I need help on number 22, please help 1051 Problems 22. The power of sunlight re

Question

I need help on number 22, please help 1051 Problems 22. The power of sunlight reaching each square meter of the Earth's surface on a clear day in the tropics is close to 1 000 W. On a winter day in Manitoba, the power concentration of sunlight can be 100 W/m2 Many human activities are described by a power per unit area on the order of 102 W/m2 or less. (a) Con- sider, for example, a family of four paying $66 to the electric company every 30 days for 600 kWh of energy carried by electrical transmission to their house, which has floor dimensions of 13.0 m by 9.50 m. Compute the power per unit area used by the family. (b) Con- sider a car 2.10 m wide and 4.90 m long traveling at 55.0 mi/h using gasoline having "heat of combustion" 44.0 MJ/kg with fuel economy 25.0 mi/gal One gallon of gasoline has a mass of 2.54 kg. Find the power per unit area used by the car (c) Explain why direct use of solar energy is not practical for running a conventional automobile. (d) What are some uses of solar energy that are more practical? 23. A community plans to build a facility to convert solar radiation to electrical power. The community requires 1.00 MW of power, and the system to be installed has an efficiency of 30.0% (that is, 30.0% of the solar energy incident on the surface is converted to useful energy that can power the community). Assuming sun- light has a constant intensity of 1 000 W/m2, what must ht the effective area of a perfectly absorbing surface

Explanation / Answer

(a) 600 kWh = 547000 Wh = 600000 W * 60 mins * 60 seconds = 216*107 W
Therefore, the power per unit area = 216*107 W / (13.0 * 9.50 m^2) = 1.749*107 W/m^2

(b) 55.0 mi / h and 25.0 mi/gal fuel economy. This means in order to go 55 miles, you need 55/25 gallons of oil: 55/25 = 2.2 Gal = 2.54 kg * 2.2 Gal = 5.59 kg.

Now the heat of combustion: 44.0 MJ / kg * 5.59 kg = 246 MJ
So 246 MJ is being used within one hour.
246 MJ / 60 minutes / 60 seconds = 6.8e-2 MJ / sec = 6.8e4 J/sec = 680 kW

Now the area of the car is 2.10 * 4.90 = 10.29 m^2
The power per unit area is 680 kW / 10.29 m^2 = 66.083 kW /m^2

c) Solar energy is almost 120 watts per square meter ,which is lot less than we needed above (66.083 kw/m2).hence solar energy cannot provide such concentrated power.Direct use is therefore impractical.

d) Solar energy in automobiles can be used to charge battery for electric components of car.And it can be used to heat car also.

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