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jolleyrancher78
16.06.2021 •
Engineering
What challenges face the beginning family as they prepare for the parenting process?
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Ответ:
Ответ:
The overall Utility delivered to customers in a year 'U' = 6.25 X 10¹⁰Kwh
However, the average power P, required for a year, t = ? Kw
Expressing their relationship, we will have
U = P x t
Given t = 1 year = 24 x 365 hours (assume a year operation is 365 days)
t = 8760 hours
P =![\frac{62500000000}{8760}](/tpl/images/0367/4960/66a59.png)
P = 7134.7Kw
Hence, the average power required during the year is 7,135Kw
Now to calculate the energy used by the power plant in a year (in quads)?
Recall, Efficiency, η = Power Output/Power Input (100)
so, we have
η = P₀/P₁, given
0.45 =![\frac{7134.7Kw}{P₁}](/tpl/images/0367/4960/c1508.png)
P₁ = 15,855Kw
the total energy E₁ used in a year = 15,855x24x365 = 138.89MJoules
So to convert this to quads, Note;
1 quads of energy = 10¹⁵ Joules
The total energy used is 0.000000139 quads
Now to find the cubic feet of natural gas required to generate this power?
Note: 0.29Kwh of Power generated = 1 cubic feet of natural gas used
Since, the power plant generated = 62500000000Kwh
The cubic feet of natural gas used =![\frac{62500000000}{0.29}](/tpl/images/0367/4960/8cd73.png)
Hence, 2.155x10²⁰cubic feet of N.gas was used to generate this much power.