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snow_lady [41]
3 years ago
6

A composite wall consists of 20 mm thick steel plate backed by insulation brick (k = 0.39 W/mK) of 50 cm thickness and overlaid

by mineral wool of 20 cm thickness (k = 0.05 W/mK) and 70 cm layer of brick of (k = 0.39 W/mK). The inside is exposed to convection at 650°C with h = 65 W/ m2K. The outside is exposed to air at 35°C with a convection coefficient of 15 W/m2K. Determine the heat loss per unit area, interface temperatures and temperature gradients in each materials.

Engineering
1 answer:
jeyben [28]3 years ago
6 0

Answer:

Heat loss=85.9W/m^2

ΔT1(Steel)=0.04C

ΔT2(Brick1)=110.13C

ΔT3(Mwood)=343.6C

ΔT1(Brick2)=154.18C

Explanation:

raise the heat transfer equation from the air inside the wall to the outside air from the wall, because that is where you have the temperature data, to find the heat.

To find the temperatures you use the heat found in the previous step, and you use the conduction and convection equations in each wall layer.

I attached the procedure

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<u>Explanation</u>:

<u>L1:</u>

         mov ah,00

         mov al,[BX]

         mov dl,al

         div dh

         cmp ah,00

         je EVEN1

         mov [DI],dl

         add OddAdd,dl

         INC DI

         INC BX

         Loop L1

         jmp CAL

    <u>EVEN1:</u>

         mov [SI],dl

         add Even Add,dl

         INC SI

         INC BX

         Loop L1

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         mov ax,0000

         mov bx,0000

         mov al,OddAdd

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         MOV  ax,4C00h

         int 21h

end

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