A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. Atmospheric air enters the gas turbine at 101 kPa and 20∘C, and the maximum gas cycle temperature is 1100∘C. The compressor pressure ratio is 8; the compressor isentropic efficiency is 85 percent; and the gas turbine isentropic efficiency is 90 percent. The gas stream leaves the heat exchanger at the saturation temperature of the steam flowing through the heat exchanger. Steam flows through the heat exchanger with a pressure of 6000 kPa and leaves at 320∘C. 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 air through the air compressor required for this system to produce 100 MW of power. Use constant specific heats for air at room temperature.

We will need Enthalpies at all points of steam and gas cycle.For the enthalpy h(1) we will use standard liquid water tables and given pressure [tex]p_{1}=20kPa[/tex].

[tex]h_{1}=251kJ/kg[/tex]

For the enthalpy of h(2) we will add the work done by the pump [tex]w_{p1}[/tex] to the enthalpy h(1).For the work [tex]w_{p1}[/tex] we will need specific volume of the water [tex]v_{1}=0.00102m^3/kg[/tex] and the pressure [tex]p_{1}[/tex] and [tex]p_{2}=6000kpa[/tex].

[tex]h_{2}= h_{1}+ v_{1}.(p_{2}- p_{1} )\\h_{2}=251kJ/kg+0.00102m^3/kg(6000kPa-20kPa)\\h_{2}=257kJ/kg[/tex]

To determine the enthalpy h(3) we will use the given pressure [tex]p_{3}=6000kPa[/tex] and the temperature [tex]T_{3}=320^0C[/tex] and we get [tex]h_{3}=2950kJ/kg[/tex].

The entropy [tex]s_{3}=6.190kJ/kg[/tex] K is equal to entropy [tex]s_{4s}[/tex] by using the pressure [tex]p_{4}=20kPa[/tex] to determine the enthalpy [tex]h_{4s}[/tex] the cycle would have if the efficiency was 100%.

[tex]h_{4s} =2040kJ/kg[/tex]

To determine the real enthalpy h(4) we will use the given turbine efficiency μ=90%.

[tex]h_{4} =h_{3}-[/tex]μ.[tex](h_{3}- h_{4s})[/tex]

[tex]h_{4}=2950kJ/kg-0.90(2950kJ/kg-2040kJ/kg)\\ h_{4}=2131kJ/kg[/tex]

To determine the enthaply h(5) we will use the given temperature [tex]T_{5} =293k[/tex] and the table of ideal gas properties of air

[tex]h_{5}=294kJ/kg[/tex]

We can also determine the Prandtl number [tex]P_{r5}=1.28[/tex] and use the given pressure ratio r=8 to determine the Prandtl number [tex]P_{r6}[/tex].

[tex]P_{r6}=r.P_{r5}=8*1.28=10.24\\[/tex]

we can now use it to determine the enthalpy h(6s) the cycle would have if the efficiency was 100%.

[tex]h_{6s}=530kJ/kg[/tex]

to determine the real enthalpy h(6) we will use the given compressor efficiency μ=85%

[tex]h_{6} =h_5}+h_{6s} -h_{5} /[/tex]μ

[tex]h_{6} =294kJ/kg+\frac{530kJ/kg-294kJ/kg}{0.85} \\h_{6} =572kJ/kg[/tex]

To determine the enthalpy h(7) we will use the given temperature [tex]T_{7}=1373K[/tex] and the table of ideal gas properties of air.

[tex]h_{7}=1483kJ/kg[/tex]

We can dertermine the prandtl number [tex]P_{r7} =364[/tex] and use the given ratio of r to determine prandtl number [tex]P_{r8}[/tex]

[tex]P_{r8}=\frac{P_{r7} }{r} =\frac{364}{8} =45.5[/tex]

we can now use it to determine the enthalpy h(8s) the cycle would have efficiency of 100%

[tex]h_{8s} =810kJ/kg[/tex]

to determine the real enthalpy h(8) we will use the given turbine efficiency μ=90%

[tex]h_{8}=h_{7}-[/tex]μ[tex](h_{7}-h_{8s})[/tex]

[tex]h_{8}=1483kJ/kg-0.90(1483kJ/kg-810kJ/kg)\\h_{8}=877kJ/kg[/tex]

for the enthalpy of h(9) we will use the given temperature of 593 K

[tex]h_{9}=600kJ/kg[/tex]

We will now use the calculated enthalpies to determine the net work output of steam [tex]w_{s}[/tex] and gas [tex]w_{g}[/tex] cycle.

[tex]w_{s} =w_{t}- w_{p1}\\ w_{s}=h_{4}- h_{3}- v_{1}.(p_{2}- p_{1})\\ w_{s}=2950kJ/kg-2131kJ/kg-0.00102m^3/kg(6000kPa-20kPa)\\ w_{s}=813kJ/kg[/tex]

now

[tex]w_{g}=w_{t}- w_{s}\\ w_{g}=(h_{7}- h_{8})-(h_{6} -h_{5} )\\w_{g}=(1483kJ/kg-877kJ/kg)-(572kJ/kg-294kJ/kg)\\w_{g}=328kJ/kg[/tex]

From the energy balance equation at the heat exchange we can determine the mass flow ratio [tex]m_{a} /m_{s}[/tex]

[tex]m_{s}.(h_{3}- h_{2})=m_{a}.(h_{8}- h_{9})\\m_{a} / m_{s} =h_{3}- h_{2}/h_{8}- h_{9} \\m_{a}/ m_{s}=2950kJ/kg-257kJ/kg/877kJ/kg-600kJ/kg\\ m_{a}/ m_{s}=9.722[/tex]

We can now use both to determine the net work output of combines cycle w

[tex]w=w_{g}+m_{a}/ m_{s}.w_{s}=328kJ/kg+1/9.722.(813kJ/kg)=411.62kJ/kg[/tex]

We will now use given power output of process P=100000KW to determine the mass flow rate of air.

[tex]m_{a} =P/w=100000kW/411.62kJ/kg=243kg/s[/tex]