APPLIED THERMODYNAMICS TUTORIAL 1 REVISION OF ISENTROPIC EFFICIENCY ADVANCED STEAM CYCLES

Transcription

1 APPLIED HERMODYNAMICS UORIAL REVISION OF ISENROPIC EFFICIENCY ADVANCED SEAM CYCLES INRODUCION his tutorial is designed for students wishing to extend their knowledge of thermodynamics to a more advanced level with ractical alications. Before you start this tutorial you should be familiar with the following. he basic rinciles of thermodynamics equivalent to level. Basic steam cycles, mainly the Rankine and Carnot cycles. Fluid roerty tables and charts mainly a set of standard thermodynamic tables and a h - s chart for steam which you must have in your ossession. he use of entroy. On comletion of the tutorial you should be able to understand isentroic efficiency for turbines and comressors. describe the use of rocess steam. describe the use of back ressure turbines. describe the use of ass out turbines. solve steam cycles involving ass out and back ressure turbines. describe the use of feed heating and suerheating in steam cycles. solve roblems involving feed heating and re-heating. You may be very familiar with all these studies in which case you should roceed directly to section. For those who wish to revise the basics, section should be comleted. his covers entroy. isentroic rocesses. roerty diagrams. isentroic efficiency. D.J.Dunn

2 . REVISION OF ENROPY. DEFINIION Entroy is a roerty which measures the usefulness of energy. It is defined most simly as ds dq/d where S is entroy is temerature Q is heat transfer he units of entroy is hence J/k. he units of secific entroy are J/kg K... ISENROPIC PROCESS ISENROPIC means constant entroy. Usually (but not always) this means a rocess with no heat transfer. his follows since if dq is zero so must be ds. A rocess with no heat transfer is called ADIABAIC. An adiabatic rocess with no friction is hence also ISENROPIC... PROPERY DIAGRAMS he two most commonly used roerty diagrams are i. Enthaly - Entroy (h - s) diagrams and ii. emerature - Entroy ( - s) diagrams. h-s diagrams are commonly used for steam work. he diagram will hence show the saturation curve. You should familiarise yourself with the h-s diagram for steam and ensure that you can use it to find values of h and s for any ressure, temerature or dryness fraction. -s diagrams are commonly used for gas..4. ISENROPIC EFFICIENCY Real exansion and comression rocesses have a degree of friction and this will generate heat which is in effect a heat transfer. increase the entroy. make the final enthaly bigger than it would otherwise be. make the final temerature bigger than it would otherwise be if it is a gas or suerheated vaour. D.J.Dunn

3 An adiabatic rocess with friction has no external heat transfer (ΦWatts or Q Joules) but the internal heat generated causes an increase in entroy. Consider the exansion and comression rocesses on fig. and. fig. fig. he ideal change in enthaly is h ' - h he actual change is h - h he isentroic efficiency is defined as h (actual) h h η is for an exansion. h (ideal) h h ' h (ideal) h ' h η is for a comression. h (actual) h h In the case of a erfect gas h c hence η is for an exansion ' ' η is for a comression Note that for an exansion this roduces a negative number on the to and bottom lines that cancels out. D.J.Dunn

4 WORKED EXAMPLE No. A steam turbine takes steam at 70 bar and 500oC and exands it to 0. bar with an isentroic efficiency 0.9. he rocess is adiabatic. he ower outut of the turbine is 5 MW. Determine the enthaly at exit and calculate the flow rate of steam in kg/s. Note you need the tables and h-s chart for steam. SOLUION h 40 kj/kg (tables) s kj/kg K for an ideal exansion ss' Assuming that the steam becomes wet during the exansion, then s' sf +x'sfg. at 0. bar x' (tables) x' Note if x' is larger than then the steam is still suerheated and the solution does not involve x. Now find h'. h ' h f + x' hfg. at 0. bar h ' 9 + (0.896)(9) 5. kj/kg. Ideal change in enthaly actual change in enthaly h' kj/kg h 0.9(-57.5) -.7 kj/kg actual change in enthaly h (h - h ) -.7 h h 78. kj/kg From the steady flow energy equation (with which you should already be familiar) we have Φ + P H/s Since there is no heat transfer then this becomes P H/s m (h - h) P m(-.7) kw hence m 0.96 kg/s (Note the sign convention used here is negative for energy leaving the system) D.J.Dunn 4

5 WORKED EXAMPLE No. A gas turbine exands gas from MPa ressure and 600 o C to 00 kpa ressure. he isentroic efficiency 0.9. he mass flow rate is kg/s. Calculate the exit temerature and the ower outut. ake cv 78 J/kg K and c 005 J/kg K SOLUION he rocess is adiabatic so the ideal temerature ' is given by ' (r ) -/ r is the ressure ratio r / 0. c /c v.005/ '87(0.) -/ K Now we use the isentroic efficiency to find the actual final temerature. η is ( - )/( ' - ) 0.9 ( - 87)/( ) K Now we use the SFEE to find the ower outut. Φ + P m c ( - ) he rocess is adiabatic Φ 0. P (.005)( ) kw (out of system) D.J.Dunn 5

6 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 he mass flow rate is 40 kg/s. Calculate the enthaly at exit and the ower outut. (Ans. 5 MW). A gas comressor comresses gas adiabatically from bar and 5oC to 0 bar with an isentroic efficiency of he gas flow rate is 5 kg/s. Calculate the temerature after comression and the ower inut. (Ans. -.5 MW) ake cv 78 J/kg K and c 005 J/kg K D.J.Dunn 6

7 . BACK-PRESSURE AND PASS-OU URBINES It is assumed that the student is already familiar with steam cycles as this is necessary for this tutorial. If an industry needs sufficient quantities of rocess steam (e.g. for sugar refining), then it becomes economical to use the steam generated to roduce ower as well. his is done with a steam turbine and generator and the rocess steam is obtained in two ways as follows. 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. 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 he diagram shows the basic circuit. he cycle could use reheat as well but this is not normal. Figure D.J.Dunn 7

8 WORKED EXAMPLE No. For a steam circuit as shown reviously, the boiler roduces suerheated steam at 50 bar and 400oC. his is exanded to bar with an isentroic efficiency of 0.9. he exhaust steam is used for a rocess. he returning feed water is at bar and 40oC. his is umed to the boiler. he water leaving the um is at 40oC and 50 bar. he net ower outut of the cycle is 60 MW. Calculate the mass flow rate of steam. SOLUION Referring to the cycle sketch revious for location oints in the cycle we can find: h 96 kj/kg s kj/kg K For an ideal exansion s s s f +x' sfg at bar x'(5.) x' 0.95 h4 hf + x'hfg at bar h (64) h kj/kg ideal change in enthaly actual change in enthaly kj/kg 0.9(-6) kj/kg he ower outut of the turbine is found from the steady flow energy equation so : P m(-550.9) kw P 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 40oC. Actual 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 m - m m 09.5 kg/s D.J.Dunn 8

9 SELF ASSESSMEN EXERCISE No. A back ressure steam cycle works as follows. he boiler roduces 8 kg/s of steam at 40 bar and 500oC. his is exanded to bar with an isentroic efficiency of he um is sulied with feed water at 0.5 bar and 0oC and delivers it to the boiler at oc and 40 bar. Calculate the net ower outut of the cycle. (Answer 5.4 MW) D.J.Dunn 9

10 .. 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. Figure 4 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. he balance of ower is as follows. Figure 5 P + P increase in enthaly er second. mchc - maha -mbhb From this the value of h C or the mass m C may be determined. his is best shown with a worked examle. D.J.Dunn 0

11 WORKED EXAMPLE No.4 he circuit below shows the information normally available for a feed um circuit. Determine the enthaly at entry to the boiler. SOLUION h A h f 9 kj/kg at 0. bar Figure 6 P(ideal) (5)(0.00)(80-)(00) 9.5 kw P (actual) 9.5/ kw P(ideal) (40)(0.00)(80-0.)(00) 9.6 kw P (actual) 9.6/ kw otal ower inut kw hb 84 kj/kg (from water tables or aroximately hf at 0 o C) hence h C - 40 h A - 5h B hc - 40(9) - 5(84) hc 90 kj/kg D.J.Dunn

12 WORKED EXAMPLE No.5 he following worked examle will show you to solve these roblems. A assout turbine lant works as shown in fig. 4. he boiler roduces steam at 60 bar and 500 o C which 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. he isentroic efficiency is 0.9 for the overall exansion. 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 0oC. he isentroic efficiency of the ums is 0.8. he net ower outut of the cycle is 40 MW. Calculate:. the flow rate of steam from the boiler.. the heat inut to the boiler.. the thermal efficiency of the cycle. SOLUION URBINE EXPANSION h 4 kj/kg from tables. h 5 ' 65 kj/kg using isentroic exansion and entroy. 0.9 (4 - h5) / (4-65) hence h5 9 kj/kg Sketching the rocess on the h - s chart as a straight line enables h4 to be icked off at bar. h4 770 kj/kg. POWER OUPU Pout m(h - h4 ) + (m - 4)(h4 - h5) P out m(4-770) + (m - 4)(770-9) P out 65 m m - 96 Figure 7 D.J.Dunn

13 POWER INPU he ower inut is to the two feed ums. Figure 8 h 6 84 kj/kg (water at bar and 0oC) h h f at 0.09 bar 8 kj/kg. P (ideal) change in flow energy 4 x 0.00 x (60 - ) x 00 kw.6 kw P (actual).6 / kw P(actual) (m-4) x 0.00 x ( ) x 00/ m kw NE POWER ENERGY BALANCE ON PUMPS kw P out - P - P m m m m - 96 hence m 7.4 kg/s hence HEA INPU P 9.5 kw P 49.4 kw (using the value of m just found) m h (m-4) h + P + P 7. h.4 x h 7 kj/kg Heat inut m(h - h) 55 kw EFFICIENCY Efficiency η 40/. % D.J.Dunn

14 SELF ASSESSMEN EXERCISE No.. A steam turbine lant is used to suly rocess steam and ower. he lant comrises an economiser, boiler, suerheater, 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 suerheater are 500oC and 70 bar, and at outlet from the turbine the ressure is 0. bar. he overall isentroic efficiency of the turbine is 0.87 and that of the feed um is 0.8. Assume that the exansion is reresented by a straight line on the h-s chart. he makeu water is at 5oC and bar and it is umed into the feed line with an isentroic efficiency 0.8 to relace the lost rocess steam. 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. Calculate the rate of steam flow leaving the suerheater and the rate of heat transfer to the boiler including the economiser and suerheater. Sketch clear - s and h-s and flow diagrams for the lant. (9.46 kg/s 95. MW). he demand for energy from an industrial lant is a steady load of 60 MW of rocess heat at 7 o C and a variable demand of u to 0 MW of ower to drive electrical generators. he steam is raised in boilers at 70 bar ressure and suerheated to 500oC. he steam is exanded in a turbine and then condensed at 0.05 bar. he rocess heat is rovided by the steam bled from the turbine at an aroriate ressure, and the steam condensed in the rocess heat exchanger is returned to the feed water line. Calculate the amount of steam that has to be raised in the boiler. Assume an overall isentroic efficiency of 0.88 in the turbine. he exansion is reresented by a straight line on the h-s diagram. Neglect the feed um work. (Answer 6 kg/s). D.J.Dunn 4

15 . ADVANCED SEAM CYCLES In this section you will extend your knowledge of steam cycles in order to show that the overall efficiency of the cycle may be otimised by the use of regenerative feed heating and steam re-heating. Regenerative feed heating is a way of raising the temerature of the feed water before it reaches the boiler. It does this by using internal heat transfer within the ower cycle. Steam is bled from the turbines at several oints and used to heat the feed water in secial heaters. In this way the temerature of the feed water is raised along with the ressure in stages so that the feed water is nearly always saturated. he heat transfers in the heaters and in the boiler are conducted aroximately isothermally. Studies of the Carnot cycle should have taught you that an isothermal heat transfer is reversible and achieves maximum efficiency. he ultimate way of conducting feed heating is to ass the feed water through a heat exchanger inside the turbine casing. In this way the temerature of the steam on one side of heat exchanger tubes is equal to the temerature of the water on the other side of the tubes. Although the temerature is changing as water and steam flow through heat exchanger, at any one oint, the heat transfer is isothermal. If no suerheating nor undercooling is used then the heat transfers in the boiler and condenser are also isothermal and efficiencies equal to those of the Carnot cycle are theoretically ossible. Figure 9 here are several reasons why this arrangement is imractical. Most of them are the same reasons why a Carnot cycle is imractical. i. he steam would be excessively wet in the turbine. ii. Placing a heat exchanger inside the turbine casing is mechanically imossible. iii. he ower outut would be small even though the cycle efficiency would be high. D.J.Dunn 5

16 Steam reheating is another way of imroving the thermodynamic efficiency by attemting to kee the steam temerature more constant during the heat transfer rocess inside the boiler. Figure 0 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. 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 which further imrove the efficiency and are necessary for ractical oeration. For examle air extraction at the condenser, steam recovery from turbine glands, de-suerheaters, de-aerators and so on. hese can be found in details in textbooks devoted to ractical steam ower lant. D.J.Dunn 6

17 4. FEED HEAING 4.. PRACICAL DESIGNS Practical feed heaters may be heat exchangers with indirect contact. 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. he tye which you should learn is the oen or direct contact mixing tye. he bled steam is mixed directly with the feed water at the aroriate ressure and condenses and mixes with the feed water. Comare a basic Rankine cycle with a similar cycle using one such feed heater. Figure Figure D.J.Dunn 7

18 4.. ENERGY BALANCE FOR MIXING FEED HEAER Consider a simle mixing tye feed heater. he bled steam at () is mixed directly with incoming feed water (6) resulting in hotter feed water (7). Figure Mass of bled steam y kg Mass of feed water entering - y kg Doing an energy balance we find y h + (-y)h 6 h 7 WORKED EXAMPLE No.6 A feed heater is sulied with condensate at 0. bar. he bled steam is taken from the turbine at 0 bar and 0.95 dry. Calculate the flow rate of bled steam needed to just roduce saturated water at outlet. SOLUION Assumtions Figure 4. Energy inut from um is negligible.. No energy is lost..he heater ressure is the same as the bled ressure. In this case h6 hf at 0. bar 9 kj/kg h 7 h f at 0 bar 008 kj/kg h hf + xhfg at 0 bar h (795) 7. kj/kg ENERGY BALANCE y(7.) (-y)(9) hence y 0.44 kg Note that it is usual to calculate these roblems initially on the basis of kg coming from the boiler and returning to it. D.J.Dunn 8

19 4.. CYCLE WIH ONE FEED HEAER Figure 5 If only one feed heater is used, the steam is bled from the turbine at the oint in the exansion where it just becomes dry saturated and the saturation temerature is estimated as follows. t s (bleed) {t s (high ressure) + t s (low ressure)}/ For examle a cycle oerating between 40 bar and 0.05 bar. t s (40 bar ) 50. oc ts (0.05 bar ) 6.7 o C ts (bleed ) ( )/ 8.5 o C he ressure corresonding to this is.5 bar so this is the bleed ressure. D.J.Dunn 9

20 WORKED EXAMPLE No.7 A Rankine cycle works between 40 bar, 400oC at the boiler exit and 0.05 bar at the condenser. Calculate the efficiency with no feed heating. Assume isentroic exansion. Ignore the energy term at the feed um. SOLUION Figure 6 h 4 kj/kg s kj/kg K ss x x h h f + x hfg (48) 04.6 kj/kg h 4 h f at 0.05 bar kj/kg Φ h - h 0 kj/kg into boiler. P h - h 89.4 kj/kg (out of turbine) η P/ Φ 8. % D.J.Dunn 0

21 WORKED EXAMPLE No.8 Reeat the last examle but this time there is one feed heater. SOLUION Figure 7 he bleed ressure was calculated in an earlier examle and was.5 bar. s s kj/kg K x x (not quite dry). h h f + x hfg (48) h 66 kj/kg h 7 h f at.5 bar 584 kj/kg Neglecting um ower h6 h5 hf kj/kg h h kj/kg Conducting an energy balance we have yh + (-y) h 6 h 7 hence y 0.85 kg Φ h - h 60 kj/kg into boiler. Rather than work out the ower from the turbine data, we may do it by calculating the heat transfer rate from the condenser as follows. Φout (-y)(h4 - h5) 0.85( ) kj/kg P Φin - Φout 07 kj/kg (out of turbine) η P/ Φin 40.8 % Note that the use of the feed heater roduced an imrovement of.5 % in the thermodynamic efficiency. D.J.Dunn

22 SELF ASSESSMEN EXERCISE No.4 A simle steam lant uses a Rankine cycle with one regenerative feed heater. he boiler roduces steam at 70 bar and 500oC. his is exanded to 0. bar isentroically. Making suitable assumtions, calculate the cycle efficiency. (4.8%) CYCLE WIH WO FEED HEAERS When two (or more) feed heaters are used, the efficiency is further increased. the rinciles are the same as those already exlained. he mass of bled steam for each heater must be determined in turn starting with the high ressure heater. It is usual to assume isentroic exansion that enables you to ick off the enthaly of the bled steam from the h-s chart at the ressures stated. D.J.Dunn

23 WORKED EXAMPLE No.9 A steam ower lant works as follows. he boiler roduces steam at 00 bar and 600oC. his is exanded isentroically to 0.04 bar and condensed. Steam is bled at 40 bar for the h.. heater and 4 bar for the l.. heater. Solve the thermodynamic efficiency. SOLUION Figure 8 Figure 9 Ignoring the energy inut from the um we find: h h0 hf 40 bar 087 kj/kg h9 h8 hf 4 bar 605 kj/kg h7 h6 hf 0.04 bar kj/kg H.P. HEAER xh + (-x)h 9 h 0 0x + 605(-x) 087 hence x 0.78 kg D.J.Dunn

24 L.P. HEAER (-x)h8 yh4 + (-x-y)h7 0.8(605) 740y + (0.8-y)() y 0.5 kg BOILER heat inut Φin h - h kj/kg CONDENSER heat outut Φout (-x-y))(h5 - h6 ) Φout 0.67(080-).5 kj/kg POWER OUPU P Φin - Φout 4.5 kj/kg η P/ Φin 48. % D.J.Dunn 4

