# DET: Mechanical Engineering Thermofluids (Higher)

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1 DET: Mechanical Engineering Thermofluids (Higher) 6485

2

3 Spring 000 HIGHER STILL DET: Mechanical Engineering Thermofluids Higher Support Materials *+,-./

4 CONTENTS Section : Thermofluids (Higher) Student Notes Section : Self-Assessment Answers Section 3: Tutorials Section 4: Tutorials Marking Scheme Section 5: Appendix, Quantities used in Thermofluids (Higher) DET: Mechanical Engineering: Thermofluids Higher

5 DET: Mechanical Engineering: Thermofluids Higher

6 Section : Thermofluids (Higher) Student Notes DET: Mechanical Engineering: Thermofluids Higher Support Notes

7 DET: Mechanical Engineering: Thermofluids Higher Support Notes

8 OUTCOME : APPLY GAS LAWS The wide use made of various gases in the field of engineering makes it necessary to predetermine their reactions when they are heated, cooled, expanded or compressed. When a process takes place, the changes which will occur in the properties of volume, absolute pressure and absolute temperature of the gases are related by the gas laws. When solving problems utilising the gas laws, both pressures and temperatures must be expressed in absolute terms and these are defined thus: Absolute pressure (Symbol p) Pressure gauges are commonly used to measure pressures in vessels and pipelines and read pressures normally above atmospheric pressure. If a gauge shows a zero reading it means the pressure is atmospheric. If the pressure in a vessel is increased above atmospheric to a gauge pressure p g, the actual or absolute pressure p in the vessel is given by: p = p g + p atm i.e. Absolute pressure p = gauge pressure + atmospheric pressure. In most practical problems, listed pressures will be in absolute terms unless otherwise stated. The unit for pressure is the N m - or Pa (PASCAL). The bar = 0 5 N m - = 00 kn m - is also commonly used. Standard atmospheric pressure = atm =.03 bar =.03 x 0 5 N m -. When working through problems, the stated or calculated values for pressure are often high numbers. In order to express multiples of SI units concisely, the undernoted prefixes are used: MULTIPLICATION FACTOR PREFIX SYMBOL,000,000 = 0 6 Mega M,000 = 0 3 Kilo k e.g. 7,00,000 N m - = 7,00 kn m - or preferably 7. MN m -. DET: Mechanical Engineering: Thermofluids Higher Support Notes

9 Absolute temperature In problems involving the gas laws, the temperature of any gas is measured from absolute zero, which has been determined to be 73 C below the zero point on the Celsius scale, i.e. absolute zero. At this point the internal energy of the substance is also zero. Absolute temperature is the temperature above absolute zero and is determined by adding 73 to the Celsius temperature scale reading. i.e. Absolute temperature = Celsius temperature scale reading Hence 7 C = 300 K. Absolute temperature takes the SI base unit, the Kelvin (K), and has the symbol T. Note: A change in temperature of C = a change in temperature of K. Other quantities encountered in our thermofluid studies are defined as follows: Mass This is usually defined as the quantity of matter in a body. Symbol: m (small letter). Unit: kg (small letters). Volume Symbol: V (capital letter). Unit: m 3. The recommended unit is the cubic metre. Subdivisions such as the cm 3 or litre (l) are also used, but as a general rule it is safer to convert data to give volumes in m 3 to avoid errors in calculations. Specific volume Symbol: v (small letter). Unit: m 3 kg -. This is the volume per unit mass and is the reciprocal of density. i.e. v = VOLUME MASS = V m Boyle s Law The Irish scientist Sir Robert Boyle investigated the behaviour of gases when expanded or compressed under constant temperature (isothermal) conditions. In essence, Boyle s Law states: For any given mass of a gas, the absolute pressure will vary inversely with the volume providing that the temperature remains constant. DET: Mechanical Engineering: Thermofluids Higher Support Notes

10 Thus p. V e.g. or pv = constant Doubling the absolute pressure gives half the volume. Three times the absolute pressure gives one-third of the volume. Boyle s Law can also be expressed algebraically in the form p V = p V = p n V n..for the mass of gas Charles Law The French scientist Jacques Charles conducted experiments on gases where the pressure of a fixed mass of gas was kept constant while variations in the volume and temperature were examined. In essence, Charles s Law states: During the change of state of any gas in which the mass and pressure remain constant, the volume varies in proportion with the absolute temperature (Kelvin). Thus V. T or V = constant T e.g. At double the absolute temperature, the volume is doubled. At three times the absolute temperature, the volume is trebled. Charles s Law can be expressed algebraically in the form: V T = V T = V T n n..for the mass of gas Constant volume process When a given mass of gas is heated at constant volume, its temperature and pressure will both increase. Conversely, if the gas is cooled, the temperature and pressure will both decrease. At any stage of either process, the ratio of the pressure p to the absolute temperature T of the gas will be a constant. Hence: PRESSURE ABSOLUTE TEMPERATURE = constant p = C T (Provided neither mass or volume of gas changes) This is known as the Pressure Law, which may be stated algebraically in the form: p T = p T = p T n n DET: Mechanical Engineering: Thermofluids Higher Support Notes 3

11 The Combined Gas Law If, during a process, the pressure, volume and absolute temperature of a gas are changed from p, V, and T to p, V and T respectively, then, provided there is no change in the mass of gas, Boyle s Law, Charles s Law and the Pressure Law may be combined to give the algebraic expression: p V T = pv T = constant This is known as the Combined Gas Law. The Characteristic Gas Equation We have seen the combined gas law stated in the form: pv = constant C T For a perfect or ideal gas, this constant C = mr where m is the mass of the gas and R is the Characteristic Gas Constant or Specific Gas Constant. Hence Or pv = mr T pv = mrt This is known as the Characteristic Gas Equation of an ideal gas. When using this equation for solving problems, it is essential to express all the terms in appropriate units which are: p = absolute pressure of the gas N m - V = volume of the gas m 3 T = absolute temperature of the gas [(t + 73)] K m = mass of the gas kg R = gas constant J kg - K -. The table below lists the gas laws, together with appropriate equations for problem solving. DET: Mechanical Engineering: Thermofluids Higher Support Notes 4

