The radiant energy incident on the earth's surface on a sunny day is roughly 900.0 W/m2. Collecting and focusing sunlight and using the focused beam to heat a fluid is an old idea, and as the environmental costs of fossil-fuel combustion mount, solar heating becomes an increasingly attractive alternative. Suppose a house is to be designed that will have a circulating forced-air central-heating unit, and solar energy is contemplated as a heat source (backed up with a conventional furnace to be used on cloudy days). If air is to be fed at a rate of 2000.0 m3/min at 30.0°C and 1 atm, and the air is to be heated to 60.0°C before being discharged into the living space, what is the required area of the solar collector plates? Assume 30.0% of the radiant energy incident on the plates is used to heat the air.

Question

Engineering

Arielle

16 May, 00:37

The radiant energy incident on the earth's surface on a sunny day is roughly 900.0 W/m2. Collecting and focusing sunlight and using the focused beam to heat a fluid is an old idea, and as the environmental costs of fossil-fuel combustion mount, solar heating becomes an increasingly attractive alternative. Suppose a house is to be designed that will have a circulating forced-air central-heating unit, and solar energy is contemplated as a heat source (backed up with a conventional furnace to be used on cloudy days). If air is to be fed at a rate of 2000.0 m3/min at 30.0°C and 1 atm, and the air is to be heated to 60.0°C before being discharged into the living space, what is the required area of the solar collector plates? Assume 30.0% of the radiant energy incident on the plates is used to heat the air.

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Kolby Leon · Answer 1

Area of solar panel required =

Required power to raise temp / useful power from irradiation

= 1228.7kw/0.27kw/m²

= 4551m²

Explanation:

Radiant energy / m² = 900w/m²

Density of air = 1.225kg/m³

Volumetric flow = 2000m³/minute = 33.33m³/s

Mass of air flow = density of air * volumetric flow = 1.225kg/m³ * 33.33m³/s = 40.83kg/s

T1 = Initial temp = 30°

T2 = Final temp = 60°

Required power to raise temp of air

= air mass flow * Cp * (T2 - T1)

Where Cp = specific heat capacity of air = 1.003kj/kg k

= 40.83kg/s * 1.003 Kj/kg k * 30k = 1228.7kw

With 30% irradiation

Useful power/m² =.9kw/m² *.3 =.27kw/m²

Area of solar panel required =

Required power to raise temp / useful power from irradiation

= 1228.7kw/0.27kw/m²

= 4551m²