12.6-1. Insulation Needed for Food Cold Storage Room. A food cold storage room is to beconstructed of an inner layer of 19.1 mm of pine, a middle layer of cork board, andan outer layer of 50.8 mm of concrete. The inside wall surface temperature is -178°Cand the outside surface temperature is 29.4°C at the outer concrete surface. Themean conductivities are for pine, 0.151; cork, 0.0433; and concrete, 0.762 W/m K.The total inside surface area of the room to use in the calculation is approximately39 m2 (neglecting corner and end effects). What thickness of cork board is needed tokeen the heat loss to 586 W?

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Asked Aug 26, 2019
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12.6-1. Insulation Needed for Food Cold Storage Room. A food cold storage room is to be
constructed of an inner layer of 19.1 mm of pine, a middle layer of cork board, and
an outer layer of 50.8 mm of concrete. The inside wall surface temperature is -178°C
and the outside surface temperature is 29.4°C at the outer concrete surface. The
mean conductivities are for pine, 0.151; cork, 0.0433; and concrete, 0.762 W/m K.
The total inside surface area of the room to use in the calculation is approximately
39 m2 (neglecting corner and end effects). What thickness of cork board is needed to
keen the heat loss to 586 W?
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12.6-1. Insulation Needed for Food Cold Storage Room. A food cold storage room is to be constructed of an inner layer of 19.1 mm of pine, a middle layer of cork board, and an outer layer of 50.8 mm of concrete. The inside wall surface temperature is -178°C and the outside surface temperature is 29.4°C at the outer concrete surface. The mean conductivities are for pine, 0.151; cork, 0.0433; and concrete, 0.762 W/m K. The total inside surface area of the room to use in the calculation is approximately 39 m2 (neglecting corner and end effects). What thickness of cork board is needed to keen the heat loss to 586 W?

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Expert Answer

Step 1

Write the expression of heat loss.

 

dT
RR,R
dT
q
dx
dx
dx,
Ak, Ak, Ak
dT
1 d,dx,dxe
A k
A (T-T)
dx2
ds dd
kg
ky
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dT RR,R dT q dx dx dx, Ak, Ak, Ak dT 1 d,dx,dxe A k A (T-T) dx2 ds dd kg ky

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Step 2

Substitute the known value...

39 m2[29.4°C-(7.8°C)]
586 W
0.0191 m
0.0508 m
0.151 W/m K
0.0433 W/m K 0.762 W/m -K
1840.8
586
dx
0.1265
0.0667
0.0433
dx
0.1932 3.1413
0.0433
dx2
= 2.9481
0.0433
dx, = 0.128 m
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39 m2[29.4°C-(7.8°C)] 586 W 0.0191 m 0.0508 m 0.151 W/m K 0.0433 W/m K 0.762 W/m -K 1840.8 586 dx 0.1265 0.0667 0.0433 dx 0.1932 3.1413 0.0433 dx2 = 2.9481 0.0433 dx, = 0.128 m

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