Sonntag, Borgnakke and van Wylen
9.44 A two-stage compressor having an interstage cooler takes in air, 300 K, 100 kPa, and compresses it to 2 MPa, as shown in Fig. P9.44. The cooler then cools the air to 340 K, after which it enters the second stage, which has an exit pressure of 15.74 MPa. Both stages are adiabatic, and reversible. Find q in the cooler, total specific work, and compare this to the work required with no intercooler. Solution:
2
1 ·
-W 1
intercooler
C1
4
3
· Q
C2
·
-W2
C.V.: Stage 1 air, Steady flow Process: adibatic: q = 0, reversible: sgen = 0 Energy Eq.6.13:
-wC1 = h2 − h1 ,
Entropy Eq.9.8:
s 2 = s1
Assume constant CP0 = 1.004 from A.5 and isentropic leads to Eq.8.32 k-1 0.286 T2 = T1(P2/P1) k = 300(2000/100) = 706.7 K
wC1 = h1 - h2 = CP0(T1 - T2) = 1.004(300 – 706.7) = -408.3 kJ/kg C.V. Intercooler, no work and no changes in kinetic or potential energy. q23 = h3 - h2 = CP0(T3 - T2) = 1.004(340 – 706.7) = -368.2 kJ/kg C.V. Stage 2. Analysis the same as stage 1. So from Eq.8.32 k-1 0.286 T4 = T3(P4/P3) k = 340(15.74/2) = 613.4 K
wC2 = h3 - h4 = CP0(T3 - T4) = 1.004(340 – 613.4) = -274.5 kJ/kg Same flow rate through both stages so the total work is the sum of the two wcomp = wC1 + wC2 = –408.3 – 274.5 = –682.8 kJ/kg For no intercooler (P2 = 15.74 MPa) same analysis as stage 1. So Eq.8.32 T2 = 300(15740/100)
0.286
= 1274.9 K
wcomp = 1.004(300 – 1274.9) = –978.8 kJ/kg