Known A combined gas turbine–vapor power plant operates at steady state with a known net power output. Operating pressures and temperatures are specified. Turbine, compressor, and pump efficiencies are also given.
Find Determine the mass flow rate of each working fluid, in kg/s; the net power developed by each cycle, in MW; and the thermal efficiency. Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor and discuss the results.
Schematic and Given Data:
Engineering Model
1. Each component on the accompanying sketch is analyzed as a control volume at steady state.
2. The turbines, compressor, pump, and interconnecting heatrecovery steam generator operate adiabatically.
3. Kinetic and potential energy effects are negligible.
4. There are no pressure drops for flow through the combustor, heat-recovery steam generator, and condenser.
5. An air-standard analysis is used for the gas turbine.
6. T_{0}=300 K , p_{0}=100 kPa.
Analysis
The property data given in the table below are determined using procedures illustrated in previously solved examples. The details are left as an exercise.
| Gas Turbine |
| State |
h(kJ/kg) |
s°(kJ/kg ⋅ K) |
| 1 |
300.19 |
1.702 |
| 2 |
669.79 |
2.5088 |
| 3 |
1515.42 |
3.362 |
| 4 |
858.02 |
2.762 |
| 5 |
400.98 |
1.9919 |
| Vapor Cycle |
| State |
h(kJ/kg) |
s(kJ/kg ⋅ K) |
| 6 |
183.96 |
0.5975 |
| 7 |
3138.3 |
6.3634 |
| 8 |
2104.74 |
6.7282 |
| 9 |
173.88 |
0.5926 |
Energy Analysis
a. To determine the mass flow rates of the vapor, \dot{m}_{ v }, and the air, \dot{m}_{ g }, begin by applying mass and energy rate balances to the interconnecting heat-recovery steam generator to obtain
0=\dot{m}_{ g }\left(h_{4}-h_{5}\right)+\dot{m}_{ v }\left(h_{6}-h_{7}\right)
or
\frac{\dot{m}_{ v }}{\dot{m}_{ g }}=\frac{h_{4}-h_{5}}{h_{7}-h_{6}}=\frac{858.02-400.98}{3138.3-183.96}=0.1547
Mass and energy rate balances applied to the gas turbine and vapor power cycles give the net power developed by each, respectively,
\begin{array}{r}\dot{W}_{ \text {gas} }=\dot{m}_{ g }\left[\left(h_{3}-h_{4}\right)-\left(h_{2}-h_{1}\right)\right] \\\dot{W}_{\text {vap }}=\dot{m}_{ v }\left[\left(h_{7}-h_{8}\right)-\left(h_{6}-h_{9}\right)\right]\end{array}
With \dot{W}_{\text {net }}=\dot{W}_{\text {gas }}+\dot{W}_{\text {vap }}
\dot{W}_{\text {net }}=\dot{m}_{ g }\left\{\left[\left(h_{3}-h_{4}\right)-\left(h_{2}-h_{1}\right)\right]+\frac{\dot{m}_{ v }}{\dot{m}_{ g }}\left[\left(h_{7}-h_{8}\right)-\left(h_{6}-h_{9}\right)\right]\right\}
Solving for \dot{m}_{ g }, and inserting \dot{W}_{\text {net }}=45 MW = 45,000 kJ/s and \dot{m}_{ v } / \dot{m}_{ a }=0.1547, we get
\dot{m}_{ g }=\frac{45,000 kJ / s }{\{[(1515.42-858.02)-(669.79-300.19)]}
+0.1547[(3138.3-2104.74)-(183.96-173.88)]\} kJ / kg
= 100.87 kg/s
and
\dot{m}_{ v }=(0.1547) \dot{m}_{ g }=15.6 kg / s
Using these mass flow rate values and specific enthalpies from the table above, the net power developed by the gas turbine and vapor power cycles, respectively, is
