Known A condenser at steady state has two streams: (1) a twophase liquid–vapor mixture entering and condensate exiting at known states and (2) a separate cooling water stream entering and exiting at known temperatures.
Find Determine the net rate at which exergy is carried from the condenser by the cooling water stream and the rate of exergy destruction for the condenser. Express both quantities in MW and as percentages of the exergy entering the plant with the fuel.
Schematic and Given Data:
Engineering Model
1. The control volume shown on the accompanying figure operates at steady state with \dot{Q}_{ cv }=\dot{W}_{ cv }=0.
2. Kinetic and potential energy effects can be ignored.
3. Only 69% of the fuel exergy remains after accounting for the stack loss and combustion exergy destruction.
4. T_{0}=22^{\circ} C \text { and } p_{0}=1 atm.
Analysis
a. The net rate at which exergy is carried out of the condenser can be evaluated by using 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)
\left[\begin{array}{c}\text { net rate at which exergy } \\\text { is carried out by the } \\\text { cooling water }\end{array}\right]=\dot{m}_{ cw }\left( e _{ fe }- e _{ fi }\right)
=\dot{m}_{ cw }\left[h_{ e }-h_{ i }-T_{0}\left(S_{ e }-S_{ i }\right)\right]
where \dot{m}_{ cw } is the mass flow rate of the cooling water from the solution to Example 8.2. With saturated liquid values for specific enthalpy and entropy from Table A-2 at the specified inlet and exit temperatures of the cooling water
\begin{aligned}\dot{m}_{ cw }\left( e _{ fe }- e _{ fi }\right)=&\left(9.39 \times 10^{6} kg / h \right)[(146.68-62.99) kJ / kg \\&-(295 K )(0.5053-0.2245) kJ / kg \cdot K ]\end{aligned}
=\frac{8.019 \times 10^{6} kJ / h }{|3600 s / h |}\left|\frac{1 MW }{10^{3} kJ / s }\right|=2.23 MW
Expressing this as a percentage of the exergy entering the plant with the fuel, we get (2.23/231.28)(69%) = 1%. This is the value listed in Table 8.4.
| TABLE 8.4 Vapor Power Plant Exergy Accounting^{a} |
| Outputs |
|
| Net power out^{b} |
30% |
| Losses |
|
| Condenser cooling water^{c} |
1% |
| Stack gases (assumed) |
1% |
| Exergy destruction |
|
| Boiler |
|
| Combustion unit (assumed) |
30% |
| Heat exchanger unit^{d} |
30% |
| Turbine^{e} |
5% |
| Pump^{f} |
– |
| Condenser^{g} |
3% |
| Total |
100% |
| ^{a} All values are expressed as a percentage of the exergy carried into the plant with the fuel. Values are rounded to the nearest full percent. Exergy losses associated with stray heat transfer from plant components are ignored. |
^{b}Example 8.8. |
| ^{c}Example 8.9. |
| ^{d}Example 8.7. |
| ^{e}Example 8.8. |
| ^{f}Example 8.8. |
| ^{g}Example 8.9. |
b. The rate of exergy destruction for the condenser can be evaluated by reducing the exergy rate balance. Alternatively, the relationship \dot{ E }_{d}=T_{0} \dot{\sigma}_{ cv } can be employed, where \dot{\sigma}_{ cv } is the time rate of entropy production for the condenser determined from an entropy rate balance. With either approach, the rate of exergy destruction can be expressed as
\dot{ E }_{ d }=T_{0}\left[\dot{m}\left(s_{3}-s_{2}\right)+\dot{m}_{ cw }\left(s_{ e }-s_{ i }\right)\right]
Substituting values
\begin{aligned}\dot{ E }_{ d }=& 295\left[\left(4.449 \times 10^{5}\right)(0.5926-6.2021)\right.\\&\left.+\left(9.39 \times 10^{6}\right)(0.5053-0.2245)\right]\end{aligned}
=\frac{416.1 \times 10^{5} kJ / h }{|3600 s / h |}\left|\frac{1 MW }{10^{3} kJ / s }\right|=11.56 MW
Expressing this as a percentage of the exergy entering the plant with the fuel, we get (11.56/231.28)(69%) = 3%. This is the value listed in Table 8.4.
Skills Developed
Ability to…
• perform exergy analysis of a power plant condenser.
Quick Quiz
Referring to data from Example 8.2, what percent of the energy supplied to the steam passing through the steam generator is carried out by the cooling water? Ans. 68.6%.
