Known Steady-state operating data are provided for a simple gas turbine power plant.
Find The net power developed, in MW, for a given fuel mass flow rate.
Schematic and Given Data:
Engineering Model
1. The control volume identified by a dashed line on the accompanying figure operates at steady state.
2. Kinetic and potential energy effects can be ignored where mass enters and exits the control volume.
3. The ideal gas model is applicable to the fuel; the combustion air and the products of combustion each form ideal gas mixtures.
4. Each mole of oxygen in the combustion air is accompanied by 3.76 moles of nitrogen, which is inert. Combustion is complete.
Analysis The balanced chemical equation for complete combustion of methane with the theoretical amount of air is given by Eq. 13.4:
CH _{4}+2\left( O _{2}+3.76 N _{2}\right) \rightarrow CO _{2}+2 H _{2} O +7.52 N _{2} (13.4)
CH _{4}+2\left( O _{2}+3.76 N _{2}\right) \rightarrow CO _{2}+2 H _{2} O +7.52 N _{2}
For combustion of fuel with 400% of theoretical air
CH _{4}+(4.0) 2\left( O _{2}+3.76 N _{2}\right) \rightarrow a CO _{2}+b H _{2} O +c O _{2}+d N _{2}
Applying conservation of mass to carbon, hydrogen, oxygen, and nitrogen, respectively,
C: 1=a
H: 4=2 b
O: (4.0)(2)(2)=2 a+b+2 c
N: (4.0)(2)(3.76)(2)=2 d
Solving these equations, a = 1, b = 2, c = 6, d = 30.08.
The balanced chemical equation for complete combustion of the fuel with 400% of theoretical air is
CH _{4}+8\left( O _{2}+3.76 N _{2}\right) \rightarrow CO _{2}+2 H _{2} O ( g )+6 O _{2}+30.08 N _{2}
The energy rate balance reduces, with assumptions 1–3, to give
1 0=\frac{\dot{Q}_{ cv }}{\dot{n}_{ F }}-\frac{\dot{W}_{ cv }}{\dot{n}_{ F }}+\bar{h}_{ R }-\bar{h}_{ P }
Since the rate of heat transfer from the power plant is 3% of the net power developed, we have \dot{Q}_{ cv }=-0.03 \dot{W}_{ cv }. Accordingly, the energy rate balance becomes
\frac{1.03 \dot{W}_{ cv }}{\dot{n}_{ F }}=\bar{h}_{ R }-\bar{h}_{ P }
Evaluating terms, we get
\frac{1.03 \dot{W}_{ cv }}{\dot{n}_{ F }}=\left\{\left[\bar{h}_{ f }^{\circ}+\Delta \bar{h}^{\nearrow0}\right]_{ C H_{4}}+8\left[\bar{h}_{ f }^{\circ \nearrow0}+\Delta \bar{h}^{\nearrow0}\right]_{ O _{2}}+30.08\left[\bar{h}_{ f }^{\nearrow0}+\Delta \bar{h}^{\nearrow0}\right]_{ N _{2}}\right\}
-\left\{\left[\bar{h}_{ f }^{\circ}+\Delta \bar{h}\right]_{ CO _{2}}+2\left[\bar{h}_{ f }^{\circ}+\Delta \bar{h}\right]_{ H _{2} O ( g )}+6\left[{h}_{ f }^{\circ \nearrow0}+\Delta \bar{h}\right]_{ O _{2}}\right.
\left.+30.08\left[\bar{h}_{f}^{\circ \nearrow0}+\Delta \bar{h}\right]_{ N _{2}}\right\}
where each coefficient is the same as the corresponding term of the balanced chemical equation and Eq. 13.9 has been used to evaluate enthalpy terms. The enthalpy of formation terms for oxygen and nitrogen are zero, and \Delta \bar{h}=0 for each of the reactants because the fuel and combustion air enter at 25°C.
\bar{h}(T, p)=\bar{h}_{ f }^{\circ}+\left[\bar{h}(T, p)-\bar{h}\left(T_{ ref }, p_{ ref }\right)\right]=\bar{h}_{ f }^{\circ}+\Delta \bar{h} (13.9)
With the enthalpy of formation for CH _{4}( g ) from Table A-25
2 \bar{h}_{ R }=\left(\bar{h}_{ f }^{\circ}\right)_{ CH _{4}( g )}=-74,850 kJ / kmol (\text { fuel })
With enthalpy of formation values for CO _{2} \text { and } H _{2} O ( g ) from Table A-25, and enthalpy values for CO _{2}, H _{2} O , O _{2}, \text { and } N _{2} at 730 K and 298 K from Table A-23
\begin{aligned}\bar{h}_{ P }=&[-393,520+28,622-9,364]+2[-241,820+25,218\\&-9,904]+6[22,177-8,682]+30.08[21,529-8,669]\end{aligned}
\bar{h}_{ P }=-359,475 kJ / kmol (\text { fuel })
Using the molecular weight of methane from Table A-1, the molar flow rate of the fuel is
\dot{n}_{ F }=\frac{\dot{m}_{ F }}{M_{ F }}=\frac{20 kg (\text { fuel }) / min }{16.04 kg ( fuel ) / kmol ( fuel )}\left|\frac{1 min }{60 s }\right|
= 0.02078 kmol(fuel)/s
Inserting values into the expression for the power
\dot{W}_{ cv }=\frac{\dot{n}_{ F }\left(\bar{h}_{ R }-\bar{h}_{ P }\right)}{1.03}
\dot{W}_{ cv }=\frac{\left(0.02078 \frac{ kmol ( fuel )}{ s }\right)[-74,850-(-359,475)] \frac{ kJ }{ kmol ( fuel )}}{1.03}
\times\left|\frac{1 MW }{10^{3} \frac{ kJ }{ s }}\right|=5.74 MW
The positive sign indicates power is from the control volume.
1 This expression corresponds to Eq. 13.12b.
\frac{\dot{Q}_{ cv }}{\dot{n}_{ F }}-\frac{\dot{W}_{ cv }}{\dot{n}_{ F }}=\bar{h}_{ P }-\bar{h}_{ R } (13.12b)
2 In the combustor, fuel is injected into air at a pressure greater than 1 atm because combustion air pressure has been increased in passing through the compressor. Still, since ideal gas behavior is assumed for the fuel, the fuel enthalpy is determined only by its temperature, 25°C.
Skills Developed
Ability to…
• balance a chemical reaction equation for complete combustion of methane with 400% of theoretical air.
• apply the control volume energy balance to a reacting system.
• evaluate enthalpy values appropriately.
Quick Quiz
Determine the net power developed, in MW, if the rate of heat transfer from the power plant is 10% of the net power developed. Ans. 5.38 MW.
