Question

Consider a 1.9 MW power plant (this is the useful output in the form of electric energy) that operates between 30∘C and 450∘C at 65% of the Carnot efficiency. This is enough electric energy for about 750 homes. One way to use energy more efficiently would be to use the 30∘C "waste" energy to heat the homes rather than releasing that heat energy into the environment. This is called cogeneration, and it is used in some parts of Europe but rarely in the United States. The average home uses 70 GJ of energy per year for heating. For estimating purposes, assume that all the power plant's exhaust energy can be transported to homes without loss and that home heating takes place at a steady rate for half a year each year. How many homes could be heated by the power plant?
I don't understand how to solve above question. Hence, please explain it in detail

Answer 1

Up to 230 homes could be heated by the performance of the power plant.

How to calculate the amount of houses benefited by an ideal co-generation plant

By definitions of energy efficiency ($\eta$), no unit, and principle of energy conservation, the amount of waste heat rate ($\dot Q_{waste}$), in megawatts, is equal to the difference between the energy input and the useful output, that is to say:

$\dot Q_{waste} = \left(\frac{1}{\eta} -1\right)\cdot \dot E_{out}$ (1)

Where $\dot E_{out}$ is the useful output, in watts.

According to the statement, an average home uses 70 GJ of energy per year for heating and home heating takes place at a steady rate for half a year, then the number of homes ($n$), no unit, benefited by the co-generation plant is calculated by this formula derived from physical definition of power:

$n = \frac{\dot Q_{waste}\cdot \Delta t}{Q_{home}}$ (2)

Where:

• $\Delta t$ - Availability time, in seconds.
• $Q_{home}$ - Yearly heating consumption, in joules.

Now we proceed to find the number of houses: ($\dot E_{out} = 1.9\times 10^{6}\,W$, $\eta = 0.65$, $\Delta t = 1.577\times 10^{7}\,s$, $Q_{home} = 7\times 10^{10}\,J$)

By (1):

$\dot Q_{waste} = \left(\frac{1}{0.65}-1 \right)\cdot (1.9\times 10^{6}\,W)$

$\dot Q_{waste} = 1.023\times 10^{6}\,W$

By (2):

$n = \frac{(1.577\times 10^{7}\,s)\cdot (1.023\times 10^{6}\,W)}{7\times 10^{10}\,J}$

$n = 230.467$

Up to 230 homes could be heated by the performance of the power plant. $\blacksquare$

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