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Mandarinka [93]

3 years ago

8

A furnace wall consisting of 0.25 m of fire clay brick, 0.20 m of kaolin, and a 0.10‐m outer layer of masonry brick is exposed t

o furnace gas at 1370 K with air at 300 K adjacent to the outside wall. The inside and outside convective heat‐transfer coefficients are 115 and 23 W/m2 ⋅ K, respectively. Determine the heat loss per square foot of wall and the temperature of the outside wall surface under these conditions.

1 answer:

dalvyx [7]3 years ago

4 0

Answer:

q=1910.71 W/m²

T=383.07 K

Explanation:

Given that

L₁=0.25 m ,K₁=1.13 W/(m.K)

L₂=0.2 m , K₂=1.45 W/(m.K)

L₃= 0.1 m , K₃=0.66 W/(m.K)

h₁=115 W/(m².K)

h₂=23 W/(m².K)

The total thermal resistance

$R=\dfrac{L_1}{K_1A}+\dfrac{L_2}{K_2A}+\dfrac{L_3}{K_3A}+\dfrac{1}{h_1A}+\dfrac{1}{h_2A}\ K/W$

Now by putting the values

Put A= 1 m² ( To find heat transfer per unit area)

$R=\dfrac{L_1}{K_1A}+\dfrac{L_2}{K_2A}+\dfrac{L_3}{K_3A}+\dfrac{1}{h_1A}+\dfrac{1}{h_2A}\ K/W$

$R=\dfrac{0.25}{1.13}+\dfrac{0.2}{1.45}+\dfrac{0.1}{0.66}+\dfrac{1}{115}+\dfrac{1}{23}\ K/W$

R= 0.56 K/W

We know that

q= ΔT/R

Here ΔT = 1370 - 300 K =1070 K

Now by putting the values

q= 1070/0.56

q=1910.71 W/m²

Lets T is the outside surface temperature

q = h₂ ( T- 300)

Now by putting the values

1910.71 = 23 ( T- 300)

T=383.07 K

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$W_{pdv} = \dfrac{100 \times 10^3 \times 523.6 -106.265 \times 10^3 \times 628.32 }{-\dfrac{1}{3} -1} = 10806697.1433 \ J$

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