12 GAS LAW PROCESS CONDITION APPROPRIATE EQUATION Boyle s Law Constant Temperature p V = p V etc Charles s Law Constant Pressure V T = V T etc Pressure Law Constant Volume p p = T T etc Combined Gas Law Pressure Volume and Temperature all Change pv = T pv T etc Characteristic Gas Equation Includes Mass and Characteristic Gas Constant R pv = mrt The gas constant R, which appears in the Characteristic Gas Equation, is identified as the Characteristic Gas Constant or the Specific Gas Constant and its value varies for different gases as will be apparent in the questions covered in the Tutorial for Outcome. Universal Gas Constant The Universal Gas Constant takes into account the concept of molecular mass of substances. This constant, symbol Ro, and also known as the Molar Gas Constant, is the product of the relative molecular mass, M, and the Characteristic Gas Constant, R, and has the same value for all gases: Thus, Ro = MR = kj kg mol - K - It follows that the value of the Characteristic Gas Constant R can be found from the relationship R = M Ro e.g. The molecular mass of nitrogen is 8. What is the value of R for nitrogen? R = Ro M = = 0.97 kj kg - The universal gas constant is frequently utilised in problems dealing with the combustion of various gases and it appears in a version of the Characteristic Gas Equation called the Ideal Gas Equation. Our studies, however, will be restricted to the use of the Characteristic Gas Equation, which employs the Characteristic or Specific Gas Constant. K - DET: Mechanical Engineering: Thermofluids Higher Support Notes 5

13 SELF-ASSESSMENT Assignment. Test your knowledge of quantities, symbols and units covered so far by completing the table below. SPECIFIC VOLUME QUANTITY SYMBOL UNIT m m 3 ABSOLUTE PRESSURE K. Convert the undernoted temperature values from one scale to the other. o C K Boyle s Law deals with the behaviour of gases under isothermal conditions. What does the condition isothermal mean? Ans: DET: Mechanical Engineering: Thermofluids Higher Support Notes 6

14 4. State Charles Law and express the law in the form of an algebraic equation. Ans: Equation: 5. When a mass of gas is cooled at constant volume, what effect has this process on its pressure and temperature? Ans: 6. State the characteristic gas equation for an ideal gas. Identify all terms in the equation and state the correct units for each. Equation: Identification of terms and units: DET: Mechanical Engineering: Thermofluids Higher Support Notes 7

15 PRACTICAL EXEMPLAR PROBLEMS Having developed the various gas law equations, we can apply these to the solution of problems dealing with the behaviour of gases when subjected to different processes. Problems may be simplified by employing a method of converting given information into symbols and units which can then be fitted into an appropriate equation. Specimen worked exemplar problems adopting this strategy now follow. Exemplar Question A fixed mass of gas is compressed isothermally from a pressure of 0 bar and volume of 3. m 3 until it occupies 5% of its original volume. Calculate the final pressure of the gas. Known Data p = 0 x 0 5 N m - V = 3. m 3 p =? 5 V = 3. x 00 = 0.8 m 3 For an isothermal process, Boyle s Law applies and equation p V = p V can be utilised. From p V = p V p = V = p x V p = 8.0 x 0 6 N m - 0 x 0 5 x FINAL PRESSURE OF GAS, p = 8 MN m - or 80 bar. Exemplar Question A quantity of gas at a pressure of 80 kn m - and temperature 8 C, occupies a volume of.43 m 3. The gas is compressed until its pressure and temperature are 670 kn m - and 7 C respectively. If there is no loss of gas, what volume will it now occupy? Known Data From the Combined Gas Law p = 80 kn m - p = 670 kn m - T = (8 + 73) K T = (7 + 73) K V = 0.43 m 3 V =? p V T = p V T 3 p V T 80 x 0 x 0.43 x 400 V = = 3 p T 670 x 0 x 9 FINAL VOLUME OF GAS IS m 3 = m 3 DET: Mechanical Engineering: Thermofluids Higher Support Notes 8

16 Exemplar Question 3 An air receiver contains a fixed mass of air at a pressure of.5 bar and temperature 84 C. After a period of time, the pressure is observed to be 7.8 bar. What will be the temperature of the air? Known Data For constant volume process, p =.5 bar p = 7.8 bar the Pressure Law applies T = ( )K T =? Constant Volume Process V = V p p = T T T p = x T p x 0 = 5 x 357 =.768K.5 x 0 FINAL TEMP. OF AIR =.8 73 = -50. C Exemplar Question 4 Gas is stored in two tanks, A and B, which are connected by a pipe fitted with a stop valve which is initially closed. Tank A has a volume of 3.0 m 3 and contains 4 kg of the gas at a pressure of 5 kn m - and a temperature of 0 C. Tank B has a volume of.0 m 3 and contains gas at a pressure of 340 kn m - and a temperature of 0 C. Determine the characteristic gas constant for the gas and the mass of gas in tank B. If the stop valve connecting the tanks is then opened, determine the final pressure of the gas, assuming the temperature remains at 0 C. V = 3.0 V =.0 m 3 m = 4 TANK m =? p = 5 k B p = 340 kn m - T = 0 + TANK T = 93 K = 93 K A Char. Gas Constant from p V = m R T R = p V m T = 3 5 x 0 x x 93 GAS CONSTANT R = 57.4 J kg - K - DET: Mechanical Engineering: Thermofluids Higher Support Notes 9

