Question

A gas-turbine power plant operates on the simple Brayton cycle with air as the working fluid and delivers 31.5 MW of power. The minimum and maximum temperatures in the cycle are 310 and 900 K, and the pressure of air at the compressor exit is eight times the value at the compressor inlet. Assuming an isentropic efficiency of 80 percent for the compressor and 86 percent for the turbine, determine the mass flow rate of air through the cycle using constant specific heats at room temperature. The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4.

Answer 1

Mass flow rate = 206.72 kg/s

Step-by-step 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

Answer 2

781.6 kg/s

Step-by-step explanation:

At the initial state, the enthalpy and pressure at the initial state from A-17 table of the steam chart is given as:

`h_1=310.24kJ/kg \\P_(r1)=1.5546`

The relative pressure at state 2 is given as:

`P_(r2)=P_(r1)(P_2)/(P_1)=1.5546*8=12.44`

The enthalpy at state 2 using interpolation from A-17 table of the steam chart is given as:

`h_(2s)=562.58kJ/kg`

The relative pressure and enthalpy at state 3 from A-17 table of the steam chart is given as:

`h_3=932.93kJ/kg\\P_(r3)=75.29`

The relative pressure at state 4 is given as:

`P_(r4)=P_(r3)(P_4)/(P_3)=75.29*(1)/(8) =9.41`

The enthalpy at state 4 using interpolation from A-17 table of the steam chart is given as:

`h_(4s)=519.3kJ/kg`

The mass flow rate (m) is given by the equation:

`m=(W)/(\eta_t(h_3-h_(4s))-(1)/(\eta_c)(h_(2s)-h_1) )`

Where ηt is isentropic efficiency of turbine = 86% = 0.86

ηc is isentropic efficiency of compressor = 80% = 0.8

W = 31.5 MW

`m=(31500)/(0.86(932.93-519.3)-(1)/(0.8)(562.58-310.24) )=781.6kg/s`