25 SELF ASSESSMEN EXERCISE No. 5. Exlain how it is theoretically ossible to arrange a regenerative steam cycle which has a cycle efficiency equal to that of a Carnot cycle. In a regenerative steam cycle steam is sulied from the boiler lant at a ressure of 60 bar and a temerature of 500 o C. Steam is extracted for feed heating uroses at ressures of 0 bar and.0 bar and the steam turbine exhausts into a condenser oerating at 0.05 bar. Calculate the aroriate quantities of steam to be bled if the feed heaters are of the oen tye, and find the cycle efficiency; base all calculations on unit mass leaving the boiler. Assume isentroic exansion in the turbine and neglect the feed um work. (Answers 0.69 kg/s, 0.45 kg/s and 45 %). he sketch shows an idealised regenerative steam cycle in which heat transfer to the feed water in the turbine from the steam is reversible and the feed um is adiabatic and reversible. he feed water enters the um as a saturated liquid at 0.0 bar, and enters the boiler as a saturated liquid at 00 bar, and leaves as saturated steam. Draw a -s diagram for the cycle and determine, not necessarily in this order, the dryness fraction in state, the cycle efficiency and the work er unit mass. (Answers 0.69 kg/s, 50% and kj/kg). Outline the ractical difficulties that are involved in realising this cycle and exlain how regenerative cycles are arranged in ractice. Figure 0 Note oint (6) is the oint in the steam exansion where the feed water enters and resumably the temeratures are equal. here is further exansion from (6) to (). D.J.Dunn 5

26 5. REHEA CYCLES We shall only examine cycles with one stage of reheating and two turbine stages, high ressure and low ressure. You should refer to text books on ractical steam turbine layouts to see how low, medium and high ressure turbines are configured and laid out in order to roduced axial force balance on the rotors. he diagram below shows a basic circuit with one stage of reheating. Figure You should be roficient at sketching the cycle on a - s diagram and a h - s diagram. hey are shown below for the cycle shown above. Figure he calculations for this cycle are not difficult. You need only take into account the extra heat transfer in the reheater. D.J.Dunn 6

27 WORKED EXAMPLE No.0 A reheat cycle works as follows. he boiler roduces 0 kg/s at 00 bar and 400oC. his is exanded isentroically to 50 bar in the h.. turbine and returned for reheating in the boiler. he steam is reheated to 400 o C. his is then exanded in the l.. turbine to the condenser which oerates at 0. bar. he condensate is returned to the boiler as feed. Calculate the net ower outut and the cycle efficiency. SOLUION h6 hf at 0. bar 5 kj/kg h 097 kj/kg at 00 bar and 400 o C. From the h-s chart we find h 90 kj/kg h 4 96 kj/kg h 5 89 kj/kg If we ignore the feed um ower then Φin at boiler 0(h -h ) + 0(h4 -h ) 9 60 kw or 9.60 MW Φout at condenser 0(h 5 -h 6 ) 58.4kW P(net) Φ in - Φ out 50 kw or 5. MW η P(net) /Φ in 7.7 % D.J.Dunn 7

28 SELF ASSESSMEN EXERCISE No. 6. Reeat worked examle No.0 but this time do not ignore the feed um term and assume an isentroic efficiency of 90% for each turbine and 80% for the um. ( Answers. MW,5%). A water-cooled nuclear reactor sulies dry saturated steam at a ressure of 50 bar to a two-cylinder steam turbine. In the first cylinder the steam exands with an isentroic efficiency of 0.85 to a ressure of 0 bar, the ower generated in this cylinder being 00 MW. he steam then asses at a constant ressure of 0 bar through a water searator from which all the water is returned to the reactor by mixing it with the feed water. he remaining dry saturated steam then flows at constant ressure through a reheater in which its temerature is raised to 50 o C before it exands in the second cylinder with an isentroic efficiency of 0.85 to a ressure of 0. bar, at which it is condensed before being returned to the reactor. Calculate the cycle efficiency and draw u an energy balance for the lant. Neglect the feed um work. (Answer 0.%). Steam is raised in a ower cycle at the suercritical ressure of 50 bar and at a temerature of 600oC. It is then exanded in a turbine to 5 bar with an overall isentroic efficiency of At that ressure some steam is bled to an oen regenerative feed heater, and the remainder of the steam is, after reheating to 600oC, exanded in a second turbine to the condenser ressure of 0.04 bar, again with an isentroic efficiency of he feed ums each have an overall isentroic efficiency of Calculate the amount of steam to be bled into the feed heater, making the usual idealising assumtions. Also calculate the cycle efficiency. Use the h-s chart wherever ossible and do not neglect feed um work. (Answers 0.79 kg/s and 47%) D.J.Dunn 8

29 APPLIED HERMODYNAMICS UORIAL GAS COMPRESSORS In order to comlete this tutorial 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 the following. Describe the working rinciles of recirocating comressors. Define and calculate swet volume. Define and calculate volumetric efficiency. Define and calculate isothermal efficiency. Define and calculate indicated ower. Dtate the benefits of cooling. Calculate the heat rejected through cooling. Define and calculate the interstage ressures for multile comressors. Define olytroic efficiency. Let s start by considering the general use of comressed air. D.J.Dunn

30 . COMPRESSED AIR. YPES 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. hese are examles. 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. he same rinciles are alied to the comression of other gasses. his tutorial is mainly about recirocating comressors. he other tyes are covered briefly. he recirocating comressor is robably the most versatile of all the tyes and is only out erformed by rotary tyes when large volumes at low ressures are required. For high ressures, the recirocating comressor is almost universal.. AMOSPHERIC VAPOUR Air and vaour mixtures are covered in detail in a later tutorial. We should note, however, the effects it has on the erformance of an air comressor. Atmosheric air contains WAER VAPOUR mixed with the other gases. When the air is cooled to the dew oint, the vaour condenses into water and we see rain or fog. 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 evaorate. 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 refers to RELAIVE HUMIDIY. For examle if the air has 40% 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. 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. D.J.Dunn

31 Water causes damage to air tools, ruins aint srays and corrodes 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 COMPRESSOR LAYOU he diagram below shows the layout of a two stage recirocating comressor tyically for sulying a worksho. Figure. 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.4 FREE AIR DELIVERY 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 DELIVERY or FAD. FAD refers to the volume the air would have if let out of the ie and returned to atmosheric ressure at the same temerature. he FAD is 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 Va (the FAD) and the comressed conditions are, V and. D.J.Dunn

32 Alying the gas law we have V V a ava V a a a F. A. D.. CYCLE FOR RECIPROCAING COMPRESSOR. HEOREICAL CYCLE he diagram shows the basic design of a recirocating comressor. he iston recirocates drawing in gas, comressing it and exelling it when the ressure inside the cylinder reaches the same level as the ressure in the delivery ie. Figure If the iston exels all the air and there is no restriction at the valves, the ressure - volume cycle is as shown below. Figure Gas is induced from 4 to at the inlet ressure. It is then traed inside the cylinder and comressed according the law Vn 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 4

33 . VOLUMERIC 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 Vn 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. Figure 4 he volumetric efficiency is defined as Induced Volume η vol Swet Volume 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 5

34 WORKED EXAMPLE No. Gas is comressed in a recirocating comressor from bar to 6 bar. he FAD is dm /s. he clearance ratio is he exansion art of the cycle follows the law V. C. he crank seed is 60 rev/min. Calculate the swet volume and the volumetric efficiency. SOLUION Swet Volume V Clearance volume 0.05 V Consider the exansion from to 4 on the -V diagram. 4 bar 6 bar. V. 4 V 4. 6(0.05V). (V 4. ) V 4 0.V or.%% of V F.A.D. 0.0 m/s. Induced volume V - V 4.05V - 0.V 0.88V Induced volume 0.0 V 0.0/ m/s Crank seed 6 rev/s so the swet volume 0.057/6.6 dm. V V V.05V η vol Induced Volume Swet Volume 0.88V η vol V 8.8 % D.J.Dunn 6

35 D.J.Dunn 7 WORKED EXAMPLE No. Show that if the clearance ratio of an ideal single stage recirocating comressor is c that the volumetric efficiency is given by ) / ( n L H vol c η where L is the inlet ressure and H the outlet ressure. SOLUION Swet volume V V Induced volume V - V 4 Clearance volume V ( ) ( ) V V V V V V V V / / / / / n L H vol n L H vol n L H vol n L H n vol vol c c c c c c c V V V c c c c V V V V V V c V V V V η η η η η

36 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 this. Figure 5 D.J.Dunn 8

37 WORKED EXAMPLE No. A single stage recirocating comressor roduces a FAD of dm /s at 40 rev/min. he inlet conditions are bar and 0 o C. he olytroic index is. for the comression and exansion. he outlet ressure is 8 bar. he clearance volume is 0 cm. Due to the restriction of the inlet valve and the warming effect of the cylinder walls, the ressure at the start of comression is 0.97 bar and the temerature is 7 o C. Determine the volumetric efficiency. SOLUION Because the induction stroke is neither at constant ressure nor constant temerature, we must solve the swet volume by using the exulsion stroke, which is assumed to be at constant ressure and temerature. he numbers of the cycle oints are as before K 0.97 x60 F. A. D. er stroke dm er stroke xaH he comressed volume exelled Exulsion volume V V n 8x0.0 V V dm n 0.96V V V.. 4 xv V n 4 n 8x dm hence V. Induced volume V V Swet volume V V η vol % 0.5 V xx dm 8x dm 0.06 dm 0.5 dm a H D.J.Dunn 9

38 D.J.Dunn 0. INDICAED POWER he indicated work er cycle is the area enclosed by the - V diagram. he easiest way to find this is by integrating with resect to the ressure axis. Figure 6 he rocesses to and to 4 are olytroic V n C. V C /n -/n C V /n Consider the exression ) ( ) ( ) / ( / / / / / / n nv n nv n C d C Vd n n n n n n Between the limits of and this becomes nv V n ( ) Between the limits 4 and this becomes nv V n 4 4 ( ) he indicated work (inut) is then W n V V n n V V n 4 4 ( ) ( ) n n n n n n n n r V n n r V n n W V n n V n n W V V V n n V V V n n W Where r is the ressure ratio. Since 4 this reduces to [ ] 4 V V r n n W n n. Where V - V4 is the swet volume.

39 If the clearance volume is neglected this becomes W n V r n n n Since V mr W n mr r n n n Note that if the rocess was isothermal and n then the integration would yield W mr l ( / ) n m is the mass comressed 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 If the clearance volume is neglected, the mass comressed is the mass exelled. In this case the actual mass flow rate delivered may be used for m and W becomes the indicated ower..4 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 η(iso) Isothermal work/actual work ( n ) ln( / ) n ( ) Note that in the ideal case, and are the inlet and outlet temeratures. D.J.Dunn

40 SELF ASSESSMEN EXERCISE No. Show how the volumetric efficiency of an ideal single stage recirocating air comressor may be reresented by the equation η vol ( / n) H c L Where c is the clearance ratio, H the delivery ressure and L the induction ressure. A recirocating air comressor following the ideal cycle has a free air delivery of 60 dm /s. he clearance ratio is 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. (4.7 kw) D.J.Dunn

41 . MULIPLE COMPRESSOR SAGES. HE EFFEC OF INERCOOLING he advantage of comressing the fluid in stages is that intercoolers may be used and the overall comression is nearer to being isothermal. Consider the - V diagram for a two stage comressor. Figure 7 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 volume entering the high ressure stage is V5 and not V. 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.. OPIMAL INERSAGE 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. mrn( - ) mrn(6-5 ) W + (n -) (n -) (-/n) (-/n) 6 Since and then assuming the same value of n for each stage n n 6 W mr + mr ( n -) ( n -) ( / n) ( / n) 6 5 D.J.Dunn

42 Since dw mr d ( / n) ( / n) ( n )/ n / n ( n) / n ( n )/ n L m M - mr H n M n6 H W mr + mr ( n -) L ( n -) M For a minimum value of W we differentiate with resect to and equate to zero. M 5 and 6 5 H and M 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 ) ½ It can further be shown that when this is the case, the work done by both stages are 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 L M D.J.Dunn 4

43 WORKED EXAMPLE No.5 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 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. he ideal interstage ressure. he ideal indicated ower er stage. he heat rejected from each cylinder. he heat rejected from each intercooler. he isothermal efficiency. he swet volume of each stage. 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 V/R x 0 5 x /(87 x 88) kg/s 88(4) 0./ K he indicated ower for each stage is the same so it will be calculated for the st. stage. - n mrn I. P. x since m is the mass comressed. n x 87 x. x 88. I. P Watts. - D.J.Dunn 5

44 CYLINDER COOLING Consider the energy balance over the first stage. Balancing the energy we have Figure 8 H A + P(in) H B + Φ (out) Φ (out) P(in) - mc( B - A ) Φ (out) x.005 ( ) 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( - 5) x.005 ( ) 7.49 kw ISOHERMAL EFFICIENCY he ideal isothermal ower mrln(/) er stage. P(isothermal) x 87 x 88 ln kw η(iso) 7.846/ % SWEP VOLUMES Consider the first stage. he F.A.D. is 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 Volume x Seed and the seed is 6 rev/s D.J.Dunn 6

45 Hence S.V. (st. Stage) / m S.V. Bore Area x Stroke π D /4 x 0. D 0.47 m. Now consider the second stage. he air is returned to atmosheric ressure 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 / m π D /4 x 0. hence D 0.7 m By the same reasoning the swet volume of the third stage is SV(rd stage) / m π D /4 x 0. D m D.J.Dunn 7

46 SELF ASSESSMEN EXERCISE No.. 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 0 C. here is an intercooler between stages which 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. Ignore the effect of the clearance volume. Calculate the following. he intermediate ressure for minimum indicated work. (6.65 bar) he theoretical indicated ower for each stage. (.85 kw) he heat rejected in each cylinder. (6. kw) he heat rejected by the intercooler. (6.5 kw) he swet volumes of both stages. (.4 dm and 5. dm ) What advantage is there in using an after-cooler? State the effect on your answers of not ignoring the clearance volume and leakages.. A single acting stage comressor draws in free air and comresses it to 8.5 bar. he comressor runs at 600 rev/min. he atmosheric conditions are.0 bar and 5 o C. he interstage ressure is bar and the intercooler cools the air back to 0 o C. he olytroic index for all comressions is.8. Due to the effect of warming from the cylinder walls and the ressure loss in the inlet valve, the ressure and temerature at the start of the low ressure comression stroke is 0.96 bar and 5 o C. he high ressure cycle may be taken as ideal. he clearance volume for each stages is 4% of the swet volume of that stage. he low ressure cylinder is 00 mm diameter and the stroke for both stages is 60 mm. Calculate the following. he free air delivery. (5.858m /min) he volumetric efficiency of the low ressure stage. (86. %) he diameter of the high ressure cylinder. (7 mm) he indicated ower for each stage. (4.6 kw and.4 kw) D.J.Dunn 8

47 . A stage recirocating air comressor has an intercooler between stages. he induction and exulsion for both stages are at constant ressure and temerature. All the comressions and exansions are olytroic. Neglecting the effect of the clearance volume show that the intermediate ressure, which gives minimum, indicated work is M ( L H ) ½ Exlain with the aid of a sketch how the delivery temerature from both cylinders varies with the intermediate ressure as it changes from L to H. 4.a. Prove that the ideal volumetric efficiency of a single stage recirocating comressor is ηvol - c(r/n-) r is the ressure ratio, n is the olytroic index and c the clearance ratio. Sketch curves of ηvol against r for tyical values of n and c. b. 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. Find the free air delivery and mass of air that can be comressed er kw h of work inut. (0.06 m/kw h.7 kg/kw h) Find the ratio of the cylinder diameters if the iston have the same stroke. Neglect the effect of the clearance volume. (d/d 0.47) D.J.Dunn 9

48 4. POLYROPIC or SMALL SAGE EFFICIENCY his is an alternative way of aroaching isentroic efficiency. In this method, the comression is suosed to be made u of many stages, each raising the ressure a small amount. he theory alies to any tye of comressor. For an adiabatic gas comression the law of comression and the gas law may be combined. d C Divide by d d C d d C d d d Since differentiate and C For an isentroic comression, let the final temerature be designated ' and the change in temerature be '. d ' + d d...() Isentroic efficiency and + d ' Figure 9 Let the change be infinitesimally smallsuch that ' d ' he ratio is designated η d η is called the POLYROPIC EFFICIENCY. ' D.J.Dunn 0

49 D.J.Dunn If we think of the comression as being made u of many tiny stes each with the same value of η. η η η η η η η ln ln ' ' d d Integrate d d d d d d is the starting temerature and is the final temerature. he overall efficiency is η 0 ' as exected. with Comaring SUBSIUE PROCESS POLYROPIC PROCESS ADIABAIC ' ' ' ' η η η η η η η r r r r r r r r r r r n n O n n his theory may be alied to exansions as well as comressions. It may also be alied to exansions in nozzles. In steam work, it is more usual to use the REHEA FACOR, which is based on the same rincile.

