A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and a simpleRankine cycle for the bottoming cycle. Atmospheric air enters the compressor at 101 kPa and 20degC, and the maximum gas cycle temperature is 1100 degC. The compressor pressure ratio is 8;the compressor isentropic efficiency is 85 percent; and the gas turbine isentropic efficiency is 90percent. The gas stream leaves the heat exchanger at the saturation temperature of the steamflowing through the heat exchanger. Steam flows through the heat exchanger with a pressure of6000 kPa and leaves at 320 degC. The steam cycle condenser operates at 20 kPa, and the isentropic efficiency of the steam turbine is 90 percent. Determine the mass flow rate of the air through the air compressor required for this system to produce 100 MW of power. Use constant specific heats for air at room temperature. (ANSWER: 279 kg/s).

Mass flow of air = 242.9kg/s

Explanation:

Given.

p1 = 20kPa

p2 = 6000kPa.

r = 8

First, we need to get the enthalpy at p1 and p2

Using saturated liquid water tables

h1 = 251kJ/Kg where p1 = 20kPa

To get, h2; we'll add the work done by the pump to h1.

To get the work done by the pump, we'll set v1 to specific volume of water

v1 = 0.00102m³/Kg

So, h2 = h1 + v1(p2 - p1)

h2 = 251 + 0.00102(6000-20)

h2 = 257.0996 kJ/Kg

Then, we need h3 and s3 (s3, represent enthropy)

Using saturated liquid water tables where p = 6000kPa and T = 320°C

h3 = 2950kj/Kg

s3 = 6.190kj/kgK

The s3 will be equal s4.

With s4 = 6.190kj/kgK and p = 20kPa, we'll get h4s

h4s = 2040kJ/Kg

Calculating the real enthalpy, h4, given that the turbine efficiency = 90% = 0.9

h4 = h3 - u(h3 - h4s)

h4 = 2950 - 0.9(2950 - 2040)

h4 = 2,131 kJ/Kg

Calculating enthalpy, h5 where T = 293k, using the table of ideal gas of air

h5 = 294kj/kg

Using pressure ratio = 8 and Prandtl number = 1.28

Pr6 = r.Pr5 = 8 * 1.28 = 10.24

Calculating the enthalpy, h6s, using Pr6 = 10.24

h6s = 530kJ/kg

Calculating the real enthalpy, h6, given that the compressor efficiency = 85% = 0.85

h6 = h5 + (h6s-h5)/u

h6= 294 + (530 - 294)/0.85

h6 = 572 kJ/K

To determine enthalpy h7, we'll use T = 1393k, using the table of ideal gas of air

h7 = 1483kj/kg

Using pressure ratio = 8 and Prandtl number = 364

Pr8 = Pr7/r = 364/8 = 45.5

Calculating the enthalpy, h8s, using Pr8 = 45.5 and the cycls efficiency = 100%

h8s = 810kJ/kg

Calculating the real enthalpy, h8, given that the turbine efficiency = 90% = 0.9

h8 = h7 - u(h7-h8s)

h8 = 1483 - (1483 - 810)0.9

h8 = 877 kJ/K

For enthalpy h9, we'll use the given temperature T = 593l

h9 = 600k

Calculating the net work output of the steam

w = wt - wp

w = h3 - h4 - v(p2 - p1)

w = 2950 - 2131 - 0.00102(6000-20)

ws = 813kj/kg

Calculating the net work output of the gas

wg = wt - wc

w = (h7 - h8) - (h6 - h5)

w = (1483 - 877) - (572 - 294 )

wg = 328kj/kg

The energy Balance equation is given as

m1(h3-h2) = m2(h8-h9)

m2/m1 = (h3-h2)/(h8-h9)

m2/m1 = (2950-257)/(877-600)

m2/m1 = 9.722

Calculating the work output of the combined cycle

w = wg + m1/m2(ws)

w = 328 + 1/9.722*(813)

w = 411.62kj/kg

Given the power output process, = 100000kW

Mass flow of air = 100000/411.62

Mass flow of air = 242.9kg/s

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