steam tables. U m u and u u f + x u fg at 9 bar u fg at 9 bar u g - u f 58-74 89 kj/kg U {74 + 0.9(89) } 454 kj U m u and u u at 9 bar and 50 o C 877 kj/kg U m u (877) 877 kj. he change in internal energy U - U 4 kj (increased) Finally deduce the heat transfer from the NFEE Q + W U hence Q U - W 4 - (-0.95) Q 56 kj (energy entering the system) D.J.Dunn www.freestudy.co.uk 46

SELF ASSESSMEN EXERCISE No.0. 0. kg of dry saturated steam at 0 bar ressure is exanded reversibly in a closed system to bar by the law. C. Calculate the following. i. he initial volume. (8.9 dm ) ii. he final volume. (64 dm ) iii. he work transfer. (-6 kj) iv. he dryness fraction. (0.779) v. he change in internal energy. (-08 kj) vi. he heat transfer. (-46 kj). Steam at 5 bar and 50 o C is exanded reversibly in a closed system to 5 bar. At this ressure the steam is just dry saturated. For a mass of kg calculate the following. i. he final volume. (0.75 m ) ii. he change in internal energy. (-65 kj) iii. he work done. (-87 kj) iv. he heat transfer. (. kj) D.J.Dunn www.freestudy.co.uk 47

EDEXCEL HIGHERS ENGINEERING HERMODYNAMICS H NQF LEEL 4 OUCOME INERNAL COMBUSION ENGINE PERFORMANCE UORIAL No. HEA ENGINE HEORY Internal combustion engine erformance Second law of thermodynamics: statement of law; schematic reresentation of a heat engine to show heat and work flow Heat engine cycles: Carnot cycle; Otto cycle; Diesel cycle; dual combustion cycle; Joule cycle; roerty diagrams; Carnot efficiency; air-standard efficiency Performance characteristics: engine trials; indicated and brake mean effective ressure; indicated and brake ower; indicated and brake thermal efficiency; mechanical efficiency; relative efficiency; secific fuel consumtion; heat balance Imrovements: turbocharging; turbocharging and intercooling; cooling system and exhaust gas heat recovery systems When you have comleted this tutorial, you should be able to do the following. Exlain the basic idea behind the Second Law of hermodynamics. Define the roerty called Entroy Define an Isentroic rocess. Solve basic roblems involving isentroic exansions. Exlain the Carnot Princile. D.J.Dunn www.freestudy.co.uk

. HE SECOND LAW OF HERMODYNAMICS he Second Law of hermodynamics is not something that can be written as a simle statement or formulae. It is a set of observations concerning the way that things flow or run as time rogresses forward. It encomasses many observations such as water normally flows from high levels to low levels and heat normally flows from hot to cold. In this module, you must concern yourself only with how the second law relates to heat engines and the efficiency of a heat engine. In the context of heat engines, the second law may be summed as : No heat engine can be 00% efficient. his should become aarent in the following sections.. HEA ENGINES Nearly all motive ower is derived from heat using some form of heat engine. Here are some examles. Steam Power Plant. Gas urbines. Jet Engines. Internal Combustion Engines. A heat engine requires a source of hot energy. We get this by burning fossil fuel or by nuclear fission. he main sources of natural heat are solar and geothermal. In order to understand the basic theory, it might hel to draw an analogy with a hydraulic motor and an electric motor. All motors require a high level source of energy and must exhaust at a low level of energy. Figure D.J.Dunn www.freestudy.co.uk

.. HYDRAULIC MOOR Fluid ower is transorted by the flow Q m /s. he energy contained in a volume Q m of liquid at a ressure is the flow energy given by the exression Q. he hydraulic motor requires a source of liquid at a high ressure and exhausts at a lower ressure.he energy sulied is Q and some of this is converted into work. he energy in the low ressure liquid is Q. For a erfect motor with no losses due to friction, the law of energy conservation gives the work outut and efficiency as follows. W out η Q - Q Q( W out Energy inut - Wout Q( - ) Q Q ) ( - ) -.. ELECRIC MOOR Electric ower is transorted by the current. Electrical energy is the roduct of the charge Q Coulombs and the electric otential olts. he energy inut at a high voltage is Q and the energy exhausted at low voltage is Q. For a erfect motor with no losses due to friction, the work outut and efficiency are found from the law of energy conservation as follows. W Q - Q Q( - ) out η W out Energy inut.. HEA MOOR Wout Q( - Q Q ) ( - ) - emerature is by analogy the equivalent of ressure and electric otential. In order to comlete the analogy, we need something that is equivalent to volume and electric charge that transorts the energy. It is not difficult to visualise a volume of liquid flowing through a hydraulic motor. It is not imossible to visualise a flow of electrons bearing electric charge through an electric motor. It is imossible to visualise something flowing through our ideal heat engine that transorts ure heat but the analogy tells us there must be something so let us suose a new roerty called ENROPY and give it a symbol S. Entroy must have units of energy er degree of temerature or Joules er Kelvin. Entroy is dealt with more fully later on. he energy sulied at temerature is S and the energy exhausted is S. For a erfect motor with no losses due to friction, the law of energy conservation gives the work outut and efficiency as follows. W out η S - S S( W out Energy inut - ) Wout S( - ) S S ( - ) - D.J.Dunn www.freestudy.co.uk

..4 EFFICIENCY In our erfect motors, the energy conversion rocess is 00% efficient but we may not have converted all the energy sulied into work and energy may be wasted in the exhaust. In the case of the electric motor, the lowest value for (so far as we know) is ground voltage zero, so theoretically we can obtain 00% efficiency by exhausting the electric charge with no residual energy. In the case of the hydraulic motor, the lowest ressure we can exhaust to is atmosheric so we always waste some energy in the exhausted liquid. In the case of the heat motor, the lowest temerature to which we can exhaust is ambient conditions, tyically 00K, so there is a lot of residual energy in the exhaust. Only by exhausting to absolute zero, can we extract all the energy. A model heat engine is usually reresented by the following diagram. (Note that the word engine is usually referred to motor). he energy transfer from the hot source is Qin Joules. he energy transfer rate from the hot source is Φin Watts. Fig. he energy transfer to the cold sink is Qout Joules. he energy transfer rate to the cold sink is Φout Watts. he work outut us W Joules. he ower outut is P Watts. By considering the total conservation of energy, it follows that the energy converted into work must be W Qin - Qout Joules or P Φin - Φout Watts he efficiency of any machine is the ratio Outut/Inut so the thermal efficiency of a heat engine may be develoed as follows. W Qin Qout Qout ηth W Qin Qout η th Q Q Q in In terms of energy transfer rates in Watts this is written as Φ ηth Φ out in in int It follows from our analogy that Q in S and Q out S and confirms η D.J.Dunn www.freestudy.co.uk 4

SELF ASSESSMEN EXERCISE No.. A heat engine is sulied with 60 MW of energy and roduces 0 MW of ower. What is the thermal efficiency and the heat lost? (Answers.% and 40 MW). A heat engine is sulied with 40 kj of energy that it converts into work with 5% efficiency. What is the work outut and the heat lost? (Answers 0 kj and 0 kj) D.J.Dunn www.freestudy.co.uk 5

.. PRACICAL HEA ENGINE CONSIDERAIONS Let us consider how we might design a ractical heat engine with a iston, connecting rod and crank shaft mechanism. Figure shows how heat may be assed to a gas inside a cylinder causing it to exand. his ushes a iston and makes it do some work. his at first looks like a good way of converting heat into work but the roblem is that it works only once and cannot convert heat into work continuously. Figure No ractical heat engine has ever been invented that continuously converts heat directly into work as suosed in our ideal model. Practical heat engines use a working fluid such as gas or steam. A cycle of thermodynamic rocesses is conducted on the fluid with the end result being a conversion of heat into work. First energy is given to the working fluid by use of a heat transfer at a hot temerature. Next we must convert as much of this energy as ossible into work by allowing the fluid to exand. Our studies of olytroic exansions tell us that the ressure, volume and temerature all change as the gas or vaour gives u its energy as work. he ressure is vitally imortant to roduce a motivating force on the iston. Having extracted as much energy as ossible from the working fluid, we must return it back to the starting condition in order to reeat the rocess. o do this, we must raise the ressure of the fluid back to the high level with some form of comression. A simle reversal of the exansion rocess would return the fluid back to the original ressure and temerature. However, this would require us to give back all the work we got out so nothing is gained. he only way we can return the fluid back to a high ressure with less work involves cooling it first. In fact, if it is to be heat engine, we must have a cooling rocess as indicated in our model. We have deduced that a ractical heat engine must meet the following criteria. It must roduce work continuously. It must return the working fluid back to the same ressure and temerature at the beginning of every cycle. D.J.Dunn www.freestudy.co.uk 6

A model of a ractical engine is shown in Fig. 4. his indicates that we need four rocesses, heating, exansion, cooling and comression. his may be achieved ractically using either closed system rocesses (as in a mechanism with a iston, connecting rod and crank shaft) or oen system rocesses such as with a steam boiler, turbine, cooler and um). Figure 4 D.J.Dunn www.freestudy.co.uk 7

.4 ENROPY We have just discovered that entroy is a roerty that governs the quantity of energy conveyed at a given temerature such that in our ideal heat engine, the energy is given by the exression Q S. Entroy is a roerty that is closely associated with the second law of thermodynamics. In thermodynamics there are two forms of energy transfer, work (W) and heat (Q). You should already be familiar with the theory of work laws in closed systems and know that the area under a ressure - volume diagram gives work transfer. By analogy there should be a roerty that can be lotted against temerature such that the area under the grah gives the heat transfer. his roerty is entroy and it is given the symbol S. his idea imlies that entroy is a roerty that can be transorted by a fluid. Consider a - and -s grah for a reversible exansion (Fig. 5). From the - grah we have W From the -S grah we have d Q ds Fig.5 his is the way entroy was develoed for thermodynamics and from the above we get the following definition ds dq/ he units of entroy are hence J/K. Secific entroy has a symbol s and the units are J/kg K It should be ointed out that there are other definitions of entroy but this one is the most meaningful for thermodynamics. A suitable integration will enable you to solve the entroy change for a fluid rocess. For those wishing to do studies in greater deth, these are shown in aendix A. Entroy values for steam may be found in your thermodynamic tables in the columns headed s f, s fg and s g. s f is the secific entroy of saturated liquid. s fg is the change in secific entroy during the latent stage. s g is the secific entroy of dry saturated vaour. D.J.Dunn www.freestudy.co.uk 8

.4. ISENROPIC PROCESSES he word ISENROPIC means constant entroy and this is a very imortant thermodynamic rocess. It occurs in articular when a rocess is reversible and adiabatic. his means that there is no heat transfer to or from the fluid and no internal heat generation due to friction. In such a rocess it follows that if dq is zero then ds must be zero. Since there is no area under the -S grah, the grah must be a vertical line as shown. Fig. 6 here are other cases where the entroy is constant. For examle, if there is friction in the rocess generating heat but this is lost through cooling, then the net result is zero heat transfer and constant entroy. You do not need to be concerned about this at this stage..4. EMPERAURE - ENROPY (-s) DIAGRAM FOR APOURS. If you lot the secific entroy for saturated liquid (s f ) and for dry saturated vaour (s g ) against temerature, you would obtain the saturation curve. Lines of constant dryness fraction and constant ressure may be shown (Figure 7). Fig. 7 D.J.Dunn www.freestudy.co.uk 9

.4. SPECIFIC ENHALPY-SPECIFIC ENROPY (h-s) DIAGRAM. his diagram is esecially useful for steady flow rocesses (figure 8). he diagram is obtained by lotting h g against s g and h f against s f to obtain the characteristic saturation curve. he two curves meet at the critical oint C. Lines of constant ressure, temerature and dryness are suerimosed on the diagram. his is an extremely useful chart and it is available commercially. If any two coordinates are known, a oint can be obtained on the chart and all other relevant values may be read off it. h s charts are esecially useful for solving isentroic rocesses because the rocess is a vertical line on this grah. Fig. 8 D.J.Dunn www.freestudy.co.uk 0

Entroy values can be used to determine the dryness fraction following a steam exansion into the wet region when the rocess is isentroic. his is a very imortant oint and you must master how to do this in order to solve steam exansion roblems, esecially in the following tutorials where steam cycles and refrigeration cycles are covered. he following examles show how this is done. WORKED EXAMPLE No. Steam at bar and 50oC is exanded reversibly and adiabatically to bar. Calculate the final dryness fraction and the enthaly change. SOLUION Let suffix () refer to the conditions before the exansion and () to the conditions after. h at bar and 50oC 770 kj/kg s at bar and 50oC is 7.80 kj/kg K. Because the rocess is adiabatic and reversible, the entroy remains the same. s at bar and assumed wet is s f + xs fg s 7.80.0 + x(6.056) x 0.987 h at bar and 0.987 dry h f + x h fg h 47 + 0.987(58) 645.6 kj/kg h 645.6 770-4.4 kj/kg Being ale to solve the changes in enthaly enable us to aly the first law of thermodynamics to solve roblems with steam turbines. he next examle shows you how to do this. D.J.Dunn www.freestudy.co.uk

WORKED EXAMPLE No. A steam turbine exands 60 kg/s from 40 bar and 00oC to 4 bar reversibly and adiabatically (isentroic). Calculate the theoretical ower outut. SOLUION Φ + P E er second (SFEE) he rocess is adiabatic. Φ 0 and the only energy term to use is enthaly. P H er second. h at 40 bar and 00oC 96 kj/kg s at 40 bar and 00oC is 6.64 kj/kg K. s at 4 bar and assumed wet is s f + xs fg s 6.64.776 + x(5.) x 0.896 h at 4 bar and 0.896 dry h f + x h fg h 605 + 0.896(4) 57 kj/kg P H er second 60(57-96) -6756 kw (out of system) D.J.Dunn www.freestudy.co.uk

SELF ASSESSMEN EXERCISE No.. A turbine exands 40 kg/s of steam from 0 bar and 50oC reversibly and adiabatically to 0.5 bar. Calculate the theoretical ower outut. (Answer 5. MW). A turbine exands 4 kg/s of steam from 50 bar and 00oC reversibly and adiabatically to 0. bar. Calculate the theoretical ower outut. (Answer.8 MW). A turbine exands 0 kg/s of steam from 800 bar and 400oC reversibly and adiabatically to 0. bar. Calculate the theoretical ower outut. (Answer. MW) 4. A turbine exands kg/s of steam reversibly and adiabatically. he inlet conditions are 0 bar and dry saturated. he outlet ressure is bar. Calculate the theoretical ower outut. (Answer 8.5 MW) D.J.Dunn www.freestudy.co.uk

. HE CARNO PRINCIPLE A man called Sadi Carnot deduced that if the heat transfers from the hot reservoir and to the cold sum were done at constant temerature (isothermal rocesses), then the efficiency of the engine would be the maximum ossible. he reasoning behind this is as follows. Consider heat being transferred from a hot body A to a slightly cooler body B. he temerature of body A falls and the temerature of body B rises until they are at the same temerature. If body B is now raised in temerature by heat transfer from the surroundings, it becomes the hotter body and the heat flow is reversed from B to A. If body A returns to its original temerature then the net heat transfers between A and B is zero. However body B is now hotter than its original temerature so there has been a net heat transfer from the surroundings. he heat transfer rocess is hence not reversible as external hel was needed to reverse the rocess. Figure 9 If it were ossible to transfer heat with no temerature difference from A to B then it could be reversed with no external hel. Such a rocess is an ISOHERMAL rocess. Isothermal heat transfer is ossible, for examle evaoration of water in a boiler is isothermal. Carnot devised a thermodynamic cycle using isothermal heat transfers only so by definition, the efficiency of this cycle is the most efficient any engine could be oerating between two temeratures. Engine cycles are covered in the next tutorial but the following shows how the Carnot cycle might be conducted. In ractice, it is not ossible to make this cycle work. D.J.Dunn www.freestudy.co.uk 4

.. CLOSED SYSEM CARNO CYCLE. he cycle could be conducted on gas or vaour in a closed or oen cycle. he cycle described here is for gas in a cylinder fitted with a iston. It consists of four closed system rocesses as follows. to. he fluid is comressed isentroically. Work is ut in and no heat transfer occurs. Fig. 0 to. he fluid is heated isothermally. his could only occur if it is heated as it exands so there is work taken out and heat ut in. Fig. to 4. he fluid continues to exand isentroically with no heat transfer. Work outut is obtained. Fig. D.J.Dunn www.freestudy.co.uk 5

