Question

9.31 The state at the beginning of compression of an air-standard Diesel cycle is fixed by pi = 100 kPa and T1 = 310 K. The compression ratio is 15. For cutoff ratios rc = 2, determine (a) the maximum temperature, in K. (b) the pressure at the end of the expansion, in kPa. (c) the net work per unit mass of air, in kJ/kg. (d) the thermal efficiency.

Answer 1

& Given that - Pressure before Compression P= 100 kPa temperature before Compression Ti= 310k - Comparession ratio Ilk = 15 - Cutt off ratio 1 c = 2 = 1 P2 = Con Y S3-con 5 ² Vu= Cons 1 12: isentropic Compression Assume Aion is an ideal gas Op = 1:005 kJ legk R = 0.287 kJ /kgk r=ly (310) = 915,79k Temperature after Compression (t = a1s. 79 k 2:3 i sobarice heat addition P2= P3. 2 1 2 2 = 3 are ) T3 = (c) (T2) T3= (2) (915.79) = 1831.5 k.

à manimum temperature in the ascle is. T3 = 1831.58k forom the process 12: Pluir = Pz V2 P2= Pi (visua) = (100) (15) "" | Pz = 4431-26 kPa = P3 heat supplied Qs = CP CTB-T2) 005 (193)-58 - 915.99) QS920.36 kJ Jkg Pressurie at the end of enpain expansion : (Bill) isentaopic Compression gato COlopp gobio * expansion socio => Y = (2) 63) e) ork= (rc) (re). i expansion natio re = 1 = 7.5 = re = 7.5 => temperature at the end of enpansion Ju = 1835.58 ) (4.5)ony Tu = 818.09 k Pay} = Puvur 2) Pu= (P3) (ane) = 4431-26 2/263.90 kPa) (as) ky"

: Pressure after expansion Pu= 263.90 kPa] Co (Tu-TI) =) heat rejected at Constant Volume QR = QR = (0.717) (818.09 – 310) QR=364.30 kJ lkg) in Network done Wret = Qg- QR = 920.36-364.30. 1 Whet= 556.05 kJ kg) in Thermal efficiency th= Whee 556.05 Qsupp 920.36 [Koh = 0obou = 1 kg = Gomy]