You leave your car out in the parking lot is Arizona at midday in July when the

Question

You leave your car out in the parking lot is Arizona at midday in July when the temperature is 116oF. The sun’s rays are normal to the windshield, area, 1m2,with an intensity of 1 kW/m2. This radiation intensity all goes into heating the car. The windshield makes an angle of 45o with the roof where the solar panel is mounted. Eventually the temperature will stabilize when conduction and radiation balance the heating from the sun, by which time the inside of the car will be unbearably hot. Could we keep the car comfortably cool at 80oF powering the air conditioner from a solar panel on the roof, area 3.0 m2 ? How much larger does the roof have to be than the windshield? What difference would it make if the sun is directly overhead so the sun’s rays are normal to the roof, and make an angle of 45o with the windshield?

The expression for the coefficient of performance is

COP=Q/W= L/(T1 -T2 +2 delta T')

To transfer heat to the outside the evaporator has to be DT’ cooler then the inside of the car and the evaporator DT’ warmer. For the airflow from the blower in the car DT’ is 20oC, L = 216 kJ/kg, cp = 1.07 kJ/kg/oC. The solar panel has an efficiency of 15% and 100 W motors are required to move air over the condenser and evaporator .

Explanation / Answer

COP = Q / W = L / [cp (T1 - T2 + 2DeltaT' )]

Where, T1 = 116 degree F , T2 = 80 degree F.

So T1 - T2 = 36 degree F = 20 degree C

So, COP = 216/[1.07 * (20 + 2*20) ] = 3.36

Case 1: Sunrays are normal to the windshield

Solar Energy in through windshield,

Qin = Area * Intensity = 1 * 1000 = 1000 W

Heat can be removed if roof surface area is 3 m^2 ,

Qout = (Area * Intensity * Cosine of Intensity on the roof * Efficiency of the solar panel - Power required by the motor) x COP

= (3 * 1000 * cos(45) * 0.15 - 100) x 3.36

= 732.98 W

As Oin > Qout , so cooling is not possible.

Let us assume A m^2 is the roof area and equate Qin = Qout

So, (A * 707 * 0.15 - 100) * 3.36 = 1000

A = 3.75

i.e. area of the roof is 3.75 times of the windshield area.

Case 2: Sunrays are normal to the roof

Qin = 1000*1 * Cos(45) = 707 W

Qout = (3 * 1000 * 0.15 - 100) * 3.36 = 1176 W

As Qout > Qin, so cooling is possible.

Let us assume, A is the area of the roof whereas windshield area is 1 m^2.

So, Qin = Qout

(A * 1000 * 0.15 -100) * 3.36 = 707

A = 2.07

i.e area of the roof is 2.07 times of the windshield area.

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