Question 8.2: The solar power plant shown in Figure 8.2 utilizes the therm...

First, draw a sketch of the system.

Here, the unknown is (\dot{W}_\text{electrical})_\text{max}, and the system is the entire power plant as shown in Figure 8.2. You must now realize that a system like this produces maximum work output when the internal losses (friction etc.) are a minimum, and the absolute maximum occurs when the system is reversible. Therefore, what we wish to find here is (\dot{W}_\text{electrical})_\text{rev} :

Since no clearly defined system states are given in the problem statement and all the given values are rate values, we recognize that this problem requires a rate analysis. The energy rate balance (ERB) equation for this system is

\dot{Q} _ \text{net} – \dot{W}_\text{net} = \underbrace{\dot{U}}_{\begin{matrix} 0 \\ \text{(steady} \\ \text{state)} \end{matrix} } + \underbrace{\dot{\dot{\text{KE}} + \dot{\text{PE}}}}_{\begin{matrix} 0 \\ \text{(assume the system} \\ \text{is stationary)} \end{matrix} }

or

(\dot{W}_\text{electrical})_\text{max} = \dot{W}_\text{net} = \dot{Q}_\text{net}

= (\dot{Q}_\text{solar} – \left|\dot{Q} \right|_\text{condenser} )

Note that there are two heat transfer surfaces in this system (the solar collector and the condenser), each at a different isothermal temperature, and one work mode (electrical). The entropy rate balance (SRB) equation for this system is

\dot{S} = \underbrace{\dot{m}s}_{\begin{matrix} 0 \\ \text{(steady} \\ \text{state)} \end{matrix} } = \int_{Σ}^{}{} (\frac{\dot{q} }{T_b} )_\text{rev}dA+ \underbrace{\dot{S}_p}_{\begin{matrix} 0 \\ \text{(reversible} \\ \text{system)} \end{matrix} }

The system surface area Σ_\text{system} is the sum of the surface areas of the collector, condenser, and the remaining surfaces where no heat transfer occurs

Σ_\text{system} = Σ_\text{collector } + Σ_\text{condenser} + Σ_\text{no heat transfer sufaces}

Since the heat transfer surfaces are all isothermal, we can set

\int_{∑}^{}{}(\frac{\dot{q} }{T_b} )dA =\int_{Σ_\text{collector }}^{}{} (\frac{\dot{q} }{T_b} )dA + \int_{Σ_\text{condenser}}^{}{}(\frac{\dot{q} }{T_b} )dA = \frac{\dot{Q} _\text{solar}}{T_\text{collector}} – \left|\frac{\dot{Q} _\text{condenser}}{T_\text{river}} \right|

Therefore,

\left|\dot{Q} _\text{condenser}\right| = \dot{Q} _\text{solar}(\frac{T_\text{river}}{T_\text{collector}} )

and the ERB becomes

(\dot{W}_\text{electrical})_\text{rev} = \dot{Q} _\text{solar}(1 -\frac{T_\text{river}}{T_\text{collector}} ) = (100. × 10³  \text{Btu/h})(1  –  \frac{40.0 + 459.67}{200. + 459.67} )

= (24,300  \text{Btu/h})(\frac{1  \text{KW}}{3412  \text{Btu/h}} ) = 7.11  \text{KW}

The positive sign indicates that the electrical power is out of the system.