Known A reheat–regenerative vapor power cycle operates with steam as the working fluid. Operating pressures and temperatures are specified, and the net power output is given.
Find Determine the thermal efficiency and the mass flow rate entering the first turbine, in kg/h.
Schematic and Given Data:
Engineering Model
1. Each component in the cycle is analyzed as a control volume at steady state. The control volumes are shown on the accompanying
sketch by dashed lines.
2. There is no stray heat transfer from any component to its surroundings.
3. The working fluid undergoes internally reversible processes as it passes through the turbines, pumps, steam generator, reheater, and condenser.
4. The expansion through the trap is a throttling process.
5. Kinetic and potential energy effects are negligible.
6. Condensate exits the closed feedwater heater as a saturated liquid at 2 MPa. Feedwater exits the open feedwater heater as a saturated liquid at 0.3 MPa. Condensate exits the condenser as a saturated liquid.
Analysis Example 8.3Let us determine the specific enthalpies at the principal states of the cycle. State 1 is the same as in , so h_{1}=3348.4 kJ / kg \text { and } s_{1}=6.6586 kJ / kg \cdot K.
State 2 is fixed by p_{2}=2.0 MPa and the specific entropy s_{2}, which is the same as that of state 1. Interpolating in Table A-4, we get h_{2}=2963.5 kJ/kg. The state at the exit of the first turbine is the same as at the exit of the first turbine of Example 8.3, so h_{3}=2741.8 kJ/kg.
State 4 is superheated vapor at 0.7 MPa, 440°C. From Table A-4, h_{4}=3353.3 kJ / kg \text { and } s_{4}=7.7571 kJ / kg \cdot K. Interpolating in Table A-4 at p_{5}=0.3 MPa \text { and } s_{5}=s_{4}=7.7571 kJ / kg \cdot K the enthalpy at state 5 is h_{5}=3101.5 kJ / kg.
Using s_{6}=s_{4}, the quality at state 6 is found to be x_{6}=0.9382. So
\begin{aligned}h_{6} &=h_{ f }+x_{6} h_{ fg } \\&=173.88+(0.9382) 2403.1=2428.5 kJ / kg\end{aligned}
At the condenser exit, h_{7}=173.88 kJ/kg. The specific enthalpy at the exit of the first pump is
\begin{aligned}h_{8} &=h_{7}+v_{7}\left(p_{8}-p_{7}\right) \\&=173.88+(1.0084)(0.3-0.008)=174.17 kJ / kg\end{aligned}
The required unit conversions were considered in previous examples. The liquid leaving the open feedwater heater at state 9 is saturated liquid at 0.3 MPa. The specific enthalpy is h_{9}=561.47.
The specific enthalpy at the exit of the second pump is
\begin{aligned}h_{10} &=h_{9}+v_{9}\left(p_{10}-p_{9}\right) \\&=561.47+(1.0732)(8.0-0.3)=569.73 kJ / kg\end{aligned}
The condensate leaving the closed heater is saturated liquid at 2 MPa. From Table A-3, h_{12}=908.79 kJ/kg. The fluid passing through the trap undergoes a throttling process, so h_{13}=908.79 kJ/kg.
The specific enthalpy of the feedwater exiting the closed heater at 8.0 MPa and 205°C is found using Eq. 3.13 as
h(T, p) \approx h_{ f }(T)+\underline{v_{ f }(T)\left[p-p_{ sat }(T)\right]} (3.13)
\begin{aligned}h_{11} &=h_{ f }+v_{ f }\left(p_{11}-p_{\text {sat }}\right) \\&=875.1+(1.1646)(8.0-1.73)=882.4 kJ / kg\end{aligned}
where h_{ f } \text { and } v_{ f } are the saturated liquid specific enthalpy and specific volume at 205°C, respectively, and p_{\text {sat }} is the saturation pressure in MPa at this temperature. Alternatively, h_{11} can be found from Table A-5.
The schematic diagram of the cycle is labeled with the fractions of the total flow into the turbine that remain at various locations. The fractions of the total flow diverted to the closed heater and open heater, respectively, are y^{\prime}=\dot{m}_{2} / \dot{m}_{1} \text { and } y^{\prime \prime}=\dot{m}_{5} / \dot{m}_{1}, \text { where } \dot{m}_{1} denotes the mass flow rate entering the first turbine.
