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larisa [96]
3 years ago
14

The use of nuclear reactors to generate electricity is?

Physics
2 answers:
SVEN [57.7K]3 years ago
6 0
The difference is that nuclear plants do not burn anything. Instead, they use uranium fuel, consisting of solid ceramic pellets, to produce electricity through a process called fission<span>. Nuclear power plants obtain the heat needed to produce steam through a physical process.

Hope this helps :)</span>
Rudik [331]3 years ago
6 0
A clean energy resource. Nuclear energy originates from the splitting of uranium atoms – a process called fission. This generates heat to produce steam, which is used by a turbine generator to generate electricity. Because nuclear power plants do not burn fuel, they do not produce greenhouse gas emissions
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Explanation:

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A single-story retail store wishes to supply all its lighting requirement with batteries charged by photovoltaic cells. The PV c
Umnica [9.8K]

Answer:

83.33% of the roof area will be occupied by the PV cells

Explanation:

Given the data in the question;

time-averaged lighting requirement P_{lighting = 10 W/m²

the annual average solar irradiance q_{solar = 150 W/m²

the PV efficiency η_{pv = 10% = 0.1

battery charging/discharging efficiency η_{battery = 80% = 0.8

we know that; Annual average power to the light = P_{lighting × A_{roof

Now, the electrical power delivered by the solar cell battery system will be;

⇒  q_{solar × A_{pv × η_{pv × η_{battery

P_{lightingA_{roof = q_{solar × A_{pv × η_{pv × η_{battery

Such that;

A_{pv = P_{lightingA_{roof / q_{solar × A_{pv × η_{pv × η_{battery

A_{pv / A_{roof = P_{lighting /  q_{solar × η_{pv × η_{battery

so we substitute

A_{pv / A_{roof = 10 W/m² / [ 150 W/m² × 0.1 × 0.8 ]

A_{pv / A_{roof = 10 W/m² / 12 W/m²

A_{pv / A_{roof = 0.8333

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A_{pv / A_{roof = 83.33%

Therefore, 83.33% of the roof area will be occupied by the PV cells.

7 0
3 years ago
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