50 WORKED EXAMPLE 6 A comressor draws in air at 5 o C and 0. bar. he air is comressed to.6 bar with a olytroic efficiency of Determine the temerature and the isentroic efficiency. ake.4 SOLUION η (0.86) 88 ' 88(5.) ηis ( 5.) K 464.5K D.J.Dunn

51 SELF ASSESSMEN EXERCISE No.. Show that for any comression rocess the overall efficiency is given by r ηo η r where η is the olytroic efficiency. Determine the index of comression for a gas with an adiabatic index of.4 and a olytroic efficiency of 0.9. (.465) Determine the overall efficiency when the ressure comression ratio is 4/ and 8/. (0.879 and 0.866). A comressor draws in air at. K temerature and 0.65 bar ressure. he comression ratio is 6. he olytroic efficiency is Determine the temerature after comression. ake.4 (405 K) D.J.Dunn

52 APPLIED HERMODYNAMICS UORIAL No. GAS URBINE POWER CYCLES In this tutorial you will do the following. Revise gas exansions in turbines. Revise the Joule cycle. Study the Joule cycle with friction. Extend the work to cycles with heat exchangers. Solve tyical exam questions. D.J.Dunn

53 . REVISION OF EXPANSION AND COMPRESSION PROCESSES. When a gas is exanded from ressure to ressure adiabatically, the temerature ratio is given by the formula he same formula may be alied to a comression rocess. Always remember that when a gas is exanded it gets colder and when it is comressed it gets hotter. he temerature change is - If there is friction the isentroic efficiency (η is) is exressed as η is (ideal)/ (actual) for a comression. η is (actual)/ (ideal) for an exansion. An alternative way of exressing this is with POLYROPIC EFFICIENCY For a comression from () to () the temerature ratio is exressed as follows. η and for an exansion from () to () ( ) η where η is called the olytroic efficiency. WORKED EXAMPLE No. A gas turbine exands 4 kg/s of air from bar and 900oC to bar adiabatically with an isentroic efficiency of 87%. Calculate the exhaust temerature and the ower outut..4 c 005 J/kg K SOLUION (/) -/.4 7 (/) K Ideal temerature change K Actual temerature change 87% x K Exhaust temerature K he steady flow energy equation states Φ + P change in enthaly/s Since it is an adiabatic rocess Φ 0 so P H/s m c 4 x 005 x (5.5) P -.5 x 06 W (Leaving the system) P(out).5 MW η D.J.Dunn

54 SELF ASSESSMEN EXERCISE No.. A gas turbine exands 6 kg/s of air from 8 bar and 700oC to bar isentroically. Calculate the exhaust temerature and the ower outut..4 c 005 J/kg K (Answers 57. K and.68 MW). A gas turbine exands kg/s of air from 0 bar and 90oC to bar adiabatically with an isentroic efficiency of 9%. Calculate the exhaust temerature and the ower outut..4 c 00 J/kg K (Answers 657. K and.6 MW). A gas turbine exands 7 kg/s of air from 9 bar and 850oC to bar adiabatically with an isentroic efficiency of 87%. Calculate the exhaust temerature and the ower outut..4 c 005 J/kg K (Answers K and.04 MW) D.J.Dunn

55 . HE BASIC GAS URBINE CYCLE he ideal and basic cycle is called the JOULE cycle and is also known as the constant ressure cycle because the heating and cooling rocesses are conducted at constant ressure. A simle layout is shown on fig.. Figure Illustrative diagram. he cycle in block diagram form is shown on fig.. Fig. Block diagram D.J.Dunn 4

56 here are 4 ideal rocesses in the cycle. - Reversible adiabatic (isentroic) comression requiring ower inut. P in H/s m c ( - ) - Constant ressure heating requiring heat inut. Φ in H/s m c ( - ) - 4 Reversible adiabatic (isentroic) exansion roducing ower outut. P out H/s m c ( - 4 ) 4 - Constant ressure cooling back to the original state requiring heat removal. Φ out H/s m c ( 4 - ) he ressure - volume, ressure - enthaly and temerature-entroy diagrams are shown on figs. a, b and c resectively.. EFFICIENCY Fig a -V diagram Fig b -h diagram Fig c -s diagram he efficiency is found by alying the first law of thermodynamics. Φ Φ η th nett in P - Φ P Φ out nett in nett P out - P Φ Φ out in in mc mc ( ( 4 - ) - ) ( ( 4 - ) - ) 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. r r r is the ressure comression ratio for the turbine and comressor. D.J.Dunn 5

57 D.J.Dunn 6.4 since ) - ( ) - ( η η 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.

58 WORKED EXAMPLE No. 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..4 c.005 kj/kg K SOLUION η r Φ η P th th in nett r mc P Φ nett in ( 6 8 x 6 - ) 0.4 x kw K or 40% 0. x.005 x ( ) 50.8 kw SELF ASSESSMEN EXERCISE No. 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.. he cycle efficiency.. he heat transfer into the heater.. the nett ower outut..4 c.005 kj/kg K (Answers 4.7 %, 06.7 kw and 88.6 kw) D.J.Dunn 7

59 . HE EFFEC OF FRICION ON HE JOULE CYCLE. URBINE he isentroic efficiency for a gas turbine is given by: ηi (Actual change in enthaly)/(ideal change in enthaly) ηi (Actual change in temerature)/(ideal change in temerature). COMPRESSOR For a comressor the isentroic efficiency is inverted and becomes as follows. ηi (Ideal change in enthaly)/(actual change in enthaly) hi (Ideal change in temerature)/(actual change in temerature) Remember that friction always roduces a smaller change in temerature than for the ideal case. his is shown on the -s diagrams (fig.4a and 4b). Fig.4a urbine exansion. Fig.4b Comression rocess. η i (( 4 )/( 4 ) ηi ( )/( ) he ower outut from the turbine is hence he ower inut to the comressor is hence P(out) m c ( 4 ) ηi P(in) m c ( )/ηi. HE CYCLE WIH FRICION It can be seen that the effect of friction on the gas turbine cycle is reduced ower outut and increased ower inut with an overall reduction in nett ower and thermal efficiency. Figs. 5a and 5b show the effect of friction on -s and -h diagrams for the Joule cycle. Fig.5a emerature - Entroy Fig.5b. Pressure - Enthaly D.J.Dunn 8

60 Note the energy balance which exists is P(in) + Φ(in) P(out) + Φ(out) P(nett) P(out) - P(in) Φ(nett) Φ(in) - Φ(out) WORKED EXAMPLE No. A Joule Cycle uses a ressure ratio of 8. Calculate the air standard efficiency. he isentroic efficiency of the turbine and comressor are both 90%. he low ressure in the cycle is 0 kpa. he coldest and hottest temeratures in the cycle are 0oC and 00oC resectively. Calculate the cycle efficiency with friction and deduce the change. Calculate the nett ower outut..4 and c.005 kj/kg K. ake the mass flow as kg/s. SOLUION No friction ηth - r / or 48.8 % With friction ' 9 x K ηi 0.9 (5-9)/(-9) 5 K 4' 47/ K ηi 0.9 (47-4)/(47-8.7) ηth - Φ(out)/Φ(in) - (4-)/(-) ηth 0.6 or 6% he change in efficiency is a reduction of 8.8% Φ(in) m c (-) x.005 x (47-557) 760 kw Nett Power Outut P(nett) ηth x Φ(in) 0.6 x kw D.J.Dunn 9

61 SELF ASSESSMEN EXERCISE No. A gas turbine uses a standard Joule cycle but there is friction in the comressor and turbine. he air is drawn into the comressor at bar 5oC and is comressed with an isentroic efficiency of 94% to a ressure of 9 bar. After heating, the gas temerature is 000oC. he isentroic efficiency of the turbine is also 94%. he mass flow rate is. kg/s. Determine the following.. he net ower outut.. he thermal efficiency of the lant..4 and c.005 kj/kg K. (Answers 6 kw and 40.4%) D.J.Dunn 0

62 4. VARIANS OF HE BASIC CYCLE In this section we will examine how ractical gas turbine engine sets vary from the basic Joule cycle. 4. 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. 4. 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.6a. shows such a layout with turbines in arallel configuration. Fig.6b shows the layout with series configuration. Fig. 6a Parallel turbines Fig.6b. Series turbines D.J.Dunn

63 4. INERCOOLING his is not art of the syllabus for the ower cycles but we will come across it later when we study comressors in detail. 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 layout is shown on fig. 7a. 4.4 REHEAING he reverse theory of intercooling alies. If several stages of exansion are used and the gas reheated between stages, the ower outut and efficiency is increased. he layout is shown on fig. 7b. Fig.7a. Intercooler Fig.7b. Reheater D.J.Dunn

64 WORKED EXAMPLE No.4 A gas turbine draws in air from atmoshere at bar and 0oC and comresses it to 5 bar with an isentroic efficiency of 80%. 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. he isentroic efficiency is 85% for both stages. Calculate the mass flow of air, the inter-stage ressure of the turbines and the thermal efficiency of the cycle. For the comressor.4 and for the turbines.. he gas constant R is 0.87 kj/kg K for both. Neglect the increase in mass due to the addition of fuel for burning. SOLUION c R and R c cv hence c c v Hence c.005 kj/kg K for the comressor and.49 kj/kg K for the turbines. COMPRESSOR 0.86 r 8 x K Power inut to comressor m c ( - ) Power outut of h.. turbine m c ( - 4 ) Since these are equal it follows that.005( ).49(00-4 ) K D.J.Dunn

65 HIGH PRESSURE URBINE 4 ηi 0.85 hence K hence LOW PRESSURE URBINE 5.9 ηi hence 4 hence K.9 bar 88.5 K NE POWER he nett ower is 80 kw hence 80 m c ( 4-5 ) m x.49( ) m 0.4 kg/s HEA INPU Φ(in) m c ( - ) 0.4 x.49 ( ) 45. kw HERMAL EFFICIENCY ηth P(nett)/Φ(in) 80/ or.% D.J.Dunn 4

66 SELF ASSESSMEN EXERCISE No. 4 A gas turbine draws in air from atmoshere at bar and 5oC and comresses it to 4.5 bar with an isentroic efficiency of 8%. 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. he isentroic efficiency is 85% for both stages. For the comressor.4 and for the turbines.. he gas constant R is 0.87 kj/kg K for both. Neglect the increase in mass due to the addition of fuel for burning. Calculate the mass flow of air, the inter-stage ressure of the turbines and the thermal efficiency of the cycle. (Answers 0.64 kg/s and 0. %) D.J.Dunn 5

67 4.5. EXHAUS HEA EXCHANGERS Because the gas leaving the turbine is hotter than the gas leaving the comressor, it is ossible to heat u the air before it enters the combustion chamber by use of an exhaust gas heat exchanger. his results in less fuel being burned in order to roduce the same temerature rior to the turbine and so makes the cycle more efficient. he layout of such a lant is shown on fig.8. Fig.8 Plant layout 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 H m g c g ( 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) H m g c g ( 5-6 ) he enthaly gained by the air is H(air) m a c a ( - ) Hence the thermal ratio is.r. m a c a ( - )/ m g c g ( 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 c a ( ). R. c g ( ) 5 D.J.Dunn 6

68 WORKED EXAMPLE No.5 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 0.88 is used. SOLUION Referring to the numbers used on fig.8 the solution is as follows. Air standard efficiency - r (-/) r 4.8% Solution with heat exchanger r (-/) 8 (7.5) K 5 4 /r (-/) 57/(7.5) K Use the thermal ratio to find..005( ).005( 50.6) ( 5 ) K ( ) In order find the thermal efficiency, it is best to solve the energy transfers. P(in) mc a ( - ) x.005 (50.6-8).7 kw P(out) mc g ( 4-5 ) x.5 ( ) 7. kw P(nett) P(out) - P(in) 97. kw Φ(in)combustion chamber) mc g ( 4 - ) Φ(in).5( ) 7. kw ηth P(nett)/Φ(in) 494./ or 69.% D.J.Dunn 7

69 SELF ASSESSMEN EXERCISE No. 5. 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. (Answers 4 kw and 57%). A gas turbine uses a ressure ratio of 6.5/. he inlet temerature and ressure are resectively 5oC and bar. 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. he isentroic efficiency is 85% for both the comression and exansion rocess. he mass flow rate is kg/s. Calculate the thermal efficiency when an exhaust heat exchanger with a thermal ratio of 0.75 is used. (Answer 48.%) D.J.Dunn 8

70 WORKED EXAMPLE No.6 A gas turbine has a free turbine in arallel with the turbine which drives the comressor. An exhaust heat exchanger is used with a thermal ratio of 0.8. he isentroic efficiency of the comressor is 80% and for both turbines is he heat transfer rate to the combustion chamber is.48 MW. he gas leaves the combustion chamber at 00oC. he air is drawn into the comressor at bar and 5oC. he ressure after comression is 7. bar. he adiabatic index is.4 for air and. for the gas roduced by combustion. he secific heat c is.005 kj/kg K for air and.5 kj/kg K for the gas. Determine the following. i. he mass flow rate in each turbine. ii. he net ower outut. iii. he thermodynamic efficiency of the cycle. SOLUION 98 K 98(7.) (-/.4) 54 K 4 7 K 5 7(/7.) (-/.) 88.5 K COMPRESSOR ηi 0.8 (54-98)/( -98) hence K URBINES reat as one exansion with gas taking arallel aths. ηi 0.85 (7-5 )/(7-88.5) hence K HEA EXCHANGER hermal ratio ( )/.5( ) hence 890. K COMBUSION CHAMBER Φ(in) mc ( 4 - ) 480 kw 480 m(.5)(7-890.) hence m.665 kg/s COMPRESSOR P(in) mc (-).665(.005)( ) kw D.J.Dunn 9

71 URBINE A P(out) kw m A c (4-5) (.5)(7-98.7) hence m A.448 kg/s Hence mass flow through the free turbine is.68 kg/s P(nett) Power from free turbine.68(.5)(7-98.7) 65.7 kw HERMODYNAMIC EFFICIENCY ηth P(nett)/Φ(in) 65.7/ or 4.8 % SELF ASSESSMEN EXERCISE No. 6. List the relative advantages of oen and closed cycle gas turbine engines. Sketch the simle gas turbine cycle on a -s diagram. Exlain how the efficiency can be imroved by the inclusion of a heat exchanger. In an oen cycle gas turbine lant, air is comressed from bar and 5oC to 4 bar. he combustion gases enter the turbine at 800oC and after exansion ass through a heat exchanger in which the comressor delivery temerature is raised by 75% of the maximum ossible rise. he exhaust gases leave the exchanger at bar. Neglecting transmission losses in the combustion chamber and heat exchanger, and differences in comressor and turbine mass flow rates, find the following. (i) he secific work outut. (ii) he work ratio (iii) he cycle efficiency he comressor and turbine olytroic efficiencies are both Comressor c.005 kj/kg K.4 urbine c.48 kj/kg K. Note for a comression and for an exansion η ( ) η D.J.Dunn 0

72 . A gas turbine for aircraft roulsion is mounted on a test bed. Air at bar and 9K enters the comressor at low velocity and is comressed through a ressure ratio of 4 with an isentroic efficiency of 85%. he air then asses to a combustion chamber where it is heated to 75 K. he hot gas then exands through a turbine which drives the comressor and has an isentroic efficiency of 87%. he gas is then further exanded isentroically through a nozzle leaving at the seed of sound. he exit area of the nozzle is 0. m. Determine the following. (i) he ressures at the turbine and nozzle outlets. (ii) he mass flow rate. (iii) he thrust on the engine mountings. Assume the roerties of air throughout. he sonic velocity of air is given by a (R) ½. he temerature ratio before and after the nozzle is given by (in)/(out) /(+) D.J.Dunn

73 . (A). A gas turbine lant oerates with a ressure ratio of 6 and a turbine inlet temerature of 97oC. he comressor inlet temerature is 7oC. he isentroic efficiency of the comressor is 84% and of the turbine 90%. Making sensible assumtions, calculate the following. (i) he thermal efficiency of the lant. (ii) he work ratio. reat the gas as air throughout. (B). If a heat exchanger is incororated in the lant, calculate the maximum ossible efficiency which could be achieved assuming no other conditions are changed. Exlain why the actual efficiency is less than that redicted. D.J.Dunn

74 APPLIED HERMODYNAMICS UORIAL 4 PISON ENGINES In this tutorial you will do a comrehensive study of iston engine cycles and all matters associated with these engines required for the examination. 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 various efficiencies of an engine. Calculate the Mean Effective Pressure of an engine. Describe the standard thermodynamic cycles for sark and comression ignition engines. Solve standard cycles. First we will examine engine testing methods.

75 . ENGINE ESING. 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.V. 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. SOICHIOMERIC RAIO Air Fuel Ratio m a /m f his is the theoretical air/fuel ratio which is required to exactly burn the fuel. RUE 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. Engines with carburettors work by restricting the air flow with a butterfly valve. his reduces the ressure to less than atmosheric at inlet to the cylinder and the restriction of the inlet valve adds to the affect. Engines with turbo charging use a comressor to deliver air to the cylinders at ressures higher than atmosheric. he actual mass of air which enters the cylinder is less than the theoretical value for various reasons such as warming from the cylinder walls, residual gas left inside and leaks from the valves and around the iston. o deal with this we use the concet of EFFICIENCY RAIO. Efficiency Ratio Actual mass/ heoretical mass D.J.Dunn

76 .. CALORIFIC VALUE 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. We can now define the fuel ower. FUEL POWER Mass of fuel/s x Calorific Value.. VOLUME FLOW RAE A two stroke engine induces the volume of air once every revolution of the crank. A 4 stroke engine does so once every two revolutions. Induced Volume Caacity x seed for a stroke engine Induced volume Caacity x seed/ for a 4 stroke engine. WORKED EXAMPLE No. A 4 stroke carburetted engine runs at 500 rev/min. he engine caacity is litres. he air is sulied at 0.5 bar and 5oC with an efficiency ratio of 0.4. he air fuel ratio is /. he calorific value is 46 MJ/kg. Calculate the heat released by combustion. SOLUION Caacity 0.00 m Volume induced 0.00 x ( 500/60)/ m/s Using the gas law V mr we have Ideal air m V/R 0.5 x 05 x 0.065/(87 x 88) 0.09 kg/s Actual air m 0.09 x kg/s. Mass of fuel mf 0.057/ 0.00 kg/s Heat released Φ calorific value x mf kj/kg x 0.00 kg/s 60. KW D.J.Dunn

77 SELF ASSESSMEN EXERCISE No. A 4 stroke carburetted engine runs at 000 rev/min. he engine caacity is 4 litres. he air is sulied at 0.7 bar and 0oC with an efficiency ratio of 0.5. he air fuel ratio is /. he calorific value is 45 MJ/kg. Calculate the heat released by combustion. (Answer 49 KW) D.J.Dunn 4

78 . BRAKE POWER 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. he second equal weight acts to cancel out the other so the torque is given by F R. Another form of dynamometer is basically an electric generator that absorbs the load and turns it into electric ower that is dissiated in a bank of resistor as heat. D.J.Dunn 5

79 . INDICAED POWER his is the ower develoed by the ressure of the gas acting on the istons. It is found by recording and lotting 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. Modern systems use electronic ressure and volume transducers and the data is gathered and stored digitally in a comuter and then dislayed and rocessed. Figure he average force on the iston throughout one cycle is F where Area of iston A F MEP x he Mean Effective Pressure is the mean ressure during the cycle. he work done during one cycle is W Force x distance moved F L AL L is the stroke and this is twice the radius of the crank shaft. 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 / the shaft seed. MEAN EFFECIVE PRESSURE 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 base 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 D.J.Dunn 6

80 .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. Hmech B.P./I.P. D.J.Dunn 7

