Consider a flat plate solar collector placed at the roof of a house. The temperatures at the inner and outer surfaces of glass cover are measured to be 28°C and 25°C, respectively. The glass cover has a surface area of 2.2. m^2 and a thickness of 0.6 cm and a thermal conductivity of 0.7 W / m · °C

Question

Consider a flat plate solar collector placed at the roof of a house. The temperatures at the inner and outer surfaces of glass cover are measured to be 28°C and 25°C, respectively. The glass cover has a surface area of [latex] 2.2. m ^{2}[/latex] and a thickness of 0.6 cm and a thermal conductivity of [latex] 0.7 W / m \cdot C [/latex]. Heat is lost from the outer surface of the cover by convection and radiation with a convection heat transfer coefficient of [latex] 10 W / m ^{2} \cdot{ }^{\circ} C [/latex] and an ambient temperature of 15°C. Determine the fraction of heat lost from the glass cover by radiation.

Accepted Answer

The glass cover of a flat plate solar collector with specified inner and outer surface temperatures is considered. The fraction of heat lost from the glass cover by radiation is to be determined.

Assumptions 1 Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values. 2 Thermal properties of the glass are constant.

Properties The thermal conductivity of the glass is given to be [latex] k=0.7 W / m \cdot{ }^{\circ} C[/latex].

Analysis Under steady conditions, the rate of heat transfer through the glass by conduction is

[latex] \dot{Q}_{ ext {cond }}=k A \frac{\Delta T}{L}=\left(0.7 W / m \cdot{ }^{\circ} C ight)\left(2.2 m ^{2} ight) \frac{(28-25)^{\circ} C }{0.006 m }=770 W [/latex]

The rate of heat transfer from the glass by convection is

[latex] \dot{Q}_{ ext {conv }}=h A \Delta T=\left(10 W / m ^{2} \cdot{ }^{\circ} C ight)\left(2.2 m ^{2} ight)(25-15)^{\circ} C = 220 W [/latex]

Under steady conditions, the heat transferred through the cover by conduction should be transferred from the outer surface by convection and radiation. That is,

[latex] \dot{Q}_{ rad }=\dot{Q}_{ ext {cond }}-\dot{Q}_{ ext {conv }}=770 - 220 = 550 W [/latex]

Then the fraction of heat transferred by radiation becomes

[latex] f = \frac{\dot{Q}_{ rad }}{\dot{Q}_{ ext {cond }}}=\frac{550}{770}=0.714 \quad( ext { or } 71.4 \%) [/latex]

Question 1.137: Consider a flat plate solar collector placed at the roof of ...

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