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EGR 334 Thermodynamics Chapter 3: Section 12-14 Lecture 10: Ideal Gas Law Quiz Today? Today’s main concepts: • • • • Be Be Be Be able to explain the Ideal Gas Law able to explain when it is appropriate to use the Ideal Gas Law able to use the Ideal Gas Law to determine State Properties able to apply the Ideal Gas Law to the solution of 1st Law problems. Reading Assignment: • Read Chap 3: Sections 15 Homework Assignment: From Chap 3: 102, 107,115, 125 3 Ideal Gas Law When the general compressibility factor, Z 1, then the following relationship between pressure, temperature, and volume of a gas applies. p ab so lu te p ressu re pv RT which can also be written pV m RT or pV nRT m = mass n = number of moles T ab so lu te tem p eratu re v sp ecific vo lu m e 8 .3 1 4 k J/k m o l K o R 1 .9 8 6 B tu /lb m o l R o 1 5 4 5 ft lb /lb R f m ol R may be found on Table 3.1 Sec 3.13 : Ideal gases and u, h, cv, cp 4 If a gas behaves as an ideal gas, then its specific internal energy, u, depends only on temperature. u u (T ) Enthalpy, h, was defined as: h u pv For an ideal gas, since pv RT then h u RT Since and u u (T ) h h (T ) u (T ) R T Therefore: If a gas can be treated as an ideal gas, its intensive properties of specific energy and enthalpy are entirely functions of temperature. Sec 3.13 : Ideal gases and u, h, cv, cp 5 If u and h are only functions of temperature, then the specific heats may be used to determine relations between temperature change and energy levels. cV u T V cP h T P For an ideal gas, the expressions for u and h can be simplified since u = f(T) and h = f(T) du T2 T1 cV dT dh T2 T1 c P dT For many cases, the specific heats will be treated as constant values over a limited temperature range and these integrals will be approximated as: u 2 u 1 cV (T 2 T1 ) h 2 h1 c p (T 2 T1 ) Sec 3.13 : Ideal gases and u, h, cv, cp 6 Another important ideal gas equation may be written as h T u T RT dh dT du R dT c P T cV T R When the specific heat ratio may also be written as cP T kR k 1 k cP cV and For monotonic gases (Ar, Ne, He) with k = 1.4 is used, this equation cV T R k 1 c P _ m o n o g a ses 5 2 R Sec 3.13 : Ideal gases and u, h, cv, cp 7 Temperature Dependence: Specific heats cv and cp are functions of temperature. du T2 T2 T1 dh T1 cV ( T ) dT c P ( T ) dT If possible, you should look up their values from tables which give the specific heat at the temperature indicated. (see Tables A-20 and A-20E) k cP cV An alternative method is to use a formula to represent cp based on cp R T T 2 T 3 T 4 where Table A-21 has values of , , , , for different gases Sec 3.13 : Ideal gases and u, h, cv, cp 8 It best to use an actual function of the specific heats to evaluate u and h, by integration du T2 T1 dh cV d T T2 T1 cP dT But, often this is simplified, evaluating cV and cP at an average T either cV a ve c v (T1 ) c v (T 2 ) 2 and c p a ve c p (T1 ) c p (T 2 ) 2 or the specific heat at the average temperature may be used. cV a ve c v T T 1 2 2 c p a ve c p T T 1 2 2 where if cV and cP are treated as constants, then u T 2 u T1 cV T 2 T1 h T 2 h T1 c P T 2 T1 Sec 3.13 : Ideal gases and u, h, cv, cp 9 Summary: 1) Decide if a substance can be treated as an ideal gas…if yes, then pv RT pV m RT pV nRT or 2) To evaluate changes in internal energy, u, and enthalpy, h: i) Integrate with cv and cp as function of T (see Table A-21) T2 T2 dh cP dT du cV d T T1 T1 or ii) Use value of cv or cp at an average temperature (see Table A-20) u 2 u 1 cV _ a ve T 2 T1 or iii) Use k and R to define cv u 2 u 1 cV T 2 T1 or h 2 h1 c P _ a ve T 2 T1 R k 1 and cP kR k 1 then h 2 h1 c P T 2 T1 iv) Look up temperature dependent values of u and h on property tables. u u (T 2 ) u (T1 ) h h (T 2 ) h (T1 ) Table A-22 has property values for Air Table A-23 has property values for CO2, C0, H20, O2, and N2. Sec 3.13 : Ideal gases and u, h, cv, cp Example: (3.111) A piston cylinder assembly contains air at 2 bar, 300K, and a volume of 2 cubic meters. the air undergoes a process to a state where the pressure is 1 bar, during which the pressure-volume relationship is pV = constant. Assuming ideal gas behavior, determine a) the mass of the air, b) the work, and c) the heat transfer. 