The temperature distribution across a wall 0.3 m thick at a certain instant of time is T(x) a bx cx2 , where T is in degrees Cel

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The temperature distribution across a wall 0.3 m thick at a certain instant of time is T(x) a bx cx2 , where T is in degrees Cel

sius and x is in meters, a 200 C, b 200 C/m, and c 30 C/m2 . The wall has a thermal conductivity of 1 W/mK. (a) On a unit surface area basis, determine the rate of heat transfer into and out of the wall and the rate of change of energy stored by the wall. (b) If the cold surface is exposed to a fluid at 100 C, what is the convection coefficient

Engineering

1 answer:

iogann1982 [279] · Answer 1 · 2026-02-22T00:28:19+03:00

Answer:

The heat transfer rate into the wall is $\mathbf{q__{in}} \mathbf{ = 200 W/m^2}$

The heat output rate is $\mathbf{q_{out} =182 \ W/m^2}$

The change in stored energy in the wall is $\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }$

The convection coefficient is h = 4.26 W/m².K

Explanation:

Considering the problem:

The temperature profile across the wall is expressed as:

$T(x) = ax+bx+cx^2$

where:

T = temperature in °C

and a, b, & c are constants.

Substituting a = 200° C, b = -200° C/m, and c = 30° C/m² results in:

$T(x) = 200x-200x+30x^2$

This follows the application of Fourier's Law of heat conduction.

$q_x = -k \dfrac{dT}{dx}$

where the heat input rate $q_{in} = q_k$ ; Then x= 0

<pThus:

$q_{in}= -k (\dfrac{d( 200x-200x+30x^2)}{dx})_{x=0}$

$q_{in}= -1 (-200+60x)_{x=0}$

$\mathbf{q__{in}} \mathbf{ = 200 W/m^2}$

Consequently, the heat transfer rate into the wall measures $\mathbf{q__{in}} \mathbf{ = 200 W/m^2}$

The heat output rate is:

$q_{out} = q_{x=L}$ ; where x = 0.3

$q_{out} = -k (\dfrac{dT}{dx})_{x=0.3}$

Replacing T with $200x-200x+30x^2$ and k with 1 W/m.K

$q_{out} = -1 (\dfrac{d(200x-200x+30x^2)}{dx})_{x=0.3}$

$q_{out} = -1 (-200+60x)_{x=0.3}$

$q_{out} = 200-60*0.3$

$\mathbf{q_{out} =182 \ W/m^2}$

Thus, the heat output rate is $\mathbf{q_{out} =182 \ W/m^2}$

Applying the energy balance to find the change in energy (internal energy) stored in the wall.

$\Delta E_{stored} = E_{in}-E_{out} \\ \\ \Delta E_{stored} = q_{in}- q_{out} \\ \\ \Delta E_{stored} = (200 - 182 ) W/m^2 \\ \\$

$\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }$

Thus, the energetic change rate stored in the wall is $\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }$

We know that in a steady state, the heat reaching the end of the plate must be convected to the surrounding fluid.

Thus:

$q_{x=L} = q_{convected}$

$q_{x=L} = h(T(L) - T _ \infty)$

where;

h represents the convective heat transfer coefficient.

Therefore:

$Replacing \ 182 W/m^2 \ for \ q_{x=L}, (200-200x +30x \ for \ T(x) \, 0.3 m \ for \ x \ and \ 100^0 C for \ T$ We find:

182 = h(200-200×0.3 + 30 ×0.3² - 100 )

182 = h (42.7)

h = 4.26 W/m².K

Thus, the convection coefficient equals h = 4.26 W/m².K