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2 months ago

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A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coeffi

cients of 25 and 1000 W/m2 ⋅ K. The wall is composed of a 12-mm-thick layer of beryllium oxide on the gas side and a 24-mm-thick slab of stainless steel (AISI 304) on the liquid side. The contact resistance between the oxide and the steel is 0.05 m2 ⋅ K/W. What is the rate of heat loss per unit surface area of the composite? Sketch the temperature distribution from the gas to the liquid.

1 answer:

Ostrovityanka [3.2K]2 months ago

3 0

Response:

$\text{heat loss} = 24864.05 \ W/m^2$

Clarification:

If

$T_1$ , $T_2$ represent the temperatures of gases and liquids in Kelvins,

$t_1$ and $t_2$ denote the thicknesses of the gas layer and steel slab in meters,
$h_1$ , $h_2$ are the convection coefficients for gas and liquid in $W/m^2 \cdot K$ ,
$R_c$ represents the contact resistance in $m^2 \cdot K/W$ ,

and $k_1, k_2$ signify thermal conductivities of gas and steel in $W/m \cdot K$ ,

then: part(a):

$\text{heat loss } = \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}$

by employing known values:

$\text {heat loss} = 2486.05 W/m^2$

part(b): Utilizing the rate equation:

$\text {heat loss} = h_1 (T_1 - T_{s1})$

the surface temperature is $T_{s1} = 1678.438 \ K$

and $T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K$

Correspondingly

$T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K$

$T_{s2} = T_{c2} - \frac {t_2 (\text{heat loss})}{ k_2} = 397.864 \ K$

The temperature profile is depicted in the image provided

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