1) Determine the thermal conductivity of the plaster board, in W/m·K. 2) Determi

Question

1) Determine the thermal conductivity of the plaster board, in W/m·K.

2) Determine the value of the inside convection resistance, in C/W.

3)Determine the value of the fiberglass blanket resistance, in C/W.

4)Determine the value of the total heat transfer resistance, in C/W.

5)Determine the value of the heat loss, in kW.

Problem 3.011 A house has a composite wall of wood, fiberglass insulation, and plaster board, as indicated in the sketch. On a cold winter day, the convection heat transfer coefficients are 60 W/m2.K and 35 W/m2.K. The total wall surface area is 350 m2. The plaster board thickness is 10 mm, the glass fiber thickness is 100 mm, and the plywood siding thickness is 20 mm. The temperature inside is 20°c and the temperature outside is -15°C. Determine the total heat loss through the wall Glass fiber blanket (28 kg/m2), k Plaster board, k Plywood siding, k Inside Outside h, To, i itt ho, Too L,

Explanation / Answer

A) from the thermal conductivity table ,ther the conductivity of plaster board or gypsum = 0.17 W/m.k

B) inside convection resistance Ri=1/ hi.A =1/(60*350 )

= 4.76 *10-5 k/w or c/w

C) fibe glass blanket resistance = l/k.A = .100/(.04*350)

= 0.00714 k/w

D) thermal resistance of plaster board = lp/kA = .01/(.17*350)

= 0.001680 k/w

And thermal resistance of plywood = l/kA = .02/(.13*350)

= 0.000476 k/w

Convection resistance of out side = 1/hA = 1/(35*350)

= 8.16*10-5 k/w

Now total resistance

= 4.76*10-5+0.00714+0.001680+0.000476+8.16*10-5

=0.0094252 k/w or c/w

E)heat loss = T / total resistance

Q = (20-(-15))/.0094252

Q= 3713 w

Q = 3.713 kw

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