17 MASS IN TANK B from p V = m R T m = p V R T x 0 x = 57.4 x 93 MASS OF GAS IN TANK B = kg When connecting valve is opened, the total volume V 3 = 5 m 3 and temperature remains at 93 K. FINAL PRESSURE from p3 V3 = m3 R T3 p 3 m3 R T = V 3 3 ( ) x 57.4 x 93 = 5 = N m - FINAL PRESSURE IN SYSTEM = 35 kn m - Exemplar Question 5 A fixed mass of gas contained in a closed system is initially at a pressure of 00 kn m -, a temperature of 5 C, and occupies a volume of 0.5 m 3. The gas is then compressed to a volume of 0.06 m 3 and a pressure of 300 kn m -. The gas is then expanded at constant pressure unit it reoccupies its original volume. If the characteristic constant for the gas is 89 J kg - K -, determine the mass of gas in the system and the temperature at the end of the compression and expansion processes. Known Data p = 00 x 0 3 N m - p = 300 x 0 3 N m - p 3 = p T = 5 C + 73 = 88 K T =? T 3 =? V = 0.5 m 3 V = 0.06 m 3 V 3 = V m =? R = 89 J kg - K - Mass of gas from p V = m R T m = p V R T 3 00 x 0 x 0.5 = = kg 89 x 88 MASS OF GAS IN SYSTEM = kg DET: Mechanical Engineering: Thermofluids Higher Support Notes 0

18 TEMPERATURE AFTER COMPRESSION from p V = m R T 3 p V 300 x 0 x 0.06 T = = = K m R x 89 TEMPERATURE AFTER COMPRESSION = = 7.56 C TEMPERATURE AFTER EXPANSION from p 3 V 3 = m R T 3 3 p3 V3 300 x 0 x 0.5 T 3 = = = K m R.756 x 89 TEMPERATURE AFTER EXPANSION T 3 = = C DET: Mechanical Engineering: Thermofluids Higher Support Notes

19 OUTCOME : SOLVE PROBLEMS USING DATA EXTRACTED FROM THERMODYNAMIC PROPERTY TABLES In this outcome we are concerned with the interpretation and extraction of data on thermodynamic properties of working fluids listed in tables as arranged by Messrs Rogers and Mayhew. These tables, commonly known as steam tables, give tabulated values for the properties of steam and refrigerants over an extensive range of pressures and temperatures. The ability to understand and extract data from the tables extends into the solution of problems in this outcome and also in Outcome 3 when the steady flow energy equation is dealt with. Before examining the tables, however, definitions need to be attached to specific thermodynamic quantities listed in the range for this outcome. Specific volume This is the volume occupied per unit mass ( kg) of a substance and is identified by the symbol v (small letter). i.e. v = VOLUME V = UNIT : m 3 kg - MASS m Thermodynamic tables give the specific volume of dry saturated steam at a particular pressure under the heading v g. e.g. SPECIFIC VOLUME OF DRY SATURATED STEAM AT.4 bar =.36 m 3 kg - For superheated steam the specific volume is read against the symbol v for different pressures and temperatures. e.g. SPECIFIC VOLUME OF SUPERHEATED STEAM AT 6 bar and 50 C = m 3 kg - Internal energy A fluid may be defined as a substance or a mixture of substances in the liquid or gaseous state. All fluids consist of large numbers of molecules that move in random directions at high speeds. Each molecule possesses a minute amount of kinetic energy and the total kinetic energy possessed by all the molecules is known as the internal energy of the fluid. When heat energy, which is a transient form of energy, is transferred to a fluid, the temperature and molecular activity of the fluid increases. These increases result in a corresponding increase in the store of internal energy within the fluid. DET: Mechanical Engineering: Thermofluids Higher Support Notes

20 As a result of experimental work on this subject, Joule concluded that the internal energy of a fluid is a function of temperature only and is independent of pressure and volume (Joule s Law). For there to be a change in the internal energy of a fluid there must be a change in temperature. The symbol used for the total internal energy in a fluid is U and its unit is the Joule (J). Generally, the internal energy of a fluid is quoted as per unit mass (per kg). This quantity is known as specific internal energy and takes the symbol u. The unit for specific internal energy is the Joule per Kilogram (J kg - ). Thermodynamic tables give three values for the specific internal energy of steam as underlisted. u f = specific internal energy of saturated water u g = specific internal energy of dry saturated steam u = specific internal energy of superheated steam. Flow or displacement energy Any volume of fluid entering or leaving a system must displace an equal volume ahead of itself in order to enter or leave the system as the case may be. Let the mass of fluid between X and Y in the figure below have a total Volume V. For flow to occur, this volume must be displaced by an equal volume from outside the system. If the pressure in the fluid is p, then the work done on the fluid inside the system by the incoming fluid = force x distance the fluid is displaced. CROSS SECTIONAL AREA A X Y p p p VOLUME DISPLACED V p S S Flow energy DET: Mechanical Engineering: Thermofluids Higher Support Notes 3

21 Now, Force = Pressure x Cross Sectional Area Therefore Work Done = p x A x s But, A s = Volume displaced V Therefore Work done on the system = p V In specific terms, i.e. per kg of mass, work done on system = p v where v equals the specific volume of the fluid. This is variously called flow energy, displacement energy or pressure energy. At entry energy is received by the system. At exit energy is lost by the system. Hence, specific flow energy = p v (J kg - ) Enthalpy In steady-flow thermodynamic systems, internal energy and flow energy are present in the moving fluid. Accordingly, it is convenient to combine these energies into a single energy quantity known as enthalpy, thus Total Enthalpy = Internal energy + flow energy The symbol used for enthalpy is H. Hence H = U + pv The unit for total enthalpy is the Joule (J). When considering kg of working fluid, then: Specific Enthalpy = Specific Internal Energy + Specific Flow Energy Hence h = u + pv (v = specific volume) Specific enthalpy h takes the unit The Joule Per kg (J kg - ). Thermodynamic property tables give four values for the specific enthalpy of steam as listed below: h f = specific enthalpy of saturated water h fg = specific enthalpy of evaporation h g = specific enthalpy of dry saturated steam h = specific enthalpy of superheated steam. DET: Mechanical Engineering: Thermofluids Higher Support Notes 4

