4版講稿例題a4512871

1、A4512871某種氣體Rg= 0.318 3 kJ/(kgK) ，cp=1.159 kJ/(kgK) 以800 ，0.6 MPa及100 m/s的參數流入一絕熱收縮噴管，若噴管背壓pb =0.2 MPa，速度系數 j = 0.92，噴管出口截面積為2 400 mm2，求： 噴管流量及摩擦引起的作功能力損失。（T0=300K）解：cV = cp - Rg= 1.159 kJ/(kgK) - 0.348 3kJ/(kgK) = 0.840 7kJ/(kgK)k = cp1.189 kJ/(kg K)= 1.3790.840 7kJ/(kg K)cVT0 = T1c2(100 m/s)2= (80

2、0 + 273)+ f 1 2cpK += 1 077.3 K2 1 159 J/(kg K)1k1.379p= p T0k -11 077.3 K1.379-1=0.6 MPa = 0.609 MPa01T1 073 K1 k1.3791.379-1 0.009 MPa= k -1= 22p=npp k +1 1+1.379 crcr00= 0.324 MPa pbp2 = pcr = 0.324 MPa若可逆：k -1k1.379-1 p2 = 1 077.3 K 0.324MPa 1.379T= T= 905.77K 0.609MPa 2s0p0 = RgT2s= 318.3 J/(kg

3、 K) 905.77K = 0.889 8 m3/kgv2s0.324106 Pa2p22 (h0 - h2 ) =2cp (T0 - T2 )=cf 2 s2 1 159 J/(kg K) (1 077.3 - 905.77) K = 630.93 m/s=2 40010-6m2 630.93 m/sA cq= 1.702 kg/s 2f 2s m0.889 8 m3/kgv2s過程不可逆：c= jc= 0.92630.93 m/s = 580.46 m/sf 2f 2sc2h2 = h0- f2 2c2(580.46 m/s)2T2 = T0- f 2 2cp= 1 077.3 K -=

4、931.94 K2 1 159J/(kg K)= RgT2= 318.3 J/(kg K) 931.94K = 0.915 5 m3/kgv20.324106Pap232 400 10-6 m2 580.46 m/sA cqm = 1.522 kg/s 2f 20.915 5 m3/kgv2絕熱熵流為零，熵產等于熵變= cln T2= 1.159kJ/(kg K) ln 931.94Ks= Ds= Ds1-22 s-2gpT905.77K2s= 0.033 kJ/(kg K)I = qmT0sg= 1.522 kg/s300 K 0.033 kJ/(kg K) =15.07kW附：-火用損=Ex,H = 0 ，所以=-4c2c2I = qm+ f 12 - ex,H,2+ f 22 ex,H,1(22cc)= qm-+-ex,H,1ex,H,2f 2f 1 22 (h - h ) -T)12(s)= q- s+- c22c12m12012f 2f 1