81 WORKED EXAMPLE No. A 4 cylinder, 4 stroke engine gave the following results on a test bed. Shaft Seed N 500 rev/min orque arm R 0.4 m Net Brake Load F 00 N Fuel consumtion mf g/s Calorific value 4 MJ/kg Area of indicator diagram Ad 00 mm Pressure scale S 80 kpa/mm Stroke L 00 mm Bore D 00 mm Base length of diagram Y 60 mm. Calculate the B.P., F.P., I.P., MEP, η Bh, η Ih,and ηmech, SOLUION BP πn π x (500/60)x (00 x 0.4) 0.94 kw FP mass/s x C.V kg/s x kj/kg 84 kw IP LAN MEP Ad/Y x S (00/60) x 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 kw η Bh 0.94/84 4.9% η Ih 6.8/84. % ηmech 0.94/6.8 80% D.J.Dunn 8

82 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. dm/min Fuel density 750 kg/m 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. (6. %) ii) he Mechanical Efficiency. (87%). iii) he Brake hermal efficiency. (.8%). D.J.Dunn 9

83 . A two stroke sark ignition 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 Net brake load 95 N orque arm 0.4 m Net indicated area 00 mm Base length of indicator diagram 40. mm Pressure scale 50 kpa/mm Calculate the following. i) he Indicated thermal efficiency. (0.5 %) ii) he Mechanical Efficiency. (8.7%). iii) he Brake hermal efficiency. (5%). 4. A four stroke sark ignition engine gave the following results during a test. Number of cylinders 4 Bore of cylinders 90 mm Stroke 80 mm Seed rev/min Fuel consumtion rate 0.09 kg/min Calorific value 44 MJ/kg Net 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%). D.J.Dunn 0

84 5. Define Indicated Mean Effective Pressure and Brake Mean Effective Pressure. he BMEP for a 4 cylinder, 4 stroke sark ignition engine is 8.4 bar. he total caacity is. dm (litres). he engine is run at 4 00 rev/min. Calculate the Brake Power. (8. kw) here are 0 kw of mechanical losses in the engine. Calculate the Indicated Mean effective Pressure. (0.6 bar). he Volumetric Efficiency is 85% and the Brake hermal Efficiency of the engine is 8%. he air drawn in to the engine is at 5oC and.0 bar. he fuel has a calorific value of 4.5 MJ/kg. Calculate the air/fuel ratio. (Answer./). Let's now have a look at the theoretical cycle for sark ignition engines. D.J.Dunn

85 . SPARK IGNIION ENGINES HE OO CYCLE he Otto cycle reresents the ideal cycle for a sark ignition engine. In an ideal sark ignition engine, there are four rocesses as follows. Fig.4 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

86 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.5 to. he air is heated at constant volume. No work is done. Q in mcv(-) Fig.6 to 4. he air exands reversibly and adiabatically with no heat transfer back to its original volume. Work outut is obtained. FFig.7 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.8 D.J.Dunn

87 If the engine is successful, then Wout he area under the to curve) is larger than Win (the area under the lower curve). he enclosed area reresents the net work obtained from the cycle. Figure 9 he Mean Effective Pressure (MEP) is the average ressure such that W(net) Enclosed Area MEP x A x L W(net) MEP x Swet Volume his is true for all cycles and for real engines. A corresonding net amount of heat must have been transferred into the cycle of: Figure 0 Alying the first law, it follows Qnet Wnet It also follows that since the heat transfer is equal to the area under a - S grah, then the area enclosed by the cycle on the - S diagram is equal to the Qnet and this is true for all cycles. D.J.Dunn 4

88 5 D.J.Dunn 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 r v V V For the rocess () to (4) we may similarly write r v V V where r v is the volume comression ratio 4 V V V V r v It follows that 4 4 and and that η η then v v r r Since this theoretical cycle is carried out on air for which.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..

89 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.4. R 87 J/kg K SOLUION Remember to use absolute temeratures throughout. Solve for a mass of kg K K r v 8 η r W ηq 0.4 ( ) V K V Q mc ( - ) x78(77 67.) J / kg kj / kg in nett v in or 56.5% 0.56x kj / kg From the gas law we have V V V V x V x 77 9 x V x 77 x MPa 9 If you have followed the rinciles used here you should be able to solve any cycle. D.J.Dunn 6

90 SELF ASSESSMEN EXERCISE No. ake c v 0.78 kj/kg K, R 87 J/kg K and.4 throughout.. In an Otto cycle air is drawn in at 0oC. he maximum cycle temerature is 500oC. he volume comression ratio is 8/. Calculate the following. i. he thermal efficiency. (56.5%) ii. he heat inut er kg of air. (789 kj/kg). iii. he net work outut er kg of air. (446 kj/kg).. 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). 4. 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). D.J.Dunn 7

91 5. 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. (Answer.5 kg). 6. he air standard cycle aroriate to the recirocating sark ignition engine internal-combustion engine is the Otto. Using this, find the efficiency and outut of a litre (dm), 4 stroke engine with a comression ratio of 9 running at 000 rev/min. he fuel is sulied with a gross calorific value of 46.8 MJ/kg and an air fuel ratio of.8. Calculate the answers for two cases. a. he engine running at full throttle with the air entering the cylinder at atmosheric conditions of.0 bar and 0oC with an efficiency ratio of (Answers 58.5% and 65 kw) b. he engine running at art throttle with the air entering the cylinder at 0.48 bar and efficiency ratio 0.8. (Answers 58.5% and 4 kw). 7. 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. (Answers 87 kj/kg and 9 kj/kg). Now let's move on to study engines with comression ignition. D.J.Dunn 8

92 . COMPRESSION IGNIION ENGINES Comression ignition is achieved by comressing the air until it is so hot that fuel srayed into it will ignite sontaneously. On modern engines the fuel is injected as a sray of drolets. Since it takes a finite time for the drolets to warm u to the air temerature, there is a time delay between injection and exlosion. he accumulation of drolets in this time cause an initial shar detonation (diesel knock) and a raid rise in ressure and temerature inside the cylinder. his is a constant volume rise in ressure similar to the sark ignition and the Otto cycle. he big difference is that with fuel injection, fuel may continue to be injected after the exlosion and controlled burning of the fuel in the air may take lace as the iston moves away from the cylinder head. his ideally maintains the ressure constant as the volume increases. When the fuel is cut off, a natural exansion occurs and the rest of the cycle is similar to the Otto cycle. he air standard cycle for this engine is the DUAL COMBUSION CYCLE. he man most credited with the invention of this engine is Rudolf Diesel but many others worked on similar ideas. Diesel's first engine used coal dust blasted into the combustion chamber with comressed air. his develoed into blasting in oil with comressed air. he air standard cycle for these old fashioned engines was deemed to be as described above but with no constant volume rocess. his cycle is called the DIESEL CYCLE. he Diesel cycle may have been a reasonable aroximation of what haened in older slow running engines but it is not reresentative of a modern engine. D.J.Dunn 9

93 . 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 he rocesses are as follows. Fig. - reversible adiabatic (isentroic) comression. - constant volume heating. - 4 constant ressure heating. 4 5 reversible adiabatic (isentroic) exansion. 5 - constant volume cooling. he analysis of the cycle is as follows. he heat is sulied in two stages hence Q in mc (4 - ) + mc v ( - ) he heat rejected is Q out mc v (5 - ) he thermal efficiency may be found as follows. Qout mcv 5 - η Q mc ( - ) + mc ( - ) ( - in v ( ) ( - ) he formula can be further develoed to show that η r v is the VOLUME COMPRESSION RAIO. r v V /V β is the CU OFF RAIO. β V 4 /V k is the ratio /. 4 5 ) + ( 4 - ) kβ - r v [( k ) + k( β ) ] D.J.Dunn 0

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

95 WORKED EXAMPLE No. 4 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 K he hottest temerature is 4 60K. β.5 r v 8.4 rv V4 4 V β η - k.7 kβ - [( k -) + k( β -)] r [(.7 -) + (.4 x.7 x (.5 -))] η 0.68 or 68% 4 9 x K K v - W nett η x Q in 0.68 x 0.68 kj er cycle. he Carnot efficiency should be higher. cold 9 η hot.7 x x 8 he figure of 0.68 is lower so the Carnot rincile has not been contravened. 0.4 D.J.Dunn

96 WORKED EXAMPLE No.5 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. he cut off ratio. ii. he cycle efficiency. iii. he nett work outut er kg of air. Assume.4 c.005 kj/kgk c v 0.78 kj/kg K. SOLUION Known data. 98 K 4 7 K 4 9 MPa 5 kpa V /V V /V 8 V V 98 x 8 (-) 947 K 6 V 9 x 0 x 98 V x V 5 x 0 V V4 4 Cut off ratio β but V 7 β V 4 4 V 5-7 x V4 but V V V 4 V x V K 5 9 x 0 5 x x 98 x 8 so β Q in mc (4 - ) + mc v ( - ) m kg Q in.005(74-96) (96-947).5 kj/kg Q out mc v ( 5 - ) Q out 0.78( ) 49 kj/g 4 96K Q η Q W nett Q out in in Q out or 65% kj/kg D.J.Dunn

97 . HE DIESEL CYCLE he Diesel Cycle roceeded 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 -V and -s cycles diagrams are shown in figure Fig. - 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 ) ( - ) 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 - r v ( β ) D.J.Dunn 4

98 WORKED EXAMPLE No.6 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 0.4 c v 78 J/kg K for air 88 K bar. he maximum temerature is and the maximum ressure is and. β - η β r η ( ).4 - η ( ) x.4 x 0 r Q Q W η Q W in in V V nett v.69 x.4 x.4 - v mc ( - ) ηq 88 x 0 β or 64.7% K 954. x 909 K.005( ) 959. kj nett in in x kj D.J.Dunn 5

99 SELF ASSESSMEN EXERCISE No.4. 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 %).. 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 %).. Draw a - V and - s diagram for the Dual Combustion Cycle. A recirocating engine oerates on the Dual Combustion Cycle. he ressure and temerature at the beginning of comression are bar and 5oC resectively. he comression ratio is 6. he heat inut is 800 kj/kg and the maximum ressure is 80 bar. Calculate the following. i. he ressure, volume and secific volume at all oints in the cycle. ii. he cycle efficiency. (6.8 %). iii. he mean effective ressure. (4.5 bar). D.J.Dunn 6

100 HERMODYNAMICS UORIAL 5 HEA PUMPS AND REFRIGERAION On comletion of this tutorial you should be able to do the following. Discuss the merits of different refrigerants. Use thermodynamic tables for common refrigerants. Define a reversed heat engine. Define a refrigerator and heat um. Define the coefficient of erformance for a refrigerator and heat um. Exlain the vaour comression cycle. Exlain modifications to the basic cycle. Sketch cycles on a ressure - enthaly diagram. Sketch cycles on a temerature - entroy diagram. Solve roblems involving isentroic efficiency. Exlain the cycle of a recirocating comressor. Define the volumetric efficiency of a recirocating comressor. Solve roblems involving recirocating comressors in refrigeration. Exlain the ammonia vaour absortion cycle.

101 . INRODUCION It is ossible to lower the temerature of a body by use of the thermo-electric affect (reversed thermo-coule or Peltier effect). his has yet to be develoed as a serious refrigeration method so refrigerators still rely on a fluid or refrigerant which is used in a reversed heat engine cycle as follows. Figure Heat is absorbed into a fluid (this is usually an evaorator) lowering the temerature of the surroundings. he fluid is then comressed and this raises the temerature and ressure. At the higher temerature the fluid is cooled to normal temerature (this is usually a condenser). he fluid then exeriences a dro in ressure which makes it go cold (this is usually a throttle valve) and able to absorb heat at a cold temerature. he cycle is then reeated. Various fluids or refrigerants are used in the reversed thermodynamic cycle. Refrigerants such as air, water and carbon dioxide are used but most refrigerants are those designed for vaour comression cycles. hese refrigerants will evaorate at cold temeratures and so the heat absorbed is in the form of latent energy. Let's look at the roerties of these and other refrigerants. D.J.Dunn

102 . REFRIGERANS Refrigerants are given R numbers. Carbon dioxide, for examle is R744. Some of them are dangerous if released because they are either exlosive or toxic. oxic refrigerants are laced in categories. Sulhur dioxide, for examle, is classed as toxic grou which means that death occurs after breathing it for 5 minutes. In the ast the most oular fluids have been ammonia (R77),fluorocarbons and halocarbons. he most oular of these is R or dichlorodifluoromethane (CFCl). he tye of refrigerant used in a cycle is largely governed by the evaoration temerature required and its latent caacity. Below is a list of some of them. Refrigerant R number Evaoration tem. oxic grou at.0 bar.(oc) C Cl F R 4 5 C ClF R -0 6 C ClF R -8 6 C F4 R4-8 6 CH ClF R 9 4 CH Cl F R CH F R C Cl F C Cl F R 47 4 C Cl F C F R4A 6 C Cl F C Cl F R4 6 C Cl F C F R5-9 All the above are Halo-Carbons and Fluro-carbons which are non-flammable and may be detected by a halide torch or electric cell sensor. Other refrigerants are shown below. Ammonia is flammable and detected by going white in the resence of sulhur dioxide. It has a strong characteristic ungent smell. Death occurs when breathed for 0 minutes. NH R77 - Carbon Dioxide is safe and non-toxic but it can suffocate. CO R Sulhur Dioxide is highly toxic and does not burn. Other refrigerants are in the Hydro-Carbon grous such as Proane, Butane and Ethane. hese are exlosive. Because of the roblems with damage to the ozone layer, new refrigerants such as R4a have been develoed and are now included in the thermodynamic tables. Now let's look at the use of thermodynamic tables for refrigerants. D.J.Dunn

103 . ABLES he section of the fluid tables devoted to refrigerants is very concise and contains only two suerheat temeratures. he layout of the tables is shown below. 5K 0 K t s vg hf hg sf sg h s h s t is the actual temerature in degrees Celsius. s is the saturation ressure corresonding to the temerature. It follows that if the refrigerant is wet or dry saturated, it must be at temerature t and ressure s. If the refrigerant has 5 degrees of suerheat, then the actual temerature is t+5 and the roerties are found under the 5 K heading. Similarly if it has 0 K of suerheat, its actual temerature is t+0. For examle, R at.9 bar and 0oC must have 0 K of suerheat since its saturation temerature would is -0oC. From the 0 K columns we find that h0.97 kj/kg and s kj/kg K. When dealing with liquid refrigerant, take the roerties as hf and sf at the given temeratures. he ressures are never very high so the ressure term will not cause much error. D.J.Dunn 4

104 4. VAPOUR COMPRESSION CYCLES 4. HE BASIC CYCLE Refrigeration/heat um cycles are similar to heat engine cycles but they work in reverse and are known as reversed heat engine cycles. A basic vaour cycle consists of isentroic comression, constant ressure cooling, isentroic exansion and constant ressure heating. You may recognise this as a reverse of the Rankine cycle or even the reverse of a Carnot cycle. he heating and cooling will involve evaoration and condensing. Let's consider the cycle first conducted entirely with wet vaour. Figure he basic rincile is that the wet vaour is comressed and becomes dryer and warmer in the rocess. It is then cooled and condensed into a wetter vaour at the higher ressure. he vaour is then exanded. Because of the cooling, the exansion back to the original ressure roduces a fluid which is much colder and wetter than it was before comression. he fluid is then able to absorb heat at the cold temerature becoming dryer in the rocess and is returned to the original state and comressed again. he net result is that heat is absorbed at a cold temerature and rejected at a higher temerature. Work is needed to drive the comressor but some of it is returned by the turbine. he thermodynamic cycle for refrigerators is often shown on a ressure enthaly diagram ( h) and rofessional charts are available but not used in the Engineering Council exams. Figure shows the basic cycle. D.J.Dunn Figure 5

105 he four thermodynamic rocesses are - Isentroic comression. P(in) mr(h - h) - Constant ressure cooling. Φ(out) mr(h - h) - 4 Isentroic exansion. P(out) mr(h - h4) 4 - Constant ressure heating. Φ(in) mr(h4 - h) mr mass flow rate of refrigerant. In ractice wet vaour is difficult to comress and exand so the refrigerant is usually dry before comression and suerheated after. he cooling rocess may roduce anything from wet vaour to undercooled liquid. he exansion of a liquid in a turbine is imractical and so a throttle is used instead. A throttle roduces no useful work but it converts the ressure into internal energy. his makes the liquid evaorate and since the saturation temerature goes down it ends u cold. he enthaly before and after a throttle are the same. he entroy increases over a throttle. Figure 4 Figure 5 D.J.Dunn 6

106 he cycle may also be drawn on a temerature entroy diagram as shown. he conditions shown are wet at (), suerheated at () and under-cooled at (). hese conditions vary. he four thermodynamic rocesses are - Isentroic comression. P(in) mr(h - h) - Constant ressure cooling. Φ(out) mr(h - h) - 4 hrottle (h h4) 4 - Constant ressure heating. Figure 6 Φ(in) mr(h4 - h) 4. COEFFICIEN OF PERFORMANCE he second law of thermodynamics tells us that no heat engine may be 00% efficient. In the reversed cycle, the reverse logic alies and it will be found that more energy is given out at the condenser and more absorbed in the evaorator, than is needed to drive the comressor. he ratio of heat transfer to work inut is not called the efficiency, but the coefficient of erformance or advantage. here are two coefficients of erformance for such a cycle, one for the refrigeration effect and one for the heat um effect. 4.. REFRIGERAOR A refrigerator is a device for removing heat at a cold temerature so we are interested in the heat absorbed in the evaorator Φ(in). he coefficient of erformance is also called the advantage and is defined as C.O.P. Φ(in)/P(in) he heat absorbed is called the refrigeration affect. 4.. HEA PUMP A heat um is a device for roducing heat so we are interested in the heat given out in the cooler Φ(out). he coefficient of erformance is defined as C.O.P. Φ(out)/P(in) It is usual to find a convenient source of low grade heat for the evaorator such as the atmoshere or a river. he heat is removed from this source and ugraded to higher temerature by the comressor. Both the work and the heat absorbed are given out at the higher temerature from the cooler. D.J.Dunn 7