\dot{W}_{ gas }=\left(100.87 \frac{ kg }{ s }\right)\left(287.8 \frac{ kJ }{ kg }\right)\left|\frac{1 MW }{10^{3} kJ / s }\right|=29.03 MW
\dot{W}_{\text {vap }}=\left(15.6 \frac{ kg }{ s }\right)\left(1023.5 \frac{ kJ }{ kg }\right)\left|\frac{1 MW }{10^{3} kJ / s }\right|=15.97 MW
The thermal efficiency is given by Eq. 9.28. The net power output is specified in the problem statement as 45 MW. Thus, only \dot{Q}_{\text {in }} must be determined. Applying mass and energy rate balances to the combustor, we get
\eta=\frac{\dot{W}_{ gas }+\dot{W}_{ vap }}{\dot{Q}_{ in }} (9.28)
\begin{aligned}\dot{Q}_{\text {in }} &=\dot{m}_{ g }\left(h_{3}-h_{2}\right) \\&=\left(100.87 \frac{ kg }{ s }\right)(1515.42-669.79) \frac{ kJ }{ kg }\left|\frac{1 MW }{10^{3} kJ / s }\right|\end{aligned}
= 85.3 MW
1 Finally, thermal efficiency is
\eta=\frac{45 MW }{85.3 MW }=0.528(52.8 \%)
b. The net rate of exergy increase of the air passing through the combustor is (Eq. 7.18)
e _{ f 1}- e _{ f 2}=\left(h_{1}-h_{2}\right)-T_{0}\left(s_{1}-s_{2}\right)+\frac{ V _{1}^{2}- V _{2}^{2}}{2}+g\left(z_{1}-z_{2}\right) (7.18)
\begin{aligned}\dot{ E }_{ f 3}-\dot{ E }_{ f 2} &=\dot{m}_{g}\left[h_{3}-h_{2}-T_{0}\left(s_{3}-s_{2}\right)\right] \\&=\dot{m}_{g}\left[h_{3}-h_{2}-T_{0}\left(s_{3}^{\circ}-s_{2}^{\circ}-R \ln p_{3} / p_{2}\right)\right]\end{aligned}
With assumption 4, we have
\dot{ E }_{ f 3}-\dot{ E }_{ f 2}=\dot{m}_{ g }\left[h_{3}-h_{2}-T_{0}\left(s_{3}^{\circ}-s_{2}^{\circ}-R \ln \frac{p_{3}}{p_{1}}^{\nearrow0}\right)\right]
=\left(100.87 \frac{ kJ }{ s }\right)\left[(1515.42-669.79) \frac{ k J }{ kg }\right.
\left.-300 K (3.3620-2.5088) \frac{ kJ }{ kg \cdot K }\right]
=59,480 \frac{ kJ }{ s }\left|\frac{1 MW }{10^{3} kJ / s }\right|=59.48 MW
The net rate exergy is carried out of the plant by the exhaust air stream at 5 is
\dot{ E }_{ f 5}-\dot{ E }_{ f 1}=\dot{m}_{ g }\left[h_{5}-h_{1}-T_{0}\left(s_{5}^{\circ}-s_{1}^{\circ}-R \ln \frac{p_{5}}{p_{1}}^{\nearrow0}\right)\right]
=\left(100.87 \frac{ kg }{ s }\right)[(400.98-300.19)
-300(1.9919-1.7020)]\left(\frac{ kJ }{ kg }\right)\left|\frac{1 MW }{10^{3} kJ / s }\right|
= 1.39 MW
The net rate exergy is carried out of the plant as water passes through the condenser is
\dot{ E }_{ f 8}-\dot{ E }_{ f 9}=\dot{m}_{ v }\left[h_{8}-h_{9}-T_{0}\left(s_{8}-s_{9}\right)\right]
=\left(15.6 \frac{ kg }{ s }\right)\left[(2104.74-173.88) \frac{ kJ }{ kg }\right.
\left.-300 K (6.7282-0.5926) \frac{ kJ }{ kg \cdot K }\right]\left|\frac{1 MW }{10^{3} kJ / s }\right|
= 1.41 MW
The rates of exergy destruction for the air turbine, compressor, steam turbine, pump, and heat-recovery steam generator are evaluated using \dot{ E }_{ d }=T_{0} \dot{\sigma}_{ cv }, respectively, as follows:
Air turbine:
2 \dot{ E }_{ d }=\dot{m}_{ g } T_{0}\left(s_{4}-s_{3}\right)
=\dot{m}_{ g } T_{0}\left(s_{4}^{\circ}-s_{3}^{\circ}-R \ln p_{4} / p_{3}\right)
=\left(100.87 \frac{ kg }{ s }\right)(300 K )\left[(2.7620-3.3620) \frac{ kJ }{ kg \cdot K }\right.
