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3 years ago

The annual average of solar photovoltaic energy in Phoenix is 6,720

Engineering

1 answer:

Dovator [93]3 years ago

3 0

Answer:

The amount of money saved is $22,075.2

Explanation:

The solar voltaic energy in Phoenix, S $_E$ = 6,720 Wh/m²/day

The area on the roof available for solar panels, A = 500 m²

The percentage of the energy received by the solar panels converted into electricity, η = 15% = 0.15

The cost of electricity, C = $0.12/kWh

The number of days in a year, t = 365 days

Therefore, the maximum area covered by the solar panel = A = 500 m²

The total solar photovoltaic energy received by the solar panels, 'E', in a year is given as follows;

E = S $_E$ × A × t

By plugging in the values of the variables, we get;

E = 6,720 Wh/m²/day × 500 m² × 365 days = 1,226,400,000 Wh

The electrical energy generated, $E_e$ = η × E

∴ $E_e$ = 0.15 × 1,226,400,000 Wh = 183960000 Wh = 183,960 kWh

The amount of money saved = The cost of electricity generated = C × $E_e$

C × $E_e$ = $0.12/kWh × 183,960 kWh = $22,075.2

∴ The amount of money saved = C × $E_e$ = $22,075.2

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Answer:

Explanation:

(a)

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(b)

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(c)

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4 years ago

The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator at a locati

Lynna [10]

Answer:

a) 0.76

b) 0.80

c) 1964 kW

Explanation:

GIVEN DATA:

$\dot m = 5000 kg/s$

Assume Mechanical energy at exist is negligible

A) Take lake bottom as reference, and then kinetic and potential energy are taken as zero.

change in mechanical energy is givrn as

$e_{in} - e_{out} = \frac{P}{\rho} - 0 = gh = 9.81 \times 50( \frac{1 kJ/kg}{1000 m^2/s^2}$

= 0.491 kJ/kg

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4 years ago

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Nitella [24]

Answer:

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3 years ago

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dybincka [34]

Answer:

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