4 to he fluid is cooled isothermally. his can only occur if it cooled as it is comressed, so work is ut in and heat is taken out. At the end of this rocess every thing is returned to the initial condition. Fig. he total work taken out is Wout and the total work ut in is Win. o be an engine, Wout must be larger than Win and a net amount of work is obtained from the cycle. It also follows that since the area under a - grah reresents the work done, then the area enclosed by the - diagram reresents the net work transfer. It also follows that since the area under the -s grah is reresents the heat transfer, and then the area enclosed on the -s diagram reresents the net heat transfer. his is true for all cycles and also for real engines. Fig.4 Alying the first law, it follows Qnett Wnett For isothermal heat transfers Q ds S since is constant. Qout cold s he efficiency would then be given by η th Q s It is aarent from the -s diagram that the change in entroy s is the same at the hot cold and cold temeratures. It follows that η th - hot his exression, which is the same as that used for the ideal model, gives the CARNO EFFICIENCY and it is used as a target figure that cannot be surassed (in fact not even attained). in hot cold hot D.J.Dunn www.freestudy.co.uk 6

WORKED EXAMPLE No. A heat engine draws heat from a combustion chamber at 00 o C and exhausts to atmoshere at 0 o C. What is the maximum ossible thermal efficiency that could be achieved? SOLUION he maximum efficiency ossible is the Carnot efficiency. Remember to use absolute temeratures. cold 7 + 0 8 η th 0.505 or 50.6% 7 + 00 57 hot SELF ASSESSMEN EXERCISE No.. A heat engine works between temeratures of 00o C and 0oC. It is claimed that it has a thermal efficiency of 75%. Is this ossible? (Answer the maximum efficiency cannot exceed 7%). Calculate the efficiency of a Carnot Engine working between temeratures of 00oC and 00o C. (Answer 67.9%) D.J.Dunn www.freestudy.co.uk 7

EDEXCEL HIGHERS ENGINEERING HERMODYNAMICS H NQF LEEL 4 OUCOME INERNAL COMBUSION ENGINE PERFORMANCE UORIAL No. 4 HEA ENGINE CYCLES Internal combustion engine erformance Second law of thermodynamics: statement of law; schematic reresentation of a heat engine to show heat and work flow Heat engine cycles: Carnot cycle; Otto cycle; Diesel cycle; dual combustion cycle; Joule cycle; roerty diagrams; Carnot efficiency; air-standard efficiency Performance characteristics: engine trials; indicated and brake mean effective ressure; indicated and brake ower; indicated and brake thermal efficiency; mechanical efficiency; relative efficiency; secific fuel consumtion; heat balance Imrovements: turbocharging; turbocharging and intercooling; cooling system and exhaust gas heat recovery systems Define AIR SANDARD CYCLES. Identify the ideal cycle for a given tye of engine. Exlain and solve roblems for the OO cycle Exlain and solve roblems for the DIESEL cycle Exlain and solve roblems for the Dual Combustion cycle Exlain and solve roblems for the JOULE cycle D.J.Dunn www.freestudy.co.uk

. HEOREICAL CYCLES FOR ENGINES Internal combustion engines fall into two grous, those that use a sarking lug to ignite the fuel (sark ignition engines) and those that use the natural temerature of the comressed air to ignite the fuel (comression ignition engines). Another way to grou engines is into those that use non-flow rocesses and those that use flow rocesses. For examle, non-flow rocesses are used in iston engines. Flow rocesses are used in gas turbine engines. heoretical cycles are made u of ideal thermodynamic rocesses to resemble those that occur in a real engine as closely as ossible. Many of these cycles are based on air as the working fluid and are called AIR SANDARD CYCLES. Before looking at air standard cycles, we should briefly revise the Carnot Cycle from tutorial.. HE CARNO CYCLE he most efficient way of transferring heat into or out of a fluid is at constant temerature. All the heat transfer in the Carnot cycle is at constant temerature so it follows that the Carnot cycle is the most efficient cycle ossible. he heat transfer into the cycle occurs at a hot temerature hot and the heat transfer out of the cycle occurs at a colder temerature cold. he thermodynamic efficiency was sown to be given as follows. cold η th hot None of the following cycles can have an efficiency greater than this when oerating between the same temeratures limits. D.J.Dunn www.freestudy.co.uk

SPARK IGNIION ENGINE. HE OO CYCLE he ideal cycle is named after Count N.A.Otto. It reresents the ideal cycle for a sark ignition engine. In an ideal sark ignition engine, there are four rocesses as follows. Fig. COMPRESSION SROKE Air and fuel are mixed and comressed so raidly that there is no time for heat to be lost. (Figure A) In other words the comression is adiabatic. Work must be done to comress the gas. IGNIION Just before the oint of maximum comression, the air is hot and a sark ignites the mixture causing an exlosion (Figure B). his roduces a raid rise in the ressure and temerature. he rocess is idealised as a constant volume rocess in the Otto cycle. EXPANSION OR WORKING SROKE he exlosion is followed by an adiabatic exansion ushing the iston and giving out work. (Figure C) EXHAUS At the end of the working stroke, there is still some ressure in the cylinder. his is released suddenly by the oening of an exhaust valve. (Figure D) his is idealised by a constant volume dro in ressure in the Otto cycle. In 4 stroke engines a second cycle is erformed to ush out the roducts of combustion and draw in fresh air and fuel. It is only the ower cycle that we are concerned with. D.J.Dunn www.freestudy.co.uk

he four ideal rocesses that make u the Otto cycle are as follows. to he air is comressed reversibly and adiabatically. Work is ut in and no heat transfer occurs. Fig. to he air is heated at constant volume. No work is done. Q in mcv(-) Fig. to 4 he air exands reversibly and adiabatically with no heat transfer back to its original volume. Work outut is obtained. Fig.4 4 to he air is cooled at constant volume back to its original ressure and temerature. No work is done Q out mcv(4-) Fig.5 D.J.Dunn www.freestudy.co.uk 4

he total heat transfer into the system during one cycle is Q nett Q in - Q out he total work outut er cycle is Wnett From the st. Law of thermodynamics Qnett Wnett EFFICIENCY ) - ( ) - ( ) - ( ) - ( 4 4 mc mc Q Q Q W v v in out in nett η For the rocess () to () we may use the rule γ γ r v For the rocess () to (4) we may similarly write - 4 4 γ γ r v where r v is the volume comression ratio 4 r v It follows that 4 4 and and that - - - - - - 4 4 4 η γ γ η 4 4 4 - - then v v r r Since this theoretical cycle is carried out on air for which γ.4 then the efficiency of an Otto Cycle is given by 0.4 v r η his shows that the thermal efficiency deends only on the comression ratio. If the comression ratio is increased, the efficiency is imroved. his in turn increases the temerature ratios between the two isentroic rocesses and exlains why the efficiency is imroved. D.J.Dunn www.freestudy.co.uk 5

WORKED EXAMPLE No. An Otto cycle is conducted as follows. Air at 00 kpa and 0oC is comressed reversibly and adiabatically. he air is then heated at constant volume to 500oC. he air then exands reversibly and adiabatically back to the original volume and is cooled at constant volume back to the original ressure and temerature. he volume comression ratio is 8. Calculate the following. i. he thermal efficiency. ii. he heat inut er kg of air. iii. he net work outut er kg of air. iv. he maximum cycle ressure. cv 78 kj/kg γ.4. R 87 J/kg K SOLUION Remember to use absolute temeratures throughout. Solve for a mass of kg. 0 +79K 500+777K r v 8 η r Q mc ( W in nett γ v ηq in 8 γ 9 8 0.565 0.4 ( ) or 56.5% 67. K - ) x78(77 67.) 789700 J / kg 789.7 kj / kg 0.4 0.56x789.7 446. kj / kg From the gas law we have 8 00000 x x 77 9 x 00000 x 77 x 8 4.84 MPa 9 If you have followed the rinciles used here you should be able to solve any cycle. D.J.Dunn www.freestudy.co.uk 6

SELF ASSESSMEN EXERCISE No. ake Cv 0.78 kj/kg K, R 87 J/kg K and γ.4 throughout.. An Otto cycle has a volume comression ratio of 9/. he heat inut is 500kJ/kg. At the start of comression the ressure and temerature are 00 kpa and 40oC resectively. Calculate the following. i. he thermal efficiency. (58.5%) ii. he maximum cycle temerature. (450 K). iii. he maximum ressure. (4.7 MPa). iv. he net work outut er kg of air. (9 kj/kg).. Calculate the volume comression ratio required of an Otto cycle which will roduce an efficiency of 60%. (9.88/) he ressure and temerature before comression are 05 kpa and 5oC resectively. he net work outut is 500 kj/kg). Calculate the following. i. he heat inut. (8 kj/kg). ii. he maximum temerature. ( 906 K) iii. he maximum ressure. (6.64 MPa).. An Otto cycle uses a volume comression ratio of 9.5/. he ressure and temerature before comression are 00 kpa and 40oC resectively. he mass of air used is.5 grams/cycle. he heat inut is 600 kj/kg. he cycle is erformed 000 times er minute. Determine the following. i. he thermal efficiency. (59.4%). ii. he net work outut. (4. kj/cycle) iii. he net ower outut. (05 kw). 4. An Otto cycle with a volume comression ratio of 9 is required to roduce a net work outut of 450 kj/cycle. Calculate the mass of air to be used if the maximum and minimum temeratures in the cycle are 00oC and 0oC resectively. (.5 kg). 5. he working of a etrol engine can be aroximated to an Otto cycle with a comression ratio of 8 using air at bar and 88 K with heat addition of MJ/kg. Calculate the heat rejected and the work done er kg of air. (87 kj/kg and 9 kj/kg). Now let's move on to study engines with comression ignition. D.J.Dunn www.freestudy.co.uk 7

COMPRESSION IGNIION ENGINES he invention of comression ignition engines, commonly known as diesel engines, was credited to Rudolf Diesel, although many other eole worked on similar engines. he basic rincile is that when high comression ratios are used, the air becomes hot enough to make the fuel detonate without a sark. Diesel's first engine used coal dust blasted into the combustion chamber with comressed air. his develoed into blasting in oil with comressed air. In modern engines the fuel oil is injected directly into the cylinder as fine drolets. here are two ideal cycles for these engines, the Diesel Cycle and the Dual Combustion Cycle.. DUAL COMBUSION CYCLE his is the air standard cycle for a modern fast running diesel engine. First the air is comressed isentroically making it hot. Fuel injection starts before the oint of maximum comression. After a short delay in which fuel accumulates in the cylinder, the fuel warms u to the air temerature and detonates causing a sudden rise in ressure. his is ideally a constant volume heating rocess. Further injection kees the fuel burning as the volume increases and roduces a constant ressure heating rocess. After cut off, the hot air exands isentroically and then at the end of the stroke, the exhaust valve oens roducing a sudden dro in ressure. his is ideally a constant volume cooling rocess. he ideal cycle is shown in figure 6. Fig. 6 he rocesses are as follows. - reversible adiabatic (isentroic) comression. - constant volume heating. - 4 constant ressure heating. 4-5 reversible adiabatic (isentroic) exansion. 5 - constant volume cooling. D.J.Dunn www.freestudy.co.uk 8

he analysis of the cycle is as follows. he heat is sulied in two stages hence Q in mc(4 - ) + mcv ( - ) he heat rejected is Q out mcv (5 - ) he thermal efficiency may be found as follows. Q η Q out in mc v ( ( - ) ( - ) mcv 5 - ) + mc ( 4 - ) ( - 5 ) + γ ( 4 - ) he formula can be further develoed to show that η kβ - γ γ r v [( k ) + γk( β ) ] r v is the OLUME COMPRESSION RAIO. r v / β is the CU OFF RAIO. β 4 / k is the ratio /. Most students will find this adequate to solve roblems concerning the dual combustion cycle. Generally, the method of solution involves finding all the temeratures by alication of the gas laws. D.J.Dunn www.freestudy.co.uk 9

hose requiring a detailed analysis of the cycle should study the following derivation. ( ) ( ) ( ) ( ) ( ) ( ) ( ) [ ] 4 5 4 4 5 4 4 5 4 4 4 4 4 4 5 - - - - - temeratures in the efficiency formula. Substitute for all Isentroic exansion 4 to 5 Constant ressure heating to 4 note note Constant volume heating to to Isentroic comression the temeratures in terms of Obtain all ) - ( ) - ( - + + + + γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ β γ β η β γ β β γ β η β β β β γ η v v r v r r v v in out r k k k k k k r r k k k k r r k r k r k k r Q Q Note that if β, the cycle becomes an Otto cycle and the efficiency formulae becomes the same as for an Otto cycle. D.J.Dunn www.freestudy.co.uk 0

WORKED EXAMPLE No. In a dual combustion cycle, the comression starts from bar and 0 o C. he comression ratio is 8/ and the cut off ratio is.5. he maximum cycle ressure is 60 K. he total heat inut is kj er cycle. Calculate the following. i. he thermal efficiency of the cycle. ii. he net work outut er cycle. Check that the efficiency does not contravene the Carnot rincile. SOLUION Known data. 0 +7 9 K he hottest temerature is 4 60K. β.5 r v 8 γ.4 r η - γ v k 4 β.7 kβ γ- [( k -) + γk( β -)] r [(.7 -) + (.4 x.7 x (.5 -))] η 0.68 or 68% 4 4 9 x 8 γ - 0.4 9K 60 8 K.5 v - W nett η x Q in 0.68 x 0.68 kj er cycle. he Carnot efficiency should be higher. cold 9 η 0.785 60 hot.7 x.5.4 - x 8 he figure of 0.68 is lower so the Carnot rincile has not been contravened. 0.4 D.J.Dunn www.freestudy.co.uk

WORKED EXAMPLE No. A dual combustion cycle has a comression ratio of 8/. he maximum ressure in the cycle is 9 MPa and the maximum temerature is 000oC. he ressure and temerature before comression is 5 kpa and 5oC resectively. Calculate the following. i. the cut off ratio. ii. the cycle efficiency. iii. the net work outut er kg of air. Assume γ.4 C.005 kj/kgk Cv 0.78 kj/kg K. SOLUION Known data. 98 K 4 7 K 4 9 MPa 5 kpa / / 8 98 x 8 (γ-) 947 K Cut off ratio β β 5 5 7.75 96 4 4 5 but 7 x 0.0974 6 9 x 0 x 98 x 5 x 0 γ- 0.4 4 4 4 5 4 4 x 895.6 K but 6 9 x 0 x 98 5 x 0 5 4 x 8 so β.75 0.0974 8 Q in mc(4 - ) + mcv ( - ) m kg Q in.005(74-96) + 0.78(96-947).5 kj/kg Q out mcv ( 5 - ) Q out 0.78(895.6-98) 49 kj/g 4 96K Q η Q W nett Q out in in Q out 49 0.65 or 65% 48.6 80.5 kj/kg D.J.Dunn www.freestudy.co.uk

. HE DIESEL CYCLE he Diesel Cycle recedes the dual combustion cycle. he Diesel cycle is a reasonable aroximation of what haens in slow running engines such as large marine diesels. he initial accumulation of fuel and shar detonation does not occur and the heat inut is idealised as a constant ressure rocess only. Again consider this cycle as being carried out inside a cylinder fitted with a iston. he - and -s cycles diagrams are shown in figure 7 Fig. 7 - reversible adiabatic (isentroic) comression. - constant ressure heating. - 4 reversible adiabatic (isentroic) exansion. 4 - constant volume cooling. Q η Q out in mc mc v ( - ) ( - ) 4 ( - 4 ) γ ( - ) he cycle is the same as the dual combustion cycle without the constant volume heating rocess. In this case since k the efficiency is given by the following formula. η β - γ γ r v ( β ) γ D.J.Dunn www.freestudy.co.uk

WORKED EXAMPLE No.4 An engine using the Diesel Cycle has a comression ratio of 0/ and a cut off ratio of. At the start of the comression stroke the air is at bar and 5 o C. Calculate the following. i. he air standard efficiency of the cycle. ii. he maximum temerature in the cycle. iii. he heat inut. iv. he net work outut. SOLUION Initial data. β r v 0 γ.4 c v 78 J/kg K for air 88 K bar. he maximum temerature is and the maximum ressure is and. γ β - η γ β γr ( ).4 - η ( ) x.4 x 0 v.69 η 0.647 or 64.7% x.4 x.4 Q Q W η Q W in in r nett γ- v mc ηq 88 x 0 β ( - ) 0.4 954.5 K 954. x 909 K.005(909-954.5) 959. kj nett in in 0.4 0.647 x 959. 60.6 kj D.J.Dunn www.freestudy.co.uk 4

