Answer: 337J/kg
Explanation:
(a) For air, take k = 1.40, R = 287 J/kg ⋅ K, and cp = 1005 J/kg ⋅K. The adiabatic steady-flow energy equation (9.23) is used to compute the downstream velocity:+ = = + = +2 2 2p 21 1 1c T V constant 1005( 260) (75) 1005(207) V or .2 2 2
Ans2mV 335 s≈2 1 p 2 1 2 1 207 273 30Meanwhile, s s c ln(T /T ) R ln( p /p ) 1005 ln 287 ln ,260 273 140+ − = − = − ÷ ÷ +or s 2 − s1 = −105 + 442 ≈ 337 J/kg ⋅ K Ans.(a)
(b) For argon, take k = 1.67, R = 208 J/kg ⋅ K, and cp = 518 J/kg ⋅ K. Repeat part (a):2 2 2p 21 1 1c T V 518(260) (75) 518(207) V , solve .2 2 2 Ans+ = + = + 2mV 246 1 207 273 30s s 518 l n 208 ln 54 320 . (b)260 273 140