The fraction y^{\prime} can be determined by application of mass and energy rate balances to a control volume enclosing the closed heater. The result is
y^{\prime}=\frac{h_{11}-h_{10}}{h_{2}-h_{12}}=\frac{882.4-569.73}{2963.5-908.79}=0.1522
The fraction y^{\prime \prime} can be determined by application of mass and energy rate balances to a control volume enclosing the open heater, resulting in
0=y^{\prime \prime} h_{5}+\left(1-y^{\prime}-y^{\prime \prime}\right) h_{8}+y^{\prime} h_{13}-h_{9}
Solving for y^{\prime \prime}
y^{\prime \prime}=\frac{\left(1-y^{\prime}\right) h_{8}+y^{\prime} h_{13}-h_{9}}{h_{8}-h_{5}}
=\frac{(0.8478) 174.17+(0.1522) 908.79-561.47}{174.17-3101.5}
= 0.0941
a. The following work and heat transfer values are expressed on the basis of a unit mass entering the first turbine. The work developed by the first turbine per unit of mass entering is the sum
\begin{aligned}\frac{\dot{W}_{ t 1}}{\dot{m}_{1}} &=\left(h_{1}-h_{2}\right)+\left(1-y^{\prime}\right)\left(h_{2}-h_{3}\right) \\&=(3348.4-2963.5)+(0.8478)(2963.5-2741.8) \\&=572.9 kJ / kg\end{aligned}
Similarly, for the second turbine
\begin{aligned}\frac{\dot{W}_{ t 2}}{\dot{m}_{1}} &=\left(1-y^{\prime}\right)\left(h_{4}-h_{5}\right)+\left(1-y^{\prime}-y^{\prime \prime}\right)\left(h_{5}-h_{6}\right) \\&=(0.8478)(3353.3-3101.5)+(0.7537)(3101.5-2428.5) \\&=720.7 kJ / kg\end{aligned}
For the first pump
\begin{aligned}\frac{\dot{W}_{ p 1}}{\dot{m}_{1}} &=\left(1-y^{\prime}-y^{\prime \prime}\right)\left(h_{8}-h_{7}\right) \\&=(0.7537)(174.17-173.88)=0.22 kJ / kg\end{aligned}
and for the second pump
\begin{aligned}\frac{\dot{W}_{ p 2}}{\dot{m}_{1}} &=\left(h_{10}-h_{9}\right) \\&=569.73-561.47=8.26 kJ / kg\end{aligned}
The total heat added is the sum of the energy added by heat transfer during boiling/superheating and reheating. When expressed on the basis of a unit of mass entering the first turbine, this is
\begin{aligned}\frac{\dot{Q}_{\text {in }}}{\dot{m}_{1}} &=\left(h_{1}-h_{11}\right)+\left(1-y^{\prime}\right)\left(h_{4}-h_{3}\right) \\&=(3348.4-882.4)+(0.8478)(3353.3-2741.8) \\&=2984.4 kJ / kg\end{aligned}
With the foregoing values, the thermal efficiency is
\eta=\frac{\dot{W}_{ t 1} / \dot{m}_{1}+\dot{W}_{ t 2} / \dot{m}_{1}-\dot{W}_{ p1 } / \dot{m}_{1}-\dot{W}_{ p 2} / \dot{m}_{1}}{\dot{Q}_{\text {in }} / \dot{m}_{1}}
=\frac{572.9+720.7-0.22-8.26}{2984.4}=0.431(43.1 \%)
b. The mass flow rate entering the first turbine can be determined using the given value of the net power output. Thus,
\dot{m}_{1}=\frac{\dot{W}_{\text {cycle }}}{\dot{W}_{ t 1} / \dot{m}_{1}+\dot{W}_{ t 2} / \dot{m}_{1}-\dot{W}_{ p 1} / \dot{m}_{1}-\dot{W}_{ p 2} / \dot{m}_{1}}
1 =\frac{(100 MW )|3600 s / h |\left|10^{3} kW / MW \right|}{1285.1 kJ / kg }=2.8 \times 10^{5} kg / h
1 Compared to the corresponding values determined for the simple Rankine cycle of Example 8.1, the thermal efficiency of the present regenerative cycle is substantially greater and the mass flow rate is considerably less.
Skills Developed
Ability to…
• sketch the T–s diagram of the reheat–regenerative vapor power cycle with one closed and one open feedwater heater.
• fix each of the principal states and retrieve necessary property data.
• apply mass, energy, and entropy principles.
• calculate performance parameters for the cycle.
Quick Quiz
If each turbine stage had an isentropic efficiency of 85%, at which numbered states would the specific enthalpy values change? Ans. 2, 3, 5, and 6.