SELF ASSESSMEN EXERCISE No. Use c v 0.78 kj/kg K, c.005 kj/kg K and γ.4 throughout.. Draw a - and - s diagram for a Diesel Cycle. he erformance of a comression ignition engine is to be comared to the Diesel cycle. he comression ratio is 6. he ressure and temerature at the beginning of comression are bar and 5oC resectively. he maximum temerature in the cycle is 00 K. Calculate the following. i. he cut off ratio.(.74) ii. he air standard efficiency. (66%). A Dual Combustion Cycle uses a comression ratio of /. he cut off ratio is /. he temerature and ressure before comression is 80 K and bar resectively. he maximum temerature 000 K. Calculate the following. i. he net work outut er cycle. (680 kj/kg). ii. he thermal efficiency. (57.6 %).. A Dual Combustion Cycle uses a comression ratio of 0/. he cut off ratio is.6/. he temerature and ressure before comression is 0oC and bar resectively. he maximum cycle ressure is 00 bar. Calculate the following. i. he maximum cycle temerature. (44 K). ii. he net work outut er cycle. (864 kj/kg). iii. he thermal efficiency. (67.5 %). D.J.Dunn www.freestudy.co.uk 5

4 GAS URBINES A gas turbine engine normally burns fuel in the air that it uses as the working fluid. From this oint of view it is an internal combustion engine that uses steady flow rocesses. Figure 8 shows a basic design. Fig.8 he air is drawn in from atmoshere and comressed. his makes it hotter. he comressed air is blown into a combustion chamber and fuel is burned in it making it even hotter. his makes the volume increase. he hot air exands out of the chamber through a turbine forcing it to revolve and roduce ower. he air becomes colder as it exands and eventually exhausts to atmoshere. he temerature dro over the turbine is larger than the temerature increase over the comressor. he turbine roduces more ower than is needed to drive the comressor. Net ower outut is the result. In the basic system, the turbine is couled directly to the comressor and the ower outut is taken from the same shaft. he ideal air standard cycle is the Joule Cycle. D.J.Dunn www.freestudy.co.uk 6

4. HE JOULE CYCLE he Joule Cycle is also known as the constant ressure cycle because the heating and cooling rocesses are conducted at constant ressure. he cycle is that used by a gas turbine engine but could conceivably be used in a closed system. We may draw the layout in block diagram form as shown in figure 9 Figure 9 here are 4 ideal rocesses in the cycle. - Reversible adiabatic (isentroic) comression requiring ower inut. P in H/s mc(-) - Constant ressure heating requiring heat inut. Φ in H/s mc(-) - 4 Reversible adiabatic (isentroic) exansion roducing ower outut. P out H/s mc(-4) 4 - Constant ressure cooling back to the original state requiring heat removal. Φ out H/s mc(4-) he ressure volume, ressure - enthaly and temerature-entroy diagrams are shown in figure 0 Fig. 0 D.J.Dunn www.freestudy.co.uk 7

he efficiency is found by alying the first law of thermodynamics. ) - ( ) - ( ) - ( ) - ( - - 4 4 mc mc P P P P in out in nett th in out out in nett nett Φ Φ Φ Φ Φ Φ η It assumed that the mass and the secific heats are the same for the heater and cooler. It is easy to show that the temerature ratio for the turbine and comressor are the same. 4 4 4 r r γ γ γ γ r is the ressure comression ratio for the turbine and comressor..4 since - - - - - - - ) - ( ) - ( -0.86 4 4 4 4 4 4 γ η η γ th th r r his shows that the efficiency deends only on the ressure ratio which in turn affects the hottest temerature in the cycle. D.J.Dunn www.freestudy.co.uk 8

WORKED EXAMPLE No. 5 A gas turbine uses the Joule cycle. he ressure ratio is 6/. he inlet temerature to the comressor is 0oC. he flow rate of air is 0. kg/s. he temerature at inlet to the turbine is 950oC. Calculate the following. i. he cycle efficiency. ii. he heat transfer into the heater. iii. he net ower outut. γ.4 C.005 kj/kg K SOLUION η r Φ η P th th in nett r -0.86 0.86 mc P Φ nett in ( 6 8 x 6 - ) 0.4 x 50.8 60. kw -0.86 0.86 0.4 47.4K or 40% 0. x.005 x ( - 47.4) 50.8 kw D.J.Dunn www.freestudy.co.uk 9

SELF ASSESSMEN EXERCISE No. γ.4 and C.005 kj/kg K throughout.. A gas turbine uses the Joule cycle. he inlet ressure and temerature to the comressor are resectively bar and -0oC. After constant ressure heating, the ressure and temerature are 7 bar and 700oC resectively. he flow rate of air is 0.4 kg/s. Calculate the following. i. he cycle efficiency. ii. he heat transfer into the heater. iii. he net ower outut. (Answers 4.7 %, 06.7 kw and 88.6 kw). A gas turbine exands draws in kg/s of air from atmoshere at bar and 0 o C. he combustion chamber ressure and temerature are 0 bar and 90oC resectively. Calculate the following. i. he Joule efficiency. ii. he exhaust temerature. iii. he net ower outut. (Answers 48. %, 67.5 K and 9 kw). A gas turbine draws in 7 kg/s of air from atmoshere at bar and 5 o C. he combustion chamber ressure and temerature are 9 bar and 850oC resectively. Calculate the following. i. he Joule efficiency. ii. he exhaust temerature. iii. he net ower outut. (Answers 46.7 %, 599 K and.96 MW) D.J.Dunn www.freestudy.co.uk 0

EDEXCEL HIGHERS ENGINEERING HERMODYNAMICS H NQF LEEL 4 OUCOME INERNAL COMBUSION ENGINE PERFORMANCE UORIAL No. 5 PERFORMANCE CHARACERISICS Internal combustion engine erformance Second law of thermodynamics: statement of law; schematic reresentation of a heat engine to show heat and work flow Heat engine cycles: Carnot cycle; Otto cycle; Diesel cycle; dual combustion cycle; Joule cycle; roerty diagrams; Carnot efficiency; air-standard efficiency Performance characteristics: engine trials; indicated and brake mean effective ressure; indicated and brake ower; indicated and brake thermal efficiency; mechanical efficiency; relative efficiency; secific fuel consumtion; heat balance Imrovements: turbocharging; turbocharging and intercooling; cooling system and exhaust gas heat recovery systems On comletion of this tutorial you should be able to do the following. Calculate the fuel ower of an engine. Calculate the brake ower of an engine. Calculate the indicated ower of an engine. Calculate the Mean Effective Pressure of an engine. Calculate the various efficiencies of an engine. Examine how the erformance of real engines may be imroved. Exlain the urose f a turbo charger. Exlain the urose of a suercharger Exlain the advantages of inter-cooling. Discuss the advantages of using waste heat recovery. Discuss the use of waster heat boilers. Discuss combined heating and ower systems. D.J.Dunn www.freestudy.co.uk

FUEL POWER (F.P.) Fuel ower is the thermal ower released by burning fuel inside the engine. F.P. mass of fuel burned er second x calorific value of the fuel. F.P. m f x C.. All engines burn fuel to roduce heat that is then artially converted into mechanical ower. he chemistry of combustion is not dealt with here. he things you need to learn at this stage follow.. AIR FUEL RAIO his is the ratio of the mass of air used to the mass of fuel burned. Air Fuel Ratio m a /m f he ideal value that just comletely burns all the fuel is called the SOICHIOMERIC RAIO. In reality, the air needed to ensure comlete combustion is greater than the ideal ratio. his deends on how efficient the engine is at getting all the oxygen to meet the combustible elements. he volume of air drawn into the engine is theoretically equal to the caacity of the engine (the swet volumes of the cylinders). he mass contained in this volume deends uon the ressure and temerature of the air. he ressure in articular deends uon the nature of any restrictions laced in the inlet flow ath.. CALORIFIC ALUE his is the heat released by burning kg of fuel. here is a higher and lower value for fuels containing hydrogen. he lower value is normally used because water vaour formed during combustion asses out of the system and takes with it the latent energy. WORKED EXAMPLE No. An engine consumes 0.057 kg/s of air. he air fuel ratio is /. he calorific value is 46 MJ/kg. Calculate the Fuel Power. SOLUION Air consumed Mass of fuel Heat released m a 0.057 kg/s. mf 0.057/ 0.00 kg/s F.P. calorific value x mf 46 000 kj/kg x 0.00 kg/s F.P. 60. KW D.J.Dunn www.freestudy.co.uk

SELF ASSESSMEN EXERCISE No.. An engine consumes 4. g/s of air with an air/fuel ratio of /. he calorific value is 45 MJ/kg. Calculate the heat released by combustion. (49 kw). An engine requires 0 kw of fuel ower by burning fuel with a calorific value of 7 MJ/kg. he air fuel/ratio required is 4/. Calculate the mass flow rate of air required. (45.4 g/s) D.J.Dunn www.freestudy.co.uk

BRAKE POWER (B.P.) Brake ower is the outut ower of an engine measured by develoing the ower into a brake dynamometer on the outut shaft. Dynamometers measure the seed and the orque of the shaft. he Brake Power is calculated with the formula B.P. πn where N is the shaft seed in rev/s is the torque in N m You may need to know how to work out the torque for different tyes of dynamometers. In all cases the torque is net brake force x radius he two main tyes are shown below. Figure Figure Figure shows a hydraulic dynamometer which absorbs the engine ower with an imeller inside a water filled casing. Basically it is a um with a restricted flow. he ower heats u the water and roduces a torque on the casing. he casing is restrained by a weight ulling down and a comression sring balance ushing down also. he torque is then (F + W) x R. Figure shows a friction drum on which a belt rubs and absorbs the ower by heating u the drum which is usually water cooled. he belt is restrained by a sring balance and one weight. the second equal weight acts to cancel out the other so the torque is given by F R. D.J.Dunn www.freestudy.co.uk 4

INDICAED POWER his is the ower develoed by the ressure of the gas acting on the istons. It is found by recording the ressure against volume inside the iston. Such diagrams are called indicator diagrams and they are taken with engine indicators. he diagram shows a tyical indicator diagram for an internal combustion engine. Figure he average force on the iston throughout one cycle is F where F MEP x Area of iston A he Mean Effective Pressure is the mean ressure during the cycle. he work done during one cycle is W F L AL L is the stroke. he number of cycles er second is N. he Indicated Power is then I.P. LAN er cylinder. Note for a 4 stroke engine N ½ the shaft seed. D.J.Dunn www.freestudy.co.uk 5

he mean effective ressure is found from the indicator diagram as follows. he area enclosed by the indicator diagram reresents the work done er cycle er cylinder. Let this area be Ad mm. he average height of the grah is H mm. he length of the diagram is Y mm. he hatched area is equal to Ad and so Ad YH H Ad/Y In order to convert H into ressure units, the ressure scale (or sring rate) of the indicator measuring system must be known. Let this be S kpa/mm. he MEP is then found from MEP S H his is also known as the Indicated Mean Effective Pressure because it is used to calculate the Indicated Power. here is also a Brake Mean Effective Pressure (BMEP) which is the mean ressure which would roduce the brake ower. BP (BMEP) LAN he BMEP may be defined from this as BMEP BP/LAN 4 EFFICIENCIES 4. BRAKE HERMAL EFFICIENCY his tells us how much of the fuel ower is converted into brake ower. η Bh B.P./F.P. 4. INDICAED HERMAL EFFICIENCY his tells us how much of the fuel ower is converted into brake ower. η Ih I.P./F.P. 4. MECHANICAL EFFICIENCY his tells us how much of the indicated ower is converted into brake ower. he difference between them is due to frictional losses between the moving arts and the energy taken to run the auxiliary equiment such as the fuel um, water um, oil um and alternator. η mech B.P./I.P. D.J.Dunn www.freestudy.co.uk 6

WORKED EXAMPLE No. A 4 cylinder, 4 stroke engine gave the following results on a test bed. Shaft Seed orque arm Net Brake Load F 00 N Fuel consumtion mf g/s Calorific value Area of indicator diagram Pressure scale Stroke Bore Base length of diagram N 500 rev/min R 0.4 m C.. 4 MJ/kg Ad 00 mm S 80 kpa/mm L 00 mm D 00 mm Y 60 mm. Calculate the B.P., F.P., I.P., MEP, η Bh, η Ih,and η mech, SOLUION BP πn π x (500/60)x (00 x 0.4) 0.94 kw FP mass/s x C.. 0.00 kg/s x 4 000 kj/kg 84 kw IP LAN MEP Ad/Y x S (00/60) x 80 400 kpa IP 400 x 0. x (π x 0./4) x (500/60)/ er cylinder IP 6.54 kw er cylinder. For 4 cylinders IP 6.54 x 4 6.8 kw η Bh 0.94/84 4.9% η Ih 6.8/84. % η mech 0.94/6.8 80% D.J.Dunn www.freestudy.co.uk 7

SELF ASSESSMEN EXERCISE No.. A 4 stroke sark ignition engine gave the following results during a test. Number of cylinders 6 Bore of cylinders 90 mm Stroke 80 mm Seed 5000 rev/min Fuel consumtion rate 0.5 kg/min Calorific value 44 MJ/kg Net brake load 80 N orque arm 0.5 m Net indicated area 70 mm Base length of indicator diagram 60 mm Pressure scale 40 kpa/mm Calculate the following. i. he Brake Power. (47. kw) ii. he Mean effective Pressure. (480 kpa) iii. he Indicated Power. (6 kw) iv. he Mechanical Efficiency. (77.%) v. he Brake hermal efficiency. (8.6%). A two stroke sark ignition engine gave the following results during a test. Number of cylinders 4 Bore of cylinders 00 mm Stroke 00 mm Seed 000 rev/min Fuel consumtion rate 5 g/s Calorific value 46 MJ/kg Net brake load 500 N orque arm 0.5 m Net indicated area 500 mm Base length of indicator diagram 66 mm Pressure scale 5 kpa/mm Calculate the following. i. he Indicated thermal efficiency. (5.9%) ii. he Mechanical Efficiency. (88%) iii. he Brake hermal efficiency. (.8%) D.J.Dunn www.freestudy.co.uk 8

. A two stroke diesel engine gave the following results during a test. Number of cylinders 4 Bore of cylinders 80 mm Stroke 80 mm Seed 00 rev/min Fuel consumtion rate.6 cm/s Fuel density 750 kg/m Calorific value 60 MJ/kg Nett brake load 95 N orque arm 0.4 m Nett indicated area 00 mm Base length of indicator diagram 40. mm Pressure scale 50 kpa/mm Calculate i. he Indicated thermal efficiency. (0.5%) ii. he Mechanical Efficiency. (8.7%) iii. he Brake hermal efficiency. (5%) 4. A four diesel engine gave the following results during a test. Number of cylinders 4 Bore of cylinders 90 mm Stroke 80 mm Seed 5 000 rev/min Fuel consumtion rate 0.09 kg/min Calorific value 44 MJ/kg Nett brake load 60 N orque arm 0.5 m MEP 80 kpa Calculate the following. i. he Mechanical Efficiency. (66.%) ii. he Brake hermal efficiency. (.8%) iii. he Indicated hermal Efficiency. (6%) 5. D.J.Dunn www.freestudy.co.uk 9

PERFORMANCE IMPROEMEN 5. INERNAL COMBUSION ENGINES In the receding work it has been shown that the efficiency of internal combustion engines deends uon the volume comression ratio and for gas turbines it deends uon the ressure comression ratio. his section debates some of the ractical roblems and solutions for imrovement of erformance. 5.. ENGINE MANAGEMEN High comression ratios in sark ignition engines leads to re-ignition as the fuel detonates without the aid of a sark before the oint of maximum comression. his roduces very high eaks of ressure and damages the engine. Reducing this roblem involves the use of fuels that are less rone to detonate (high octane ratings). iming of the sark ignition is also vital. he correct timing deends uon many factors such as air/fuel ratio and engine load. Modern engines use fuel management systems in which the timing and the air/fuel ratio are controlled by a comuter connected to sensors. his allows greater comression ratios and hence efficiency. Fuel injection gives a measure of control over the combustion rocess and this is now ossible with etrol engines as well as diesel engines. 5.. URBO CHARGING he ower roduced by an engine basically deends on the amount of fuel burned. his is limited by the mass of air in the cylinder. o burn more fuel requires more air. Blowing air into the cylinders under ressure may do this and requires an air blower. his is a successful rocess in comression ignition engines but increases the roblem of re-ignition on sark ignition engines. he blower may be driven by a mechanical connection direct to the engine crankshaft. he Lobe comressor shown in figure 4 is commonly used. his arrangement is called SUPERCHARGING. On large engines, the blower is driven by a small gas turbine that uses the exhaust from the engine to ower it. his is called URBO CHARGING. Figure 5 shows a turbo charger. Fig. 4 D.J.Dunn www.freestudy.co.uk 0

