I know this is not the right section but i have posted 3 questions in the physics section but didnt get any answers. So iam expecting few people iin maths section might know the answer.
I have a question on physics,
Q: A solar panel of area 2.5m^2 is fitted to a satellite in orbit above the earth. The panel produces a current of 2.4A at a pottential difference of 20V when solar radiation is incident normally on it.
(1) Calculate the electrical power output of the panel.
I used power=current*voltage
p=2.4*20=48W ..........I this Correct?
(2) Solar radiation on the satellite has an intensity of 1.4kWm-2s. Calculate the efficiency.
I looked at 1.4kWm-2s, and i timed 1.4 by 1000 and got 1400Wm-2. Because it is /2 sec i divided 1400 by 2 and got 700Wm-1s. Now I am confused what to do next. How to find the efficiency? please help!
I have a question on physics,
Q: A solar panel of area 2.5m^2 is fitted to a satellite in orbit above the earth. The panel produces a current of 2.4A at a pottential difference of 20V when solar radiation is incident normally on it.
(1) Calculate the electrical power output of the panel.
I used power=current*voltage
p=2.4*20=48W ..........I this Correct?
(2) Solar radiation on the satellite has an intensity of 1.4kWm-2s. Calculate the efficiency.
I looked at 1.4kWm-2s, and i timed 1.4 by 1000 and got 1400Wm-2. Because it is /2 sec i divided 1400 by 2 and got 700Wm-1s. Now I am confused what to do next. How to find the efficiency? please help!
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Hi Hammad,
Of course, my physics is not sound, but if they are not answering over there, I'll do what I can to help.
(1)
I think you are correct.
Power = Current * Voltage = 2.4A * 20V = 48W.
(2)
1.4 kWm⁻² reads as "1.4 kilowatts per square metre".
Since the panel has 2.5m² of area, multiply by 2.5 to get the total power input:
2.5 m² * 1.4 kWm⁻² = 3.5 kW.
So, 3.5 kW = 3500 W is input, and only 48 W is output. This is an efficiency of:
48 / 3500
≈ 0.0137...
≈ 1.37% to 3 significant figures.
I hope this is correct, or at least, it makes sense (=
Of course, my physics is not sound, but if they are not answering over there, I'll do what I can to help.
(1)
I think you are correct.
Power = Current * Voltage = 2.4A * 20V = 48W.
(2)
1.4 kWm⁻² reads as "1.4 kilowatts per square metre".
Since the panel has 2.5m² of area, multiply by 2.5 to get the total power input:
2.5 m² * 1.4 kWm⁻² = 3.5 kW.
So, 3.5 kW = 3500 W is input, and only 48 W is output. This is an efficiency of:
48 / 3500
≈ 0.0137...
≈ 1.37% to 3 significant figures.
I hope this is correct, or at least, it makes sense (=