Please help me in***THERMODYNAMICS OF TURBINE AND HEAT EXCHANGER*** problem in the link given below please?
Let's consider the line 5 - 6, Heat Exchanger, side hot: Before some tips about specific heats. If I deal the air like a perfect gas I have: Gas Constant: R = 8.3145*10^3/(0.75*28+0.25*32) = 286,7 J/Kg*K and: Cv = 5/2*R = 716.75 J/Kg*K Cp = 7/2*R = 1003.5 J/Kg*K (gamma) = g g = Cp/Cv = 1.40 But that's true for a temperature range between 0 and 100°C. The exchanger works at a range of about 927°C and 1207°C so I prefer to use the Kern's data ( se my answer at the same your question) I consider a transformation at Volume = const (delta)U = Q - L L = 0 Mean specific heat of air: Cair = 1.19*10^3 J/Kg*K - Q56 = 1.19*10^3 * (1480 - 1200)*1200/60 = 6.66*10^6 J/sec. (negative because it's given heat) Turbine - 1 Wt1 = 10000*10^3 Watt = 10*10^6 J/sec. Let's consider an adiabatic transformation: (delta)U = Q - L Q = 0 L = - (delta)U Poisson equation: - (delta)U = [Mair*R*T1/(g-1)]* *[1-(P2/P1)^(g-1)/g] L =[ Mair*286.7*1400/(1.4-1)]* *[1-(0.5*10^6/2.0*10^6)^(1.4-1)/1.4] L = Mair * 328178.2 J Wt1 = L/t = Mair * 328178.2/t 10*10^6 = Mair * 328178.2/t Mair/t = 10*10^6/328178.2 = 30.5 Kg/sec. = 1828 Kg/min. Line 2 - 3 Just as 5 - 6. But in this case the unknown value is: T3 and Q23 = Q56 (L23 = 0) ........ Bye