5.. INER-COOLING Fig. 5 he mass contained in a volume of air deends uon the temerature. he colder the air, the more mass it contains. Comressed air is naturally hot so if it can be cooled after comression, a greater mass of air may be sulied to the cylinder. urbo charging and inter-cooling on large comression ignition engines leads to imroved efficiency as well as increased ower. Fig.6 shows an intercooler designed to fit under a car radiator. Fig. 6 D.J.Dunn www.freestudy.co.uk

6 EXHAUS GAS HEA RECOERY When large amounts of hot exhaust gas is roduced, by either gas turbines or large diesel engines, the heat in the exhaust gas may be recovered for useful alications such as using it to roduce hot water or steam in a boiler. A factory might well use a gas turbine to roduce electric ower and hot water or steam. his is more economical than buying electricity. Fig. 7 When the generation of electric ower is combined with a waste heat system, it is called COMBINED POWER AND HEAING. On a large scale, this is sometimes done on major ower stations. he enormous quantities of waste heat roduced in the form of hot water from the condensers may be umed through a hot water ie system to heat buildings or large greenhouses. D.J.Dunn www.freestudy.co.uk

SELF ASSESSMEN EXERCISE No. A factory is to be built that uses both electricity and steam. here are two roosals to be considered. PROPOSAL Produce steam in an oil fired boiler and urchase electricity. PROPOSAL Generate electric ower with a gas turbine and roduce steam in a waste heat boiler using the exhaust gas. OPERAING DAA FOR SEAM BOILER Mass Flow rate kg/s Steam condition 5 bar and dry saturated. Feed water temerature 5 o C. When burning fuel, the combustion efficiency is tyically 85% When using exhaust gas, the heat transfer from the gas may be assumed to be equal to the heat gained by the water and steam. he exhaust gas is cooled to 00 o C before leaving the boiler. GAS URBINE DAA Pressure ratio 7 Inlet air ressure bar Inlet air temerature 5 o C Combustion chamber temerature 500 o C FUEL DAA Any fuel to be burned in either the gas turbine or the boiler will be light oil with a calorific value of 4 MJ/kg. he cost of fuel is.7 ence er kg. Electricity cost.5 ence er kwhr ( kwhr 600 kj) PROPERIES AIR C.005 kj/kg K γ.4 BURNED GAS C. kj/kg K γ. D.J.Dunn www.freestudy.co.uk

Produce a reort comaring the costs for both schemes. You will need to do the following tasks. GUIDANCE In the course of your work you will need to do the following. SEAM BOILER You will need to determine the following. i. he energy required to make the steam. ii. he fuel required in kg/s. iii. he mass of exhaust gas required to roduce the same steam in kg/s. GAS URBINE You will need to equate the heat transfer from burning fuel to the energy required to raise the temerature in the combustion chamber. You will need to determine the following. vi. he mass flow of air. v. he fuel burned in kg/s. vi. he Power inut of the comressor. vii. he ower outut of the turbine. viii. he net ower for generating electricity. COSING Base t he cost of otion on the cost of fuel lus the cost of buying the same electricity as for otion. Base the cost on the cost of fuel only. What other factors would you consider when making a decision on which otion take? D.J.Dunn www.freestudy.co.uk 4

EDEXCEL HIGHERS ENGINEERING HERMODYNAMICS H NQF LEEL 4 OUCOME UORIAL No. 6 - RECIPROCAING AIR COMPRESSORS Air comressors Proerty diagrams: theoretical ressure-volume diagrams for single and multistage comressors; actual indicator diagrams; actual, isothermal and adiabatic comression curves; induction and delivery lines; effects of clearance volume Performance characteristics: free air delivery; volumetric efficiency; actual and isothermal work done er cycle; isothermal efficiency First law of thermodynamics: inut ower; air ower; heat transfer to intercooler and aftercooler; energy balance Faults and hazards: effects of water in comressed air; causes of comressor fires and exlosions In order to comlete this section you should be familiar with gas laws and olytroic gas rocesses. You will study the rinciles of recirocating comressors in detail and some rinciles of rotary comressors. On comletion you should be able to do the following. Describe the working rinciles of recirocating comressors. Describe the basic design of various other comressors. Define and calculate swet volume. Define and calculate volumetric efficiency. Define and calculate isothermal efficiency. Define and calculate indicated ower. State the benefits of cooling. Calculate the heat rejected through cooling. Define and calculate the inter-stage ressures for multile comressors. D.J.Dunn www.freestudy.co.uk

INRODUCION Air is an exansive substance and dangerous when used at high ressures. For this reason, most alications are confined to things requiring low ressures (0 bar or lower) but there are industrial uses for high ressure air u to 00 bar. he common source of the air is the comressor. here are many tyes of comressors with different working rinciles and working conditions. his is a list of the main tyes. Recirocating. Sliding vane comressors. Lobe comressors. Helical screws. Centrifugal. Axial turbine comressors. he function of all of them is to draw in air from the atmoshere and roduce air at ressures substantially higher. Usually a storage vessel or receiver is used with the comressor. Comressed air has many alications. It is also used for owering neumatically oerated machines. It is used as a ower medium for worksho tools such as shown in Fig... COMPRESSED AIR Fig.. AMOSPHERIC APOUR Water vaour in the atmoshere has imortant consequences on comressors. Atmosheric air contains WAER APOUR mixed with the other gases. he ratio of the mass of water vaour in the air to the mass of the air is called the ABSOLUE HUMIDIY. he quantity of water that can be absorbed into the air at a given ressure deends uon the temerature. he hotter the air, the more water it can absorb. When the air contains the maximum ossible amount of vaour it is at its dew oint and rain or fog will aear. he air is then said to have 00% humidity. When the air contains no water vaour at all (dry air), it has 0% humidity. his is called the RELAIE HUMIDIY. For examle if the air has 40% relative humidity it means that it contains 40% of the maximum that it could contain. here are various ways to determine the humidity of air and instruments for doing this are called HYGROMEERS. D.J.Dunn www.freestudy.co.uk

he imortance of humidity to air comressors is as follows. When air is sucked into the comressor, it brings with it water vaour. When the air is comressed the ressure and the temerature of the air goes u and the result is that the comressed air will have a relative humidity of about 00% and it will be warm. When the air leaves the comressor it will cool down and the water vaour will condense. Water will then clog the comressor, the receiver and the ies. Water damages air tools; ruins aint srays, and corrode ies and equiment. For this reason the water must be removed and the best way is to use a well designed comressor installation and distribution network.. YPICAL RECIPROCAING COMPRESSOR LAYOU Figure shows the layout of a two stage recirocating comressor tyically for sulying a worksho. Fig.. Induction box and silencer on outside of building with course screen.. Induction filter.. Low ressure stage. 4. Intercooler. 5. High ressure stage. 6. Silencer. 7. Drain tra. 8. After cooler 9. Pressure gauge. 0. Air receiver.. Safety ressure relief valve.. Sto valve D.J.Dunn www.freestudy.co.uk

. HAZARDS AND SAFEY Dangers associated with air comressors are as follows. Pressure vessels may ruture. Oil leaks may burn or cause other accidents. Oil in the comressed air may exlode. Water in the comressed air may damage equiment. here are many regulations concerning the use, maintenance and insection of ressure vessels. essels must have the safe working ressure marked on them. hey must have a ressure gauge and be fitted with an isolating valve. hey must also be fitted with a ressure release valve to revent overressure. In articular, with recirocating comressors, if water accumulates in the cylinder, it may fill the sace so comletely that it revents the iston reaching the end of its travel and cause damage to the iston and head. Oil in the cylinder can exlode during comression. Normally the oerating ressure is not high enough to roduce the temerature required. However, if the outlet becomes blocked (e.g. the valve sticks or the outlet ie is closed with an isolating valve), then the danger exists. Oil or water in the air can also cause damage when sulied to some kinds of tools. For this reason a good installation fully conditions the air to remove water, dirt and oil. he following is a list of recautions to be taken against fires and exlosions. Avoid overheating. Kee discharge temeratures within the recommended limits. Kee the deosit formation to a minimum by using the correct lubricant. Ensure efficient filtration of the air. his reduces wear on the valves and istons and reduces deosits of carbonised articles. Avoid over-feeding oil to the cylinders. Minimise the carry over of oil between stages. Avoid high temeratures and low airflow when idling. Kee the coolers in good condition. Do not use solvents for cleaning or use anywhere near to an installation as the vaour given off can ignite. Do not allow naked flames (e.g. smoking) near to an installation when it is oened. D.J.Dunn www.freestudy.co.uk 4

.4 FREE AIR DELIERY When a gas such as air flows in a ie, the mass of the air deends uon the ressure and temerature. It would be meaningless to talk about the volume of the air unless the ressure and temerature are considered. For this reason the volume of air is usually stated as FREE AIR DELIERY or FAD. In other words FAD refers to the volume the air would have if let out of the ie and returned to atmosheric ressure and temerature. he FAD is also the volume of air drawn into a comressor from the atmoshere. After comression and cooling the air is returned to the original temerature but it is at a higher ressure. Suose atmosheric conditions are aa and a (the FAD) and the comressed conditions are, and. Alying the gas law we have a a a a a a F. A. D D.J.Dunn www.freestudy.co.uk 5

. CYCLE FOR RECIPROCAING COMPRESSOR. HEOREICAL CYCLE Figure shows the basic design of a recirocating comressor. Fig. Figure 4 shows the ressure volume diagram for an ideal recirocating comressor. Fig.4 he iston recirocates drawing in gas, comressing it and exelling it. If the iston exels all the air and there is no restriction at the valves, the ressure - volume cycle is as shown. Gas is induced from 4 to at the inlet ressure. It is then traed inside the cylinder and comressed according the law n C. At oint the ressure reaches the same level as that in the delivery ie and the outlet valve os oen. Air is then exelled at the delivery ressure. he delivery ressure might rise very slightly during exulsion if the gas is being comacted into a fixed storage volume. his is how ressure builds u from switch on. D.J.Dunn www.freestudy.co.uk 6

. OLUMERIC EFFICIENCY In reality, the iston cannot exel all the gas and a clearance volume is needed between the iston and the cylinder head. his means that a small volume of comressed gas is traed in the cylinder at oint. When the iston moves away from the cylinder head, the comressed gas exands by the law n C until the ressure falls to the level of the inlet ressure. At oint 4 the inlet valve oens and gas is drawn in. he volume drawn in from 4 to is smaller than the swet volume because of this exansion. Fig. 5 he volumetric efficiency is defined by the following equation. η vol Induced olume Swet olume It may be shown that this reduces to η vol ( / n) c his efficiency is made worse if leaks occur ast the valves or iston. he clearance ratio is defined as c Clearance volume/swet volume. Ideally the rocess to and 4 to are isothermal. hat is to say, there is no temerature change during induction and exulsion. D.J.Dunn www.freestudy.co.uk 7

WORKED EXAMPLE No. Gas is comressed in a recirocating comressor from bar to 6 bar. he FAD is dm /s. he clearance ratio is 0.05. he exansion art of the cycle follows the law. C. he crank seed is 60 rev/min. Calculate the swet volume and the volumetric efficiency. SOLUION Swet olume Clearance volume 0.05 Consider the exansion from to 4 on the - diagram. 4 bar 6 bar.. 4 4. 6(0.05). (4. ) 4 0. or.%% of F.A.D. 0.0 m/s. + 0.05.05 Induced volume - 4.05-0. 0.88 Induced volume 0.0 m/s 0.0/0.88 0.057 m/s Crank seed 6 rev/s so the swet volume 0.057/6.6 dm. η vol Induced olume Swet olume η vol 0.88 8.8 % D.J.Dunn www.freestudy.co.uk 8

WORKED EXAMPLE No. Show that the volumetric efficiency of an ideal single stage recirocating comressor with a clearance ratio is c is given by the exression below. ) / ( n L H vol c η L is the inlet ressure and H the outlet ressure. SOLUION Swet volume - Induced volume - 4 Clearance volume ( ) ( ) ( ) + + + + + Substitute (B) Substitute (A) for the bottom line. )...( )...( / / / / 4 4 4 4 4 4 4 n L H vol n L H vol n L H n vol vol vol c c c c c c c c c c B c c c c A c c η η η η η D.J.Dunn www.freestudy.co.uk 9

WORKED EXAMPLE No. A single stage recirocating comressor has a clearance volume of 0 cm. he bore and stroke are 00 mm and 80 mm resectively. he comression and exansion rocesses have a olytroic index of.5. he inlet and outlet ressures are and 6 bar resectively. Determine the volumetric efficiency. SOLUION he swet volume is the roduct of stroke and bore area. π x 00 S.. 80 x 68. x 0 mm or 68. cm 4 he Clearance volume is 0 cm. c η η vol vol 0 68. 0.08 H c L 0.08 / n 6 0.08 /.5 { 4.9 } 0.06 0.898 or 89.8%. REAL - DIAGRAMS In real comressors the warm cylinder causes a slight temerature rise over the induction from 4 to. he gas is restricted by the valves and is slightly less than 4. he valves also tend to move so the real cycle looks more like figure 6 Fig. 6 D.J.Dunn www.freestudy.co.uk 0

WORKED EXAMPLE No.4 A single stage recirocating comressor roduces a FAD of dm /s at 40 rev/min. he inlet ressure is bar. he olytroic index is. for the comression and exansion. he outlet ressure is 8 bar. he clearance volume is 0 cm. Determine the volumetric efficiency. SOLUION First find the induced volume.his is the free air drawn in for each revoution. x60 F. A. D. er rev. 0.857 dm /rev 40 his is the induced volume he clearance volume is Next we need to find 8 x 0.0 0.857 + 0.0566 0.4m Swet volume η vol n. x 4 4 n 4. 4 4 hence 0 cm 0. 0.0566 dm dm or 0.0dm 0.857 0.859 or 85.9% 0. 4 4..4 INDICAED POWER he indicated work er cycle is the area enclosed by the - diagram. he easiest way to find this is by integrating with resect to the ressure axis. Fig.7 D.J.Dunn www.freestudy.co.uk

.4. POLYROPIC PROCESSES If the rocesses to and to 4 are olytroic n C. C /n -/n he work done is given by d W Consider the exression ( ) ) ( ) ( substitute in and c C ) / ( / / n n / / / / / / n n n n d C n n n C d C d n n n n n n n n n Between the limits of and this becomes [ ] ) (n n W - Between the limits 4 and this becomes [ ] ) ( W 4 4 4- n n Subtract one from the other to find the indicated work. [ ] [ ] ) ( ) ( 4 4 n n n n W [ ] [ ] [ ] r n n mr W R m he corresonding induced mass is the induced volume. is where r n n W r n n W since r n n W r r n n W n n W and the relationshis Substitute n n W n n n n 4 n n 4 4 4 n n n n 4 4 n n n n 4 4 4 n n 4 4 4 4 4 4 n n D.J.Dunn www.freestudy.co.uk

If the clearance volume is ignored. n n n W r or n W n mr r n n n.4. ISOHERMAL PROCESSES If the rocesses to and to 4 are isothermal then C. C - he work done is given by W d Consider the exression d C d C ln Between the limits of and this becomes Between the limits 4 and this becomes ln 44 ln he indicated work (inut) is then W ln W ln since W ln W W ln ( r ) ln( r ) ( r )( ) 44ln 4 4 ln( r )( 4 ) ln( r ) ( r ) mr 4 4 4 we get the following. W ln( r )( ) If the clearance volume is neglected W ln ( r ) mr m is the mass induced and exelled each cycle and W is the indicated work er cycle. he indicated ower is found by multilying W by the strokes er second. I.P. W x N where N is the shaft seed in Rev/s D.J.Dunn www.freestudy.co.uk