107 5. MODEL REVERSED HEA ENGINE Figure 7 he ideal model ums heat from a cold source to a hot lace. he st. Law of hermodynamics alies so Φ(in) + P(in) Φ(out) C.O.P. (refrigerator) Φ(in)/P(in) Φ(in)/{Φ(out) - Φ(in)} If the heat transfers are reversible and isothermal at temeratures (hot) and (cold) then C.O.P. (refrigerator) (cold) /{ (hot) - (cold) } C.O.P. (heat um) Φ(out))/P(in) Again for reversible isothermal heat transfers this reduces to C.O.P. (heat um) (hot) /{ (hot) - (cold) } his is the inverse of the Carnot efficiency exression for heat engines. C.O.P. (heat um) {Φ(in)+ P(in)}/P(in) C.O.P. (heat um) Φ(in)/P(in) + C.O.P. (heat um) C.O.P. (refrigerator) + D.J.Dunn 8

108 WORKED EXAMPLE No. A heat um uses a vaour comression cycle with refrigerant. he comressor is driven by a heat engine with a thermal efficiency of 40%. Heat removed from the engine in the cooling system is recovered. his amounts to 40% of the energy sulied in the fuel. he heat um cycle uses an ideal cycle with an evaorator at 5oC and a condenser at.9 bar. he vaour is dry saturated at inlet to the comressor. he condenser roduces liquid at 45oC. Calculate the thermal advantage (Coefficient of Performance) for the heat um. Comare it with a boiler running at 90% thermal efficiency. he lant is to deliver 40 kw of heat. Determine the mass flow rate of refrigerant. SOLUION Figure 8 From the R tables we find h 5oC kj/kg s 5oC kj/kg K bar If a - h chart was available, h could be found easily. We must use the tables and we can see that occurs between 0 K of suerheat and 5 K of suerheat. Using linear interolation we may find the enthaly as follows. 0 K θ 5 K s h h h (0.046/0.069).9.7 kj/kg h h4 hf at 45oC 79.7 kj/kg h kj/kg D.J.Dunn 9

109 P(in) h - h.67 kj/kg 40% of fuel energy Φ(out) h - h.56 kj/kg Φ(in) h - h kj/kg C.O.P (condenser)..56/ C.O.P (evaorator) / Φ(out) 40% x % of fuel ower otal heat from system 4.8 % + 40% 8.8 % of fuel energy. Comared to a boiler which gives 90% this is 9.8 % more. otal heat outut 40 kw 8.8 % of fuel ower. Fuel Power 4.4 kw 40 kw Φ(out) + energy recovered from cooling water 40 mr(h- h) + 40% x 4.4 mr(.56) mr0.6 kg/s Φ(in) mr(h- h4) 8.7 kw Φ(out) mr(h- h) 4. kw D.J.Dunn 0

110 6. ISENROPIC EFFICIENCY When the comression is not reversible and isentroic then the isentroic efficiency is used in the usual way for a comression rocess. ηis ideal enthaly change/actual enthaly change WORKED EXAMPLE No. he ower inut to the comressor of an ammonia vaour comression lant is 8. kw. he mechanical efficiency is 85%. he ammonia is dry saturated at -6oC at inlet to the comressor. After comression the vaour is at.67 bar. he comression has an isentroic efficiency of 90%. he condenser roduces saturated liquid. Calculate the following. i. he flow rate. ii. he coefficient of erformance for the refrigerator. iii. he coefficient of erformance for the heat um. SOLUION Figure 9 h hf at -6oC 47.6 kj/kg h h4 bar. kj/kg s sf at -6oC 5.49 kj/kg K Ideally s s 5.49 kj/kg K From the tables at.67 bar we see that the secific entroy is 5.47 kj/kg K when there is 50 K of suerheat. his is the ideal condition after comression and the corresonding enthaly is 60.5 kj/kg K. Ideal change in enthaly kj/kg. Actual change 7.9/90% 9. kj/kg. Actual value of h kj/kg. Power inut to cycle8. kw x 85% 6.97 kw mr( ) mr 6.97/ kg/s Heat inut to evaorator mr(h - h4) kw Coefficient of erformance (refrigerator) 40.44/ Heat outut at condenser mr(h - h) 470 kw Coefficient of erformance (heat um) 47.4/ D.J.Dunn

111 SELF ASSESSMEN EXERCISE No.. A simle vaour comression refrigerator comrises an evaorator, comressor, condenser and throttle. he condition at the 4 oints in the cycle are as shown. Point Pressure emerature After evaorator bar -0oC After comressor 5.67 bar 50oC After condenser 5.67 bar 5oC After throttle bar -5oC he refrigerant is R which flows at 0.05 kg/s. he ower inut to the comressor is kw. Comression is reversible and adiabatic. Calculate the following. i. he theoretical ower inut to the comressor. (.85 kw) ii. he heat transfer to the evaorator. (6.57 kw) iii. he coefficient of erformance based answer (i.)(.59) iv. he mechanical efficiency of the comressor. (90.7%) v. he coefficient of erformance based on the true ower inut. (.6) Is the comression rocess isentroic?. A vaour comression cycle uses R. the vaour is saturated at -0oC at entry to the comressor. At exit from the comressor it is at 0.84 bar and 75oC. he condenser roduces saturated liquid at 0.84 bar. he liquid is throttled, evaorated and returned to the comressor. Sketch the circuit and show the cycle on a -h diagram. Calculate the coefficient of erformance of the refrigerator. (.0) Calculate the isentroic efficiency of the comressor. (7%) D.J.Dunn

112 7 MODIFIED CYCLES An imrovement to the basic comression cycle is the use of a flash chamber instead of a throttle valve. he condensed high ressure liquid at oint 7 is srayed into the low ressure flash chamber. he dro in ressure has the same effect as throttling and the liquid artially evaorates and dros in temerature. he dry saturated vaour is drawn into the comressor and the saturated liquid is umed to the evaorator. he rincial difference is that the evaorator now oerates at a higher ressure and so the liquid at oint is below the saturation temerature. Figure 0 Further modifications may be made by comressing the vaour in two stages and mixing the vaour from the flash chamber at the inter-stage oint. he outut of the evaorator then goes to the inut of the low ressure stage as shown in fig.0 hese modifications require more hardware than the basic cycle so the extra cost must be justified by savings and increased caacity to refrigerate. D.J.Dunn

113 WORKED EXAMPLE No. A refrigeration lant uses R in the cycle below. he evaorator temerature is - 50oC and the condenser ressure is 50oC. he flash chamber is maintained at 0oC. Saturated vaour from the chamber is mixed with the comressed vaour at the inter-stage oint (). he liquid in the chamber is further throttled to -50oC in the evaorator. he vaour leaving the evaorator is dry saturated and comression is isentroic. Find the coefficient of erformance for the refrigerator assuming kg/s flow rate. SOLUION Figure At oint () the vaour is dry saturated so h -50oC kj/kg Similarly s sg kj/kg K he ressure at oint / must be s at 0oC.08 bar s s kj/kg bar. From the tables this is seen to be suerheated so the degree of suerheat and the enthaly must be estimated by interolation as follows. / ( )/( ) 4.09 K Now the enthaly at oint may be estimated as follows. 4.09/5 (h )/( ) h 99.9 kj/kg Energy balance on Flash Chamber Assume saturated liquid at oint (6). h6 50oC kj/kg Since the enthaly is the same after throttling then h kj/k h4 hg at 0oC 87.5 kj/kg D.J.Dunn 4

114 Let the flow rate be kg/s at (7) and (y) at oint (4). Balancing enthaly we have h7 y(h4) + (-y)h8 h8 0oC 6.05 kj/kg K y(87.5) + (-y)(6.05) y 0.7 kg/s Energy balance at mixing oint. (-y)h + yh4 h ((-0.7)(99.9) + (0.7)(87.5) h 95.9 kj/kg Power of LP urbine P (-y) kg/s (h - h) (-0.7)( ).67 kw he vaour at oint is clearly suerheated between 0 and 5 K. Interolation gives the degree of suerheat as follows. /5 ( )/( ).9 K We may now interolate to find s.9/5 (s )/( ) s 0.76 kj/kg K bar (s at 50oC) Again we must interolate to find the degree of suerheat at oint 5 which is between 0 and 5 K again. /5 ( )/( ) 4. K We may now interolate with this degree of suerheat to find h5. 4./5 (h5-8.64)/( ) h5 kj/kg HP urbine Power P kg/s(h5 - h) (-95) 6.9 kw otal Power inut kw Evaorator Heat Inut h9 h8 hf at 0oC 6.05 kj/kg Φ(in) kg/s (h - h9) Φ(in) 0.677( ) 87. kw Coefficient of Performance 87./6.9.7 D.J.Dunn 5

115 SELF ASSESSMEN EXERCISE No.. A refrigerator oerates with ammonia. he lant circuit is shown below. he conditions at the relevant oints of the cycle are as follows. saturated liquid at -0oC,4 and 7 saturated liquid at 0oC 5 saturated vaour at -0oC he um and comressor have an isentroic efficiency of 80%. there are no heat losses. he secific volume of ammonia liquid is m/kg. Determine the coefficient of erformance and the mass flow rate if the refrigeration effect is 0 kw. (Ans..964 and kg/s) Figure D.J.Dunn 6

116 . A heat um consists of a comressor, condenser, throttle, and evaorator. he refrigerant is R. he refrigerant is at 0oC at entry to the comressor and 80oC at exit. he condenser roduces saturated liquid at 50oC. he throttle roduces wet vaour at -0oC. he mass flow rate is 0.0 kg/s. he indicated ower to the comressor is kw. Sketch the - s diagram and - h diagram for the cycle. Calculate the coefficient of erformance for the heat um (.9 based on I.P.) Calculate the rate of heat loss from the comressor. (0. kw) Calculate the coefficient of erformance again for when the refrigerant is sub cooled to 45oC at exit from the condenser. ( based on I.P.) Calculate the temerature at exit from the comressor if the comression is reversible and adiabatic. (68.7oC). A refrigeration cycle uses R. he evaorator ressure is.86 bar and the condenser ressure is 0.84 bar. here is 5K of suerheat at inlet to the comressor. he comressor has an isentroic efficiency of 90%. the condensed liquid is undercooled by 5K and is throttled back to the evaorator. Sketch the cycle on a -s and -h diagram. Calculate the coefficient of erformance. (.04) Exlain why throttles are used rather than an exansion engine. D.J.Dunn 7

117 8. RECIPROCAING COMPRESSORS his is covered in detail in tutorial. he knowledge of comressors is often required in refrigeration and heat um studies so the basics are covered here along with examle questions on comressors used in this area. he diagram shows the basic design of a recirocating comressor. he iston recirocates drawing in gas, comressing it and exelling it when the ressure inside the cylinder reaches the same level as the ressure in the delivery ie. Figure If the iston exels all the air and there is no restriction at the valves, the ressure - volume cycle is as shown below. Figure 4 Gas is induced from 4 to at the inlet ressure. It is then traed inside the cylinder and comressed according the law Vn 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 8

118 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 Vn 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. he volumetric efficiency is defined as Figure 5 η vol Induced Volume/Swet volume. his efficiency is made worse if leaks occur ast the valves or iston. In real comressors, the gas is restricted by the valves and the valves tend to move so the real cycle looks more like this. Figure 6 D.J.Dunn 9

119 WORKED EXAMPLE No.4 Dry saturated Refrigerant vaour at 5oC is comressed in a recirocating comressor to.9 bar at a rate of 0.74 kg/s. he clearance volume is 5% of the swet volume. he exansion art of the cycle follows the law V. C. he crank seed is 60 rev/min. Calculate the swet volume and the volumetric efficiency. SOLUION Swet Volume V Clearance volume 0.05 V Consider the exansion from to 4 on the -V diagram. he inlet ressure must be s at 5oC hence 4.66 bar. V. 4V4..96(0.05V)..66 (V4. ) V4 0.7V or.7% of V V V V.05V Induced volume V - V4.05V - 0.7V 0.9V mr 0.74 kg/s At inlet v vg at 5oC m/kg. Volume flow rate required 0.74 x m/s. Induced volume V V 0.0/ m/s Crank seed 6 rev/s so the swet volume 0.04/ m. η vol Induced Volume/Swet volume. η vol 0.9V/V 9.% D.J.Dunn 0

120 SELF ASSESSMEN EXERCISE No.. Why is it referable that vaour entering a comressor suerheated? A vaour comression refrigerator uses R. he vaour is evaorated at -0oC and condensed at 0oC. he vaour has 5 K of suerheat at entry to the comressor. Comression is isentroic. he condenser roduces saturated liquid. he comressor is a recirocating tye with double action. he bore is 50 mm and the stroke is 00 mm. he seed is 00 rev/min. he volumetric efficiency is 85%. You may treat suerheated vaour as a erfect gas. Determine i. the mass flow rate (0.956 kg/s) ii. the coefficient of erformance. (5.5) iii. the refrigeration effect. (.7 kw) (Note that double acting means it ums twice for each revolution. he molecular mass for R is given in the tables.) D.J.Dunn

121 9. AMMONIA ABSORPION CYCLE he basic rincile of the ammonia absortion cycle is similar to that of the vaour comression cycle. An evaorator is used to absorb heat at a low temerature and a condenser is used to reject the heat at a higher temerature. he difference is in the way the ammonia is assed from the evaorator to the condenser. In a comression cycle this is done with a comressor. In the absortion cycle it is done by absorbing the ammonia into water at the lower temerature. he water and ammonia is then umed to a heater raising the ressure and temerature. he heater also searates the ammonia from the water and the ammonia vaour is driven off is at a higher ressure and temerature than it started at. he vaour is then condensed and throttled back to the evaorator. Figure 7 he advantage of this system is that a water um relaces the vaour comressor. he um may be done away with altogether by making use of the rincile of artial ressures. When hydrogen is mixed with ammonia vaour, the total ressure of the mixture '' is the sum of the artial ressures such that (ammonia) + (hydrogen) he mixing is done in the evaorator but the total ressure stays the same. he ammonia vaour hence exeriences a dro in ressure when mixing occurs and the effect is the same as throttling so a throttle valve is not needed either. Mixing causes the vaour to cool and condense so that a cold wet vaour results. D.J.Dunn

122 Evaoration dries it out and a mixture of ammonia and hydrogen gases leaves the evaorator. he mixture goes to the absorber where the ammonia is absorbed into the water leaving the hydrogen behind. he hydrogen goes back to the evaorator in a continuous cycle. When the ammonia and hydrogen searate out in the absorber, they both exeriences a ressure rise back to which is also the water ressure. Figure 8 Since no ressure difference exists between the evaorator and condenser, circulation may be caused by a thermo-sihon which is induced by heating the water and ammonia. he -h cycle is similar to that of a vaour comression cycle. Process to 4 is due to the mixing in the evaorator. Process to is due to the absortion. Figure 9 D.J.Dunn

123 Heat which is ut into the searator in order to make the ammonia leave the water, is carried with the water back to the absorber. his heat can be used to oerate the thermosihon by use of a heat exchanger. A schematic of a comlete lant is shown below. he heat required to oerate the system may be obtained from anywhere and is commonly a gas flame (Electrolux refrigeration system). his system is oular in caravan refrigerators. Figure 0 D.J.Dunn 4

124 APPLIED HERMODYNAMICS UORIAL 6 AIR-VAPOUR MIXURES In this tutorial you will do the following. Revise the UNIVERSAL GAS LAW Learn DALONS LAW OF PARIAL PRESSURES Aly these laws to mixtures of water vaour and air. Solve roblems involving air conditioning lant. Solve roblems involving cooling towers. Solve roblems involving steam condensers. Let's start by revising the Universal gas law and the law of artial ressures.

125 . UNIVERSAL GAS LAW mro V N ~ where Ro is the universal constant 84.4 J/kmol K Ñ is the relative molecular mass which is 8 for water vaour treated as a gas and 8.96 for dry air treated as a single gas.. PARIAL PRESSURES he ressure exerted by a gas on the surface of containment is due to the bombardment of the surface by the molecules. he relative distance between molecules is very large so if two or more gases exist in the same sace, their behaviour is unaffected by the others and so each gas roduces a ressure on the surface according to the gas law above. Each gas occuies the total volume V and has the same temerature. If two gases A and B are considered, the ressure due to each is : a m a N ~Ro V a b m b N ~Ro V b he total ressure on the surface of containment is a + b his is Daltons Law of artial ressures. Now let s see how these laws are alied to mixtures of vaour and air. D.J.DUNN

126 . AIR - VAPOUR MIXURES In the following work, water vaour is treated as a gas. Consider a mixture of dry air and vaour. If the temerature of the mixture is cooled until the vaour starts to condense, the temerature must be the saturation temerature (dew oint) and the artial ressure of the vaour s must be the value of s in the fluids tables at the mixture temerature. If the mixture is warmed u at constant ressure so that the temerature rises, the vaour must become suerheated. It can be shown that the artial ressure of the vaour and the dry air remains the same as at the saturation temerature. Let condition () be at the saturation condition and condition () be at the higher temerature. is constant so it follows that : V V he initial artial ressure of the vaour is: s m s N ~Ro s V he final ressure of the vaour is : s m s N ~Ro s V Since V V then s s By the same rocess it can be shown that a a If is constant then the artial ressures are constant and the artial ressure of the vaour may easily be found by looking u the saturation ressure at the dew oint if it is known. When the air is contact with water, it will evaorate the water and the water will cool down until it is at the saturation temerature or dew oint. his idea is used in wet bulb thermometers for examle, which measure the dew oint. When stable conditions are reached, the air becomes saturated and equal to the temerature of the water and so its temerature is the dew oint (t s ) in fluids tables. D.J.DUNN

127 WORKED EXAMPLE No. Moist air at bar and 5oC asses over water and emerges at bar and 8oC. Calculate the artial ressure of the air and vaour before cooling. SOLUION When cooled 8oC must be the saturation temerature so the artial ressure of the vaour is s in the fluids tables and is bar. he artial ressure of the vaour was the same before cooling so the artial ressure of the air must be bar. Now let's look at the definitions and use of humidity.. HUMIDIY here are two ways to exress humidity SPECIFIC AND RELAIVE... SPECIFIC HUMIDIY ω ω mass of water vaour/mass of dry air Starting with the gas law VN ~ m Ro svron ~ ω VRoN ~ a s ω 0.6 s a s sn ~ N ~ a s a s a x s a his derivation is often requested in Examinations. D.J.DUNN 4

128 .. RELAIVE HUMIDIY φ φ mass of vaour/maximum ossible mass of vaour he maximum ossible mass of water vaour which can be held by air is when the vaour is saturated and the temerature of the mixture is the saturation temerature. mass Volume/secific volume V/v When saturated, v vg at the mixture temerature. m m s φ g V v s V v a v v a s Alternatively v V/m N ss vs Ro s φ g and v g N g g Ro s artial ressure of the actual vaour g artial ressure when saturated. s ω 0.6 ( ) ω φ 0.6 s g s and φ s g D.J.DUNN 5