\left.-\left(\frac{8.314}{28.97} \frac{ k J }{ kg \cdot K }\right) \ln \left(\frac{100}{1200}\right)\right]\left|\frac{1 MW }{10^{3} kJ / s }\right|
= 3.42 MW
Compressor:
\dot{ E }_{ d }=\dot{m}_{ g } T_{0}\left(s_{2}-s_{1}\right)
=\dot{m}_{ g } T_{0}\left(s_{2}^{\circ}-s_{1}^{\circ}-R \ln p_{2} / p_{1}\right)
=(100.87)(300)\left[(2.5088-1.7020)-\frac{8.314}{28.97} \ln \left(\frac{1200}{100}\right)\right]\left|\frac{1}{10^{3}}\right|
= 2.83 MW
Steam turbine:
\dot{ E }_{ d }=\dot{m}_{ v } T_{0}\left(s_{8}-s_{7}\right)
=(15.6)(300)(6.7282-6.3634)\left|\frac{1}{10^{3}}\right|
= 1.71 MW
Pump:
\dot{ E }_{ d }=\dot{m}_{ v } T_{0}\left(s_{6}-s_{9}\right)
=(15.6)(300)(0.5975-0.5926)\left|\frac{1}{10^{3}}\right|
= 0.02 MW
Heat-recovery steam generator:
\dot{ E }_{ d }=T_{0}\left[\dot{m}_{ g }\left(s_{5}-s_{4}\right)+\dot{m}_{ v }\left(s_{7}-s_{6}\right)\right]
=(300 K )\left[\left(100.87 \frac{ kg }{ s }\right)(1.9919-2.7620) \frac{ kJ }{ kg \cdot K }\right.
\left.+\left(15.6 \frac{ kg }{ s }\right)(6.3634-0.5975) \frac{ kJ }{ kg \cdot K }\right]\left|\frac{1 MW }{10^{3} kJ / s }\right|
= 3.68 MW
3 The results are summarized by the following exergy rate balance sheet in terms of exergy magnitudes on a rate basis:
| Net exergy increase of the gas passing through the combustor: |
|
|
|
|
|
|
| 59.48 MW |
100% |
(70%)* |
| Disposition of the exergy: |
|
|
|
| • Net power developed |
|
|
|
| gas turbine cycle |
29.03 MW |
48.80% |
(34.2 \%) |
| vapor cycle |
15.97 MW |
26.80% |
(18.8\%) |
| Subtotal |
45.00 MW |
75.60% |
(53.0\%) |
| • Net exergy lost |
|
|
|
| with exhaust gas at state 5 |
1.39 MW |
2.30% |
(1.6\%) |
| from water passing |
1.41 MW |
2.40% |
(1.7\%) |
| through condenser |
|
|
|
| • Exergy destruction |
|
|
|
| air turbine |
3.42 MW |
5.70% |
(4.0\%) |
| compressor |
2.83 MW |
4.80% |
(3.4\%) |
| steam turbine |
1.71 MW |
2.90% |
(2.0\%) |
| pump |
0.02 MW |
– |
– |
| heat-recovery steam |
3.68 MW |
6.20% |
(4.3\%) |
| generator |
|
| *Estimation based on fuel exergy. For discussion, see note 3. |
The subtotals given in the table under the net power developed heading indicate that the combined cycle is effective in generating power from the exergy supplied. The table also indicates the relative significance of the exergy destructions in the turbines, compressor, pump, and heat-recovery steam generator, as well as the relative significance of the exergy losses. Finally, the table indicates that the total of the exergy destructions overshadows the losses. While the energy analysis of part (a) yields valuable results about combined-cycle performance, the exergy analysis of part (b) provides insights about the effects of irreversibilities and true magnitudes of the losses that cannot be obtained using just energy.
1 For comparison, note that the combined-cycle thermal efficiency in this case is much greater than those of the standalone regenerative vapor and gas cycles considered in Examples 8.5 and 9.11, respectively.
2 The development of the appropriate expressions for the rates of entropy production in the turbines, compressor, pump, and heat-recovery steam generator is left as an exercise.
3 In this exergy balance sheet, the percentages shown in parentheses are estimates based on the fuel exergy. Although combustion is the most significant source of irreversibility, the exergy destruction due to combustion cannot be evaluated using an air-standard analysis. Calculations of exergy destruction due to combustion (Chap. 13) reveal that approximately 30% of the exergy entering the combustor with the fuel would be destroyed, leaving about 70% of the fuel exergy for subsequent use. Accordingly, the value 59.48 MW for the net exergy increase of the air passing through the combustor is assumed to be 70% of the fuel exergy supplied. All other percentages in parentheses are obtained by multiplying the corresponding percentages, based on the exergy increase of the air passing through the combustor, by the factor 0.7. Since they account for combustion irreversibility, the table values in parentheses give the more accurate picture of combined cycle performance.
Skills Developed
Ability to…
• apply mass and energy balances.
• determine thermal efficiency.
• evaluate exergy quantities.
• develop an exergy accounting.
Quick Quiz
Determine the net rate energy is carried out of the plant by heat transfer as water passes through the condenser, in MW, and comment. Ans. 30.12 MW. The significance of this energy loss is far less than indicated by the answer. In terms of exergy, the loss at the condenser is 1.41 MW [see part (b)], which better measures the limited utility of the relatively low-temperature water flowing through the condenser.