22 The formation of steam Consider a quantity of water initially at 0 C being heated in a vessel fitted with a movable piston such that a constant atmospheric pressure can be maintained in the vessel. If the water is heated until it has all been converted to steam then the temperature/time graph would be as illustrated. During the stage A to B sensible heat energy flows to the water accompanied by a rise of temperature. At B the water boils at a temperature referred to as saturation temperature. This temperature depends on the pressure in the vessel and is 00 C at atmospheric pressure. The energy required to produce this temperature rise is called the liquid enthalpy. During stage B to C steam is being formed whilst the temperature remains constant and the contents of the vessel will be a mixture of water and steam known as wet steam. At point C the steam will have received all the heat energy required to convert the water completely to dry steam. The energy required to produce the total change from all water to all dry steam is called the enthalpy of evaporation. When completely dry saturated steam has been formed at saturation temperature, further transfer of heat energy will produce superheated steam which will be accompanied by a rise in temperature. The amount of heat energy in the superheat phase is called the superheat enthalpy. Steam, therefore, can exist in three states: wet, dry, or superheated. Values for specific enthalpy, specific internal energy, and specific volume may be obtained directly from thermodynamic property tables for dry and superheated steam. For wet steam, it is necessary to know the degree of dryness, or the dryness fraction, of the steam before the various properties can be calculated. DET: Mechanical Engineering: Thermofluids Higher Support Notes 5

23 Dryness fraction The degree of dryness, or dryness fraction, of steam is that proportion of a given mass of water which has been evaporated to form steam. The dryness fraction may have any value from 0 (corresponding to boiling water) to (corresponding to dry saturated steam). For example, steam with a dryness fraction 0.6 means that for each kg of water, 0.6 will be steam and 0.4 kg of saturated liquid. The symbol x is used to represent dryness fraction. Mass of Dry Steam Dryness fraction, x = Total mass of Steam and Moisture Layout and use of thermodynamic tables for water and steam The figure below duplicates the information given at the top of page 4 in the Rogers & Mayhew tables. The s figures in the right hand columns of the actual tables relate to entropy values and these are not required for this unit. P T vg u s f ug hf hfg h g 0 3 [ bar] [ C] m /kg kj/kg kj/kg The various symbols in the eight columns are identified with their quantities and units as below. SYMBOL QUANTITY UNIT p T s v g u f u g h f h fg h g Absolute pressure Saturation temperature relating to value of p Specific volume of dry saturated steam Specific internal energy of saturated water Specific internal energy of dry saturated steam Specific enthalpy of saturated water Specific enthalpy of evaporation Specific enthalpy of dry saturated steam bar C m 3 /kg kj/kg kj/kg kj/kg kj/kg kj/kg DET: Mechanical Engineering: Thermofluids Higher Support Notes 6

24 Thermodynamic property tables With reference to pages 3, 4 and 5 of the Rogers & Mayhew tables: The first column headed p is the absolute pressure measured in bar, where bar = x 0 5 N m - or 00 kn m -. The second column headed T s is the temperature in C at which the water boils (saturation temperature). Note how T s changes relative to pressure. The third column v g is the specific volume in m 3 of kg of completely dry saturated steam. That is at pressure of bar, saturation temperature is 0. C and kg of dry steam occupies a volume of m 3. The fourth column u f is termed the specific internal energy of saturated liquid. That is the heat energy required to raise the temperature of kg of water from 0 C to saturation temperature (kj kg - ). This is a constant volume operation. The fifth column u g is the specific internal energy of kg of completely dry saturated steam. That is the heat energy required to raise the temperature of kg of water at 0 C to saturation temperature plus the heat energy required to completely evaporate it (specific enthalpy of evaporation) as if the operation were carried out at constant volume. The sixth column h f is the specific enthalpy of saturated liquid i.e. the enthalpy of kg of water from 0 C to saturation temperature at constant pressure. Note that u f and h f are identical down to 4.5 bar and then they gradually drift apart since h f increases slightly faster than u f. The seventh column h fg is the specific enthalpy of evaporation i.e. the heat energy required to completely evaporate kg of water at saturation temperature to kg of dry saturated steam at constant pressure and at same temperature. The eighth column h g is the specific enthalpy of saturated vapour and is the enthalpy of kg of dry saturated steam at constant pressure measured from water at 0 C. On page of the Rogers & Mayhew tables the same properties are given but are set out against the reference of saturated water temperature (T C) in the first column. The graph below illustrates three phases of steam formation and incorporates enthalpy values from tables. DET: Mechanical Engineering: Thermofluids Higher Support Notes 7

25 GRAPH OF HEAT ADDED AGAINST TEMPERATURE AT bar ABSOLUTE TEMP O C SENSIBLE HEAT PROCESS EVAPORATION PROCESS SUPERHEAT PROCESS D WATER WATER & STEAM SUPERHEATED STEAM SATURATED WATER DRY SATURATED STEAM Ts 9.6 O C B SATURATION TEMPERATURE C 0 C O A - hf hfg = 58 kj kg = 47 kj kg - hg = 675 kj kg - HEAT ADDED kj kg - From A to B, a heat transfer of 47 kj kg - raises the temperature from 0 C to saturation temperature (boiling point) of 99.6 C at the pressure of bar. From point B, if more heat is added, the boiling water will evaporate to form steam at the same temperature and pressure. At point C, the enthalpy of evaporation process is completed by a further heat energy transfer of 58 kj kg -. At point C the steam is in a completely dry saturated state. In the region C to D, further heat addition produces superheated steam. Superheated steam So far we have considered pages to 5 of the thermodynamic tables. These pages set out the properties and heat energy requirements for steam to be raised from water at 0 C to dry saturated steam at different pressures. DET: Mechanical Engineering: Thermofluids Higher Support Notes 8