129 WORKED EXAMPLE No. Moist air at bar and 5oC is cooled to 8oC by assing it over water at 8oC. It emerges at 8oC and bar with a relative humidity of.0. Assuming that there is no net water absorbed nor lost, calculate the relative and secific humidity before cooling. SOLUION his is the same as the revious roblem so the dew oint must be 8oC and the artial ressure of the vaour is s at 8oC and is bar. a bar g at 5oC bar It follows that if no net water is gained nor lost then the secific humidity must be the same before and after and is : ω 0.6(s/a) 0.0 φ s/g 0.65 hese are the humidity values which will result in no evaoration nor condensation. If φ<0.65 then there will have been evaoration. If φ>0.65 then condensation will have taken lace on contact with the water and cooling also. D.J.DUNN 6

130 MASS BALANCE Consider the worked examle again only this time suose the relative humidity at inlet is 0.5. his means that water is evaorated. Consider kg of dry air assing through from inlet to outlet. At outlet φ and s g at 8oC bar. a bar ω 0.6s/g 0.0 ms/ma hence for kg of dry air there must be 0.0 kg of saturated vaour. At inlet φ 0.5 his time the mass of the vaour at inlet and outlet are not the same so the secific humidity is different at inlet. φ 0.5 ω (-s)/(0.6g) at 5oC. Remember s is the saturation ressure at the dew oint (8oC) and g is the saturation ressure at the actual temerature (5oC). 0.5 ω (-0.006)/(0.6 x 0.066) 49.7 ω Hence ω 0.0 m s /m a Since the air mass is kg throughout, then the mass of vaour at inlet is 0.0 kg. It follows that the mass of water evaorated is kg D.J.DUNN 7

131 SELF ASSESSMEN EXERCISE No.. Reeat the worked examle but this time the relative humidity 0.8 at inlet. Is water condensed or evaorated? (m s kg so water is condensed). Define secific humidity ω and rove that ω sñs/ña( - s) Humid air at bar flows through an insulated vessel over a ool of water and emerges saturated. he temeratures are 5oC and 8oC at inlet and outlet resectively. he mass of water is maintained constant at 8oC all the time. Calculate the relative humidity at inlet assuming constant ressure throughout. (Ans. 0.65) D.J.DUNN 8

132 4. ENERGY BALANCE Consider a simle air conditioner. Moist air is drawn in and cooled so that water condenses out. he air at this oint must be at the dew oint. Figure Alying the law of energy conservation we get : Ma caθ + m s h s m w c w θ + m a c a θ + m s h s + energy removed he suffixes a, s and w refer to air, vaour and water resectively. reating vaour as a gas, the acceted value of the secific heat caacity is.864 kj/kg K. he enthaly of steam relative to 0oC is then hs hg θs - ts) ts is the saturation temerature of the vaour. It is robably best to use the thermodynamic tables or the h - s chart sulied in the exam whenever ossible to find the enthaly of vaour at low ressures and temeratures. WORKED EXAMPLE No. Moist air enters a conditioning unit at 5oC and bar with a relative humidity of 0.7. It is assed through a cooler causing the temerature to fall to 8oC and condensate is formed. Calculate the mass of condensate formed er kg of dry air and the energy removed er kg. SOLUION Following the same method as in the revious examles, the mass of vaour at exit is m s 0.0 kg er kg of dry air. he vaour ressure at inlet is.0 bar. At inlet the relative humidity is 0.7 φ ω ( - s)/(0.6g) at 5oC. ω 0.6 x x 0.7/( ) he mass of vaour at inlet is then m s kg/kg D.J.DUNN 9

133 he condensate formed mw kg Conducting an energy balance: x.005 x h s x.005 x h s + mwθw + Φ h s hg (θs - ts) at bar. h s (5-8)(.864) 546.9kJ/kg h s hg at 8oC since it is saturated. h s 5.9 kj/kg he balance becomes x.005 x (546.9) x.005 x (5.9) x4.68x8 + Φ Φ 9.87 kj er kg of dry air. In air conditioning it is normal to heat the air to the required temerature before it leaves the unit. Figure Suose the air is heated to oc before leaving. What is the heat transfer required in the heater? he masses are unchanged so we only need an energy balance between and. x.005 x h s + Φ x.005 x h s h s hg (θs - ts) at bar. ts 8oC h s ) 54. kj/kg h s 5.9 kj/kg x.005 x x Φ x.005 x x Φ 55.7 Φ 4. kj D.J.DUNN 0

134 SELF ASSESSMEN EXERCISE No. Air having a ressure, temerature and relative humidity of bar, 6oC and 0.65 resectively, flows into an air conditioner at a steady rate and is dehumidified by cooling and removing water from it. he air is then heated to roduce an outlet temerature and relative humidity of 4oC and 0.59 resectively. he ressure is constant throughout. Determine the heat transfers in the cooler and heater er kg of conditioned air at exit. Draw u a comlete mass balance. (6. kj/kg and 6.4 kj/kg) D.J.DUNN

135 5. COOLING OWERS Cooling towers fall into two tyes, dry and wet. Dry cooling towers are no more than very large air conditioners and the theory is the same as already outlined. Wet cooling towers work on the rincile of sraying warm water downwards so that heat and vaour is assed to the air which rises and carries away latent heat leaving the water at a lower temerature to collect in a ool at the bottom of the tower. he water is then recycled from the ool. he moist air leaves the to of the tower as a lume. he tower has a venturi shae to assist the rocess by causing a slight ressure reduction in the sray area followed by resurgence as the to widens. his causes condensation to form and make the lume visible. Some of the condensate rains down into the ool. We can say with certainty that the air leaves the tower with 00% humidity. he best way to understand the roblem is to do a worked examle as follows. WORKED EXAMPLE No.4 A cooling tower must cool 40 kg of water er minute. he water is sulied at 4oC and it is srayed down into the column of air which enters the bottom of the tower at a rate of 540m/min with a temerature of 8oC and relative humidity of 60%. he moist air leaves the to of the tower saturated at 7oC. he whole rocess occurs at a constant ressure of.0 bar. Determine the temerature of the cooled water in the ool and the rate at which make u water must be sulied to relace that evaorated. Figure D.J.DUNN

136 SOLUION R 87 J/kg K for air and 46 J/kg K for vaour. INLE AIR g bar at 8oC φ 0.6 s / g s bar hence a bar ma V/R x 05 x 540/(87 x 9) kg/min m s V/R x 05 x 540/(46 x 9 ) 4.97 kg/min OULE AIR φ s g bar hence a bar φ 0.6 x / m s x kg/min Water evaorated kg/min Make u water 9.69 kg/min m w ENERGY BALANCE In this examle enthaly values from the steam tables and chart will be used. h w 4oC 75.8 kj/kg h w is unknown h a.005 x 7 kj/kg h s bar & 8oC 50 kj/kg (from h-s chart) h s 7oC 550. kj/kg Balancing energy we get (40 x75.8) + {646.6 x.005 x (8-7)} + {4.97 x 50) (4.66 x 550.) 0. h w h w.5 kj/kg and from the tables the temerature must be 9.5 oc. he temerature of the cooled water is 9.5 oc. D.J.DUNN

137 SELF ASSESSMEN EXERCISE No.. Derive the exression for secific humidity ω 0.6(s/a) Water flows at kg/h and 40oC into a cooling tower and is cooled to 6oC. he unsaturated air enters the tower at 0oC with a relative humidity of 0.4. It leaves as saturated air at 0oC. he ressure is constant at bar throughout. Calculate i. the mass flow of air er hour. (4 66 kg/h) ii. the mass of water evaorated er hour. (00.5 kg/h). he cooling water for a small condenser is sent to a small cooling tower. 7 m/s of air enters the tower with a ressure, temerature and relative humidity of.0 bar, 5oC and 0.55 resectively. It leaves saturated at oc. he water flows out of the tower at 7.5 kg/s at oc. Using a mass and energy balance, determine the temerature of the water entering the tower. (Answer.9oC). A fan sulies 600 dm/s of air with a relative humidity of 0.85, temerature 0oC and ressure.04 bar into an air conditioner. Moisture is removed from the air by cooling and both the air and condensate leave at the same temerature. he air is then heated to 0oC and has a relative humidity of 0.6. Determine the following. i. he mass of dry air and water at entrance to the conditioner. (0.697 kg/s and kg/s) ii. he mass of water vaour delivered at exit. ( kg/s) iii. he mass of water extracted from the cooler. ( kg/s) iv. he temerature at exit from the cooler. (oc) v. he heat transfer in the cooler. D.J.DUNN 4

138 6. CONDENSERS It is inevitable that air will be drawn into steam condensers oerating with a vacuum. he effect of this is to reduce the saturation ressure and temerature of the steam resulting in a colder condensate that would otherwise be obtained. his in turn means more heat required to turn it back into steam in the boiler and a reduced thermal efficiency for the ower lant. he air must be removed from the condenser in order to kee the artial ressure as small as ossible. his is done with an extractor um. Some vaour will be removed with the air but this loss is tolerable because of the energy saved. Figure 4 he solution to roblems on condensers is similar to that for cooling towers and requires mass and energy balances. It is normal to neglect the artial ressure of the air at inlet as it makes little difference to the answers. SELF ASSESSMEN EXERCISE No. 4 a. Discuss the reasons why air mixed with steam in a condenser is not desirable. b. Wet steam with a dryness fraction of 0.9 enters a condenser at 0.05 bar ressure at a rate of kg/h. he condensate leaves at 5oC. Air also enters with the steam at a rate of 40 kg/h. he air is extracted and cooled to 0oC. he artial ressure of the air at inlet is negligible and the rocess is at constant ressure. he cooling water is at 0oC at inlet and oc at outlet. i. Determine the mass of vaour extracted with the air. (50 kg/h) ii. Calculate the flow rate of the cooling water required. ( kg/h) D.J.DUNN 5

139 HERMODYNAMICS UORIAL 7 COMPRESSIBLE FLOW On comletion of this tutorial you should be able to do the following. Define entroy Derive exressions for entroy changes in fluids Derive Bernoulli's equation for gas Derive equations for comressible ISENROPIC flow Solve roblems involving comressible flow Note that more work on comressible flow may be found under FLUID MECHANICS. Let's start by revising entroy.

140 . ENROPY. DEFINIION You should already be familiar with the theory of work laws in closed systems. You should know that the area under a ressure-volume diagram for a reversible exansion or comression gives the work done during the rocess. In thermodynamics there are two forms of energy transfer, work (W) and heat (Q). By analogy to work, there should be a roerty which if lotted against temerature, then the area under the grah would give the heat transfer. his roerty is entroy and it is given the symbol S. Consider a -V and -s grah for a reversible exansion. Figure From the -V grah we have W dv From the -S grah we have Q ds his is the way entroy was develoed for thermodynamics and from the above we get the 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. D.J.Dunn

141 . ISENROPIC PROCESSES he word Isentroic 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, then the grah must be a vertical line as shown. Figure 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 nett result is zero heat transfer and constant entroy. You do not need to be concerned about this at this stage. Entroy is used in the solution of gas and vaour roblems. We should now look at ractical alications of this roerty and study the entroy changes which occur in closed and steady flow systems for erfect gases and vaours. hese derivations should be learned for the examination. D.J.Dunn

142 . ENROPY CHANGES FOR A PERFEC GAS IN A CLOSED SYSEMS Consider a closed system exansion of a fluid against a iston with heat and work transfer taking lace. Figure Alying the non-flow energy equation we have Q + W U Differentiating we have Since dq ds and dw -dv then dq + dw du ds - dv du ds du + dv his exression is the starting oint for all derivations of entroy changes for any fluid (gas or vaour) in closed systems. It is normal to use secific roerties so the equation becomes ds du + dv but from the gas law v R we may substitute for and the equation becomes ds du + Rdv/v rearranging and substituting du c v d we have s is secific entroy v is secific volume. ds c v d/ + Rdv/v...() u is secific internal energy and later on is also used for velocity. D.J.Dunn 4

143 . ISOHERMAL PROCESS Figure 4 In this case temerature is constant. Starting with equation () ds c v d/ + Rdv/v. since d 0 then s - s s R ln(v /v ) A quicker alternative derivation for those familiar with the work laws is: V Q + W U but U 0 then Q - W and W -mr ln V Q ds S but is constant. Q W S V S mr ln V v s R ln v s R ln V mr ln V v and since v D.J.Dunn 5

144 . CONSAN VOLUME PROCESS Figure 5 Starting again with equation () we have ds c v d/ + Rdv/v In this case dv0 so ds c v d/ Integrating between limits () and () s c v ln( / ). CONSAN PRESSURE PROCESS Figure 6 Starting again with equation () we have d dv ds Cv + R In this case we integrate and obtain v v s Cv ln R ln For a constant ressure rocess, v/ constant v v so the exression becomes s Cv ln + R ln ( Cv + R) v ln It was shown in an earlier tutorial that R c - c v hence s C ln D.J.Dunn 6

145 .4 POLYROPIC PROCESS his is the most difficult of all the derivations here. Since all the forgoing are articular examles of the olytroic rocess then the resulting formula should aly to them also. Figure 7 he olytroic exansion is from () to () on the -s diagram with different ressures, volumes and temeratures at the two oints. he derivation is done in two stages by suosing the change takes lace first at constant temerature from () to (A) and then at constant ressure from (A) to (). You could use a constant volume rocess instead of constant ressure if you wish. s -s (s A -s ) - (s A -s ) s -s (s A -s ) + (s -s A ) For the constant temerature rocess (s A -s ) R ln( / A ) For the constant ressure rocess (s -s A ) (c ) ln( / A ) Hence s R ln + C ln + s -s Since A and A A A hen s s -s R ln + C ln Divide through by R s C ln + ln R R From the relationshi between c, c v, R and we have c /R /(-) Hence s R ln s + ln ln R his formula is for a olytroic rocess and should work for isothermal, constant ressure, constant volume and adiabatic rocesses also. In other words, it must be the derivation for the entroy change of a erfect gas for any closed system rocess. his derivation is often requested in the exam. D.J.Dunn 7

146 WORKED EXAMPLE No. A erfect gas is exanded from 5 bar to bar by the law V. C. he initial temerature is 00 o C. Calculate the change in secific entroy. R 87 J/kg K.4. SOLUION 47 5 s ln R. 6.7K.5 s 6.7 (ln5) 0.67 R 47 s 0.67x J/kgK SELF ASSESSMEN EXERCISE No.. Calculate the secific entroy change when a erfect gas undergoes a reversible isothermal exansion from 500 kpa to 00 kpa. R 87 J/kg K. (Answer J/kg K).. Calculate the total entroy change when kg of erfect gas is comressed reversibly and isothermally from 9 dm to dm. R00 J/kg K. (Answer -. kj/k). Calculate the change in entroy when.5 kg of erfect gas is heated from 0oC to 00oC at constant volume. ake c v 780 J/kg K (Answer 470 J/K) 4. Calculate the total entroy change when 5 kg of gas is exanded at constant ressure from 0oC to 00oC. R 00 J/kg K c v 800 J/kg K (Answer.45 kj/k) 5. Derive the formula for the secific change in entroy during a olytroic rocess using a constant volume rocess from (A) to (). 6. A erfect gas is exanded from 5 bar to bar by the law V.6 C. he initial temerature is 00 o C. Calculate the change in secific entroy. R 87 J/kg K.4. (Answer -44 J/kg K) 7. A erfect gas is exanded reversibly and adiabatically from 5 bar to bar by the law V C. he initial temerature is 00oC. Calculate the change in secific entroy using the formula for a olytroic rocess. R 87 J/kg K.4. (he answer should be zero since the rocess is constant entroy). D.J.Dunn 8

147 Let's go on to aly the knowledge of entroy to the flow of comressible fluids starting with isentroic flow. 4. ISENROPIC FLOW Isentroic means constant entroy. In this case we will consider the flow to be ADIABAIC also, that is, with no heat transfer. Consider gas flowing in a duct which varies in size. he ressure and temerature of the gas may change. Figure 8 Alying the steady flow energy equation between () and () we have : Φ - P U + F.E. + K.E. + P.E. For Adiabatic Flow, Φ 0 and if no work is done then P 0 hence : U + F.E. H 0 H + K.E.+ P.E. In secific energy terms this becomes : rewriting we get: 0 h + k.e. +.e. h + u / + g z h + u / + g z For a gas, h C so we get Bernoulli's equation for gas which is : C + u / + g z C + u / + g z Note that is absolute temerature in Kelvins oc + 7 D.J.Dunn 9

148 4. SAGNAION CONDIIONS If a stream of gas is brought to rest, it is said to SAGNAE. his occurs on leading edges of any obstacle laced in the flow and in instruments such as a Pitot ube. Consider such a case for horizontal flow in which P.E. may be neglected. Figure 9 u 0 and z z so C + u / C + 0 u /C + is the stagnation temerature for this case. Let - u /C u /C Now C - C v R and C / C v is the adiabatic index. hence C R / ( - ) and so : u ( - ) / ( R) It can be shown elsewhere that the seed of sound a is given by : hence at oint : a R / u ( - ) / ( R ) u ( - ) /a he ratio u/a is the Mach Number M a so this may be written as : / M a ( - ) / If M a is less than 0. then M a is less than 0.04 and so / is less than It follows that for low velocities, the rise in temerature is negligible under stagnation conditions. D.J.Dunn 0

149 D.J.Dunn he equation may be written as : ( ) ( ) + a a M M Since V/ constant and V constant then : Hence : ( ) ( ) a a M M + + is the stagnation ressure. If we now exand the equation using the binomial theorem we get : Ma 4 Ma Ma 4 If M a is less than 0.4 then : Ma + Now comare the equations for gas and liquids : LIQUIDS u ( /ρ)0.5 GAS Ma + Put + so : u R v Ma ρ where ρ / R and M a u / ( R) hence u ( /ρ)0.5 which is the same as for liquids.