10 Sec 3.13 : Ideal gases and u, h, cv, cp 11 Example: (3.111) A piston cylinder assembly contains air at 2 bar, 300K, and a volume of 2 cubic meters. the air undergoes a process to a state where the pressure is 1 bar, during which the pressure-volume relationship is pV = constant. Assuming ideal gas behavior, determine a) the mass of the air, b) the work, and c) the heat transfer. State 1: p1 = 2 bar T1 = 300 K V1 = 2 m3 pV m RT For Ideal Gas: m p1V1 R T1 State 2: p2 = 1 bar T2 = ? V2 = ? 3 5 ( 2 b a r )( 2 m ) 10 N / m (0 .2 8 7 0 kJ / kg K )(3 0 0 K ) 1b a r For constant pV: p1V1 p 2V 2 2 1kJ J 1000 J N m 4 .6 5 kg 2bar 3 3 V2 V1 2 m 4 m p2 1 b a r p1 Sec 3.13 : Ideal gases and u, h, cv, cp 12 Example: (3.111) continued... State 2: p2 = 1 bar T2 = ? V2 = 4 m3 State 1: p1 = 2 bar T1 = 300 K V1 = 2 m3 find T2: p 2V 2 T2 R p1V1 T1 T2 p 2V 2 p1V1 T1 1b a r 4 m 3 2bar 2 m 2 (3 0 0 K ) 3 0 0 K 1st Law of Thermodynamics: U c v (T 2 T 1 ) 0 U Q W W pdV C V dV V2 V2 W C ln p1V1 ln V V 1 5 1 3 p1V1 p 2V 2 C 4m 10 N / m 2 J 1kJ ( 2 b a r )( 2 m ) ln 2 7 7 .3 kJ 3 2 m 1 b a r N m 1 0 0 0 J 3 Q U W 0 2 7 7 .3 kJ Sec 3.13 : Ideal gases and u, h, cv, cp Example: (3.124) Two kilograms (2 kg) of air, initially at 5 bar, 350 K and 4 kg of CO initially at 2 bar, 450 K are confined to opposite sides of a rigid, well-insulated container by a partition. The partition is free to move and allows conduction from one gas to the other without energy storage in the partition itself. The air and CO each behave as ideal gases with constant specific heat ratio, k = 1.395. Determine at equilibrium (a) the temperature in K, (b) the pressure, in bar, and (c) the volume occupied by each gas, in m3. 13 Sec 3.13 : Ideal gases and u, h, cv, cp Example: (3.124) k = 1.395. CO State 1: Air: State 1: 14 TC O _ 1 4 5 0 K T a ir _ 1 3 5 0 K pCO _1 2b a r p a ir _ 1 5 b a r m C O 4 kg m a ir 2 kg State 2: TC O _ 2 T a ir _ 2 T 2 p C O _ 1 p air _ 2 p 2 m C O m air m total 1st Law of Thermo: U system Q system W system U C O U air 0 or U air U C O since system is isolated Sec 3.13 : Ideal gases and u, h, cv, cp 15 CO State 1: Example: (3.124) continued… U m u 2 u1 m cV T 2 T1 For CO: Air: State 1: TC O _ 1 4 5 0 K T a ir _ 1 3 5 0 K pCO _1 2b a r p a ir _ 1 5 b a r mCO 4 kg m a ir 2 k g U C O m C O cV C O T 2 T C O _ 1 cV , C O RC O k 1 0 .2 9 6 8 kJ / kg K 1 .3 9 5 1 0 .7 5 1 4 kJ kg K U C O (4 kg )(0.7514 kJ / kg K ) T 2 450 K For Air: U air m air cV air T 2 T air _ 1 cV _ a ir R a ir k 1 0 .2 8 7 0 kJ / kg K 1 .3 9 5 1 0 .7 2 6 6 kJ kg K U air (2 kg )(0.7266 kJ / kg K ) T 2 350 K Sec 3.13 : Ideal gases and u, h, cv, cp Example: (3.124) continued… U C O U air 0 16 CO State 1: Air: State 1: TC O _ 1 4 5 0 K T a ir _ 1 3 5 0 K pCO _1 2b a r p a ir _ 1 5 b a r mCO 4 kg m a ir 2 k g (4 kg )(0.7514 kJ / kg K ) T 2 450 K (2 kg )(0.7266 kJ / kg K ) T 2 350 K 0 (3.006 kJ / K ) T 2 450 K (1.453 kJ / K ) T 2 350 K 0 (3.006 kJ / K )T 2 1352.7 kJ (1.453 kJ / K ) T 2 508.55 kJ 0 (4.459 kJ / K )T 2 1858.3 kJ T 2 416.8 K Sec 3.13 : Ideal gases and u, h, cv, cp 17 Example: (3.124) continued Find Vtotal V T V C O V A ir mRT mRT P C O ,i P A ir,i 2 0.2870 350 4 0.2968 450 5 5 2 10 5 10 C O ,i A ir,i 2.67 0.40 = 3.07 m Then find pfinal ( kg )( kJ / kg K )( K ) 1000 N m 2 N /m kJ 3 pf mRT m R T CO , f V to ta l mRT A ir , f V to ta l 4 0.2968 416.8 2 0.2870 416.8 ( kg )( kJ / kg K )( K ) 1000 J N m 3.07 m 3 kJ J 1bar 5 10 N / m 2.39 b a r Sec 3.13 : Ideal gases and u, h, cv, cp 18 Example: (3.124) continued… The final volumes are then, mRT P C O ,f VC O VC O 4 0.2968 416.8 5 2.39 10 V A ir V A ir ( kg )( kJ / kg K )( K ) 1000 N m N /m 2 2.079 m 3 kJ C O ,f mRT P A ir,f 2 0.2870 416.8 5 2.39 10 ( kg )( kJ / kg K )( K ) 1000 N m N /m A ir,f 2 kJ 1.001 m 3 19 Solution using IT: Note: The results are slightly different as the cv and cp values that IT pulled out slightly different. 20 End of Slides for Lecture 10