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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 ([tex]\eta[/tex]), no unit, and principle of energy conservation, the amount of waste heat rate ([tex]\dot Q_{waste}[/tex]), in megawatts, is equal to the difference between the energy input and the useful output, that is to say:
[tex]\dot Q_{waste} = \left(\frac{1}{\eta} -1\right)\cdot \dot E_{out}[/tex] (1)
Where [tex]\dot E_{out}[/tex] 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 ([tex]n[/tex]), no unit, benefited by the co-generation plant is calculated by this formula derived from physical definition of power:
[tex]n = \frac{\dot Q_{waste}\cdot \Delta t}{Q_{home}}[/tex] (2)
Where:
- [tex]\Delta t[/tex] - Availability time, in seconds.
- [tex]Q_{home}[/tex] - Yearly heating consumption, in joules.
Now we proceed to find the number of houses: ([tex]\dot E_{out} = 1.9\times 10^{6}\,W[/tex], [tex]\eta = 0.65[/tex], [tex]\Delta t = 1.577\times 10^{7}\,s[/tex], [tex]Q_{home} = 7\times 10^{10}\,J[/tex])
By (1):
[tex]\dot Q_{waste} = \left(\frac{1}{0.65}-1 \right)\cdot (1.9\times 10^{6}\,W)[/tex]
[tex]\dot Q_{waste} = 1.023\times 10^{6}\,W[/tex]
By (2):
[tex]n = \frac{(1.577\times 10^{7}\,s)\cdot (1.023\times 10^{6}\,W)}{7\times 10^{10}\,J}[/tex]
[tex]n = 230.467[/tex]
Up to 230 homes could be heated by the performance of the power plant. [tex]\blacksquare[/tex]
To learn more on co-generation, we kindly invite to check this verified question: https://brainly.com/question/1344306
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