.5 ISOHERMAL EFFICIENCY he minimum indicated ower is obtained when the index n is a minimum. he ideal comression is hence isothermal with n. he isothermal efficiency is defined as ( )( ) [ ] ( ) ( ) r r n - ln r n n r ln POLYROPIC WORK ISOHERMAL WORK n n n n n iso iso iso η η η D.J.Dunn www.freestudy.co.uk 4

WORKED EXAMPLE No.5 A single stage recirocating comressor draws in air at atmosheric ressure of.0 bar and delivers it at 9.5 bar. he olytroic index is.8 for the comression and exansion. he swet volume is.5 dm and the clearance volume is 0.0 dm. he seed is 500 rev/min. Determine the following. i. he volumetric efficiency. ii. he free air delivery. iii. he indicated ower. iv. he isothermal efficiency. SOLUION clearance volume 0. he clearance ratio c 0.0667 swet volume.5 Pressure ratio r η v - c r n η - 0.0667 v 9.406 Induced olume η x Swet olume 0.6x.5 0.95 dm FAD er stroke Induced volume 0.98 dm FAD er minute 0.95 x 500 465.8 dm n n Indicated Power r n 5.8 I.P..0x 0 9.406 0.8 I. P. 54. 9.406 0.55 [ ] [ ] 9.5.0 9.406-0.6 n 0.8.8 I. P. 54..407-096 Watt v.8 x FAD x / min 0.4658 60 or 0.4658 m / s η iso ( n -) ln( r ) n r n n 0.8ln9.406 0.8.8.8 9.406 0.49 0.407 0.84 D.J.Dunn www.freestudy.co.uk 5

SELF ASSESSMEN EXERCISE No.. A recirocating air comressor oerates between bar and 8 bar. he clearance volume is 5 cm and the swet volume is 900 cm. he index of comression and exansion is.. Calculate the following. i. he ideal volumetric efficiency. (9.4%) ii. he ideal indicated work er cycle. (08. J) iii. he Isothermal work er cycle. (7.9 J) iv. he Isothermal efficiency. (8%). A recirocating air comressor following the ideal cycle has a free air delivery of 60 dm /s. he clearance ratio is 0.05. he inlet is at atmosheric ressure of bar. he delivery ressure is 7 bar and the comression is olytroic with an index of.. Calculate the following. i. he ideal volumetric efficiency. (8.7%) ii. he ideal indicated ower. (4.74 KW) iii. he Isothermal efficiency. (79. %). A single stage recirocating comressor draws in air at atmosheric ressure of.0 bar and delivers it at bar. he olytroic index is. for the comression and exansion. he swet volume is.0 dm and the clearance volume is 0.6 dm. he seed is 600 rev/min. Determine the following. i. he ideal volumetric efficiency. (45.6%) ii. iii. he free air delivery. (547.6 dm /min) he ideal indicated ower. (.8 kw) iv. he Isothermal efficiency. (80%) D.J.Dunn www.freestudy.co.uk 6

MULIPLE SAGE COMPRESSORS he main advantage to comressing the air in stages is that the air may be cooled between each stage and the overall comression is nearer to being isothermal. his reduces the ower requirement and allows removal of water from the air. wo stage comressions are common but when very high ressure is required, more stages may be used.. HE EFFEC OF INER-COOLING ON HE INDICAED WORK Consider the - diagram for a comressor with two stages. Fig.8 he cycle to 4 is a normal cycle conducted between L and M. he air is exelled during rocess to 4 at M and constant temerature. he air is then cooled at the intermediate ressure and this causes a contraction in the volume so that the induced volume 8 to 5 is smaller than the exelled volume to. he high ressure cycle is then a normal cycle conducted between M and H. he shaded area of the diagram reresents the work saved by using the intercooler. he otimal saving is obtained by choosing the correct intermediate ressure. his may be found as follows. D.J.Dunn www.freestudy.co.uk 7

. OPIMAL INER-SAGE PRESSURE W W + W where W is the work done in the low ressure stage and W is the work done in the high ressure stage. ( ) ( ) + + -) ( n -) ( n n for each stage then assuming the same value of Since (n -) ) - mrn( (n -) ) - mrn( W / 5 6 6 / (-/n) 5 6 5 6 (-/n) 5 6 n n n mr n mr W and ( ) ( ) ( ) ( ) ( ) n n M n n H n M n n L n M H n L M L H m mr mr dw n mr n mr W / / 5 / / M M / 6 / 6 5 - d and equate to zero. W we differentiate with resect to minimum value of For a -) ( n -) ( n and and Since + If the intercooler returns the air to the original inlet temerature so that 5, then equating to zero reveals that for minimum work M ( L H ) / It can further be shown that when this is the case, the work done by both stages is equal. When K stages are used, the same rocess reveals that the minimum work is done when the ressure ratio for each stage is ( L / H ) /K D.J.Dunn www.freestudy.co.uk 8

4 ROARY COMPRESSORS Figure 9 shows three tyes of rotary comressors. From left to right - the vane tye, centrifugal tye and axial flow tye. Figure 0 shows a screw comressor. Fig. 9 4. ANE YPE he vanes fit in slots in the rotor. he rotor is eccentric to the bore of the cylinder. When the rotor is turned, centrifugal force throws the vanes out against the wall of the cylinder. he sace between the vanes grows and shrinks as the rotor turns so if inlet and outlet assages are cut in the cylinder at the aroriate oint, air is drawn in, squeezed and exelled. his tye of comressor is suitable for small ortable alications and is relatively chea. ane comressors often use oil to lubricate and cool the air and a system similar to that shown for the screw comressor (figure 0) is used. 4. CENRIFUGAL YPE he rotor has a set of vanes shaed as shown in the diagram. When the rotor sins, the air between the vanes is thrown outwards by centrifugal force and gathered inside the casing. As the air slows down in the casing, the kinetic energy is converted into ressure. he shae of the casing is imortant and is basically an eccentric assage surrounding the rotor edge. Fresh air is drawn in from the front of the rotor. hese tyes of comressor are suitable for medium and large flow rates. Pressure u to 5 bar may be obtained by using several stages or using them as the second stage of an axial flow tye. ery large comressors are used to suly large volumes of air at low ressure to combustion chambers and blast furnaces. 4. AXIAL FLOW YPE he axial flow comressor is basically many rows of fan blades arranged along the axis. Each row gives the air kinetic energy. Not shown on the diagram are fixed vanes in between each row that slows the air down again and raises the ressure. In this way the ressure gradually increases as the air flows along the axis from inlet to outlet. Often a centrifugal stage is situated at the end in order to give it a boost and change the direction of flow to the side. yical industrial comressors can rovide 70 m /s at 5 bar. hey are not suitable for small flow rates (below 5 m /s). Axial flow comressors are commonly used in jet and gas turbine engines but they have many alications where large flow rates and medium to high ressure are required. D.J.Dunn www.freestudy.co.uk 9

4.4 SCREW YPES wo rotors have helical lobes cut on them in such a way that when they mesh and rotate in oosite directions, air is drawn along the face of the lobes from inut to outut. Oil is used liberally to seal the air. he oil also acts as a coolant and the diagram shows how the oil and air are searated and then cooled in a radiator. he oil is re-circulated. 4.5 LOBE YPES Fig. 0 Lobe comressors are commonly used as suerchargers on large engines. Figure shows the basic design. Air is carried around between the lobes and the outer wall and is exelled when the lobes come together. hese comressors are not suitable for high ressures but flow rates around 0 000 m /hr are achievable. Figure D.J.Dunn www.freestudy.co.uk 0

5. COOLERS Coolers are used for the following reasons. o reduce the indicated work in multile stage comressors. o condense water from the air. For recirocating comressors, the cooling takes lace in the following laces. he cylinder. Between stages. After final comression. 5. CYLINDER COOLING he cylinders may be cooled with air and are designed with cooling fins on the outside. Circulating water through a cooling jacket roduces more effective cooling. 5. INER COOLING hese are usually simle heat exchangers with water-cooling. Drain tras are fitted to them to remove the water that condenses out of the air. Indicated work is saved as exlained in section. Assuming that no heat is lost to the surroundings, the st. Law may be alied to the air and water side to roduce the following heat balance. Φ m a c a a m w c w w m a is the mass flow rate of air. m w is the mass flow rate of water. a is the temerature change of the air. w is the temerature change of the water. c a is the secific heat caacity of air. c w is the secific heat caacity of water. 5. AFER COOLING he only urose of an after cooler is to cool the air to around ambient conditions and condense water from the air. his is usually another water-cooled heat exchanger and the same heat balance may be alied. D.J.Dunn www.freestudy.co.uk

WORKED EXAMPLE No.6 A single acting recirocating comressor runs at 60 rev/min and takes in air at bar and 5 o C and comresses it in stages to 64 bar. he free air delivery is 0.0566 m /s. here is an intercooler between each stage, which returns the air to 5 o C. Each stage has one iston with a stoke of 00 mm. Calculate the following. i. he ideal ressure between each stage. ii he ideal indicated ower er stage. iii he heat rejected from each cylinder. iv he heat rejected from each intercooler. v he isothermal efficiency. vi he swet volume of each stage. vii he bore of each cylinder. Ignore leakage and the effect of the clearance volume. he index of comression is. for all stages. SOLUION Pressure ratio for each stage (64/) / 4 Hence the ressure after stage is x 4 4 bar. he ressure after the second stage is 4 x 4 6 bar he final ressure is 6 x 4 64 bar. 88 K. m /R x 0 5 x 0.0566/(87 x 88) 0.06847 kg/s 88(4) 0./. 96.5 K he indicated ower for each stage is the same so it will be calculated for the st. stage. I.P. mrn {(/) (-/n) -}/(n-) since m is the mass comressed. I.P. 0.06847 x 87 x. x 88{4 0./. - } /(.-) 9.46 kw D.J.Dunn www.freestudy.co.uk

CYLINDER COOLING Consider the energy balance over the first stage. Balancing the energy we have Fig. HA + P(in) HB + Φ (out) Φ (out) P(in) - mc(b - A) Φ (out) 9.46-0.06847 x.005 (96.5-88) Φ (out).78 kw (rejected from each cylinder) INERCOOLER Now consider the Intercooler. No work is done and the temerature is cooled from to 5. Φ (out) mc( C - D ) 0.0687 x.005 (96.5-88) 7.49 kw Fig. D.J.Dunn www.freestudy.co.uk

ISOHERMAL EFFICIENCY he ideal isothermal ower mrln(/) er stage. P(isothermal) 0.06847 x 87 x 88 ln4 7.846 kw η (iso) 7.846/9.46 84.9 % SWEP OLUMES Consider the first stage. he F.A.D. is 0.0566 m /s. In the ideal case where the air is drawn in at constant temerature and ressure from the atmoshere, the FAD is given by FAD Swet olume x Seed and the seed is 6 rev/s If the clearance volume is ignored, the FAD gives the swet volume. S.. (st. Stage) 0.0566/6 0.0094 m S.. Bore Area x Stroke 0.0094 π D /4 x 0. D 0.47 m. Now consider the second stage. he air is returned to atmosheric temerature at inlet with a ressure of 4 bar. he volume drawn is hence /4 of the original FAD. he swet volume of the second stage is hence 0.0094/4 0.006 m. 0.006 π D /4 x 0. hence D 0.7 m By the same reasoning the swet volume of the third stage is S(rd stage) 0.0094/6 0.000589 m. 0.000589 π D /4 x 0. D 0.0866 m D.J.Dunn www.freestudy.co.uk 4

SELF ASSESSMEN EXERCISE No.. A two stage comressor draws in 8 m /min from atmoshere at 5 o C and.0 bar. he air is comressed with an index of comression of.7 to the interstage ressure of 6 bar. he intercooler must cool the air back to 5 o C. Look u the aroriate mean value of c in your tables. Calculate the heat that must be extracted from the air in kw. (.8 kw). A single acting stage comressor draws in 8.5 m /min of free air and comresses it to 40 bar. he comressor runs at 00 rev/min. he atmosheric conditions are.0 bar and 5 o C. here is an intercooler between stages that cools the air back to 5 o C. he olytroic index for all comressions is.. he volumetric efficiency is 90% for the low ressure stage and 85% for the high ressure stage. Calculate the following. i. he intermediate ressure for minimum indicated work. (6.7 bar) ii. he theoretical indicated ower for each stage. (.8 kw) iii. he heat rejected in each cylinder. (6. kw for both) iv. he heat rejected by the intercooler. (6.5 kw) v. he swet volumes of both stages. (. and 5.4 dm ). A two stage recirocating air comressor works between ressure limits of and 0 bar. he inlet temerature is 5oC and the olytroic index is.. Intercooling between stages reduces the air temerature back to 5oC. Both stages have the same stroke. Neglect the effect of the clearance volume. Calculate he following. i. he free air delivery for each kwh of indicated work. (0.06 m ) ii. he mass of air that can be comressed for each kw h of indicated work. (.7 kg) iii. he ratio of the cylinder diameters. (.5) (Note kw h is.6 MJ) D.J.Dunn www.freestudy.co.uk 5

4. A single acting comressor draws in atmosheric air at bar and 5 o C. he air is comressed in two stages to 9 bar. he comressor runs at 600 rev/min. he installation has an inter-cooler that reduces the air temerature to 0 o C at inlet to the second stage. he olytroic index for all comressions is.8. he clearance volume for each stage is 4% of the swet volume. he low ressure cylinder is 00 mm diameter and the stroke for both stages is 60 mm. Calculate the following. i. he otimal inter-stage ressure. ( bar) ii. he volumetric efficiency of each stage. (98.9)% iii. he free air delivery. (7.77 m /min) iv. he induced volume of the high ressure stage. (4.545 dm ) v. he diameter of the high ressure cylinder. (9. mm) vi. he indicated ower for each stage. (6 and 6.9 kw) D.J.Dunn www.freestudy.co.uk 6

EDEXCEL HIGHERS ENGINEERING HERMODYNAMICS H NQF LEEL 4 OUCOME 4 SEAM AND GAS URBINE POWER PLAN UORIAL No. 7 - URBINE HEORY Steam and gas turbine Princiles of oeration: imulse and reaction turbines; condensing; ass-out and back ressure steam turbines; single and double shaft gas turbines; regeneration and re-heat in gas turbines; combined heat and ower lants Circuit and roerty diagrams: circuit diagrams to show boiler/heat exchanger; suerheater; turbine; condenser; condenser cooling water circuit; hot well; economiser/feedwater heater; condensate extraction and boiler feed ums; temerature - entroy diagram of Rankine cycle Performance characteristics: Carnot, Rankine and actual cycle efficiencies; turbine isentroic efficiency; ower outut; use of roerty tables and enthaly-entroy diagram for steam When you have comleted this short tutorial, you should be able to exlain the basic theory and design rinciles of turbines used in steam and gas turbine cycles. D.J.Dunn www.freestudy.co.uk

. URBINES DESIGN. A BRIEF HISORY 0 B.C. Hero of Alexandria constructs a simle reaction turbine. his was constructed from a sherical vessel with two souts as shown. Heat turned the water inside into steam that escaed through the souts and made the vessel rotate. Figure Hero s urbine 69 Branca, an Italian, created the first imulse turbine. Steam issuing from a nozzle struck the vanes on a wheel and made it revolve. Figure Branca s urbine Windmills, develoed in medieval times formed the main source of ower for centuries. 884 Charles Parsons develoed the first ractical reaction turbine. his machine develoed around 7 kw of ower. 889 De Laval develoed the first ractical imulse turbine caable of roducing around kw of ower. Others who develoed the imulse turbine were Rateau in France and Curtis in the U.S.A. D.J.Dunn www.freestudy.co.uk

