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mote1985 [20]
3 years ago
14

A photovoltaic panel of dimension 2m×4m is installed on the

Engineering
1 answer:
blsea [12.9K]3 years ago
6 0

Answer:

Explanation:

Simply put, a solar panel works by allowing photons, or particles of light, to knock electrons free from atoms, generating a flow of electricity. Solar panels actually comprise many, smaller units called photovoltaic cells. (Photovoltaic simply means they convert sunlight into electricity. The attached diagram give an ilustsration of the photovotaic pannel mounted on a roof top.

Solution

To Determine the electric power generated for

a) A still summer day.

E = A * r * H * PR

E = Total Amount of Energy in kilowatt

A = Total Surface Area

r = efficiency Rating

H = global radiation value

PR = Performance Ratio

kwh = watt * Time/1000

kwh = 100 * 35/1000

3.5

b)

kwh = watt * Time/1000

kwh = 30 *15/1000

4.5

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2 years ago
Calculate the unit cell edge length for an 80 wt% Ag−20 wt% Pd alloy. All of the palladium is in solid solution, the crystal str
alisha [4.7K]

Answer:

λ^3 = 4.37

Explanation:

first let us to calculate the average density of the alloy

for simplicity of calculation assume a 100g alloy

80g --> Ag

20g --> Pd

ρ_avg= 100/(20/ρ_Pd+80/ρ_avg)

         = 100*10^-3/(20/11.9*10^6+80/10.44*10^6)

         = 10744.62 kg/m^3

now Ag forms FCC and Pd is the impurity in one unit cell there is 4 atoms of Ag since Pd is the impurity we can not how many atom of Pd in one unit cell let us calculate

total no of unit cell in 100g of allow = 80 g/4*107.87*1.66*10^-27

                                                          = 1.12*10^23 unit cells

mass of Pd in 1 unit cell = 20/1.12*10^23

Now,

                      ρ_avg= mass of unit cell/volume of unit cell

                      ρ_avg= (4*107.87*1.66*10^-27+20/1.12*10^23)/λ^3

                          λ^3 = 4.37

6 0
3 years ago
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