Question

Problem 3 In modern high-speed compression ignition engines, fuel is injected into the combustion chamber much sooner than in early diesel engines. That means, heat is added partly at constant volume and partly at constant pressure. The advantage of this modification is that more time is available for the fuel to completely combust. The ideal dual cycle consists of the following operations: • Process -: Isentropic compression • Process -3: Addition of heat at constant volume. • Process - Addition of heat at constant pressure. • Process O-6: Isentropic expansion. • Process -0: Rejection of heat at constant volume. Consider an ideal dual cycle process using air as the working fluid. The compression ratio is ry = = 15 and the cut-off ratio is rc = = 1.1. At the beginning of the compression, the the state is defined by P = 90 kPa and Ti = 290 K. The maximum pressure is 7.5 MPa. Furthermore assume a total specific heat intake Gin = 42-3 + 93-4 = 1800 kJ .kg . Your tasks: 1. Sketch this process in P-v and T-s diagrams. 2. Determine the properties P, v, T for all endpoints of the five processes. 3. Calculate the fraction of heat transferred at constant volume. 4. Determine the net specific work Wnet,out 5. Determine the thermal efficiency Nth of the process. 6. Find the corresponding thermal efficiency Nth.c of a Carnot cycle operating between the same minimum and maximum temperatures. Remark: You can assume air to be an ideal gas with constant specific heats at room temperature. The air- standard assumptions are applicable. Changes of the kinetic and the potential energies can be neglected.

Answer 1

Givendata: compression rello e You = 15 cut off ratio = rc = 24 = 1) Pi= go ipa & T 1 = 290k Maximum Pressure Pmax= 7.5 mpa total specific heat at intake = q in = q 2-3 + q 3-4 = 1800 15/169 sketch this process on pan & T-s diagram Heat addition at 4 const-press. AT is ontropic L expansion 1. A 3 9 Sv Add. const vol - Heat Rejected isentropic compression PWP C Dual cyce. PV & T-S diagram.

2. Determine the properties PNT for all endpoints of the five processes. for Process 1-2 P2=(0-9) (15) Pro 9bar for air 1 P2 = 39.88 bar I T2 = 11 (2) = (290) (15) 4-1 1. T2 = 856. Yi KL using ProMRT PIVI= RT m. lis. R= 8.314 Slik

Glees VIERT pi Vi = (8.314) (290) (0.9) INI = 26 78.95 m3 V2 =R Tz P2 V2 = (8.314) (858-71) . (39.88) V2 = 178.60 m3

for processes 2-3 & 3-4 = 7.5mpa 73 - (856.9)/15) 13 = 1611.16 k J cut off ruho : Pas Pmaxely = 75 bar 11 THE we know that Vc=1-1= T4 T4 = T3. (1-1) (1611. 16)x(1.1) Ty = 1742.27 1.)

for process 4-5: VA N Ps= (75) (0.35) i P5= 26-37 borr

Expansion ratio = 15 = 5 x V 2 - VW us = 1 1 1 1 NS- V4 - 13.63 V5 Vu [ V3 = 1785.6 m2 v = vs / 1368 = 200.52 m3) 1/3=V2 [/3.178.60m3 = 200.52 me

3) Heat transferred at const- volum & Fraction of HT at u-) Ratio = q 2-3 q in 9 2-3 = CVCT3 - T2) : Co.o. Yig is leg for air. 92-32 (0.4183 754.95) = 542.05 Ratio = 0.301) Iratio = 30:lly

5-) Thermal efficiency nth mth = What 9, in - 1560. 43 1800 = 10.86 Mth=86/ 6) thermal efficieney of connot cyde for same max & min temperaturas. nth Carnot = Tu - Ti = 1772-27-290 1772- 27 = 0.83 on the = 834.