. IMPULSE HEORY urbines are generally classified as either imulse or reaction. his refers to the tye of force acting on it and causing it to rotate. IMPULSIE FORCES are exerted on an object when it diverts or changes the flow of a fluid assing over it. A very basic imulse turbine is the windmill and this converts the kinetic energy of the wind into mechanical ower. Consider a rotor with vanes arranged around the edge. Fluid is directed at the vanes by a set of nozzles. Fig. If there is no ressure dro in the rocess, the resulting force on the vane is entirely due to the change in the momentum of the fluid and the force is entirely imulsive. It is of interest to note that the name imulsive comes from Newton s second law of motion. Imulse change in momentum Imulsive force rate of change in momentum. F m v m is the mass flow rate in kg/s and v is the change in velocity of the fluid. his is a vector quantity and may be alied to any direction. If we make v the change in velocity in the direction of motion we obtain the force making the rotor turn. his direction is usually called the whirl direction and v w means the change in velocity in the whirl direction. F m v w Suose the vanes to be rotating on a mean circle of diameter D at N rev/s. he linear velocity of the vanes is u m/s. his is given by the following equation. u πdn he ower roduced by any moving force is the roduct of force and velocity. he ower of the ideal rotor is given by the following equation. P m v w u m v w πnd D.J.Dunn www.freestudy.co.uk

his is the fundamental way of finding the ower roduced by fluids assing over moving vanes. In order to find the vector quantity v w, we draw vector diagrams for the velocities. For this reason, the ower is called DIAGRAM POWER. DIAGRAM POWER m v w πnd he construction of the vector diagrams for fluids flowing over vanes is not covered in this book and you should refer to more advanced text if you wish to study it at a deeer level. WORKED EXAMPLE No. he vanes on a simle steam turbine are mounted on a rotor with a mean diameter of 0.6 m. he steam flows at a rate of 0.8 kg/s and the velocity in the whirl direction is changed by 80 m/s. he turbine rotates at 600 rev/min. Calculate the diagram ower. SOLUION Rotor Seed N 600/60 0 rev/s elocity of the vanes u πnd π x 0 x 0.6 8.85 m/s Diagram Power DP m u v w 0.8 x 8.85 x 80 06.5 W A ractical imulse turbine needs several sets of moving vanes and fixed vanes as shown in figure 4. he fixed vanes act as nozzles that convert ressure into velocity. he steam from the nozzles is deflected by the moving row. here is a ressure dro over each fixed row. Fig.4 D.J.Dunn www.freestudy.co.uk 4

. REACION HEORY REACION FORCES are exerted on an object when it causes the velocity of the fluid to change. Consider a simle nozzle in which the fluid accelerates due to the change in the cross sectional area. he kinetic energy of the fluid increases and since energy is conserved, the ressure of the fluid dros. In other words, the ressure behind the fluid forces it through the nozzle causing it to seed u. he force required to accelerate the fluid is in the direction of the acceleration. Every force has an equal and oosite reaction so an equal and oosite force is exerted on the nozzle. his is the rincile used in rockets. Fig. 5 It should be borne in mind that steam and gas, unlike liquids, undergoes a volume increase when the ressure falls. It is thus ossible to accelerate steam and gas without narrowing the flow assage. he force required to accelerate the fluid is given by the following equation. F m v he reaction force acting on the nozzle is equal and oosite in direction. Figure 6 shows the layout of the blades for a turbine that uses both reaction and imulse. he fixed rows accelerate the steam and there is a ressure dro over the row. he moving row also accelerates the steam and there is a further ressure dro over the moving row. he moving blades are thus moved by both imulse and reaction forces. If the rows of blades are identical, the ressure dro over each is the same and there is 50% imulse and 50% reaction. Fig.6 D.J.Dunn www.freestudy.co.uk 5

he moving vanes exerience both reaction and imulsive forces and the two together is given by the change in momentum. he formula develoed in section. alies to any kind of turbine. AXIAL FORCE DIAGRAM POWER m v w πnd he change in momentum that roduces the force on the blade is not only in the direction of rotation. here is also a change of velocity and hence momentum in the direction of the axis of rotation and this ushes the turbine rotor in that direction. his would require a large thrust bearing in the turbine design. his can be avoided by lacing two identical rotors back to back so the axial thrust cancels out. Figure 7 shows the schematic for such an arrangement. Fig. 7 Because the volume of the steam or gas increases greatly as it rogresses along the axis, the height of the blades increases in order to accommodate it. Figure 8 shows a turbine with the casing removed. here are three sets or cylinders each with double flow. he exhaust steam has such a large volume that entry to the condenser is through the large assages underneath. he condenser occuies the sace below the turbine hall. D.J.Dunn www.freestudy.co.uk 6

Fig. 8 D.J.Dunn www.freestudy.co.uk 7

Figure 9 is another icture showing the rotor of a large steam turbine. Fig. 9 D.J.Dunn www.freestudy.co.uk 8

SELF ASSESSMEN EXERCISE No.. A steam turbine has its vanes on a mean diameter of. m and rotates at 500 rev/min. he change in the velocity of whirl is 65 m/s and the change in the axial velocity is 0 m/s. he flow rate is kg/s. Calculate the following. i. he diagram ower. (6. kw) ii. he axial force. (0 N). A steam turbine is to be designed to rotate at 000 rev/min and roduce 5 kw of ower when kg/s is used. he vanes will be laced on a mean diameter of.4 m. Calculate the change in the velocity of whirl that will have to be roduced. (.7 m/s). A gas turbine has rotor blades on a mean diameter of 0.5 m and the rotor turns at 000 rev/min. he change in the whirl velocity is 0 m/s and the diagram ower is MW. Calculate the mass flow rate of gas. (7.6 kg/s) D.J.Dunn www.freestudy.co.uk 9

EDEXCEL HIGHERS ENGINEERING HERMODYNAMICS H NQF LEEL 4 OUCOME 4 SEAM AND GAS URBINE POWER PLAN UORIAL No. 8 SEAM CYCLES Steam and gas turbine Princiles of oeration: imulse and reaction turbines; condensing; ass-out and back ressure steam turbines; single and double shaft gas turbines; regeneration and re-heat in gas turbines; combined heat and ower lants Circuit and roerty diagrams: circuit diagrams to show boiler/heat exchanger; suerheater; turbine; condenser; condenser cooling water circuit; hot well; economiser/feedwater heater; condensate extraction and boiler feed ums; temerature entroy diagram of Rankine cycle Performance characteristics: Carnot, Rankine and actual cycle efficiencies; turbine isentroic efficiency; ower outut; use of roerty tables and enthaly-entroy diagram for steam When you have comleted tutorial 8 you should be able to do the following. Exlain the Carnot steam cycle. Exlain the Rankine steam ower cycle. Describe imrovements to the Rankine cycle. D.J.Dunn www.freestudy.co.uk

. SEAM CYCLES. HE CARNO SEAM CYCLE In revious tutorials you learned that a Carnot cycle gave the highest thermal efficiency ossible for an engine working between two temeratures. he cycle consisted of isothermal heating and cooling and reversible adiabatic exansion and comression. Consider a cycle that uses vaour throughout. Evaoration and condensation at constant ressure is also constant temerature. Isothermal heating and cooling is theoretically ossible. he cycle would consist of the same 4 rocesses as before only this time each rocess would be carried out in a searate steady flow lant item with the vaour flowing from one to the other in a closed loo as shown below. he four rocesses are: Fig. - Evaoration at constant ressure and temerature requiring heat inut. - Reversible adiabatic exansion in the turbine giving ower outut. - 4 Cooling and condensing at constant ressure and temerature in the condenser requiring heat outut. 4 - Reversible adiabatic comression requiring ower inut. In order that no temerature changes occur in the evaorator and condenser, the vaour must be wet at inlet and outlet. Over-cooling will roduce liquid at temeratures below the saturation temerature and over-heating will suerheat it beyond the saturation temerature. he cycle will be a rectangle on the -s diagram and as shown on the h-s diagram. D.J.Dunn www.freestudy.co.uk

Fig. he limits are that at oint () it may be dry saturated vaour but not suerheated. At oint it may be saturated water but not under-cooled. If these limits are not used, then the vaour has a dryness fraction at each oint. Since heat transfer only occurs at the evaorator and condenser the heat transfer rates are given by the following exressions. Φ in m(h - h) h S (Boiler) Φ out m(h - h4) c S (Condenser) h is the boiler temerature and c is the condenser temerature. he thermal efficiency may be found from the st. Law. ηth - Φ out /Φ in - c / h his exression is the same as for the gas version. WORKED EXAMPLE No. A Carnot cycle is conducted on steam as follows. he evaorator roduces dry saturated steam at 0 bar. he steam is exanded reversibly and adiabatically in a turbine to bar. he exhaust steam is artially condensed and then comressed back to 0 bar. As a result of the comression, the wet steam is changed comletely into saturated water. Assuming a flow rate of kg/s throughout determine the condition and secific enthaly at each oint in the cycle. Calculate the energy transfers for each stage. Show that the efficiency is correctly redicted by the exression ηth (cold)/(hot) D.J.Dunn www.freestudy.co.uk

SOLUION We will refer to the revious diagrams throughout. EAPORAOR h hg at 0 bar (since it is dry saturated) 778 kj/kg. s sg at 0 bar (since it is dry saturated) 6.586 kj/kg K. h hf at 0 bar (since it is saturated water) 76 kj/kg. Φ in (778-76) 05 kw URBINE Since the exansion is isentroic then s s 6.586 kj/kg K s 6.586 s f + x s f g at bar 6.586.0 + x (6.056) hence x 0.87 h h f + x h f g at bar 47 + (0.87)(58) 87 kj/kg P(outut) (778-87) 9. kw COMPRESSOR Since the comression is isentroic then s4 s s sf at 0 bar (since it is saturated water).8 kj/kg K. s4 s.8 sf + x4s f g at bar.8.0 + x4(6.056) hence x4 0.8 h4 hf + x4h f g at bar 47 + (0.9)(58) 78. kj/kg Power Inut (76-78.) 4.7 kw CONDENSER Heat outut (87-78.) 658.7 kw Energy Balances rounded off to nearest kw. otal energy inut 4.7 + 05 050 kw otal energy outut 9. + 658.7 050 kw Net Power outut 9. - 4.7 56 kw Net Heat inut 05-658.7 56 kw hermal efficiency P nett /Φ in 56/05 7.7% hermal Efficiency - Φ out / Φ in - 658.7/05 7.7% he hottest temerature in the cycle is ts at 0 bar 79.9 oc or 45.9 K he coldest temerature in the cycle is ts at bar 99.6 oc or 7.6 K he Carnot efficiency - 7.6/45.9 7.7% D.J.Dunn www.freestudy.co.uk 4

SELF ASSESSMEN EXERCISE No.. A steam ower lant uses the Carnot cycle. he boiler uts 5 kw of heat into the cycle and roduces wet steam at 00 o C. he condenser roduces wet steam at 50 o C. Calculate the following. i. he efficiency of the lant. (4.6%) ii. he net ower outut. (0.9 kw) iii. he heat removed by the condenser. (4 kw). A steam ower lant is based on the Carnot cycle. he boiler is sulied with saturated water at 0 bar and roduces dry saturated steam at 0 bar. he condenser oerates at 0. bar. Assuming a mass flow rate of kg/s calculate the following. i. he thermal efficiency. (4.%) ii. he ower outut of the turbine. (79 kw) iii. he heat transfer rate into the boiler. (.89 MW) D.J.Dunn www.freestudy.co.uk 5

. HE RANKINE CYCLE he Rankine Cycle is a ractical cycle and most steam ower lants are based on it. he roblems with the Carnot Cycle are as follows. It roduces only small net ower oututs for the lant size because dry saturated steam is used at inlet to the turbine. It is imractical to comress wet steam because the water content searates out and fills the comressor. It is imractical to control the condenser to roduce wet steam of the correct dryness fraction. In order to get around these roblems, the Rankine Cycle uses suerheated steam from the boiler to the turbine. he condenser comletely condenses the exhaust steam into saturated water. he comressor is relaced with a water (feed) um to return the water to the boiler. he result of this is reduced efficiency but greater quantities of ower. Fig. he lant layout is shown above. First let s briefly examine the boiler. D.J.Dunn www.freestudy.co.uk 6

BOILER For reasons of combustion efficiency (which you do not have to study), a ractical boiler is made u of three sections. a) Economiser his is a water heater inside the boiler that raises the water temerature at the boiler ressure to just below the saturation temerature at that ressure. b) Evaorator his is a unit usually consisting of a drum and tubes in which the water is evaorated and the steam driven off. c) Suer-heater his is a heater laced in the hottest art of the boiler that raises the temerature of the steam well beyond the saturation temerature. here are many boiler designs and not all of them have these features. he main oint is that a heat transfer rate is needed into the boiler unit in order to heat u the water, evaorate it and suerheat it. he overall heat transfer is Φ in m (h - h) Next let s look at some other ractical asects of a steam ower lant. EXRACION PUMP AND HOWELL. In a ractical steam cycle the condensate in the condenser is extracted with an extraction um and the water roduced is the coldest oint in the steam cycle. his is usually laced into a vessel where it can be treated and extra added to make u for leaks. his oint is called the HOWELL because it contains hot water. he main feed um returns this water to the boiler at high ressure. In the following work, extraction ums and hotwells are not shown. Fig.4 D.J.Dunn www.freestudy.co.uk 7

Now let s examine the cycle with the aid of roerty diagrams. Fig.5 he rocess 4 to is cramed into the corner of the h-s diagram and is not clear. BOILER PROCESS ( ) to () HEA INPU he water at oint is below the saturation temerature at the boiler ressure. he economiser first heats it u raising the temerature, enthaly and entroy until it reached the saturation curve. he water is then evaorated and finally, the temerature is raised by suerheating the steam to oint. Φ in m (h - h) URBINE PROCESS () to () POWER OUPU he second rocess is the exansion in the turbine and this is ideally reversible and adiabatic and is reresented by a vertical line on the diagrams. P out m(h - h) urbines in real lant are often in several stages and the last stage is secially designed to coe with water drolets in the steam that becomes wet as it gives u its energy. You must use the isentroic exansion theory in order to calculate the dryness fraction and enthaly of the exhaust steam. CONDENSER PROCESS () to (4) HEA OUPU he third rocess is the condenser where the wet steam at oint is ideally turned into saturated water at the lower ressure (oint 4). Condensers usually work at very low ressures (vacuums) in order to make the turbine give maximum ower. he heat removed is given by Φ out m (h - h4) Since the condenser roduces condensate (saturated water) then h4 hf at the condenser ressure. D.J.Dunn www.freestudy.co.uk 8

PUMP PROCESS (4) to () POWER INPU he final rocess which comletes the cycle is the uming of the water (oint 4) from the low condenser ressure to the boiler at high ressure (oint ). In reality there are many things which are done to the feed water before it goes back into the boiler and the ressure is often raised in several stages. For the Rankine Cycle we assume one stage of uming which is adiabatic and the ower inut to the um is P in m (h - h4) he ower required to um the water is much less than that required to comress the vaour (if it was ossible). he ower inut to the feed um is very small comared to the ower outut of the turbine and you can often neglect it altogether. In this case we assume hh4. If you are not ignoring the ower inut, then you need to find h. If you know the exact temerature of the water at inlet to the boiler (outlet from the um) then you may be able to look it u in tables. he nearest aroximation is to look u hf at the water temerature. Since the water is at high ressure, this figure will not be very accurate and you may correct it by adding the flow energy. We will look at this in greater detail later. Lets first do a simle examle with no great comlications. D.J.Dunn www.freestudy.co.uk 9

WORKED EXAMPLE No. A steam ower lant is based on the Rankine cycle. he steam roduced by the boiler is at 40 bar and 400oC. he condenser ressure is 0.05 bar. Assume isentroic exansion. Ignore the energy term at the feed um. Calculate the Rankine cycle efficiency and comare it to the Carnot efficiency for the same uer and lower temerature limits. SOLUION urbine Figure 6 h 4 kj/kg at 40 bar and 400oC. Since the exansion is isentroic then s 6.769 kj/kg K s 0.9 + 8. x x 0.785 h hf + x hfg + 0.785(48) 04.6 kj/kg Condenser h4 hf at 0.05 bar kj/kg Boiler If the ower inut to the um is neglected then h4 h kj/kg Φ in h - h 0 kj/kg. P(outut) h - h 89.4 kj/kg η P/ Φ in 8. % D.J.Dunn www.freestudy.co.uk 0

Carnot Efficiency he hottest temerature in the cycle is 400oC (67 K) and the coldest temerature is ts at 0.05 bar and this is 6.7 oc(99.7 K). he Carnot efficiency is - 99.7/67 55.5 % which is higher as exected. Now let s examine the feed um in more detail. FEED PUMP When water is comressed its volume hardly changes. his is the imortant factor that is different from the comression of a gas. Because the volume hardly changes, the temerature should not increase and the internal energy does not increase. he Steady flow Energy equation would then tell us that the ower inut to the um is virtually equal to the increase in flow energy. We may write P in m v Since the volume of water in nearly all cases is 0.00 m/kg then this becomes If we use ressure units of bars then Exressed in kilowatts this is P in 0.00 m 0.00 m ( - ) P in 0.00 m( - ) x 05 Watts P in m( - ) x 0- kw From this we may also deduce the enthaly of the water after the um. Hence h may be deduced. P in m (h - h4) D.J.Dunn www.freestudy.co.uk

