Question 8.7: CONTINUATION OF EXAMPLE 5.3, WITH THE ADDITION SHOWN IN ITAL...
A CONTINUATION OF EXAMPLE 5.3, WITH THE ADDITION SHOWN IN ITALIC TYPE
A basic vapor cycle power plant consists of the following four parts:
a. The boiler, where high-pressure vapor is produced.
b. The turbine, where energy is removed from the high-pressure vapor as shaft work.
c. The condenser, where the low-pressure vapor leaving the turbine is condensed into a liquid.
d. The boiler feed pump, which pumps the condensed liquid back into the high-pressure boiler for reheating.
In such a power plant, the boiler receives 950. × 10^{5 } \text{kJ/h} from the burning fuel, and the condenser rejects 600. × 10^{5 } \text{kJ/h} to the environment. The boiler feed pump requires 23.0 \text{kW} input, which it receives directly from the turbine. Assuming that the turbine, pump, and connecting pipes are all insulated, determine the net power of the turbine and the rate of entropy production of the plant if the boiler temperature is 500.°\text{C } and the condenser temperature is 10.0°\text{C }.
First, draw a sketch of the system (Figure 8.7).
The unknowns here are (\dot{W}_T)_\text{net} and \dot{S}_P for the entire power plant. Therefore, the system is the ntire ower plant. Since we have few specific details on the internal operation of the plant, the ermodynamic properties within the power plant are apparently not needed in the solution.
The net turbine work output is determined in Example 5.3 to be
(\dot{W}_T)_\text{net} = 9720 \text{kW}
The answer to the second part of this problem can be obtained by the indirect method from Eq. (8.2) as
\dot{S}_P = \dot{S} – \frac{d}{dt} \int_{Σ}(\frac{\overline{d} Q}{T_b} )_\text{act}
The surface area of our system can be divided into three major parts: the boiler’s surface area, the condenser’s surface area, and all the remaining surface areas. Therefore, the surface area Σ_\text{system} of the system is composed of the boiler, the condenser, and everything else, or
Σ_\text{system} = Σ_\text{boiler} + Σ_\text{condenser} + Σ_\text{everything else}
Now, both the boiling and the condensing processes are isothermal phase changes; and since no heat transfer occurs at any other point in the system, we can write
\frac{d}{dt} \int_{∑}\frac{\overline{d} Q}{T_b} = \frac{d}{dt}(\int_{Σ_\text{boiler}}\frac{\overline{d} Q}{T_b}+\int_{Σ_\text{condenser}}\frac{\overline{d} Q}{T_b}+\int_{Σ_\text{remainder}}\frac{\overline{d} Q}{T_b})
where
\frac{d}{dt}\int_{Σ_\text{boiler}}\frac{\overline{d} Q}{T_b} = \frac{\dot{Q}_\text{boiler}}{T_\text{boiler}}
\frac{d}{dt}\int_{Σ_\text{condenser}}\frac{\overline{d} Q}{T_b} = \frac{\dot{Q}_\text{condenser}}{T_\text{condenser}}
and
\frac{d}{dt}\int_{Σ_\text{condenser}}\frac{\overline{d} Q}{T_b} = 0 \text{no heat transfer across the remaining surface area}
Then, we have
\dot{S} _P = \dot{S} – \frac{\dot{Q}_\text{boiler}}{T_\text{boiler}} – \frac{\dot{Q}_\text{condenser}}{T_\text{condenser}}
and, for steady state operation (\dot{S}= 0), this reduces to
\dot{S} _P = -(\frac{\dot{Q}_\text{boiler}}{T_\text{boiler}} + \frac{\dot{Q}_\text{condenser}}{T_\text{condenser}})
= -(\frac{950. × 10^{5 }}{500.+273.15} + \frac{−600. × 10^{5 }}{10.0 + 273.15} )(\frac{\text{kJ/h}}{\text{K}} ) = 89.0 × 10^{3} \text{kJ/(h.K)}
Note that the actual thermal efficiency of this power plant is given by Eq. (7.9) as
(η_T)_\text{act} = 1− \frac{\left|\dot{Q}_\text{act}\right| }{\dot{Q}_\text{in}} = 1-\frac{600. × 10^{5 }}{950. × 10^{5 }} = 0.368 = 36.8\%
whereas its theoretical reversible (Carnot) efficiency is given by Eq. (7.16) as
(η_T)_\text{rev} = 1−\frac{T_\text{condenser}}{T_\text{boiler}} = 1 – \frac{10.0 + 273.15}{500. + 273.15} = 0.634 = 63.4\%
Therefore, the actual efficiency is less than the theoretical maximum (reversible) efficiency, as it should be.
