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Question 1 (25 marks) a. A small house needs 2.4kWh per day, all of which is used in the evening between the hours of 7 to 11pm. Using only 12V batteries, calculate the minimum rated Ahr of battery storage required to supply the load if the battery maximum depth of discharge is 60%. Assume battery temperature is 25°C. b. With the aid of a well labeled diagram, explain the principle of operation, from resource to electrical power, of a PV module. c. How does shading impact on I-V curves of a PV module with and without a bypass and blocking diode. d. Imagine a space station with a solar array measuring 10 meters by 10 meters. In full sunlight, how much solar energy is impinging on the array? If it can be converted to electrical energy at 28% efficiency, how much power is generated (in W)? [8 + 5 + 5 + 7 marks]

          Question 1 (25 marks)
a. A small house needs 2.4kWh per day, all of which is used in the evening between
the hours of 7 to 11pm. Using only 12V batteries, calculate the minimum rated
Ahr of battery storage required to supply the load if the battery maximum depth of
discharge is 60%. Assume battery temperature is 25°C.
b. With the aid of a well labeled diagram, explain the principle of operation, from
resource to electrical power, of a PV module.
c. How does shading impact on I-V curves of a PV module with and without a
bypass and blocking diode.
d. Imagine a space station with a solar array measuring 10 meters by 10 meters. In
full sunlight, how much solar energy is impinging on the array? If it can be
converted to electrical energy at 28% efficiency, how much power is generated
(in W)?
[8 + 5 + 5 + 7 marks]

Show more…

Question 1 (25 marks)
a. A small house needs 2.4kWh per day, all of which is used in the evening between
the hours of 7 to 11pm. Using only 12V batteries, calculate the minimum rated
Ahr of battery storage required to supply the load if the battery maximum depth of
discharge is 60%. Assume battery temperature is 25°C.
b. With the aid of a well labeled diagram, explain the principle of operation, from
resource to electrical power, of a PV module.
c. How does shading impact on I-V curves of a PV module with and without a
bypass and blocking diode.
d. Imagine a space station with a solar array measuring 10 meters by 10 meters. In
full sunlight, how much solar energy is impinging on the array? If it can be
converted to electrical energy at 28% efficiency, how much power is generated
(in W)?
[8 + 5 + 5 + 7 marks]

Added by Christina M.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Umar Sohail Qureshi

11/10/2021

Step 1

4 kWh/day Battery voltage = 12V Depth of discharge = 60% Show more…

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Question 1 (25 marks) a. A small house needs 2.4kWh per day, all of which is used in the evening between the hours of 7 to l lpm.Using only 12V batteries,calculate the minimum rated Ahr of battery storage required to supply the load if the battery maximum depth of discharge is 60%. Assume battery temperature is 25C. b. With the aid of a well labeled diagram, explain the principle of operation, from resource to electrical power, of a PV module. c. How does shading impact on I-V curves of a PV module with and without a bypass and blocking diode. d. Imagine a space station with a solar array measuring 10 meters by 10 meters. In full sunlight, how much solar energy is impinging on the array? If it can be converted to electrical energy at 28% efficiency, how much power is generated (in W)? [8 + 5 + 5 +7 marks]

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00:01 For this question, we're considering the energy coming from the sun to have an intensity i of 1 ,000 watts per meter squared.

00:08 But our solar panel can only convert about 10 % of that.

00:10 So the intensity for our solar panel i is 0 .1 times that 1 ,000 watt per meter squared or 100 watts per meter squared.

00:17 And we're breaking this into part a, b, and c.

00:20 For part a, we consider the hair dryer that needs a power of 50 kilowatts.

00:26 So that's 50 times 10 of the minus 3 watts.

00:28 I call that piece of a for the power in part a.

00:31 Part b needs to power piece of b of 1 ,500 watts.

00:35 Excuse me, so a was the calculator.

00:37 Sorry, not the hair dryer.

00:38 Part b is the hairdriar.

00:39 It needs 1 ,500 watts.

00:41 And part c, we're powering a car, p sub c, of 20 horsepower, which i convert to watts.

00:46 There's 746 watts per horsepower.

00:48 So this is 14 ,920 watts.

00:51 Okay.

00:51 So for part a, let's go ahead and calculate the area.

00:54 So in general, area here is equal to power divided by intensity.

00:57 So this is the power for part a divided by the intensity i.

01:02 The intensity is the same for all of them.

01:03 So this comes out to equal 5 times 10 to the minus 4 meters squared times 10 to minus 4 square meters...

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