Question

Question 2: (13 Marks) A quantity of carbon dioxide occupies a volume of 0.14 m'at 975 kPa and 370 °C. It is heated at constant volume to a pressure of 4.2 MPa, after which it is expanded in an insulated cylinder to a pressure of 270 kPa. Sketch these processes on a P-V diagram. Assuming Cp = 1.075 kJ/kg K for carbon dioxide, determine: A. The temperature at the start of the expansion B. The volume after expansion C. The temperature after expansion D. The nett work transfer E. The nett heat transfer F. The nett change in internal energy The molar mass for carbon dioxide 44 kg/kmol.

Answer 1

Solution → 11 initial rolome V = 0.14 m3 pressure Pi 975KP9 Temperale Ti = 3700 2 P2 = 4200 kPa P P₂ = 270 kPq 3 P = '975189 V (A) Temperature before expansion will be Tz PIVX - Pav2 T2 975 4200 (376+273) Tz Tz = T2=2496.84°C 27 69.84K B) N=1.33 For Co2 or sa 1.4 or s= 1.4 use any volume After expension 3 (2-3) Adi batic expensiom. P2 렇 P3 allot ( 2 Ver = Pz Vz?

V3 = 1.lom3 process (2-3) Adibatic prous dw23 4200 13 270 V2 V3 Von = 7.837 V2 V₂ = 7.8378V2 = 7.837X0.14 Tz Z P₃ Tz 1.988 2769.8 1393.24k 1.988 T3 = 1120.24°C work done procus (1-2) V2 = v dw=o 0.333 1.333 = (©) Temperature After Expension. T = ( 132 ) 7 품: 4200 270 T3 = = D dWaz P3 V₃ - P2 V2 270X1.10 - 4200% 0-1 23= 7) 0.33 dwa3==881.81 kJ /kg

(E) Net Healt transfer. do d01.2= d.2 + duiz du-2= U2U = Cv (T2-T) 0.85 10/1gzCv = 0.85 (2769.8-643) = 1807.78 kJ /igk. do=0= 1807.78 KJ 1 kg k de1-2- (6) Heelt frans tes in adibatic procey Oco Total or Net = 1807.78 kJ / kg K. (F) internal Energy (1-2) process. dw=o 4,2 = Cv (T2-T ) = 1807.78 Kgligk. For proces (2-3) Adibatic procus. doz du edw do=0 dy = dw- - (-88181 KJ 1kg K) Net U = U1-2 + U23 U 1807.78 + 881.8) = G41 Ų 2689.59 kJ 59 kJ /kg