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slamgirl [31]
2 years ago
8

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

Engineering
1 answer:
Dovator [93]2 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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A welding rod with κ = 30 (Btu/hr)/(ft ⋅ °F) is 20 cm long and has a diameter of 4 mm. The two ends of the rod are held at 500 °
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Answer:

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

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Thermal Conductivity is SI units:

k=30(Btu/hr)/(ft.F) * \frac{1055.06}{3600*0.3048*0.556} \\k=51.88 W/m.K

Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft

Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft

T_1=500 C=932 F

T_2=50 C= 122 F

Part A:

In Joules (J)

A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J

In Btu:

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Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu

PArt B:

At midpoint Length=L/2=0.1 m

Q=\frac{k*A*(T_1-T_2)}{L}

On rearranging:

T_2=T_1-\frac{Q*L}{KA}

T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C

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