Answer:
flow(m) = 7.941 lbm/s
Q_in = 90.5184 Btu/lbm
Q_out = 56.01856 Btu/lbm
Explanation:
Given:
- T_1 = 60 F = 520 R
- T_6 = 940 = 1400 R
- Heat ratio for air k = 1.4
- Compression ratio r = 3
- W_net,out = 1000 hp
Find:
mass flow rate of the air
rates of heat addition and rejection
Solution:
- Using ideal gas relation compute T_2, T_4, T_10:
T_2 = T_1 * r^(k-1/k)
T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R
- Using ideal gas relation compute T_7, T_5, T_9:
T_7 = T_6 * r^(-k-1/k)
T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R
- The mass flow rate is obtained by:
flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)
flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)
flow(m) = 7.941 lbm/s
- The heat input is as follows:
Q_in = c_p*(T_6 - T_5)
Q_in = 0.24*(1400 - 1022.84)
Q_in = 90.5184 Btu/lbm
- The heat output is as follows:
Q_out = c_p*(T_10 - T_1)
Q_out = 0.24*(711.744 - 520)
Q_out = 56.01856 Btu/lbm