26 When steam has a temperature higher than its saturation temperature for a stated pressure, then the steam is in a superheat state in which case we use pages 6 to 9 of the steam tables. The following explanation of the columns on these pages will enable their use: Column Column as before, states the pressure (p) in bar but the figure in brackets under each pressure is the saturation temperature corresponding to that pressure. lists the properties still of dry saturated steam i.e. v g - Specific volume of dry saturated steam u g - Specific internal energy of dry saturated steam h g - Specific enthalpy of dry saturated steam. The remaining columns list these same properties corresponding to various degrees of superheat. Rule In order to define the condition of superheated steam it is necessary to state both the pressure and temperature of the steam. Thus, if a temperature is quoted for steam in a problem, check it against the tables and if it is higher than (T s ) for the corresponding pressure then superheated tables must be used. The difference between the superheated temperature (T) and the saturation temperature (T s ) is called the degree of superheat. Units in superheated steam tables: v in m 3 kg - u and h in kj kg - Where exact values of the condition of steam are not listed in the tables, linear interpolation for both pressure and temperature may therefore be required. Class exemplars and tutorials on the use of steam tables cover this aspect. DET: Mechanical Engineering: Thermofluids Higher Support Notes 9

27 Specific volume of wet steam The specific volume of steam with a dryness fraction x is given by: v x = v f + x v fg This is illustrated on the graph of temperature against specific volume shown below: TEMP Vx = Vf + X Vfg X Vfg Vf Vfg = Vg - Vf SPECIFIC VOLUME Vg Referring to above graph: v fg = v g - v f Therefore v x = v f + x(v g v f ) = v f + xv g x v f = ( x) v f + x v g Since v f is extremely small compared with v g, the term ( x) v f may be ignored. Hence, v x = x v g ** Example Determine the specific volume of wet steam having a pressure of.5 MN m - and dryness fraction MN m - =.5 bar, v x = x v g bar 3 bar v x = 0.9 = m 3 kg - DET: Mechanical Engineering: Thermofluids Higher Support Notes 0

28 Specific internal energy of wet steam The specific internal energy of steam with a dryness fraction of x is given by: u x = u f + x u fg This is illustrated in the graph of temperature against specific internal energy shown below: TEMP Ux = Uf + X Ufg X Ufg Uf Ufg = Ug - Uf SPECIFIC INTERNAL ENERGY Ug Temperature against specific internal energy Referring to above graph: u fg = u g - u f Therefore u x = u f + x(u g u f ) = u f + xu g x u f Therefore u x = ( x) u f + x u g Example Determine the specific internal energy in wet steam at a pressure of 4 bar when it has a dryness fraction of Specific internal energy u x = ( x)u f + xu g u x = ( 0.87) x 554 u x = u x = kj kg - DET: Mechanical Engineering: Thermofluids Higher Support Notes

29 Specific enthalpy of wet steam The specific enthalpy of steam with a dryness fraction x is given by: h x = h f + x h fg ** This is illustrated on the graph of temperature against specific enthalpy shown below: TEMP hx = hf + Xhfg Xhfg X hfg STEAM ( - x) kg WATER hf hg hfg = hg - hf SPECIFIC ENTHALPY Temperature against specific enthalpy Example Determine the specific enthalpy of wet steam at a pressure of 70 kn m - and having a dryness fraction of Pressure of 70 kn m - = 0.7 bar Specific enthalpy h x = h f + x h fg = x 83 = SPECIFIC ENTHALPY, h x = kj kg - DET: Mechanical Engineering: Thermofluids Higher Support Notes

30 Using steam tables and wet steam formulae Example Determine the specific internal energy, specific enthalpy and specific volume of steam at a pressure of 3 bar (300 kn m - ) when it is.8 dry. What is the saturate temperature? From steam tables at a pressure of 3 bar:- Saturation temperature T s = 33.5 C Specific volume v g =.6057 m 3 kg - Specific internal energy of sat. liquid u f = 56 kj kg - Specific internal energy of sat. vapour u g = 544 kj kg - Specific enthalpy of sat. liquid h f = 56 kj kg - Specific enthalpy of evaporation h fg = 64 kj kg - Specific internal energy of steam at 3 bar and.8 dry. u x = (- x) u f + xu g = ( -.8) x x 544 = u x = kj kg - Specific enthalpy of steam at 3 bar and.8 dry. h x = h f + xh fg = x 64 h x = kj kg - Specific volume at 3 bar and.8 dry. v x = xv g =.8 x.6057 v x =.4967 m 3 kg - DET: Mechanical Engineering: Thermofluids Higher Support Notes 3

31 Interpolation of steam tables When quantities cannot be extracted directly from tables, intermediate values need to be interpolated between the nearest listed values above and below the required value. Examples. Determine the specific volume of wet steam at 68 bar. at 65 bar, v g = m 3 kg - at 70 bar, v g = m 3 kg - Difference = at 68 bar, v g = (3/5 x.0035) = m 3 kg - or v g = (/5 x.0035) = m 3 kg -. Determine the specific enthalpy of dry saturated steam at 5 bar. at 50 bar, h g = 794 kj kg - at 55 bar, h g = 790 kj kg - Difference = 4 at 5 bar, h g = 794 (/5 x 4) = 79.4 kj kg - or h g = (3/5 x 4) = 79.4 kj kg - 3. Determine the specific internal energy of superheated steam at 4 bar and 35 C. at 0 bar and 35 C, u = at 5 bar and 35 C, u = = kj kg - = 86.0 kj kg - Difference = 8.5 kj kg - at 4 bar and 35 C, u = (4/5 x 8.5) = 87 kj kg - or u = (/5 x 8.5) = 87 kj kg - Refrigeration In general, refrigeration may be defined as any process of heat removal. More specifically, refrigeration is defined as that branch of science that deals with the process of reducing and maintaining the temperature of a space or body below the temperature of its surroundings. DET: Mechanical Engineering: Thermofluids Higher Support Notes 4

