Known An ideal Rankine cycle operates with steam as the working fluid. The boiler and condenser pressures are specified, and the net power output is given.
Find Determine the thermal efficiency, the back work ratio, the mass flow rate of the steam, in kg/h, the rate of heat transfer to the working fluid as it passes through the boiler, in MW, the rate of heat transfer from the condensing steam as it passes through the condenser, in MW, the mass flow rate of the condenser cooling water, which enters at 15°C and exits at 35°C.
Schematic and Given Data:
Engineering Model
1. Each component of the cycle is analyzed as a control volume at steady state. The control volumes are shown on the accompanying sketch by dashed lines.
2. All processes of the working fluid are internally reversible.
3. The turbine and pump operate adiabatically.
4. Kinetic and potential energy effects are negligible.
5. Saturated vapor enters the turbine. Condensate exits the condenser as saturated liquid.
1 Analysis Table A-3To begin the analysis, we fix each of the principal states located on the accompanying schematic and T–s diagrams. Starting at the inlet to the turbine, the pressure is 8.0 MPa and the steam is a saturated vapor, so from , h_{1}=2758.0 kJ/kg and s_{1}=5.7432 kJ / kg \cdot K.
State 2 is fixed by p_{2}=0.008 MPa and the fact that the specific entropy is constant for the adiabatic, internally reversible expansion through the turbine. Using saturated liquid and saturated vapor data from Table A-3, we find that the quality at state 2 is
x_{2}=\frac{s_{2}-s_{ f }}{s_{ g }-s_{ f }}=\frac{5.7432-0.5926}{7.6361}=0.6745
The enthalpy is then
h_{2}=h_{ f }+x_{2} h_{ fg }=173.88+(0.6745) 2403.1
= 1794.8 kJ/kg
State 3 is saturated liquid at 0.008 MPa, so h_{3}=173.88 kJ/kg.
State 4 is fixed by the boiler pressure p_{4} and the specific entropy S_{4}=S_{3}. The specific enthalpy h_{4} can be found by interpolation in the compressed liquid tables. However, because compressed liquid data are relatively sparse, it is more convenient to solve Eq. 8.3 for h_{4}, using Eq. 8.7b to approximate the pump work. With this approach
\frac{\dot{W}_{ p }}{\dot{m}}=h_{4}-h_{3} (8.3)
\left(\frac{\dot{W}_{ p }}{\dot{m}}\right)_{ s } \approx v_{3}\left(p_{4}-p_{3}\right) (8.7b)
h_{4}=h_{3}+\dot{W}_{ p } / \dot{m}=h_{3}+v_{3}\left(p_{4}-p_{3}\right)
By inserting property values from Table A-3
h_{4}=173.88 kJ / kg +\left(1.0084 \times 10^{-3} m ^{3} / kg \right)
\times(8.0-0.008) MPa \left|\frac{10^{6} N / m ^{2}}{1 MPa }\right|\left|\frac{1 kJ }{10^{3} N \cdot m }\right|
=173.88+8.06=181.94 kJ / kg
a. The net power developed by the cycle is
\dot{W}_{\text {cycle }}=\dot{W}_{ t }-\dot{W}_{ p }
Mass and energy rate balances for control volumes around the turbine and pump give, respectively,
\frac{\dot{W}_{ t }}{\dot{m}}=h_{1}-h_{2} \quad \text { and } \quad \frac{\dot{W}_{ p }}{\dot{m}}=h_{4}-h_{3}
where \dot{m} is the mass flow rate of the steam. The rate of heat transfer to the working fluid as it passes through the boiler is determined using mass and energy rate balances as
\frac{\dot{Q}_{ in }}{\dot{m}}=h_{1}-h_{4}
The thermal efficiency is then
\eta=\frac{\dot{W}_{ t }-\dot{W}_{ p }}{\dot{Q}_{\text {in }}}=\frac{\left(h_{1}-h_{2}\right)-\left(h_{4}-h_{3}\right)}{h_{1}-h_{4}}
=\frac{[(2758.0-1794.8)-(181.94-173.88)] kJ / kg }{(2758.0-181.94) kJ / kg }
= 0.371(37.1%)
b. The back work ratio is
2 bwr =\frac{\dot{W}_{ p }}{\dot{W}_{ t }}=\frac{h_{4}-h_{3}}{h_{1}-h_{2}}=\frac{(181.94-173.88) kJ / kg }{(2758.0-1794.8) kJ / kg }
