Answer:
781.6 kg/s
Explanation:
At the initial state, the enthalpy and pressure at the initial state from A-17 table of the steam chart is given as:
[tex]h_1=310.24kJ/kg \\P_{r1}=1.5546[/tex]
The relative pressure at state 2 is given as:
[tex]P_{r2}=P_{r1}\frac{P_2}{P_1}=1.5546*8=12.44[/tex]
The enthalpy at state 2 using interpolation from A-17 table of the steam chart is given as:
[tex]h_{2s}=562.58kJ/kg[/tex]
The relative pressure and enthalpy at state 3 from A-17 table of the steam chart is given as:
[tex]h_3=932.93kJ/kg\\P_{r3}=75.29[/tex]
The relative pressure at state 4 is given as:
[tex]P_{r4}=P_{r3}\frac{P_4}{P_3}=75.29*\frac{1}{8} =9.41[/tex]
The enthalpy at state 4 using interpolation from A-17 table of the steam chart is given as:
[tex]h_{4s}=519.3kJ/kg[/tex]
The mass flow rate (m) is given by the equation:
[tex]m=\frac{W}{\eta_t(h_3-h_{4s})-\frac{1}{\eta_c}(h_{2s}-h_1) }[/tex]
Where ηt is isentropic efficiency of turbine = 86% = 0.86
ηc is isentropic efficiency of compressor = 80% = 0.8
W = 31.5 MW
[tex]m=\frac{31500}{0.86(932.93-519.3)-\frac{1}{0.8}(562.58-310.24) }=781.6kg/s[/tex]
Answer:
Mass flow rate = 206.72 kg/s
Explanation:
Given Data:
T1 = 310 K
T3 = 900 K
P1 = 1 atm
P2 = 8 atm
k = 1.4
cp = 1.005 kJ/kg·K
Calculating the temperature at state 2 using the isetropic relation, we have;
T2/T1 = (P2/P1)^k-1/k
T2 = T1 * (P2/P1)^k-1/k
= 310 * (8/1)^1.4-1/1.4
= 310 * 8^0.4/1.4
= 310 * 1.8114
= 561.55 K
Calculating the temperature at stage 4 using the isetropic relation, we have
T3/T4 = (P2/P1)^k-1/k
900/T4 = (8/1)^1.4-1/1.4
900/T4 = 8^0.4/1.4
900/T4 = 1.8114
T4 = 900/1.8114
= 496.83 K
Finding the net work output, we have;
Wnet = cp(T3-T4) -cp(T2-T1)
= 1.005(900-496.83) - 1.005*( 561.55 -310)
1.005*403.17 - 1.005*251.55
= 405.18585 - 252.80775
= 152.37 kJ/kg
Calculating the mass flow rate using the formula;
Wnet = Mair*wnet
31.5*10^6 = 152.37*10^3*Mair
Mair = 31.5*10^6/152.37*10^3
= 206.72 kg/s
Therefore, mass flow rate = 206.72 kg/s