32 If a space or body is to be maintained at a temperature lower than its surrounding ambient temperature, heat must be removed from the space or body being refrigerated and transferred to another body or substance whose temperature is below that of the refrigerated body. Mechanical refrigeration is primarily an application of thermodynamics wherein the cooling substance goes through a cycle in which it is recovered for re-use. A thermodynamic cycle can be operated in the forward direction to produce mechanical power from heat energy, or it can be operated in the reverse direction to produce heat energy from mechanical power. The reversed cycle is essentially utilised for the cooling effect that it produces during a portion of the cycle and is thus called a refrigeration cycle. Vapour-compression refrigeration cycles The most widely used domestic refrigerators function on a vapour-compression cycle which operates between two pressure levels using a two-phase working substance or refrigerant which alternates cyclically between the liquid and vapour phases in continuos circulation. In order to convert a liquid into a vapour, an energy transfer is required. This energy is acquired by the vapour molecules in the evaporation process and is termed the enthalpy of evaporation. If this energy is subsequently transferred from the vapour, the energy of the molecules is diminished and liquid is formed during the process of condensation. The evaporation and condensation processes take place when the refrigerant is absorbing and rejecting heat, and these are essentially constant temperature and constant pressure processes. Commonly, the vapour compression cycle system within a refrigerator comprises four main devices, viz: Evaporator Compressor Condenser Expansion or throttle valve. These individual elements are illustrated and have their functions examined with reference to the following systems diagram of a refrigeration unit. Referring to the diagram below, a wet low-pressure, low temperature refrigerant enters the evaporator at point and boils (evaporates) to a nearly dry state at point by absorbing heat from a controlled refrigerated space thereby producing the refrigerating effect. The vaporised refrigerant then enters the compressor in which it is compressed, by a work input, ideally to a dry saturated state at a higher pressure and temperature to point 3. The refrigerant next passes through a condenser at constant pressure and temperature until it is completely liquid at point 4 by transferring heat to the surroundings. DET: Mechanical Engineering: Thermofluids Higher Support Notes 5

33 Work Input Heat Rejection Q 3 Compressor Condenser Evaporator Heat Absorption Q Throttle Valve 4 High Pressure Side Low Pressure Side Systems Diagram for Vapour Compression Cycle The cycle is completed when the refrigerant is expanded through a throttle valve back to its original low pressure, low temperature, wet state at point. The enthalpy at point 4 being equal to the enthalpy at point. This cycle of operations is repeated on a continuous basis in order to maintain a predetermined sub-zero temperature within the controlled space. Refrigerant The working fluid that circulates in a refrigeration system is called a refrigerant and may be defined as a substance that, by undergoing a change in phase (liquid to gas, gas to liquid), absorbs or releases a large quantity of heat in relation to its volume, thereby producing a considerable cooling effect. A refrigerant is a fluid that absorbs heat during evaporation at a low temperature and pressure, and rejects heat by condensing at a higher temperature and pressure. Examples of refrigerants are ammonia, sulphur dioxide, and methyl chloride, although these are no longer widely used, having been largely replaced by fluorocarbons such as Freon (refrigerants R and R). The Freon refrigerants R and R are general purpose fluids specially manufactured for refrigeration and these are non-toxic and non-flammable. DET: Mechanical Engineering: Thermofluids Higher Support Notes 6

34 Apart from the ability to boil (evaporate) at a low temperature, refrigerants should possess other desirable characteristics such as: low cost and commercially available in quantity chemical stability non-explosive suitable working pressures and temperatures low specific volume in order to keep pipe sizes small the liquid enthalpy should be low and evaporation enthalpy high in order to achieve a high refrigeration effect per kilogram of refrigerant. There is no refrigerant with all these properties, so the choice of a suitable fluid for any particular application must represent some form of compromise. The R family of refrigerants is the safest group and most widely used. All new refrigerants in the R family should have zero ozone depletion potential and be user friendly. Property tables and charts are produced for the various refrigerants similar to those produced for water and steam. The behaviour of refrigerants is akin to that of water when subjected to heat. Water boils at 00 C when heat energy is supplied at atmospheric pressure. Evaporation then takes place at constant temperature until the vapour is completely dry and in the gaseous state. Further heating raises the temperature and the fluid is in the superheated condition. When the temperature and pressure of a refrigerant bear a `natural stable relationship to each other, the refrigerant is regarded as being in its saturated state. A refrigerant liquid in its saturated state can be further cooled at the same pressure. It will then become subcooled or undercooled. A refrigerant vapour in its saturated state can be heated further at the same pressure. It will then become superheated. Saturated Liquid Refrigerant at 40 C and 9.6 bar - Sensible Heat Undercooled Liquid Refrigerant at 35 C and 9.6 bar Sat. Refrigerant Vapour at 40 C and 9.6 bar + Sensible Heat Superheated Refrigerant Vapour at 45 C and 9.6 bar DET: Mechanical Engineering: Thermofluids Higher Support Notes 7

35 Use of thermodynamic tables for refrigerants In this outcome we have already dealt with the use of property tables for water and steam. With the exception of entropy (s values), our studies now extend into the interpretation and extraction of information covering the ammonia and fluorocarbon refrigerants R77 and R as listed on pages and 3 of the Rogers and Mayhew tables. Refrigerant quantities together with appropriate symbols and units are identified in the table below: SYMBOL QUANTITY UNIT T p s v g h f h g h Saturation Temperature Corresponding Saturation Pressure Specific Volume of Saturated Vapour Specific Enthalpy of Saturated Liquid Specific Enthalpy of Saturated Vapour Specific Enthalpy of Superheated Vapour C bar m 3 kg - kj kg - kj kg - kj kg - As previously stated, a refrigerant is regarded as being in its saturated state when its saturation temperature, T, and its corresponding pressure, p s bear a `natural stable relationship to one another. The specific volume, v g, of a saturated refrigerant vapour, (i.e. completely dry) can be read directly from the tables against any given pressure. For a wet vapour, the total volume of the mixture is given by the volume of liquid present plus the volume of dry vapour present. The volume of liquid is usually negligibly small compared to the volume of dry saturated vapour, hence for most practical problems, v x = xv g. e.g. Spec. Volume of Refrigerant R at.004 bar and.96 dry v x =.96 x.594 =.530 m 3 kg - The heat energy required to change kg of saturated liquid refrigerant to a completely dry saturated vapour (gas) is called the enthalpy of evaporation. i.e. e.g. Enthalpy of evaporation h fg = h g - h f Enthalpy of evaporation of refrigerant R77 at.680 bar h fg = = kj kg - DET: Mechanical Engineering: Thermofluids Higher Support Notes 8

