4版讲稿例题courseware4版讲稿例题a470299

1、A470299已知柴油机混合加热理想循环p1=0.17 MPa、t1=60，压缩比=14.5，气缸中气体最大压力10.3 MPa，循环加热量q1 = 900 kJ/kg。设工质为空气，比热容为定值并取cp = 1004 J/(kgK)，cV= 718 J/(kgK) ，=1.4；环境温度t0= 20，压力p0=0.1 MPa。试解：由已知条件 p1=0.17 kPa、T1=333.15 K= RgT1= 287J/(kg K) 333.15 K = 0.562m3/kg点1：v10.17 106Pap10.562 m3/kgv点2： v= 0.038 7 m /kg3 1e214.511-2是

2、定熵过程，有k v= p e kp= p= 0.17MPa 14.51.4 = 7.18 MPa 1 21 v12 7.18106Pa 0.038 7m3/kgp v= 968 KT2 22Rg287J/(kg K)点3：v= v= 0.038 7m3/kgp= 10.3 MPa32310.3106Pa 0.038 7m3/kgp vT= 1 389 33K3287J/(kg K)Rgp3= 10.3 MPal = 1.43p27.18MPa= cV (T3 -T2 ) = 0.718 kJ/(kgK) (1 389 - 968)K = 302kJ/kgq1V= q1 - q1V= 900kJ

3、/kg - 302kJ/kg = 598kJ/kgq1 p2= q1 - q2 = 900 kJ/kg - 320 kJ/kg = 580 kJ/kgwnetlrk -1e k -1 (l -1) + kl(r -1)h= 1-t1.431.421.4 -114.51.4-1 (1.43 -1) +1.41.43 (1.42 -1)= 1-= 0.639= wnet= 580kJ/kg = 0.644或htq900kJ/kg1在吸热过程中空气熵增为Ds= cln T4 - Rln p424pgTp22= 1.004 kJ/(kg K) ln 1 985 K - 0.287kJ/(kg K) l

4、n 10.3 MPa968 K7.18 MPa= 0.617 4 kJ/(kg K)3K=p3 =10.3 MPa，p4q1 p+ q1 p点4：= cp (T4 - T3 )所以598 kJ/kgT= T= 1 389 K += 1 985431.004 kJ/(kg K)cp= RgT4= 287J/(kg K) 1 985K = 0.055m3/kgv410.3106pPa4点5：v= v= 0.562m3/kg51k 0.055 m3/kg 1.4 v= p4 4 = 10.3 MPa = 0.398p5MPa0.562m3/kgv5 0.398106 Pa 0.562 m3/kgp vT= 779 K 555287 J/(kg K)Rgq= c(T-T ) = 0.718 kJ/(kgK)(779 - 333)K = 320 kJ/kg2V514所以平均吸热温度为q1900kJ/kgT= 1 457.6K1mDs0.617 4kJ/(kg K)24循环吸热量中的可用能：q= 1- 293.15 K 900T=1-kJ/kg = 7