Known A heat exchanger at steady state has a water stream entering and exiting at known states and a separate gas stream entering and exiting at known states.
Find Determine the net rate at which exergy is carried into the heat exchanger by the gas stream, in MW, the net rate at which exergy is carried from the heat exchanger by the water stream, in MW, the rate of exergy destruction, in MW, and the exergetic efficiency.
Schematic and Given Data:
Engineering Model
1. The control volume shown in 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. The gaseous combustion products are modeled as air as an ideal gas.
4. The air and the water each pass through the steam generator at constant pressure.
5. Only 69% of the exergy entering the plant with the fuel remains after accounting for the stack loss and combustion exergy destruction.
6. T_{0}=22^{\circ} C , p_{0}=1 atm.
Analysis The analysis begins by evaluating the mass flow rate of the air in terms of the mass flow rate of the water. The air and water pass through the boiler in separate streams. Hence, at steady state the conservation of mass principle requires
\begin{array}{ll}\dot{m}_{ i }=\dot{m}_{ e } & (\text { air }) \\\dot{m}_{4}=\dot{m}_{1} & \text { (water) }\end{array}
Using these relations, an energy rate balance for the overall control volume reduces at steady state to
0=\dot{ Q }_{ cv }^{\nearrow0}-\dot{ W }_{ cv }^{\nearrow0}+\dot{m}_{ a }\left(h_{ i }-h_{ e }\right)+\dot{m}\left(h_{4}-h_{1}\right)
where \dot{Q}_{ cv }=\dot{W}_{ cv }=0 by assumption 1, and the kinetic and potential energy terms are dropped by assumption 2. In this equation \dot{m}_{ a } \text { and } \dot{m} denote, respectively, the mass flow rates of the air and water. On solving
\frac{\dot{m}_{ a }}{\dot{m}}=\frac{h_{1}-h_{4}}{h_{ i }-h_{ e }}
The solution to Example 8.2 gives h_{1}=2758 kJ / kg \text { and } h_{4}=183.36 kJ / kg. From Table A-22, h_{ i }=1491.44 kJ / kg \text { and } h_{ e }=843.98 kJ/kg. Hence,
\frac{\dot{m}_{ a }}{\dot{m}}=\frac{2758-183.36}{1491.44-843.98}=3.977 \frac{ kg \text { (air) }}{ kg (\text { steam })}
From Example 8.2, \dot{m}=4.449 \times 10^{5} kg / h \text { . Thus, } \quad \dot{m}_{ a }=17.694 \times 10^{5} kg / h
a. The net rate at which exergy is carried into the heat exchanger unit by the gaseous stream can be evaluated 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 in by the } \\\text { gaseous stream }\end{array}\right]=\dot{m}_{ a }\left( e _{ fi }- e _{ fe }\right)
=\dot{m}_{ a }\left[h_{ i }-h_{ e }-T_{0}\left(s_{ i }-s_{ e }\right)\right]
Since the gas pressure remains constant, Eq. 6.20a giving the change in specific entropy of an ideal gas reduces to s_{ i }-s_{ e }=s_{ i }^{\circ}-s_{ e }^{\circ}. Thus, with h and s° values from Table A-22
s\left(T_{2}, p_{2}\right)-s\left(T_{1}, p_{1}\right)=s^{\circ}\left(T_{2}\right)-s^{\circ}\left(T_{1}\right)-R \ln \frac{p_{2}}{p_{1}} (6.20a)
\begin{aligned}\dot{m}_{ a }\left( e _{ fi }- e _{ fe }\right)=&\left(17.694 \times 10^{5} kg / h \right)[(1491.44-843.98) kJ / kg \\&-(295 K )(3.34474-2.74504) kJ / kg \cdot K ]\end{aligned}
=\frac{8.326 \times 10^{8} kJ / h }{|3600 s / h |}\left|\frac{1 MW }{10^{3} kJ / s }\right|=231.28 MW
b. The net rate at which exergy is carried out of the boiler by the water stream is determined similarly:
\left[\begin{array}{c}\text { net rate at which exergy } \\\text { is carried out by the } \\\text { water stream }\end{array}\right]=\dot{m}\left( e _{ f 1}- e _{ f 4}\right)
=\dot{m}\left[h_{1}-h_{4}-T_{0}\left(S_{1}-S_{4}\right)\right]
From Table A-3, s_{1}=5.7432 kJ / kg \cdot K. Double interpolation in Table A-5 at 8.0 MPa and h_{4}=183.36 kJ / kg \text { gives } s_{4}=0.5957 kJ/kg ⋅ K. Substituting known values
\begin{aligned}\dot{m}\left( e _{ f 1}- e _{ f 4}\right)=&\left(4.449 \times 10^{5}\right)[(2758-183.36)\\&-295(5.7432-0.5957)]\end{aligned}
1 =\frac{4.699 \times 10^{8} kJ / h }{|3600 s / h |}\left|\frac{1 MW }{10^{3} kJ / s }\right|=130.53 MW
c. The rate of exergy destruction can be evaluated by reducing the exergy rate balance to obtain
2 \dot{ E }_{ d }=\dot{m}_{ a }\left( e _{ fi }- e _{ fe }\right)+\dot{m}\left( e _{ f 4}- e _{ f 1}\right)
With the results of parts (a) and (b)
3 \dot{ E }_{ d }=231.28 MW -130.53 MW =100.75 MW
d. The exergetic efficiency given by Eq. 7.27 is
\varepsilon=\frac{\dot{m}_{ c }\left( e _{ f 4}- e _{ f 3}\right)}{\dot{m}_{ h }\left( e _{ f 1}- e _{ f 2}\right)} (7.27)
\varepsilon=\frac{\dot{m}\left( e _{ f 1}- e _{ f 4}\right)}{\dot{m}_{ a }\left( e _{ fi }- e _{ fe }\right)}=\frac{130.53 MW }{231.28 MW }=0.564(56.4 \%)
This calculation indicates that 43.6% of the exergy supplied to the heat exchanger unit by the cooling combustion products is destroyed. However, since only 69% of the exergy entering the plant with the fuel is assumed to remain after the stack loss and combustion exergy destruction are accounted for (assumption 5), it can be concluded that 0.69 × 43.6% = 30% of the exergy entering the plant with the fuel is destroyed within the heat exchanger. 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. |
1 Since energy is conserved, the rate at which energy is transferred to the water as it flows through the heat exchanger equals the rate at which energy is transferred from the gas passing through the heat exchanger. By contrast, the rate at which exergy is transferred to the water is less than the rate at which exergy is transferred from the gas by the rate at which exergy is destroyed within the heat exchanger.
2 The rate of exergy destruction can be determined alternatively by evaluating the rate of entropy production, \dot{\sigma}_{ cv }, from an entropy rate balance and multiplying by T_{0} \text { to obtain } \dot{ E }_{ d }=T_{0} \dot{\sigma}_{ cv }.
3 Underlying the assumption that each stream passes through the heat exchanger at constant pressure is the neglect of friction as an irreversibility. Thus, the only contributor to exergy destruction in this case is heat transfer from the higher-temperature combustion products to the vaporizing water.
Skills Developed
Ability to…
• perform exergy analysis of a power plant steam generator.
Quick Quiz
If the gaseous products of combustion are cooled to 517°C (h _{ e }=810.99 kJ/kg), what is the mass flow rate of the gaseous products, in kg/h? Ans. 16.83 \times 10^{5} kg/h.
