Engineering, 29.10.2019 06:31 madi1820

The roof of a house consists of a 22-cm-thick concrete slab (k = 2 w/m•k) that is 16.8 m wide and 20 m long. the emissivity of the outer surface of the roof is 0.9, and the convection heat transfer coefficient on that surface is estimated to be 15 w/m2•k. the inner surface of the roof is maintained at 15°c. on a clear winter night, the ambient air is reported to be at 10°c, while the night sky temperature for radiation heat transfer is 255 k

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Answer from: brooklynmikestovgphx

Roof's outer surface is at a temperature of 7.67 °C (Celsius degrees)

Explanation:

The only variable left to be found is the roof's outer temperature, which we shall name . For this purpuse, we must assume the following, related to the law of continuity of energy: all heat through the roof's thickness as a solid body (conduction) is later released to the ambient as a sum of radiated heat and convected heat, by their respective heat transfer processes.

NOTE: We will assume here that air's temperature is lower than the roof's outer temperature, so the heat flux is given from the roof towards the air ambient.

Please note that, for heat transfer calculus involving radiation, all temperatures must be expressed in Kelvin degrees, T(K) =T(C)+273.15 .

Thus,

$Q_{conduction}=Q_{convection}+Q_{radiation}$ (1)

Let us write down all these terms:

$Q_{conduction} =\frac{(T_{in}-T_{out})}{R_{thermal}} =\frac{(T_{in}-T_{out})}{\frac{t}{kA}}}$ (2)

where we know the roof's inner temperature, the area A (m2), the thickness t(m) and the roof's material's conductivity k (W/m*K).

$Q_{convection} = hA(T_{out}-T_{air})$ (3)

where we know convection coefficient h, area A and temperature of air.

$Q_{radiation} = \varepsilon \sigma A ((T_{out})^4-(T_{sky})^4)$ (4)

where we know emissivity $\varepsilon$ , Steffan-Boltzmann constant $\Sigma$ , and, sky radiative temperature, and, again, the area A.

Placing equation 2-4 into eq. (1), and considering that the area value A can be simplified from all three terms, we find the following equation, whose only unknown is the roof's outer temperature, named T:

$\frac{2}{0.22}(288.15-T)=15(T-283.15)+0.9*5.6704E-8(T^4-255^4)$

The result of this equations fourth degree, involving fourth power of unkown T and linear T) is:

T=280.82K=7.67degC

NOTE: the resulting roof's outer temperature is lower than the air's temperature, which implies that, against our initial assumption, the heat flux is not from roof's outer surface towards the air, but the air is heating up the roof, since the air is warmer than the roof's outer surface.

Answer from: Quest

answer: eco and eco system

explanation:

Answer from: Quest

answer:

an example of solubility is salt or sugar dissolving in water.

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The roof of a house consists of a 22-cm-thick concrete slab (k = 2 w/m•k) that is 16.8 m wide and 20...