Solution for Homework 6.19, 6.21, 6.23, 6.25(a-c).pdf - CC

6.19Find: Determine the change in specific entropy, in kJ/kg∙K, between the specifiedstates.(a) Water p 1 =10 MPa, T 1 =400°C , p 2 =10 MPa, T 2 =100°C .Table A-4, s 1 =6.2120 kJ/kg∙KTable A-5, s 2 =1.2992 kJ/kg∙Ks 2 -s 1 =1.2992-6.2120=-4.9128 kJ/kg∙K.(b) Refrigerant 134a, h 1 =111.44 kJ/kg, T 1 =-40°C , saturated vapor at p 2 =5 bar.Table A-10 at -40°C h f =0, h fq =222.88, h g =232.88kJ/kgThus,x 1 =(111.44-0)/222.88=0.5Thens 1 =s f +x 1 (s g -s f )=0+0.5(0.9560-0)=0.478 kJ/kg∙K.Then , with s 2 from Table A-11s 2 -s 1 =0.9117-0.478=0.4337 kJ/kg∙K.(c) Air as an ideal gas, T 1 =7°C , p 1 =2 bar, T 2 =327°C , p 2 =1 bar.With Eq.6.21a and so data from Table A-22o o R p2 ( ) ( ) ln2 1 2 1s s s T s TMp1kJ 8.314 kJ 1= (2.40902-1.63279) - lnkg K 28.89 kg K 2=0.97515kJkg K

s =1.007ln 373 -0=0.2431 kJ/kg∙K293(b) air, p 1 =1 bar, T 1 =27°C , p 2 =3 bar, T 2 =377°C2Ideal gas Table s s ( T ) s ( T ) R ln with2 1p1,o o pos data from Table A-22 s =2.49264-1.70203-(8.314/28.97)ln3=0.47632 kJ/kg∙KConstant specific heatT ps c ln R lnpT p2 21 1With c p at 475 K from Table A-20 s =1.0245ln(650/800)- (8.314/28.97)ln3=0.47684 kJ/kg∙K(c) carbon dioxide, p 1 =150 MPa, T 1 =30°C , p 2 =300 MPa, T 2 =300°C2Ideal gas Table s s ( T ) s ( T ) R ln with2 1p1,from Table A-1o o pos data from Table A-23 and M800241.033 214.284 8.314 lns 150 =0.4769kJ/kg K44.01Constant specific heatT ps c ln R lnpT p2 21 1With c p at 438 K from Table A-20573 8.314 300s 0.9636 ln ln 0.4862 kJ/kg∙K303 44.01 150(d) cardon monoxide, T 1 =300 K, v 1 =1.1 m 3 /kg, T 2 =500 K, v 2 =0.75 m 3 /kg.2Ideal gas Table s s ( T ) s ( T ) R ln2 1po o p1With T2c ( T)pps s dT R ln2 1 T1Tp21,pv RTP T V 500 1.102 2 1 2.444P T V 300 0.751 1 2 Withos data from Table A-23 and M from Table A-1

212.719 197.723 8.314 ln 2.444s 0.2701 kJ/kg∙K28.01Constant specific heatT ps c ln R lnpT p2 21 1With c p at 400 K from Table A-20500 8.314s 1.047 ln ln 2.444 0.2696 kJ/kg∙K300 28.01(e)2Ideal gas Table s s ( T ) s ( T ) R ln2 1po o p1withos data from Table A-23 and M from Table A-11191.682 220.907 8.314 lns 2 0.8373 kJ/kg∙K28.02Constant specific heatT ps c ln R lnpT p2 21 1With c p at 550 K from Table A-20300 8.314 1s 1.065 ln ln 0.8389 kJ/kg∙K800 28.01 26.23Given: One kg of O 2 undergoes a process between specified states.Find: Using three methods, determine the change in specific entropy: (a) Equation6.19 with cp(T) from Table A-21.(b) Equation 6.21b withEquation 6.23 with c p at 900 K from Table A-20.Schematic and given data:0s from Table A-23.(c)

Assumption: The system consist of one kg of O 2 , which behaves as an ideal gas.(a) Using Eq.6.19Table A-21 givesT2c ( T)pps s dT R ln2 1 T1Tp21c pR2 3 4 T T T Twhere for O 2 =3.626, =-1.878 10 3, =7.056 10 6, =-6.764 10 9, =2.156Thus2 3 4c ( T ) R T T T T dT dTT M TT2 Tp2T1 T1R T T T ln T T M 2 3 42 3 410 12Finally8.3141500 1.878 7.056 3.626 ln 1500 300 1500 30032 300 10 2 10 2 23 6 3 3 4 4 1500 300 1500 300 9 126.764 2.1563 10 4 10=1.657 kJ/kg∙Kc ( T)ps s dT R2 1T2p2ln =T1Tp18.314 1.5=1.7317 kJ/kg∙K1.657 ln32 2(b) Using Eq.6.21b withos from Table A-23oo ( ) ( ) ln /2 1 2 1 2 1s s s T s T R p ps s 2 1MM

257.965 205.213 8.314 ln 1.5 / 2= =1.7235 kJ/kg∙K32(c) Using Eq.6.23 with c p at 900 K from Table A-20T ps s c ln R ln2 1 pT p2 21 11500 8.314 1.5= 1.074 ln ln =1.8033 kJ/kg∙K300 32 26.25Given: Water undergoes process from a liquid state at 80 °C , 5MPa, to a sat.liq at40°C .Find : determine the change in specific entropy using three alternative methods.Analysis :(a) Table A-2 and A-5. s 1 =1.0720 kJ/kg∙K, s 2 =0.5725 kJ/kg∙K,(b) Sat .lid data from Table A-2. s s ( T ) 1.0753 kJ/kg∙K, s1 f 12 as above, s=-0.4995 kJ/kg∙K s =-0.5028 kJ/kg∙K(c) incompressible liquid model with c= 4.18 kJ/kg∙K from Table A-19: withEq.6.242s c =4.18 (kJ/kg∙K) lnln TT1313=0.5027 kJ/kg∙K353