9 kilowatt hours of power per day? Assume ideal conditions for a flat panel arra

Question

9 kilowatt hours of power per day? Assume ideal conditions for a flat panel array matching Insolation Map data for month and location that provides 3.2 kwh/ml/day How many square feet ofphoto voltaic panels (15% efficiency) in a single amy will be required to generate 2s at solar noon. How many degrees Fahrenheit will 10 gallons of water be raised in a 30 square foot collector system? (One gallon of water weighs 8.35 pounds). 10. In ideal solar thermal conditions a flat plate collector (consider 100 % efficient) can capture 300 BTU's per hour

Explanation / Answer

9)

Efficiency eff= 15%

Energy per day required is E = 2.5 kWh/day

Insolation at given location per day per square meter I = 3.2 kWh/m2 / day

1 m2 = 10.7639 ft2

I = 0.2973 kWh/ft2 / day

Area of photo panels required in square feet = E / (I*eff) = 2.5 / (0.15 * 0.2973) = 56.06 ft2

10)

Area of solar thermal collector = 30 ft2

No.of gallons of water = 10 gallons

1 gallon = 8.35 pounds

1 BTU = 262.164 calories

300 BTU = 75.6493 kcalories

By definition 1 calorie is amount of heat (energy) required to rise 1o of 1 gram of water.

1 pound = 453.592 grams

10 gallons of water = 83.5 pouns = 37874.932 grams of water

1 c - 1o rise of 1 gram of water

to rise 1o of 37874.932 grams of water requires calories = ?

37874.932 grams of water requires 37874.932 calories of heat (energy required)

Rise of temperature in 0C of water by 75649.3 calories of heat

= 75649.3/37874.932 = 1.9973 0C

In Fahrenheit = (9/5)*T 0C + 32 = 35.5952 0F rise in temperature of 10 gallons of water.

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