WORKED EXAMPLE No. Reeat examle, but this time do not ignore the feed um and assume the boiler inlet condition is unknown. SOLUION P in kg/s(40-0.05) x 0-4 kw 4 kg/s(h - h4) (h - ) h 6 kj/kg Reworking the energy transfers gives Φ in h - h 4-6 098 kj/kg. P nett P out - P in 89.4-4 85.4 kj/kg η P nett /Φ in 85.4/098 8. % Notice that the answers are not noticeably different from those obtained by ignoring the feed um. WORKED EXAMPLE No.4 A steam ower lant uses the Rankine Cycle. he details are as follows. Boiler ressure Condenser ressure emerature of steam leaving the boiler Mass flow rate 00 bar 0.07 bar 400oC 55 kg/s Calculate the cycle efficiency, the net ower outut and the secific steam consumtion. D.J.Dunn www.freestudy.co.uk

SOLUION urbine h 097 kj/kg at 00 bar and 400oC. For an isentroic exansion we find the ideal condition at oint as follows. s 6. kj/kg K s 0.559 + 7.75 x x 0.7 h hf + x hfg 6 + 0.7(409) 98 kj/kg P out m(h-h) 55(097-98) 64. MW Condenser h4 hf at 0.07 bar 6 kj/kg Φ out m(h - h4) 55(98-6) 97. MW PUMP Ideal ower inut Flow Energy change mv( ) P in 55(0.00)(00-0.07) x 05 550 kw P in m(h - h4) 55(h-6) hence h 7 kj/kg Boiler Φ in m(h - h) 55(097-7) 60.8 MW EFFICIENCY P nett P out - P in 64. - 0.55 6.7 MW η P nett / Φ in 6.7/60.8 9.6 % Alternatively P nett Φ in - Φ out 60.8-97. 6.7 MW his should be the same as P nett since the net energy entering the cycle must equal the net energy leaving. η - Φ out / Φ in - 97./60.8 9.6% SPECIFIC SEAM CONSUMPION his is given by S.S.C. P nett /mass flow 6.78/55.59 MW/kg/s or MJ/kg D.J.Dunn www.freestudy.co.uk

SELF ASSESSMEN EXERCISE No.. A simle steam lant uses the Rankine Cycle and the data for it is as follows. Flow rate 45 kg/s Boiler ressure 50 bar Steam temerature from boiler 00oC Condenser ressure 0.07 bar Assuming isentroic exansion and uming, determine the following. i. he ower outut of the turbine. (44.9 MW) ii. he ower inut to the um. (5 kw) iii. he heat inut to the boiler. (4 MW) iv. he heat rejected in the condenser. (79 MW) v. he thermal efficiency of the cycle. (6%). A simle steam ower lant uses the Rankine Cycle. he data for it is as follows. Flow rate kg/s Boiler ressure 00 bar Steam temerature from boiler 600oC Condenser ressure 0.04 bar Assuming isentroic exansion and uming, determine the following. i. he ower outut of the turbine. (4.6 MW) ii. he ower inut to the um. (0 kw) iii. he heat inut to the boiler. (0.5 MW) iv. he heat rejected in the condenser. (5.9 MW) v. he thermal efficiency of the cycle. (44%). a) Exlain why ractical steam ower lants are based on the Rankine Cycle rather than the Carnot Cycle. b) A simle steam ower lant uses the Rankine Cycle. he data for it is Boiler ressure Steam temerature from boiler Condenser ressure Net Power Outut 5 bar 00oC 0. bar. MW Calculate the following. i. he cycle efficiency. (9.7 %) ii. he steam flow rate. (. kg/s) D.J.Dunn www.freestudy.co.uk 4

. BACK-PRESSURE AND PASS-OU URBINES If an industry needs sufficient quantities of rocess steam (e.g. for sugar refining) and electric ower, it becomes economical to use the steam for both uroses. his is done with a steam turbine and generator and the rocess steam is obtained in two ways as follows. A) By exhausting the steam at the required ressure (tyically bar) to the rocess instead of to the condenser. A turbine designed to do this is called a BACK-PRESSURE URBINE. B) By bleeding steam from an intermediate stage in the exansion rocess. A turbine designed to do this is called a PASS-OU URBINE. he steam cycle is standard excet for these modifications... BACK-PRESSURE URBINES Back-ressure turbines are designed to oerate with a back ressure, unlike normal turbines that oerate with a vacuum at the exhaust. he diagram shows the basic circuit. Fig.7 D.J.Dunn www.freestudy.co.uk 5

WORKED EXAMPLE No.5 For a steam circuit as shown reviously, the boiler roduces suerheated steam at 50 bar and 400 o C. his is exanded isentroically to bar. he exhaust steam is used for a rocess. he returning feed water is at bar and 40 o C. his is umed to the boiler. he water leaving the um is at 40 o C and 50 bar. he net ower outut of the cycle is 60 MW. Calculate the mass flow rate of the steam. SOLUION Referring to the revious cycle sketch for location oints we find h 96 kj/kg s 6.646 kj/kg K For an ideal exansion s s 6.646 s f +x sfg at bar 6.646.67 + x (5.) x 0.95 h 4 h f + x hfg at bar h 4 56 + 0.95(64) h 4 584 kj/kg Change in enthaly 584-96 -6 kj/kg he ower outut of the turbine is found from the steady flow energy equation so P m(-6) kw P -6 m kw (outut) Next we examine the enthaly change at the um. h 68 kj/kg at bar and 40 o C h 7 kj/kg at 50 bar and 40 o C. Change in enthaly 7-69 kj/kg he ower inut to the um is found from the steady flow energy equation so : P -m() kw P - m kw(inut) Net Power outut of the cycle 60 MW hence 60 000 6 m - m m 98.5 kg/s D.J.Dunn www.freestudy.co.uk 6

SELF ASSESSMEN EXERCISE No.. A steam cycle is erformed as follows. he boiler roduces kg/s of suerheated steam at 60 bar and 400 o C. he steam is sulied to a turbine that it exands it isentroically and with no friction to.5 bar. he exhaust steam is sulied to a rocess. he feed water is sulied to the um at.0 bar and 00 o C and delivered to the boiler at 60 bar. he um may be considered as ideal. Calculate the following. i. he ower outut of the turbine. (. MW) ii. he heat inut to the boiler. (8. MW) iii. he ower inut to the um. (7.7 kw) iv. he thermal efficiency of the cycle. (6.%). A back ressure steam cycle works as follows. he boiler roduces 8 kg/s of steam at 40 bar and 500 o C. his is exanded isentroically to bar. he um is sulied with feed water at 0.5 bar and 0 o C and delivers it to the boiler at o C and 40 bar. Calculate the following. i. he net ower outut. (6 MW) ii. he heat inut to the boiler. (6.5 MW) iii. he thermal efficiency of the cycle. (.5%) D.J.Dunn www.freestudy.co.uk 7

. PASS-OU URBINES he circuit of a simle ass-out turbine lant is shown below. Steam is extracted between stages of the turbine for rocess use. he steam removed must be relaced by make u water at oint 6. Fig. 8 In order to solve roblems you need to study the energy balance at the feed ums more closely so that the enthaly at inlet to the boiler can be determined. Consider the ums on their own as below. Fig. 9 he balance of ower is as follows. P + P increase in enthaly er second. P + P m h - m 6 h 6 -m 5 h 5 From this the value of h or the mass m may be determined. his is best shown with a worked examle. D.J.Dunn www.freestudy.co.uk 8

WORKED EXAMPLE No.6 he circuit below shows the information normally available for a feed um circuit. Determine the enthaly at entry to the boiler. Fig. 0 SOLUION If no other details are available the ower inut to the um is m v Assume v 0.00 m /kg P m v m x 0.00 x (bar) x 0 5 x 0 - m x 0 - kw PUMP PUMP P m v (40)(80-0.)(0 - ) 9.6 kw P m v (5)(80-)(0 - ) 9.5 kw h 5 h f 9 kj/kg at 0. bar otal ower inut 9.6 + 9.5 59. kw h 6 84 kj/kg (from water tables or aroximately h f at 0o C) Balancing energy we have the following. 59. 45 h - 40 h 5-5h 6 59. 45 h - 40(9) - 5(84) h 88 kj/kg D.J.Dunn www.freestudy.co.uk 9

WORKED EXAMPLE No.7 A ass-out turbine lant works as shown in fig. he boiler roduces steam at 60 bar and 500 o C and this is exanded through two stages of turbines. he first stage exands to bar where 4 kg/s of steam is removed. he second stage exands to 0.09 bar. Assume that the exansion is a straight line on the h - s chart. he condenser roduces saturated water. he make u water is sulied at bar and 0 o C. he net ower outut of the cycle is 40 MW. Calculate the following. i. he flow rate of steam from the boiler. ii he heat inut to the boiler. iii. he thermal efficiency of the cycle. SOLUION URBINE EXPANSION h 4 kj/kg from tables. h 678 kj/kg using isentroic exansion and entroy. h 4 66 kj/kg using isentroic exansion and entroy. hese results may be obtained from the h-s chart. POWER OUPU Fig. P out m(h - h ) + (m - 4)(h - h 4 ) P out m(4-678) + (m - 4)(678-66) P out 74 m + 5 m - 048 D.J.Dunn www.freestudy.co.uk 0

POWER INPU he ower inut is to the two feed ums. Fig. h 6 84 kj/kg (water at bar and 0 o C) h 5 hf at 0.09 bar 8 kj/kg. P mv 4 x (60 - ) x 0 -.6 kw P mv (m-4) x (60-0.09) x 0-5.99m -.96 kw NE POWER 40 000 kw P out - P - P 40 000 74 m + 5 m - 048 -.6-5.99 m +.96 40000 49 m - 048 hence m.66 kg/s ENERGY BALANCE ON PUMPS P.6 kw P 77.7 kw (using the value of m just found) m h (m - 4) h 5 + P + P.66 h 9.66 x 8 +.6 + 77.7 h 67. kj/kg HEA INPU Φ in m(h - h ) 095 kw EFFICIENCY Efficiency η 40/09.5 6.5 % D.J.Dunn www.freestudy.co.uk

SELF ASSESSMEN EXERCISE No.4. A steam turbine lant is used to suly rocess steam and ower. he lant comrises an economiser, boiler, suer-heater, turbine, condenser and feed um. he rocess steam is extracted between intermediate stages in the turbine at bar ressure. he steam temerature and ressure at outlet from the suer-heater are 500 o C and 70 bar resectively. he turbine exhausts at 0. bar. he make-u water is at 5 o C and bar and it is umed into the feed line to relace the lost rocess steam. Assume that the exansion is isentroic. If due allowance is made for the feed um-work, the net mechanical ower delivered by the lant is 0 MW when the rocess steam load is 5 kg/s. Sketch clear - s and h-s and flow diagrams for the lant. Calculate the following. i. he flow rate of steam flow leaving the suer-heater. (5.6 kg/s) ii. he rate of heat transfer to the boiler. (8. MW). An industrial lant requires 60 MW of rocess heat using steam at.6 bar which it condenses to saturated water. his steam is derived from the intermediate stage of a turbine that roduces 0 MW of ower to drive the electric generators. he steam is raised in boilers and sulied to the turbines at 80 bar ressure and 600 o C. he steam is exanded isentroically to the condenser ressure of 0.05 bar. Calculate the following. i. he flow rate of the steam bled to the rocess. (7.7 kg/s) ii. he flow rate of the steam from the boiler. (0.5 kg/s) D.J.Dunn www.freestudy.co.uk

. REHEAING Steam reheating is another way of imroving the thermodynamic efficiency by attemting to kee the steam temerature more constant during the heating rocess. Fig. Suerheated steam is first assed through a high ressure turbine. he exhaust steam is then returned to the boiler to be reheated almost back to its original temerature. he steam is then exanded in a low ressure turbine. In theory, many stages of turbines and reheating could be done thus making the heat transfer in the boiler more isothermal and hence more reversible and efficient..4. FEED HEAING Feed heating is another way of imroving the efficiency of a steam ower lant. he feed water returning to the boiler is heated nearer to the saturation temerature with steam bled from the turbines at aroriate oints. here are two tyes. Indirect contact. Direct contact. INDIREC CONAC he steam does not mix with the feed water at the oint of heat exchange. he steam is condensed through giving u its energy and the hot water resulting may be inserted into the feed system at the aroriate ressure. D.J.Dunn www.freestudy.co.uk

DIREC CONAC he bled steam is mixed directly with the feed water at the aroriate ressure and condenses and mixes with the feed water. he diagram shows a basic lant with a single feed heater added. Fig.4 If a steam cycle used many stages of regenerative feed heating and many stages of reheating, the result would be an efficiency similar to that of the Carnot cycle. Although racticalities revent this haening, it is quite normal for an industrial steam ower lant to use several stages of regenerative feed heating and one or two stages of reheating. his roduces a significant imrovement in the cycle efficiency. here are other features in advanced steam cycles that further imrove the efficiency and are necessary for ractical oeration. For examle air extraction at the condenser, steam recovery from turbine glands, de-suer-heaters, de-aerators and so on. hese can be found in detail in textbooks devoted to ractical steam ower lant. SELF ASSESSMEN EXERCISE No.5 Draw the lant cycle of a steam lant with the following features.. Economiser.. Evaorator.. Suer-heater 4. Re-heater. 5. High ressure turbine. 6. Low ressure turbine. 7. Condenser. 8. Direct contact feed heaters fed from the l.. turbine. 9. Direct contact feed heaters fed from the h.. turbine. 0. Hot well.. High ressure feed um.. Low ressure feed um.. Extraction um. D.J.Dunn www.freestudy.co.uk 4

EDEXCEL HIGHERS ENGINEERING HERMODYNAMICS H NQF LEEL 4 OUCOME 4 SEAM AND GAS URBINE POWER PLAN UORIAL No. 9 GAS URBINE HEORY Steam and gas turbine Princiles of oeration: imulse and reaction turbines; condensing; ass-out and back ressure steam turbines; single and double shaft gas turbines; regeneration and re-heat in gas turbines; combined heat and ower lants Circuit and roerty diagrams: circuit diagrams to show boiler/heat exchanger; suerheater; turbine; condenser; condenser cooling water circuit; hot well; economiser/feedwater heater; condensate extraction and boiler feed ums; temerature entroy diagram of Rankine cycle Performance characteristics: Carnot, Rankine and actual cycle efficiencies; turbine isentroic efficiency; ower outut; use of roerty tables and enthaly-entroy diagram for steam When you have comleted tutorial 9 you should be able to do the following. Describe the basic gas turbine ower cycle. Describe advanced gas turbine ower cycles. Solve roblems concerning gas turbine ower lant. D.J.Dunn www.freestudy.co.uk

GAS URBINE ENGINES In this section we will examine how ractical gas turbine engine sets vary from the basic Joule cycle. he efficiency of gas turbine engines increases with ressure comression ratio. In ractice this is limited, as the tye of comressor needed to roduce very large flows of air cannot do so at high ressures. 6 bar is a tyical ressure for the combustion chamber. he efficiency of gas turbines may be imroved by the use of inter-cooling and heat exchangers.. GAS CONSANS he first oint is that in reality, although air is used in the comressor, the gas going through the turbine contains roducts of combustion so the adiabatic index and secific heat caacity is different in the turbine and comressor.. FREE URBINES Most designs used for gas turbine sets use two turbines, one to drive the comressor and a free turbine. he free turbine drives the load and it is not connected directly to the comressor. It may also run at a different seed to the comressor. Fig. shows the layouts for arallel turbines. Figure shows the layout for series turbines. Fig. PARALLEL URBINES D.J.Dunn www.freestudy.co.uk

Figure SERIES URBINES. INER-COOLING AND REHEAING Basically, if the air is comressed in stages and cooled between each stage, then the work of comression is reduced and the efficiency increased. he reverse theory also alies. If several stages of turbine exansions are used and the gas reheated between stages, the ower outut and efficiency is increased. Figure shows the layout for intercoolers and re-heaters. Fig. INER-COOLER and RE-HEAER D.J.Dunn www.freestudy.co.uk