150 SELF ASSESSMEN EXERCISE No. ake.4 and R 8 J/kg K in all the following questions.. An aerolane flies at Mach 0.8 in air at 5o C and 00 kpa ressure. Calculate the stagnation ressure and temerature. (Answers 4.9 K and 5.4 kpa). Reeat roblem if the aerolane flies at Mach. (Answers 58.4 K and 78.4 kpa). he ressure on the leading edges of an aircraft is 4.5 kpa more than the surrounding atmoshere. he aerolane flies at an altitude of metres. Calculate the seed of the aerolane.( Answer m/s) Note from fluids tables, find that a 0.5 m/s kpa.4 4. An air comressor delivers air with a stagnation temerature 5 K above the ambient temerature. Determine the velocity of the air. (Answer 00. m/s) Let's now extend the work to itot tubes. D.J.Dunn

151 5. PIO SAIC UBE A Pitot Static ube is used to measure the velocity of a fluid. It is ointed into the stream and the differential ressure obtained gives the stagnation ressure. + Figure 0 Using the formula in the last section, the velocity v may be found. WORKED EXAMPLE No. A itot tube is ointed into an air stream which has a ressure of 05 kpa. he differential ressure is 0 kpa and the air temerature is 0oC. Calculate the air seed. SOLUION kpa Ma ( ) ( ) + 5 Ma + hence Ma a (R)0.5 (.4 x 87 x 9 )0.5 4 m/s M a u/a hence u 7.7 m/s Let's further extend the work now to venturi meters and nozzles. D.J.Dunn

152 4 D.J.Dunn 6. VENURI MEERS AND NOZZLES Consider the diagrams below and aly Isentroic theory between the inlet and the throat. Figure u - u h - h If the Kinetic energy at inlet is ignored this gives us u h - h For a gas h C so: [ ] C u P Using C R/(-) we get [ ] R u R V/m /ρ so ρ ρ u so it follows that ρ ρ ρ ρ ρ u V V ρ u he mass flow rate m ρ A u C d where C d is the coefficient of discharge which for a well designed nozzle or Venturi is the same as the coefficient of velocity since there is no contraction and only friction reduces the velocity. ρ ρ hence [ ] + ρ A C m d

153 If a grah of mass flow rate is lotted against ressure ratio ( / ) we get: Figure Aarently the mass flow rate starts from zero and reached a maximum and then declined to zero. he left half of the grah is not ossible as this contravenes the nd law and in reality the mass flow rate stays constant over this half. What this means is that if you started with a ressure ratio of, no flow would occur. If you gradually lowered the ressure, the flow rate would increase u to a maximum and not beyond. he ressure ratio at which this occurs is the CRIICAL RAIO and the nozzle or Venturi is said to be choked when assing maximum flow rate. Let r dm For maximum flow rate, 0 dr he student should differentiate the mass formula above and show that at the maximum condition the critical ressure ratio is : 6. MAXIMUM VELOCIY r + If the formula for the critical ressure ratio is substituted into the formula for velocity, then the velocity at the throat of a choked nozzle/venturi is : u R a ρ Hence the maximum velocity obtainable at the throat is the local seed of sound. D.J.Dunn 5

154 6. CORRECION FOR INLE VELOCIY In the receding derivations, the inlet velocity was assumed negligible. his is not always the case and esecially in Venturi Meters, the inlet and throat diameters are not very different and the inlet velocity should not be neglected. he student should go through the derivation again from the beginning but this time kee v in the formula and show that the mass flow rate is m C d A ρ A A + he critical ressure ratio can be shown to be the same as before. 6. MORE ON ISENROPIC FLOW When flow is isentroic it can be shown that all the stagnation roerties are constant. Consider the conservation of energy for a horizontal duct : h + u/ constant h secific enthaly If the fluid is brought to rest the total energy must stay the same so the stagnation enthaly h o is given by : h o h + u/ and will have the same value at any oint in the duct. since h o C o then o (the stagnation temerature) must be the same at all oints. It follows that the stagnation ressure o is the same at all oints also. his knowledge is very useful in solving questions. D.J.Dunn 6

155 6.4 ISENROPIC EFFICIENCY (NOZZLE EFFICIENCY) If there is friction resent but the flow remains adiabatic, then the entroy is not constant and the nozzle efficiency is defined as : η actual enthaly dro/ideal enthaly dro For a gas this becomes : ( - )/( - ' ) ' is the ideal temerature following exansion. Now aly the conservation of energy between the two oints for isentroic and non isentroic flow : C + u / C + u /... for isentroic flow Hence C + u / C ' + u ' /...for non isentroic η ( - )/( - ' ) (u - u )/(u ' - u ) If v is zero (for examle Rockets) then this becomes : η u /u ' D.J.Dunn 7

156 SELF ASSESSMEN EXERCISE No.. A Venturi Meter must ass 00g/s of air. he inlet ressure is bar and the inlet temerature is 0oC. Ignoring the inlet velocity, determine the throat area. ake Cd as ake.4 and R 87 J/kg K (assume choked flow) (Answer m). Reeat roblem given that the inlet is 60 mm diameter and the inlet velocity must not be neglected. (Answer m). A nozzle must ass 0.5 kg/s of steam with inlet conditions of 0 bar and 400oC. Calculate the throat diameter that causes choking at this condition. he density of the steam at inlet is.6 kg/m. ake for steam as. and Cd as (Answer. mm) 4. A Venturi Meter has a throat area of 500 mm. Steam flows through it, and the inlet ressure is 7 bar and the throat ressure is 5 bar. he inlet temerature is 400oC. Calculate the flow rate. he density of the steam at inlet is.74 kg/m. ake.. R 46 J/kg K. Cd0.97. (Answer 8 g/s) 5. A itot tube is ointed into an air stream which has an ambient ressure of 00 kpa and temerature of 0oC. he ressure rise measured is kpa. Calculate the air velocity. ake.4 and R 87 J/kg K. (Answer 89.4 m/s) 6. A fast moving stream of gas has a temerature of 5oC. A thermometer is laced into it in front of a small barrier to record the stagnation temerature. he stagnation temerature is 8oC. Calculate the velocity of the gas. ake.5 and R 00 J/kg K. (Answer 7.5 m/s) Let's do some further study of nozzles of venturi shaes now. D.J.Dunn 8

157 9 D.J.Dunn 7. CONVERGEN - DIVERGEN NOZZLES A nozzle fitted with a divergent section is in effect a Venturi shae. he divergent section is known as a diffuser. Figure If is constant and is reduced in stages, at some oint will reach the critical value which causes the nozzle to choke. At this oint the velocity in the throat is sonic. If is further reduced, will remain at the choked value but there will be a further ressure dro from the throat to the outlet. he ressure dro will cause the volume of the gas to exand. he increase in area will tend to slow down the velocity but the decrease in volume will tend to increase the velocity. If the nozzle is so designed, the velocity may increase and become suersonic at exit. In rocket and jet designs, the diffuser is imortant to make the exit velocity suersonic and so increase the thrust of the engine. 7. NOZZLE AREAS When the nozzle is choked, the velocity at the throat is the sonic velocity and the Mach number is. If the Mach number at exit is M e then the ratio of the throat and exit area may be found easily as follows. u t (R t )0.5 u e M e (R e )0.5 mass/s ρ t A t v t ρ e A e v e. ( ) ( ) + ρ ρ ρ ρ t e e e t 0.5 t e e t e e t t e t e 0.5 t 0.5 e e t e e t t e t e t t e e e t M A A M A A was also shown earlier that It R R M A A but earlier it was shown that u u A A here is much more which can be said about nozzle design for gas and steam with imlications to turbine designs. his should be studied in advanced text books.

158 WORKED EXAMPLE No. Solve the exit velocity for the nozzle shown assuming isentroic flow: Figure 4 50 K P MPa 00 kpa he nozzle is fully exanded (choked). Hence M t (the Mach No.) he adiabatic index.4 SOLUION he critical ressure t {/( - )} /(-) 0.58 MPa t / ( t / ) ( -)/ hence t 9.7 K o / t { + M( -)/ } hence o 50 K It makes sense that the initial ressure and temerature are the stagnation values since the initial velocity is zero. t ( / t ) ( -)/ 8. K a (R ) m/s o / { + M ( - )/ } /(-) Hence M.57 and u.57 x m/s D.J.Dunn 0

159 SELF ASSESSMEN EXERCISE No. 4. A nozzle is used with a rocket roulsion system. he gas is exanded from comlete stagnation conditions inside the combustion chamber of 0 bar and 000K. Exansion is isentroic to bar at exit. he molar mass of the gas is kg/kmol. he adiabatic index is.. he throat area is 0. m. Calculate the thrust and area at exit. (Answers 0.6 m and 8.5 kn) Recalculate the thrust for an isentroic efficiency of 95%. (Answer 74. kn) Note that exansion may not be comlete at the exit area. You may assume t o +. A erfect gas flows through a convergent-divergent nozzle at kg/s. At inlet the gas ressure is 7 bar, temerature 900 K and velocity 78 m/s. At exit the velocity is 80m/s. he overall isentroic efficiency is 85%. he flow may be assumed to be adiabatic with irreversibility's only in the divergent section. C. kj/kg K R 87 J/kg K. Calculate the cross sectional areas at the inlet, throat and exit. (Answers 0.8 cm, 0. cm and.69 cm) Calculate the net force acting on the nozzle if it is stationary. he surrounding ressure is bar. (-57 N) You may assume t o +. Dry saturated steam flows at kg/s with a ressure of 4 bar. It is exanded in a convergent-divergent nozzle to 0.4 bar. Due to irreversibility's in the divergent section only, the isentroic efficiency 96%. he critical ressure ratio may be assumed to be Calculate the following. he dryness fraction, secific volume and secific enthaly at the throat. (Answers 0.958, 0. m/kg and 68 kj/kg) he velocity and cross sectional area at the throat and exit. (Answers 46.6 m/s, 497 mm, 6 m/s and 7. cm.) he overall isentroic efficiency. (Answer 96.6%) D.J.Dunn

160 4. A jet engine is tested on a test bed. At inlet to the comressor the air is at bar and 9 K and has negligible velocity. he air is comressed adiabatically to 4 bar with an isentroic efficiency of 85%. he comressed air is heated in a combustion chamber to 75 K. It is then exanded adiabatically in a turbine with an isentroic efficiency of 87%. he turbine drives the comressor. he gas leaving the turbine is exanded further reversibly and adiabatically through a convergent nozzle. he flow is choked at exit. he exit area is 0. m. Determine the following. he ressures at the outlets of the turbine and nozzle. (Answers.8 bar and.9 bar) he mass flow rate. (Answer 7. kg/s) he thrust roduced. (Answer 7 kn) It may be assumed that t o + and a R 5. Dry saturated steam exands through a convergent-divergent nozzle. he inlet and outlet ressures are 7 bar and bar resectively at a rate of kg/s. he overall isentroic efficiency is 90% with all the losses occurring in the divergent section. It may be assumed that.5 and t o + Calculate the areas at the throat and exit. (Answers 9.6 cm and 8.8 cm). he nozzle is horizontal and the entry is connected directly to a large vessel containing steam at 7 bar. he vessel is connected to a vertical flexible tube and is free to move in all directions. Calculate the force required to hold the receiver static if the ambient ressure is.0 bar. (Answer.868 kn) D.J.Dunn

161 HERMODYNAMICS UORIAL No.8 COMBUSION OF FUELS On comletion of this tutorial you should be able to do the following.. Let's start by revising the basics. Write down combustion equations. Solve the oxygen and air requirements for the combustion of solid, liquid and gaseous fuels. Determine the roducts of combustion. Determine the air/fuel ratio from the roducts of combustion. Solve roblems involving energy released by combustion. Solve roblems involving dissociation of roducts. Solve ast aer questions.

162 . INRODUCION Combustion is the rocess of chemical reaction between fuel and oxygen(reactants). he rocess releases heat and roduces roducts of combustion. he main elements which burn are : CARBON HYDROGEN SULPHUR he heat released by kg or m of fuel is called the calorific value. he oxygen used in combustion rocesses normally comes from the atmoshere and this brings nitrogen in with it which normally does nothing in the rocess but makes u the bulk of the gases remaining after combustion. he main elements in combustion are then : Symbol Atomic Mass Molecular Mass Product Carbon C CO Hydrogen H H O Sulhur S SO Oxygen O 6 Nitrogen N 4 8 If the water formed during combustion leaves as vaour, it takes with it the latent heat of evaoration and thus reduces the energy available from the rocess. In this case the calorific value is called the lower Calorific value (LCV). If the roducts cool down after combustion so that the vaour condenses, the latent heat is given u and the calorific value is then the higher calorific value (HCV). Solid and liquid fuels are normally analysed by mass to give the content of carbon, hydrogen, sulhur and any other elements resent. Often there is silica, moisture and oxygen resent in small quantities which have some effect on rocess. he silica leaves slaggy deosits on the heat transfer surfaces in boilers. Gaseous fuels are normally analysed by volumetric content and are in the main hydrocarbon fuels. For uroses of calculation, the content of air is considered to be : VOLUMERIC GRAVIMERIC Oxygen % % Nitrogen 79% 77% D.J.Dunn

163 he sulhur content of the fuel is considered to be a ollutant and so undesirable. he theoretically correct quantity of air or oxygen required to just exactly burn the fuel exressed as a ratio to the fuel burned, is called the SOICHIOMERIC RAIO. In ractice it is found that not all the oxygen in the reactant reaches the fuel elements and that excess air is required in order to ensure comlete combustion. his results in oxygen aearing in the roducts. If too little air or oxygen is sulied, the result is incomlete combustion resulting in the formation of carbon monoxide CO instead of carbon dioxide CO. he resulting roducts contain water HO. Industrial equiment for measuring the contents of the roducts usually remove the water from the samle and the roducts are then called the dry roducts.. COMBUSION CHEMISRY. SOLID AND LIQUID FUELS In the case of solid and liquid fuels, we do the combustion of each element searately. he imortant rule is that you must have the same number of atoms of each substance before and after the rocess. his may be obtained by juggling with the number of molecules. CARBON C + O CO Mass ratio + 44 Hence kg of C needs /kg of O and makes 44/kg of CO HYDROGEN H + O H O Mass ratio Hence kg of H needs 8kg of O and makes 9 kg of H O SULPHUR S + O SO + 64 Hence kg of S needs kg of O and makes kg of SO. D.J.Dunn

164 .. GASEOUS FUELS yical hydrocarbons are : Methane Ethane Proane Butane Pentane Hexane Hetane Octane Ethene Proene Ethyne Benzenol Cyclohexane CH4 CH6 CH8 C4H0 C5H C6H4 C7H6 C8H8 CH4(Ethylene) CH6 (Proylene) CH (Acetylene) C6H6 (Benzene) C6H he combustion equation follows the following rule : CaHb + (a+b/4)o (a)co + (b/)ho If this results in fractional numbers of molecules, then the whole equation may be multilied u. WORKED EXAMPLE No. Write out the combustion equation for C8H8 SOLUION C8H8 + (8+8/4)O 8CO + (8/)HO C8H8 + ½O 8CO + 9HO C8H8 + 5O 6CO +8HO here are other gases which burn and the main one to know about is Carbon Monoxide (CO) which is artially burned carbon. he equation for the combustion of CO is : CO + O CO D.J.Dunn 4

165 . COMBUSION BY MASS he only rule to be observed in deducing the quantities of each substance is the law of conservation of mass. he roortions of the mass is that of the molecular masses. his is shown in the following examle. WORKED EXAMPLE No. A fuel contains by mass 88% C, 8%H, %S and % ash (silica). Calculate the stoichiometric air. SOLUION CARBON C + O CO Mass ratio + 44 Hence 0.88kg of C need (/)x kg of oxygen. It makes (44/) x kg of carbon dioxide. HYDROGEN H + O HO Mass ratio hence 0.08kg of hydrogen needs (/4) x kg of oxygen. SULPHUR S + O SO Mass ratio + 64 Hence 0.0kg of sulhur needs 0.0kh of oxygen and makes 0.0kg of sulhur dioxide. OAL OXYGEN needed is kg OAL AIR needed is.997/%.0kg he SOICHIOMERIC air/fuel ratio is.0/ D.J.Dunn 5

166 WORKED EXAMPLE No. If the air sulied is 0% more than the stoichiometric value, find the analysis of the dry roducts by mass. SOLUION If 0% excess air is sulied then the air sulied is: 0% x kg Oxygen is also 0% excess so 0. x kg is left over. Nitrogen in the air is 77% x kg List of roducts : Nitrogen.04kg 75.8% Carbon dioxide.7kg 0.% Sulhur dioxide 0.0kg 0.% Oxygen 0.599kg.8% otal dry roduct 5.886kg 00% It is of interest to note that for a given fuel, the % of any roduct is a direct indication of the excess air and in ractice the carbon dioxide and/or oxygen is used to indicate this. his is imortant in obtaining otimal efficiency in a combustion rocess. D.J.Dunn 6

167 SELF ASSESSMEN EXERCISE No.. A boiler burns fuel oil with the following analysis by mass : 80% C 8% H %S 0% excess air is sulied to the rocess. Calculate the stoichiometric ratio by mass and the % Carbon Dioxide resent in the dry roducts. (5.6/ 4.9% CO). A boiler burns coal with the following analysis by mass : 75% C 5% H 7%S remainder ash Calculate the % Carbon Dioxide resent in the dry roducts if 0% excess air is sulied. (6.5% CO). Calculate the % of each dry roduct when coal is burned stoichiometrically in air. he analysis of the coal is: 80% C 0% H 5% S and 5% ash. (76.7%N,.5% CO 0.8% SO) D.J.Dunn 7

168 4.COMBUSION BY VOLUME First we need to revise gas mixtures and understand the meaning of VOLUMERIC CONEN. o do this we must understand Dalton's law of artial ressures and Avagadro's Law. First let us define the kmol. A kmol of substance is the number of kg numerically equal to the aarent molecular mass. For examle kg of Carbon is a kmol, so is kg of O and kg of H and 8 kg of N. he molecular mass of a substance is exressed as kg/kmol so the molecular mass of O, for examle, is kg/kmol. Avagadro's Law states : m of any gas at the same ressure and temerature contains the same number of molecules. It follows that the volume of a gas at the same and is directly roortional to the number of molecules. From this we find that the volume of a kmol of any gas is the same if and are the same. Dalton's law states: he total ressure of a mixture is the sum of the artial ressures. he artial ressure is the ressure each gas would exert if it alone occuied the same volume at the same temerature. Consider two gases A and B occuying a volume V at temerature. Using the Universal gas law for each : AVA maro/ña BVB mbro/ñb where Ñ is the relative molecular mass. A/B mañb/mbña ratio of the kmol fractions. A and B are the artial ressures. VA and VB are the artial volumes. D.J.Dunn 8

