Known Steady-state operating data are provided for a system consisting of a heat-recovery steam generator and a turbine.
Find Determine the power developed by the turbine and the turbine inlet temperature. Evaluate the annual value of the power developed.
Schematic and Given Data:
Engineering Model
1. The control volume shown on the accompanying figure is at steady state.
2. Heat transfer is negligible, and changes in kinetic and potential energy can be ignored.
3. There is no pressure drop for water flowing through the steam generator.
4. The combustion products are modeled as air as an ideal gas.
Analysis:
a. The power developed by the turbine is determined from a control volume enclosing both the steam generator and the turbine. Since the gas and water streams do not mix, mass rate balances for each of the streams reduce, respectively, to give
\dot{m}_{1}=\dot{m}_{2}, \quad \dot{m}_{3}=\dot{m}_{5}
For this control volume, the appropriate form of the steady-state energy rate balance is Eq. 4.18, which reads
0=\dot{Q}_{ cv }-\dot{W}_{ cv }+\sum_{i} \dot{m}_{i}\left(h_{i}+\frac{ V _{i}^{2}}{2}+g z_{i}\right)-\sum_{e} \dot{m}_{e}\left(h_{e}+\frac{ V _{e}^{2}}{2}+g z_{e}\right) (4.18)
0=\underline{\dot{Q}_{ cv }}-\dot{W}_{ cv }+\dot{m}_{1}\left(h_{1}+\underline{\frac{ V _{1}^{2}}{2}+g z_{1}}\right)+\dot{m}_{3}\left(h_{3}\underline{+\frac{ V _{3}^{2}}{2}+g z_{3}}\right)
-\dot{m}_{2}\left(h_{2}+\underline{\frac{ V _{2}^{2}}{2}+g z_{2}}\right)-\dot{m}_{5}\left(h_{5}+\underline{\frac{ V _{5}^{2}}{2}+g z_{5}}\right)
The underlined terms drop out by assumption 2. With these simplifications, together with the above mass flow rate relations, the energy rate balance becomes
\dot{W}_{ cv }=\dot{m}_{1}\left(h_{1}-h_{2}\right)+\dot{m}_{3}\left(h_{3}-h_{5}\right)
The mass flow rate \dot{m}_{1} can be evaluated with given data at inlet 1 and the ideal gas equation of state
\dot{m}_{1}=\frac{( AV )_{1}}{v_{1}}=\frac{( AV )_{1} p_{1}}{(\bar{R} / M) T_{1}}=\frac{\left(2 \times 10^{5} ft ^{3} / min \right)\left(14.7 lbf / in .^{2}\right)}{\left(\frac{1545 ft \cdot lbf }{28.97 lb \cdot{ }^{\circ} R }\right)\left(860^{\circ} R \right)}\left|\frac{144 in .^{2}}{1 ft ^{2}}\right|
= 9230.6 lb/min
The specific enthalpies h_{1} \text { and } h_{2} can be found from Table A-22E: At 860°R, h_{1}=206.46 Btu/lb, and at 720°R h_{2}=172.39 Btu/lb. At state 3, water is a liquid. Using Eq. 3.14 and saturated liquid data from Table A-2E, h_{3} \approx h_{ f }\left(T_{3}\right)=70 But/lb. State 5 is a two-phase liquid–vapor mixture. With data from Table A-3E and the given quality
h(T, p) \approx h_{ f }(T) (3.14)
h_{5}=h_{ f 5}+x_{5}\left(h_{ g 5}-h_{ f 5}\right)
=69.74+0.93(1036.0)=1033.2 Btu / lb
Substituting values into the expression for \dot{W}_{ cv }
\dot{W}_{ cv }=\left(9230.6 \frac{ lb }{ min }\right)(206.46-172.39) \frac{ Btu }{ lb }
+\left(275 \frac{ lb }{ min }\right)(70-1033.2) \frac{ Btu }{ lb }
=49610 \frac{ Btu }{min }
b. To determine T_{4}, it is necessary to fix the state at 4. This requires two independent property values. With assumption 3, one of these properties is pressure, p_{4}=40 lbf / in .^{2} The other is the specific enthalpy h_{4}, which can be found from an energy rate balance for a control volume enclosing just the steam generator. Mass rate balances for each of the two streams give \dot{m}_{1}=\dot{m}_{2} \text { and } \dot{m}_{3}=\dot{m}_{4} With assumption 2 and these mass flow rate relations, the steady-state form of the energy rate balance reduces to
0=\dot{m}_{1}\left(h_{1}-h_{2}\right)+\dot{m}_{3}\left(h_{3}-h_{4}\right)
Solving for h_{4}
1 h_{4}=h_{3}+\frac{\dot{m}_{1}}{\dot{m}_{3}}\left(h_{1}-h_{2}\right)
=70 \frac{ Btu }{ lb }+\left(\frac{9230.6 lb / min }{275 lb / min }\right)(206.46-172.39) \frac{ Btu }{ lb }
=1213.6 \frac{ Btu }{ lb }
Interpolating in Table A-4E at p_{4}=40 lbf / in .^{2} \text { with } h_{4}, we get T_{4}=354^{\circ} F.
c. Using the result of part (a), together with the given economic data and appropriate conversion factors, the value of the power developed for 8000 hours of operation annually is
annual value =\left(49610 \frac{ Btu }{\min }\left|\frac{60 min }{1 h } \right| \left|\frac{1 kW }{3413 Btu / h }\right|\right)
\times\left(8000 \frac{ h }{\text { year }}\right)\left(0.115 \frac{\$}{ kW \cdot h }\right)
2 =802,000 \frac{\$}{\text { year }}
1 Alternatively, to determine h_{4} a control volume enclosing just the turbine can be considered.
2 The decision about implementing this solution to the problem of utilizing the hot combustion products discharged from an industrial process would necessarily rest on the outcome of a detailed economic evaluation, including the cost of purchasing and operating the steam generator, turbine, and auxiliary equipment.
Skills Developed
Ability to…
• apply the steady-state mass and energy rate balances to a control volume.
• apply the mass flow rate expression, Eq. 4.4b.
\dot{m}=\frac{ AV }{v} (4.4b)
• develop an engineering model.
• retrieve property data for water and for air modeled as an ideal gas.
• conduct an elementary economic evaluation.
Quick Quiz
Taking a control volume enclosing just the turbine, evaluate the turbine inlet temperature, in °F. Ans. 354°F.