=\frac{8.06}{963.2}=8.37 \times 10^{-3}(0.84 \%)
c. The mass flow rate of the steam can be obtained from the expression for the net power given in part (a). Thus,
\dot{m}=\frac{\dot{W}_{\text {cycle }}}{\left(h_{1}-h_{2}\right)-\left(h_{4}-h_{3}\right)}
=\frac{(100 MW )\left|10^{3} kW / MW \| 3600 s / h \right|}{(963.2-8.06) kJ / kg }
=3.77 \times 10^{5} kg / h
d. With the expression for \dot{Q}_{\text {in }} from part (a) and previously determined specific enthalpy values
\dot{Q}_{\text {in }}=\dot{m}\left(h_{1}-h_{4}\right)
=\frac{\left(3.77 \times 10^{5} kg / h \right)(2758.0-181.94) kJ / kg }{\left|3600 s / h \| 10^{3} kW / MW \right|}
= 269.77 MW
e. Mass and energy rate balances applied to a control volume enclosing the steam side of the condenser give
\dot{Q}_{\text {out }}=\dot{m}\left(h_{2}-h_{3}\right)
=\frac{\left(3.77 \times 10^{5} kg / h \right)(1794.8-173.88) kJ / kg }{\left|3600 s / h \| 10^{3} kW / MW \right|}
= 169.75 MW
3 Note that the ratio of \dot{Q}_{\text {out }} \text { to } \dot{Q}_{\text {in }} is 0.629 (62.9%).
Alternatively, \dot{Q}_{\text {out }} can be determined from an energy rate balance on the overall vapor power plant. At steady state, the net power developed equals the net rate of heat transfer to the plant
\dot{W}_{\text {cycle }}=\dot{Q}_{\text {in }}-\dot{Q}_{\text {out }}
Rearranging this expression and inserting values
\dot{Q}_{\text {out }}=\dot{Q}_{\text {in }}-\dot{W}_{\text {cycle }}=269.77 MW -100 MW =169.77 MW
The slight difference from the above value is due to round-off.
f. Taking a control volume around the condenser, the mass and energy rate balances give at steady state
0=\dot{Q}_{ cv }^{\nearrow0}-\dot{W}_{ cv }^{\nearrow0}+\dot{m}_{ cw }\left(h_{ cw , \text { in }}-h_{ cw , out }\right)+\dot{m}\left(h_{2}-h_{3}\right)
where \dot{m}_{ cw } is the mass flow rate of the cooling water. Solving for \dot{m}_{ cw }
\dot{m}_{ cw }=\frac{\dot{m}\left(h_{2}-h_{3}\right)}{\left(h_{ cw , \text { out }}-h_{ cw , in }\right)}
The numerator in this expression is evaluated in part (e). For the cooling water, h \approx h_{ f }(T), so with saturated liquid enthalpy values from Table A-2 at the entering and exiting temperatures of the cooling water
\dot{m}_{ cw }=\frac{(169.75 MW )\left|10^{3} kW / MW \| 3600 s / h \right|}{(146.68-62.99) kJ / kg }=7.3 \times 10^{6} kg / h
1 Note that a slightly revised problem-solving methodology is used in this example problem: We begin with a systematic evaluation of the specific enthalpy at each numbered state.
2 Note that the back work ratio is relatively low for the Rankine cycle. In the present case, the work required to operate the pump is less than 1% of the turbine output.
3 In this example, 62.9% of the energy added to the working fluid by heat transfer is subsequently discharged to the cooling water. Although considerable energy is carried away by the cooling water, its exergy is small because the water exits at a temperature only a few degrees greater than that of the surroundings. See Sec. 8.6 for further discussion.
Skills Developed
Ability to…
• sketch the T–s diagram of the basic Rankine cycle.
• fix each of the principal states and retrieve necessary property data.
• apply mass and energy balances.
• calculate performance parameters for the cycle.
Quick Quiz
If the mass flow rate of steam were 150 kg/s, what would be the net power, in MW, and the thermal efficiency? Ans. 143.2 MW, 37.1%.