36 The specific enthalpy, h, of a refrigerant in the superheat condition can be extracted directly from the tables from either of two column headings (50 K and 00 K) dependent on the degree of superheat which is obtained by the difference between the superheat temperature and the saturation temperature at the specified pressure. Example Determine the specific enthalpy of refrigerant R77 at a pressure of.908 bar and temperature of 0 C. From tables, the saturation temperature at.908 bar is 0 C. Hence (T T s ) = 30 C or 30 K. Specific Enthalpy, h = 55.7 kj kg - (From 50 K column). If the degree of superheat had been, say, 84K at the same pressure, then Specific Enthalpy, h = kj kg - (from 00 K column). DET: Mechanical Engineering: Thermofluids Higher Support Notes 9

37 OUTCOME 3: SOLVE PROBLEMS ASSOCIATED WITH STEADY FLOW ENERGY EQUATION APPLICATIONS FOR GASES AND VAPOURS In this outcome we are concerned with the solution of problems utilising the steady flow energy equation as applied to typical thermodynamic devices such as boilers, steam turbines, compressors, etc. Thermodynamics deals with the relationships between energy transfers within such devices/systems in the form of heat and work, and the related changes in the properties of the working fluid. Steady flow thermodynamic systems The steady flow energy equation (SFEE) is applicable to open two-flow systems where the working fluid may be a gas or vapour. Steady flow conditions prevail when an equal mass of fluid per unit time is both entering and leaving the system. In order to analyse specific situations where thermodynamic principles are involved, we adopt a systems approach and make use of diagrams to illustrate the system, its boundary, its surroundings, together with input, output and process data. The figure below identifies these elements of a two-flow open system. WORKING FLUID ENTERING SURROUNDINGS (USUALLY THE ATMOSPHERE) WORK, IN OR OUT SYSTEM CONTAINING THE WORKING FLUID WORKING FLUID LEAVING HEAT, IN OR OUT BOUNDARY, e.g. THE WALL OF Only three things can cross the system boundary: a) energy in the form of heat b) energy in the form of work c) a mass of fluid which will possess certain forms of energy. Heat and work transfers across the system boundary are shown by two-way arrows since both quantities can either enter or leave the system. DET: Mechanical Engineering: Thermofluids Higher Support Notes 30

38 Heat received or rejected In any system a fluid can have a direct reception or rejection of heat energy transferred through the system boundary. This is designated by Q (Unit J), or if the rate of heat energy is given, by Q (Unit J s - or Watt). Thus if heat is received, then Q is positive heat is rejected, then Q is negative. If heat is neither received or rejected, then Q = 0. External work done In any system a fluid can do external work or have external work done on it transferred through the system boundary. This is designated by W (Unit J) or if the rate of work done is given by W (Unit J s - or W). Thus if external work is done by the fluid, then W is positive external work is done on the fluid, then W is negative. If no external work is done on or by the fluid, then W = 0. In order to satisfy performance criteria (a) in this outcome, students are required to convert common thermodynamic devices into representative input/output sub-system diagrams. Examples of these follow in pages 3 to 7 of the notes and also in the tutorial questions. Steam power plant An important industrial application for vapours is the steam power plant as represented in the system diagram shown below: BOILER TURBINE WORK OUT Q IN ELECTRIC GENERATOR CONDENSER COOLING WATER PUMP Q OUT STEAM POWER PLANT -- SYSTEM DIAGRAM DET: Mechanical Engineering: Thermofluids Higher Support Notes 3

39 Feed water from the pump enters the boiler which is supplied with fuel to provide heat input Q +ve. Wet or superheated steam from the boiler rotates the turbine and the work output W +ve drives an electric generator via the turbine s output shaft. Exhaust steam from the turbine flows to the condenser where heat energy Q ve is removed by the cooling water. The steam becomes water again (condensate) then returns to the feed pump where the cycle is continued. Each of the devices identified in the above diagram can be categorised as sub-systems of the integrated whole steam power plant. Each of the items is an example of a steady-flow system to which the steady flow energy equation can be applied. Boiler HEAT LOSSQ -VE WET STEAM SPACE STEAM OUTPUT FEED WATER INPUT WATER HEAT INPUTQ +VE MASS FLOW AT = MASS FLOW AT Input/output system In a steam power plant facility, the boiler is the device/sub-system in which steam is generated. In essence, a conventional boiler consists of a water container together with some heating device. The boiler is supplied with a steady flow of water which is converted into wet steam using the heat released by burning a fuel such as coal, oil or gas. If superheated steam is required, the wet steam is removed from the steam space in the boiler and piped into an integrated superheater where it is further converted into dry or superheated steam by the addition of more heat energy. DET: Mechanical Engineering: Thermofluids Higher Support Notes 3