169 hese are the volumes each gas would occuy if they were searated and ket at the original and. his concet is very useful in roblems involving the combustion of gases. It also follows that the artial volumes are directly related to the artial ressures so that VA/VB A/B Figure When not mixed the ressure is and the volumes are VA and VB. Hence : VA/ mro/ña ` maro/ñava...() VB/ mro/ñb mbro/ñbvb...() Since () () then : ma/ñavamb/ñbvb and so VA/VB(mA/ÑA)(mB/ÑB) which shows that in a mixture, the artial volumes are in the same ratio as the kmol fractions which in turn are in roortion to the number of molecules of each gas. When mixed they both have volume V, hence: A maro/ña V...() B mbro/ñb V...(4) ()/() gives A/ VA/V and (4)/() gives B/ VB/V hence VA/VB A/B Consider the combustion of Methane. CH4 + O CO + HO Since the volumetric content of each gas is in the same ratio as the kmol fractions then volumetric content is in the same roortion as the molecules. Hence it needs volumes of oxygen to burn volume of methane. he volume of air needed is /% 9.5 volumes. Hence it burn m of methane we need 9.5 m of air for stoichiometric combustion. If the roducts are at the same and as the original reactants, we would obtain m of carbon dioxide and m of water vaour which would robably condense and cause a reduction in volume and/or ressure. D.J.Dunn 9

170 WORKED EXAMPLE No.4 Calculate the % CO in the dry roducts when methane is burned with 5% excess air by volume. SOLUION CH4 + O CO + HO Volume ratio he stoichiometric air is /% 9.54 m he actual air is 9.54 x 5% 0.95 m Analysis of dry roducts : Nitrogen 79% x m Carbon Dioxide.00 m Oxygen 5% x 0.0 m otal 9.95 m he % Carbon Dioxide (/9.95) x 00 0% When the fuel is a mixture of gases, the rocedure outlined must be reeated for each combustible gas and the oxygen deduced for the volume of each in m of total fuel. WORKED EXAMPLE No. 5 A fuel is a mixture of 60% Methane and 0% carbon monoxide and 0% oxygen by volume. Calculate the stoichiometric oxygen needed. SOLUION As before, the volume of oxygen required to burn m of methane is m.o burn 0.6m needs.m of oxygen. For carbon monoxide we use the combustion equation : CO + O CO Hence to burn m of CO need 0.5 m of oxygen, so to burn 0. m needs 0.5 m of oxygen. he total oxygen needed is m. However there is already 0. m in the fuel so the stoichiometric oxygen needed.5m D.J.Dunn 0

171 SELF ASSESSMEN EXERCISE No.. Find the air fuel ratio for stoichiometric combustion of Ethene by volume. (6.9/). Find the air fuel ratio for stoichiometric combustion of Butane by volume.(0.95/). Calculate the % carbon dioxide resent in the dry flue gas if 0% excess air is used. (0.6%). Find the air fuel ratio for stoichiometric combustion of Proane by volume.(.8/). Calculate the % oxygen resent in the dry flue gas if 0% excess air is used. (.8%) 4. A gaseous fuel contains by volume : 5% CO, 40% H, 40% CH4, 5% N Determine the stoichiometric air and the % content of each dry roduct. (4.76 m, 89.7%,N 0.% CO). D.J.Dunn

172 5. RELAIONSHIP BEWEEN PRODUC AND EXCESS AIR. It follows that if we can deduce the % roduct then given the figure, we can work backwards to determine the air or oxygen that was used. WORKED EXAMPLE No.6 consider the combustion of methane again. CH4 + O CO + HO vol vol vol vols SOLUION Let the excess air be x (as a decimal) he stoichiometric air is 9.5 vols. Actual air is 9.5( + x) Dry Products: Nitrogen 0.79 x 9.5( + x) 7.54x Oxygen.000x Carbon Dioxide.000 otal 9.54x % Carbon monoxide 00 {/(9.54x )} % Oxygen 00{/(9.54x )} For examle if the % CO is 0% then the excess air is found as follows : 0% 00 {/(9.54x )} 0. /(9.54x ) (9.54x ) x.476 x0.55 or 5.5% Similarly if the O is 0% then the excess air is 8% (show this for yourself) If the analysis of the fuel is by mass, then a different aroach is needed as follows : D.J.Dunn

173 WORKED EXAMPLE No.7 An analysis of the dry exhaust gas from an engine burning Benzole shows 5% Carbon Dioxide resent by volume. he Benzole contains 90% C and 0% H by mass. Assuming comlete combustion, determine the air/fuel ratio used. SOLUION kg of fuel contains 0.9kg of C and 0.kg of H. Converting these into kmol we have 0.9/ kmol of C and 0./ kmol of H. For kmol of dry exhaust gas we have : 0.5 kmol of CO Y kmol of excess O Y Y kmol of N kmol of CO is 44 kg kmol of N is 8 kg kmol of O is kg 0.5 kmol of CO is 0.5 x 44kg his contains (/44) carbon so the carbon resent is 0.5 x kg he carbon in the fuel is 0.9 kmol er kmol of fuel. Hence the number of kmols of DEG must be 0.9/(.5 x ) 0.5 here are 0.5 kmol of DEG for each kmol of fuel burned. he Nitrogen resent in the DEG is Y kmol er kmol of DEG. his has a mass of 8( Y) er kmol of DEG he oxygen sulied to the rocess must be : (./76.7) x 8 x ( Y) Y kg er kmol of DEG. (using recise roortions of air for accuracy). he oxygen contained within the carbon dioxide is: (/44) x 0.5 x kg er kmol DEG kmol of CO contains 44 kg and /44 of this is oxygen. he oxygen in the CO is hence x 0.5 kg er kmol DEG. he excess oxygen is Y kg er kmol DEG otal oxygen in the roducts excluding that used to make HO is : x Y D.J.Dunn

174 he oxygen used to burn hydrogen is hence : Y - x Y O used to burn H is For 0.5 kmol this is y kg er kmol DEG Y kg o burn hydrogen requires oxygen in a ratio of 8/. here is 0. kg of H in each kmol of fuel so 0.8 kg of O is needed. Hence : Y Y 0.08kmol er kmol DEG he nitrogen in the DEG is Y 0.64 kmol er kmol DEG he actual Nitrogen 0.64 x 0.5 x 8.6 kg he air sulied must be.6/ kg er kg of fuel. A simle calculation shows the stoichiometric mass of air is.7 so there is 0.% excess air. SELF ASSESSMEN EXERCISE No.. CH6 is burned in a boiler and the dry roducts are found to contain 8% CO by volume. Determine the excess air sulied. (59%). he analysis of the dry exhaust gas from a boiler shows 0% carbon dioxide. Assuming the rest is oxygen and nitrogen determine the excess air sulied to the rocess and the % excess air. he fuel contains 85% C and 5% H (.5 kg, 44.5%) Now we will look at a comlete examle involving all the rinciles so far covered. D.J.Dunn 4

175 6. ENERGY RELEASED BY HE REACION he contents of the fuel and air or oxygen rior to combustion are called the reactants. he resulting material is called the roducts. he rocess releases energy but the amount of energy deends uon the temerature before and after the reaction. Consider a mixture of reactants at condition () which is burned and the resulting roducts are at condition (). In order to solve roblems we consider that the reactants are first cooled to a referance condition (0) by removing energy Q. he reaction then takes lace and energy is released. he roducts are then brought back to the same referance conditions (0) by removing energy Q. he energy Q and Q are then returned so that the final condition of the roducts is reached (). Figure For constant volume combustion (closed system), we use Internal Energy. Balancing we have U - UR (URo - UR) + (Uo - URo) + (U - Uo) he energy released by combustion is in this case the Internal Energy of combustion and this occurs at standard conditions of bar and 5oC. his ressure is designated θ and the internal energy of combustion is designated Uθ. When this is based on kmol it is designated uθ U - UR (URo - UR) + Uo θ + (U - Uo) he standard conditions chosen for the combustion are bar and 5oC. At this temerature the internal energy of all gases is the same (- 479 kj/kmol). he figure is negative because the zero value of internal energy arbitrarily occurs at a higher temerature. If the rocess is conducted in a steady flow system, enthaly is used instead of internal energy. he reasoning is the same but U is relaced by H. H - HR (HRo - HR) + Ho θ + (H - Ho) ho θ may be found in the thermodynamic tables for some fuels. he figures are quoted in kj er kmol of substance. For the roducts In terms of kmol fractions ho uo + nroo D.J.Dunn 5

176 For the reactants In terms of kmol fractions hro uro + nrroo where n is the kmols. ho θ (uo + nroo) - (uro + nrroo ) ho θ (uo - uro ) - nrroo + nroo ho θ (uo - uro ) + (n - nr)roo uo θ ho θ + (n - nr)roo If the combustion roduces equal numbers of kmols before and after, the ressure would be constant (assuming constant volume and no condensation). n n R so ho θ is the same as the internal energy of reaction νo θ. For examle consider the combustion of ethylene. CH4 + O CO + HO In this case there are 4 kmols before and after. When this occurs, we may use the secific heat c to solve the roblems. uo θ ho θ c he secific heats are listed in the thermodynamic tables. Note that in order to make the method of solution conform to standard data, the combustion equations should always be based on kmol of fuel. he heat transfer Q is found either by use of the mean secific heat or by looking u the enthaly of the gas at the required temeratures (enthaly of formation) and deducing the change. In general for a constant volume we should use uo θ and c v to solve roblems. For constant ressure with no work being done (e.g. a combustion chamber) we should use ho θ and c. Since tables only list ho θ and c we may find uo θ ho θ + (n - nr)roo where is 98. K and R o is 8.4 kj/kmol K and n is the number of kmols of roduct. c v c R c R o /molecular mass D.J.Dunn 6

177 WORKED EXAMPLE No.8. A vessel contains 0. m of CH4O and oxygen in its stoichiometric ratio. he mixture is at bar ressure and 5oC. he mixture is ignited and allowed to cool back to 5oC. Determine the following. i. he final ressure. ii. he amount of condensate formed. iii. he heat transfer. iv. he enthaly of reaction er kmol of CH4O. he enthaly of formation ( h o θ ) for the gases involved is shown below for a temerature of 98 K. Molecular mass. kg/kmol Enthaly of reaction (kj/kmol) CO(gas) HO(gas) HO(liquid) O(gas) 0 CH4O D.J.Dunn 7

178 SOLUION CH4O + ½O CO + HO mass ratio kmol ratio ½ bar V 0. m 98 K V0.m Vf + Vox ½ Volumes bar f + ox f x (/.5) bar ox x (.5/.5) 0.74 bar mass of fuel VÑ/Ro0.857 x 05 x 0. x 44/(84.4 x 98) mass of fuel 0.05 kg mass of oxygen VÑ/Ro 0.74 x 05 x 0. x /(84.4 x 98) mass of oxygen kg total mass 0.86 kg Mass of CO (88/4) x kg Mass of HO (6/4) x kg total mass 0.86 kg (CO) mro/ñv 0.0 x 84.4 x 98/(44 x 0.) x bar Since condensate forms, the gas is saturated with water vaour so (vaour) 8oC from the steam tables. (vaour) bar otal ressure bar Answer (i) Mass of vaour V/vg where vg 4.4 8oC from tables. Mass of vaour 0./ kg Mass of condensate formed kg Answer (ii) Now consider the reaction. Since it starts and finishes at 5oC there is no initial cooling required (Q 0). D.J.Dunn 8

179 REACAN Fuel Mass 0.05 kg kmol 0.05/ kmol Oxygen Mass kg kmol 0.845/ kmol Enthaly of formation for oxygen 0 Enthaly of formation for CH4O -560 kj/kmol Hf 0.00(-560) -.4 kj (Minus relative to higher oint of reference) PRODUCS CO kmol 0.0/ kmol Hf x (-9 50) kj HO (gas) kmol / kmol Hf x (-480) kj HO (water) kmol / kmol Hf x (-8580) kj otal -608 kj he change in enthaly (-.4) kj Ho Answer (iv) ho θ / GJ/kmol Answer (v) D.J.Dunn 9

180 WORKED EXAMPLE No.9 Air and Ethane (CH6) are mixed with twice the stoichiometric ratio at 600K in a vessel at bar ressure. Determine the temerature and ressure after combustion assuming no energy losses. he enthaly of combustion at 5oC is Ho kj/kmol SOLUION CH6 +.5 O HO + CO kmol.5 he air required.5/ kmol Actual air. kmol Nitrogen 0.79 x. 6. kmol Excess oxygen.5 kmol he equation may be rewritten as CH6 + 7O + 6. N HO + CO +.5 O + 6. N he rocess may be idealised as follows First find the enthaly of the reactants. he mean temerature of the reactants relative to 5oC is {(5+7) + 600}/ 449K near enough 450K for the tables. We look u secific heats in the thermodynamic tables at 450 K. he temerature change from 5oC to 600 K is 0 K. We roceed to work out the heat transfer based on kmol of fuel, Q as follows. able of values CH6 c.40 kj/kg K mass kmol x 0 0 kg Q 0 x.40 x 0 76 kj O c kj/kg K mass 7 kmol x 4 kg Q 4 x x kj N c.049 kj/kg K mass 6. kmol x kg Q 77.4 x.049 x kj otal Q kj er kmol of fuel (negative leaving system) Next we reeat the rocess for the roducts to find Q + Q D.J.Dunn 0

181 In order to use a mean secific heat we must guess the aroximate final temerature of the roducts. A good guess is always 000 K so the mean of 5oC and 000 K is near enough 50 K. Using this we work out the heat transfer to the roducts with an unknown temerature change from 5oC to of. HO c.9 kj/kg K mass kmol x 6 48 kg Q 48 x.9 x 4.8 O c.09 kj/kg K mass.5 kmol x kg Q x.09 x 4. N c.96 kj/kg K mass 6. kmol x kg Q 77.4 x.96 x 88.7 CO c.70 kj/kg K mass kmol x kg Q 88 x.70 x.76 otal Q + Q.5 kj er kmol of fuel (ositive entering system) Q kj/kmol of fuel (from question). Conserving energy we have hence.5 K and 76 K which is different from our original guess of 000 K but more accurate. Next we must reeat the last stage with a more accurate mean temerature. Mean temerature ( )/ 007 K. Say 000 K. HO c.88 kj/kg K Q 48 x.88 x 09.8 O c.090 kj/kg K Q x.09 x. N c.67 kj/kg K Q 77.4 x.67 x CO c.4 kj/kg K Q 88 x.4 x 08.6 otal Q + Q 0 kj er kmol of fuel (ositive entering system) Conserving energy we have hence 455 K and 75 K which is different from our original guess of 76K but more accurate. the true answer is between 76 and 75 K and may be narrowed down by making more stes but two is usually sufficient in the exam. D.J.Dunn

182 Finally the ressure. V / N Ro V / N and the volumes are equal. bar 600 K N 4.5 kmol 75 K N 5 kmol x 5 x 75/(4.5 x 600) 8.9 bar SELF ASSESSMEN EXERCISE No. 4. he gravimetric analysis of a fuel is Carbon 78%, hydrogen %, oxygen 5% and ash 5%. he fuel is burned with 0% excess air. Assuming comlete combustion, determine the following. i. he comosition of the roducts. (7.6% N, 7.% CO, 6.5%HO and.6% O) ii. he dew oint of the roducts.47oc) iii. he mass of water condensed when the roducts are cooled to (0.67 kg) 0oC.. Carbon monoxide is burned with 5% excess oxygen in a fixed volume of 0. m. he initial and final temerature is 5oC. he initial ressure is bar. Calculate the following. i. he final ressure.(0.874 bar) ii. he heat transfer. (574.5 kj) Use your thermodynamic tables for enthalies of reaction. D.J.Dunn

183 . Prove that the enthaly and the internal energy of reaction are related by Ho Uo + Roo(nP - nr) where np and nr are the kmols of roducts and reactants resectively. Ethylene (CH4) and 0% excess air at 77oC are mixed in a vessel and burned at constant volume. Determine the temerature of the roducts. You should use your thermodynamic tables to find Uo or Ho and the table below. (Answer 6 K) U (kj/kmol) CH4 O N CO HO (K) An engine burns hexane (C6H4) in air. At a articular running condition the volumetric analysis of the dry roducts are CO 8.7% CO 7.8 % N 8.5% Calculate the air-fuel ratio by mass and find the stoichiometric ratio. (Answer.9 and 5.7) D.J.Dunn

184 7. DISSOCIAION At the high temeratures and ressures exerienced in combustion, dissociation occurs. his results in some of the fuel not burning. CO is roduced and in the case of hydrogen, some of it remains as hydrogen after the rocess even though oxygen is resent. he reasons for this will not be covered here other than to say it is redicted by the nd law of thermodynamics and involves equilibrium in the chemical rocess. When dissociation occurs, the energy released is reduced accordingly and if the amount of unburned fuel is known the revious examles may easily be modified to take account of it. When hydrogen is burned, it can be shown that the artial ressures of the hydrogen, oxygen and water vaour resent in the roducts are related by θ H O k θ H ( ) O he roerties tables list values of ln K θ. Similarly when dissociation occurs in the formation of carbon dioxide, the relationshi between the artial ressures of CO,CO and O is given by θ CO k θ C CO Other similar equations for other combinations of roducts may be found in the tables. D.J.Dunn 4

185 SELF ASSESSMEN EXERCISE No. 5. Hydrogen is mixed with stoichiometric air at 5oC and burned adiabatically at constant volume. After combustion 6% of the hydrogen remains unburned. Determine the temerature and ressure of the roducts. (Answer the temerature is 44K after two aroximations) You need to find K θ in the tables. Also find Ho4800 kj/kmol. Deduce the artial ressures of the roducts as a fraction of and then use K θ to solve.. A mixture of air and CO is burned adiabatically at constant volume. he air is 90% of the stoichiometric requirement. he mixture is initially at 5 bar and 400K. he only dissociation that occurs is CO CO + ½O. Show that the equilibrium constant at the final temerature is K.a/({(-a)(0.9-a)½(/)½} where a is the amount of CO kmol in the roducts er kmol of CO in the reactants. If it assumed that initially 900 K for which log K 0.649, the solution of the above equation gives a Check the assumed value of given that the internal energy of reaction at K is kj/kmol. (K) U kj/kmol CO O N CO D.J.Dunn 5