Question

Do problem 9.128E using variable specific heat assumptions instead of constant.

Solution 9.128E Problem Statement Reconsider Prob. 9–127E. Determine the change in the rate of heat addition to the cycle when the isentropic efficiency of each compressor is 88 percent and that of each turbine is 93 percent. Problem 9.127E A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60°F; the pressure ratio for each stage of compression is 3; the air temperature when entering a turbine is 940°F; and the regenerator operates perfectly. Determine the mass flow rate of the air passing through this engine and the rates of heat addition and rejection when this engine produces 1000 hp. Assume isentropic operations for all compressor and the turbine stages and use constant specific heats at room temperature. intercool Preheat - Pin FIGURE P9-127E

Please don't use constant specific heat assumption

Answer 1

Is diagram for the given problem 9. 127€ AT A For air (p = 1.005k/18k N=1.4 P2 = 14 psi = 0.965266 bar, Ti = 60°F = 288.7K Rp =3 (for each stage), To = Tg = 94056 = 777.6K Ideal Regenerator (E = Effectiveness-1), ma =? Power produced Whet = 1000 hp = 745.7kW. Assuming isentropic operations for all compressors and turbines D T2 = T (2) t = 288.7 X (3) 1.4 _ 3. Assuming perfect intercooling To = T1=238.7 of condition - T4=T2=395.15k and min compression work 777.6 Now, T7= 20.44 - 568./7 R. Assuming maximum temp. limit for both turbine to be Yame , To=78 = 777,6K And for max. power output Ty=Tg = 568.11K.

Now, for ideal regenerator, t=1 T- Tg-Tg * To = Tg = 568.11K Also, for HEX Tg-To = Ts-Tn Đ To = Tu = 395.15k Now, it is given that what = 745.7 kW map{ (T6 T7) + (T3 - Tg) - (T2-T)) - (Ts Ta} = 745.7 kW. max 1.005 { 2x (777.6 - 568.11) - 2 (395.15–288.77 = 745.7 745.7 207.11 - = 3.6 kg/sl. Aus Now, for problem 9. 1285 TS diagram -0.88 T = 0.93 Ta = Tit T2-T1 = 288.7 + 395.15-288.7 0.88 To = Ta' = 408.66K

and + - T6 T7' = 2 T6 T7) TH' = To - 91 (To-Tx) = #AGAZIN T = TG' = 777.6 - 0.93(77716-568.11) - 582.77K. Now, for perfect Regenerator t 1 Ts-Tal Tg'-T4 Ts=Tg = 582.77K. - For problem 9. 127€ Rate of Heat addition Qadd = ma Cp ( (To-Ts) + Tp =T=)) = 3.621.005 (777.6–568.11) +(7776 -568.11) - 1515. 8 7 KW. And Rate of Heat Rejection Spej z nalek (To-T, ) + (Tg -f3)} = 3.621.005 { (395.15–288.7) x2] = 770.27 kW For problem 9.128€ , Rate of HEAT Addition & add = malp { (T6 - Ts) + (Tg = Tə)} = 31,6x1.6054 (77706-582177) + (777-6-582.17 = 1409,79 kW Change in rate of heat Additon = dado - Qady = 1409.79-1515-87=-106 KW