40 In a boiler no work is done, hence W = 0. Heat input Q is required to generate steam in a boiler which can also have a heat loss through the boiler casing to the surroundings. In a steam power plant the boiler provides wet or superheated steam to the turbine in the system. The steam turbine Q FLUID INPUT TURBINE SYSTEM WORK OUTPUT FLUID OUTPUT MASS FLOW AT = MASS FLOW AT Input/output system In the steam turbine, inlet steam is supplied to the system with a high energy level and impinges across curved blades causing the turbine to rotate. An output shaft coupled to the blade mechanism delivers external work. The exhaust steam exits from the system with a low energy level. Heat may be lost from the system to the surroundings or additional heat may be transferred into the system. In this case work is done by the system. In an integrated steam power plant the turbine element may be used to drive an electric generator. In a steam power plant, the boiler supplies high energy steam to the turbine element. A simple integrated systems diagram for these two devices is shown below. DET: Mechanical Engineering: Thermofluids Higher Support Notes 33

41 Heat exchanger/condenser COOLING WATER STEAM INPUT CONDENSER OUTPUT 3 4 MASS FLOW AT 3 = MASS FLOW AT 4 MASS FLOW OF COOLING WATER IN = MASS FLOW OF COOLING WATER OUT Input/output system A heat exchanger is a device that transfers heat energy from a hot fluid to a colder fluid, e.g. oil coolers in engines and turbines where hot oil is cooled by a flow of cold water condensers in steam power plants. Exhaust steam from the turbine is cooled and condensed by cold water. Normally the two fluids interacting are separated by tube walls. In a condenser the work transfer is zero, i.e. W = 0. In a steam power plant, exhaust steam from a turbine is fed into a condenser for cooling into condensate. A simple integrated systems diagram for these devices is shown below. COOLING WATER TURBINE STEAM CONDENSER 3 4 DET: Mechanical Engineering: Thermofluids Higher Support Notes 34

42 Rotary air compressor HEAT LOSS Q TO SURROUNDINGS LOW PRESSURE INTAKE WORK INPUT COMPRESSOR SYSTEM HIGH PRESSURE OUTPUT Input/output system In the rotary type compressor, atmospheric air is induced to a cylinder where it is compressed by an offset rotor and blade mechanism or rotary screw type arrangement. The high-pressure air is subsequently delivered to a storage tank from where it can be tapped off and used to operate pneumatic tools such as rock drills, demolition tools and riveting hammers. Portable compressors usually have a diesel engine as the power source and an input shaft drives the rotor. In this case work is done on the system. Having developed system diagrams for various thermodynamic devices, we extend our studies in this outcome into the solution of practical problems involving the steady flow energy equation for both gases and vapours. In outcome one, we defined and generated formulae for internal energy, flow energy and enthalpy. Two other energy forms, potential energy and kinetic energy are also present in a moving fluid and these are dealt with below. Potential energy This is the energy possessed by a mass of fluid, m, by virtue of its height Z above a given datum position, thus: Total potential energy = mgz (kg x m s - x m) = Nm = (J) and for unit mass of the fluid Specific potential energy = gz (J kg - ) DET: Mechanical Engineering: Thermofluids Higher Support Notes 35

43 Kinetic energy If a fluid is in motion then it possesses kinetic energy. Thus, for a mass of fluid m, flowing with velocity C. Total Kinetic Energy = ½ mc (kg x m s - x m s - ) = Nm = (J) and for unit mass of the fluid Specific Kinetic Energy = C (J kg - ) Various energy forms exist in thermodynamic systems. In certain systems they may all be present. In other systems only some may be present. Not infrequently, energy forms of insignificant value may be ignored in the solution of problems. The steady flow energy equation The figure below represents an open system in which a steady-flow process is taking place. At entry to the system, the working fluid possesses potential, kinetic and internal energy and entry flow work is done. During its passage through the system the working fluid is considered to take in a quantity of heat Q and do external work W. At exit from the system the working fluid will again possess potential, kinetic and internal energy and will do flow work to leave the system. ENERGY IN FLUID ENTERING SYSTEM E FLUID IN SYSTEM W OUT Z HEAT Q IN ENERGY IN FLUID LEAVING SYSTEM E FLUID OUT DET: Mechanical Engineering: Thermofluids Higher Support Notes 36

44 The forms of energy associated with the moving fluid mass entering the system are: Potential energy = mgz (J) Kinetic energy = m C (J) Internal energy = U (J) Flow energy = p V (J) Hence, total energy of the moving fluid mass entering the system = mgz + m C + U + p V Also, total energy of the moving fluid mass leaving the system = mgz + m C + U + p V In a steady-flow system it is considered that the mass flow rate and the total energy of the working fluid remains constant throughout the process. Applying the principle of conservation of energy to the steady-flow open system then: Initial energy + Energy entering = Final energy + Energy leaving of the system the system of the system the system mgz + m C + U + p V + Q = mgz + m C + U + p V + W This is known as the steady flow energy equation. For unit mass ( kg) of working fluid the equation becomes: gz + C + u + p v + Q = gz + C + u + p v + W Also, the combination of properties of internal and flow energies is called enthalpy and these may be combined and designated by the specific enthalpy symbol h. Hence, the steady flow energy equation can be expressed in the form: gz + C + h + Q = gz + C + h + W for unit mass of working fluid. DET: Mechanical Engineering: Thermofluids Higher Support Notes 37

45 When the mass flow rate of working fluid (m) and rates of heat input (Q) and work output (W) are given then the steady-flow energy equation can be rearranged as follows:... C - C Q = W + m[g(z Z ) + + (u u ) + (p v p v )] OR Q = W + m[g(z Z ) + C - C + (h h )] Frequently in thermodynamic problems, changes in potential energy are small compared with other energy changes or even non-existent when there is no difference between entry and exit datum levels. The gz terms can therefore be neglected or dropped and the equation shortens to:... C - C Q = W + m [ + (h h )] It is important to note that, in thermofluids, the symbol H represents total enthalpy and h represents specific enthalpy. For this reason we identify height in the PE formula by the symbol Z. Similarly, for the KE formula, the symbol C is used for fluid velocity in order to distinguish velocity from total volume, V or specific volume, v. DET: Mechanical Engineering: Thermofluids Higher Support Notes 38

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