WORKED EXAMPLE No. A gas turbine draws in air from atmoshere at bar and 0oC and comresses it to 5 bar. he air is heated to 00 K at constant ressure and then exanded through two stages in series back to bar. he high ressure turbine is connected to the comressor and roduces just enough ower to drive it. he low ressure stage is connected to an external load and roduces 80 kw of ower. Draw the circuit. Calculate the mass flow of air, the inter-stage ressure of the turbines and the thermal efficiency of the cycle. Assume γ.4 and c.005 kj/kg K for both the turbines and the comressor. Neglect the increase in mass due to the addition of fuel for burning. Comare the efficiency to the air standard efficiency. SOLUION COMPRESSOR γ 0.86 r 8 x 5 448.4 K Fig. 4 Power inut to comressor mc(-) Power outut of h.. turbine mc(-4) Since these are equal we may equate them..005(448.4-8).005(00-4) 404.6 K D.J.Dunn www.freestudy.co.uk 4

HIGH PRESSURE URBINE 4 4 5 4 5 - γ 04.6 4 0.595 00 5 0.86 4 5 x 0.595 0.86.977 bar LOW PRESSURE URBINE 5 4.977 - γ 0.86 ( 0.6) 0.7 757. K NE POWER he net ower is 80 kw. 80 mc(4-5) m x.005(04.6-757.) m 0.88 kg/s HEA INPU Φ(in) mc(-) 0.88 x.005 (00-448.4) 7.5 kw HERMAL EFFICIENCY ηth P(nett)/Φ(in) 80/9.9 0.67or 6.7% he air standard efficiency is the Joule efficiency. η -r -0.86-5 -0.86 0.69 or 6.9% 5 D.J.Dunn www.freestudy.co.uk 5

SELF ASSESSMEN EXERCISE No.. A gas turbine draws in air from atmoshere at bar and 5oC and comresses it to 4.5 bar. he air is heated to 00 K at constant ressure and then exanded through two stages in series back to bar. he high ressure turbine is connected to the comressor and roduces just enough ower to drive it. he low ressure stage is connected to an external load and roduces 00 kw of ower. For the comressor γ.4 and for the turbines γ.. he gas constant is 0.87 kj/kg K for both. Neglect the increase in mass due to the addition of fuel for burning. Assume the secific heat of the gas in the combustion chamber is the same as that for the turbines. Calculate the following. i. he secific heat c of the air and the burned mixture. (.005 and.4) ii. he mass flow of air. (0.407 kg/s) iii. he inter-stage ressure of the turbines. (.67 bar) iv. he thermal efficiency of the cycle. (0%). A gas turbine draws in air from atmoshere at bar and 00 K and comresses it to 6 bar. he air is heated to 00 K at constant ressure and then exanded through two stages in series back to bar. he high ressure turbine is connected to the comressor and roduces just enough ower to drive it. γ.4 and R 0.87 kj/kg K for both the turbine and comressor. Neglect the increase in mass due to the addition of fuel for burning. Calculate the following for a mass flow of kg/s. i. he inter-stage ressure. (. bar) ii. he net ower outut. ( kw) iii. he thermal efficiency of the cycle. (40%) D.J.Dunn www.freestudy.co.uk 6

.4 EXHAUS GAS HEA EXCHANGER he exhaust gas from a turbine is hotter than the air leaving the comressor. If heat is assed to the air from the exhaust gas, then less fuel is needed in the combustion chamber to raise the air to the oerating temerature. his requires an exhaust heat exchanger. Fig.5 shows the layout required. Fig.5 In order to solve roblems associated with this cycle, it is necessary to determine the temerature rior to the combustion chamber (). A erfect heat exchanger would heat u the air so that is the same as 5. It would also cool down the exhaust gas so that 6 becomes. In reality this is not ossible so the concet of HERMAL RAIO is used. his is defined as the ratio of the enthaly given to the air to the maximum ossible enthaly lost by the exhaust gas. he enthaly lost by the exhaust gas is Hmgcg(5-6) his would be a maximum if the gas is cooled down such that 6. Of course in reality this does not occur and the maximum is not achieved and the gas turbine does not erform as well as redicted by this idealisation. H (maximum) mgcg(5-) he enthaly gained by the air is H (air) maca(-) Hence the thermal ratio is.r. maca(-)/mgcg(5-) he suffix a refers to the air and g to the exhaust gas. Since the mass of fuel added in the combustion chamber is small comared to the air flow we often neglect the difference in mass and the equation becomes D.J.Dunn www.freestudy.co.uk 7.R. ca(-)/cg(5-)

WORKED EXAMPLE No. A gas turbine uses a ressure ratio of 7.5/. he inlet temerature and ressure are resectively 0oC and 05 kpa. he temerature after heating in the combustion chamber is 00 oc. he secific heat caacity c for air is.005 kj/kg K and for the exhaust gas is.5 kj/kg K. he adiabatic index is.4 for air and. for the gas. Assume isentroic comression and exansion. he mass flow rate is kg/s. Calculate the air standard efficiency if no heat exchanger is used and comare it to the thermal efficiency when an exhaust heat exchanger with a thermal ratio of 0.8 is used. SOLUION Referring to the numbers used on fig.6 the solution is as follows. Air standard efficiency - r (-/γ) -7.5-0.86 0.48 0r 4.8% Solution with heat exchanger r (-/γ) 8 (7.5)0.86 50.6 K 54/r (-/γ) 57/(7.5)0.48 954. K Use the thermal ratio to find. 0.8.005(-)/.5(5-) 0.8.005(-50.6)/.5(954.-50.6) 96 K In order find the thermal efficiency, it is best to solve the energy transfers. P(in) mca(-) x.005 (50.6-8).7 kw P(out) mcg(4-5) x.5 (57-954.) 7.7 kw P(net) P(out) - P(in) 490 kw Φ(in)combustion chamber) mcg(4-) Φ(in).5(57-96) 755.5 kw ηth P(net)/Φ(in) 490/755.5 0.65 or 65% D.J.Dunn www.freestudy.co.uk 8

SELF ASSESSMEN EXERCISE No.. A gas turbine uses a ressure ratio of 7/. he inlet temerature and ressure are resectively 0oC and 00 kpa. he temerature after heating in the combustion chamber is 000 oc. he secific heat caacity C is.005 kj/kg K and the adiabatic index is.4 for air and gas. Assume isentroic comression and exansion. he mass flow rate is 0.7 kg/s. Calculate the net ower outut and the thermal efficiency when an exhaust heat exchanger with a thermal ratio of 0.8 is used. (4 kw and 57%). A gas turbine draws in air from the atmoshere at.0 bar and 00 K. he air is comressed to 6.4 bar isentroically. he air entering the turbine is at 500 K and it exands isentroically to.0 bar. Assume the secific heat C is.005 kj/kg K and γ is.4 for both the turbine and comressor. Ignore the addition of mass in the burner. Calculate the following. i. he air standard efficiency. (40.8%) ii. he efficiency when an exhaust heat exchanger with a thermal ratio of 0.75 is added. (70.7%) D.J.Dunn www.freestudy.co.uk 9

EDEXCEL HIGHERS ENGINEERING HERMODYNAMICS H NQF LEEL 4 OUCOME 4 SEAM AND GAS URBINE POWER PLAN UORIAL No. 0 ISENROPIC EFFICIENCY Steam and gas turbine Princiles of oeration: imulse and reaction turbines; condensing; ass-out and back ressure steam turbines; single and double shaft gas turbines; regeneration and re-heat in gas turbines; combined heat and ower lants Circuit and roerty diagrams: circuit diagrams to show boiler/heat exchanger; suerheater; turbine; condenser; condenser cooling water circuit; hot well; economiser/feedwater heater; condensate extraction and boiler feed ums; temerature entroy diagram of Rankine cycle Performance characteristics: Carnot, Rankine and actual cycle efficiencies; turbine isentroic efficiency; ower outut; use of roerty tables and enthaly-entroy diagram for steam On comletion of this tutorial you should be able to do the following. Exlain the effect of friction on steam and gas exansions in turbines. Solve steam cycle roblems taking into account friction. Solve gas turbine cycle roblems taking into account friction. D.J.Dunn www.freestudy.co.uk

. ISENROPIC EFFICIENCY. HE EFFEC OF FRICION When a fluid is exanded or comressed with fluid friction occurring, a degree of irreversibility is resent. he result is the generation of internal heat equivalent to a heat transfer. his always results in an increase in entroy. Figure shows exansion and comression rocesses on a -s diagram. In the case of vaour, the line crosses the saturation curve. In the case of gas, the rocess takes lace well away from the saturation curve and indeed the saturation curve would not normally be shown for gas rocesses. Note that in every case, the ideal rocess is from () to (') but the real rocess is from () to (). Friction does the following. Increases the entroy. Increases the enthaly. he true rocess ath on roerty diagrams is always to the right of the ideal rocess. When the final oint () is in the gas (suerheat) region, the result is a hotter temerature. When the final oint () is in the wet region, the result is a dryer vaour. Gas and vaour rocesses should be described by sketching them on an aroriate roerty diagram and these effects of friction clearly shown. Figure D.J.Dunn www.freestudy.co.uk

he same oints are also aarent on the h-s diagram. Figure shows a vaour exansion from () to () with the ideal being from () to ('). Note how it ends u dryer at the same ressure with an increase in entroy. aour is not normally comressed so this is not shown. Figure D.J.Dunn www.freestudy.co.uk

. ISENROPIC EFFICIENCY An ideal reversible adiabatic rocess would be constant entroy as shown on the diagrams from () to ('). When friction is resent, the rocess is () to (). he ideal change in enthaly is he actual change is h (ideal) h' - h h(actual) h - h he isentroic efficiency is defined as follows. EXPANSION η is h (actual) h (ideal) h h ' h h COMPRESSION η is h (ideal) h (actual) h h ' h h For gas only h c EXPANSION COMPRESSION η η is is ' ' Note that for an exansion negative changes are obtained on the to and bottom lines that cancel. If the work transfer rate is only due to the change in enthaly we may also define isentroic efficiency as follows. η η is is Actual Power Outut Ideal Power Outut Ideal Power Inut Actual Power Inut for a turbine for a comressor D.J.Dunn www.freestudy.co.uk 4

WORKED EXAMPLE No. A turbine exands steam adiabatically from 70 bar and 500oC to 0. bar with an isentroic efficiency of 0.9. he ower outut is 5 MW. Determine the steam flow rate. SOLUION he solution is easier with a h-s chart but we will do it with tables only. h 40 kj/kg at 70 bar and 500oC. s 6.796 kj/kg K at 70 bar and 500oC. For an ideal exansion from () to (') we calculate the dryness fraction as follows. s s s f + x's fg at 0. bar. 6.796 0.649 + x'(7.5) x' 0.896 Note that you can never be certain if the steam will go wet. It may still be suerheated after exansion. If x' came out to be larger than unity, then because this is imossible, it must be suerheated and you need to deduce its temerature by referring to the suerheat tables. Now we find the ideal enthaly h ' h ' h f + x' h fg at 0. bar. h ' 9 + 0.896(9) 5. kj/kg Now we use the isentroic efficiency to find the actual enthaly h. h( ideal) η is h( actual) 5. 40 0.9 h 40 h 78. kj/kg Now we may use the SFEE to find the mass flow rate. D.J.Dunn www.freestudy.co.uk 5

Φ + P m(h - h ) Φ 0 since it is an adiabatic rocess. P -5 000 kw (out of system) m(78.-40) m 0.96 kg/s D.J.Dunn www.freestudy.co.uk 6

WORKED EXAMPLE No. A turbine exands gas adiabatically from MPa and 600oC to 00 kpa. he isentroic efficiency is 0.9. he mass flow rate is kg/s. Calculate the ower outut. c.005 kj/kg K c v 0.78 kj/kg K. SOLUION he rocess is adiabatic so the ideal temerature ' is given by ' (r ) -/γ r is the ressure ratio r / 0. γ c /c v.005/0.78.4 '87(0.) -/.4 45.9 K Now we use the isentroic efficiency to find the actual final temerature. η is ( - )/( ' - ) 0.9 ( - 87)/(45.9-87) 485.6 K Now we use the SFEE to find the ower outut. Φ + P m c ( - ) he rocess is adiabatic Φ 0. P (.005)(485.6-87) - 467 kw (out of system) D.J.Dunn www.freestudy.co.uk 7

WORKED EXAMPLE No. A simle steam ower lant uses the Rankine cycle. he boiler sulies suerheated steam to the turbine at 40 bar and 400 o C. he condenser oerates at 0. bar and roduces saturated water. he ower inut to the um is negligible. i. Calculate the thermal efficiency of the ideal cycle. ii. Calculate the thermal efficiency when the turbine has an isentroic efficiency of 89%. D.J.Dunn www.freestudy.co.uk 8

SOLUION he solution is easier with a h-s chart. Figure IDEAL CONDIIONS From the chart h 0 kj/kg and h 0 kj/kg/k he ideal work outut 0 0 980 kj/kg When the ower inut to the um is ignored, the ower out is the net ower and the enthaly at inlet to the boiler is h f at 0. bar D.J.Dunn www.freestudy.co.uk 9

he heat inut to the boiler 0 5 959 kj/kg 980 η th 0. 0r.% 959 AKING ACCOUN OF ISENROPIC EFFICIENCY η 0.89 Actual work outut Ideal work outut Actual work outut 980 Actual work outut η is th 87. 959 0.95 980 x 0.89 87. kj/kg or 9.5% Actual work outut 980 D.J.Dunn www.freestudy.co.uk 0

WORKED EXAMPLE No.4 A simle gas turbine uses the Joule cycle. he ressure ratio is 6.5. he air temerature is 00 K at inlet to the comressor and 7 K at inlet to the turbine. he adiabatic index is.4 throughout and the secific heat caacities may be considered constant. i. Calculate the thermal efficiency of the ideal cycle. ii. Calculate the thermal efficiency when the turbine and comressor has an isentroic efficiency of 90%. iii. Sketch the cycle on a -s diagram showing the effect of friction. SOLUION IDEAL CYCLE Comressor urbine 4 ηth 4 00 7 ( 6.5) ( 6.5) 0.86 0.86 5.4 K 80.8 K 80.8 00 0.45 or 4.5% 7 5.4 INCLUDING ISENROPIC EFFICIENCY Comressor urbine η 4 th 56 K 860.7 K 4 η η is is 5.4 00 0.9 00 7 4 0.9 7 80.8 860.7 00 0. or % 7 56 D.J.Dunn www.freestudy.co.uk

D.J.Dunn www.freestudy.co.uk Fig. 4

SELF ASSESSMEN EXERCISE No.. Steam is exanded adiabatically in a turbine from 00 bar and 600oC to 0.09 bar with an isentroic efficiency of 0.88. he mass flow rate is 40 kg/s. Calculate the ower outut. (5 mw). A comressor takes in gas at bar and 5oC and comresses it adiabatically to 0 bar with an isentroic efficiency of 0.89. he mass flow rate is 5 kg/s. Calculate the final temerature and the ower inut. c.005 kj/kg K γ.4 (590 K and.5 MW). A turbine is sulied with kg/s of hot gas at 0 bar and 90oC. It exands adiabatically to bar with an isentroic efficiency of 0.9. Calculate the final temerature and the ower outut. c.005 kj/kg K γ-.4 (66 K and.6 MW) 4. A turbine is sulied with 7 kg/s of hot gas at 9 bar and 850oC that it exands adiabatically to bar with an isentroic efficiency of 0.87. Calculate the final temerature and the ower outut. c.005 kj/kg K γ.4 (667 K and. MW) 5. A simle steam ower lant uses the Rankine cycle. he boiler sulies suerheated steam to the turbine at 00 bar and 550 o C. he condenser oerates at 0.05 bar and roduces saturated water. he ower inut to the um is negligible. i. Calculate the thermal efficiency of the ideal cycle. (4.5%) ii. Calculate the thermal efficiency when the turbine has an isentroic efficiency of 85%. (6.%) 6. A simle gas turbine uses the Joule cycle. he ressure ratio is 7.5. he air temerature is 88 K at inlet to the comressor and 400 K at inlet to the turbine. he adiabatic index is.4 throughout and the secific heat caacities may be considered constant. i. Calculate the thermal efficiency of the ideal cycle. (4.8%) ii. Calculate the thermal efficiency when the turbine and comressor has an isentroic efficiency of 9%. (6.9%) D.J.Dunn www.freestudy.co.uk