Answer :
Answer:
the rate of heat transfer into the wall is [tex]\mathbf{q__{in}} \mathbf{ = 200 W/m^2}[/tex]
the rate of heat output is [tex]\mathbf{q_{out} =182 \ W/m^2}[/tex]
the rate of change of energy stored by the wall is [tex]\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }[/tex]
the convection coefficient is h = 4.26 W/m².K
Explanation:
From the question:
The temperature distribution across the wall is given by :
[tex]T(x) = ax+bx+cx^2[/tex]
where;
T = temperature in ° C
and a, b, & c are constants.
replacing 200° C for a, - 200° C/m for b and 30° C/m² for c ; we have :
[tex]T(x) = 200x-200x+30x^2[/tex]
According to the application of Fourier's Law of heat conduction.
[tex]q_x = -k \dfrac{dT}{dx}[/tex]
where the rate of heat input [tex]q_{in} = q_k[/tex] ; Then x= 0
So:
[tex]q_{in}= -k (\dfrac{d( 200x-200x+30x^2)}{dx})_{x=0}[/tex]
[tex]q_{in}= -1 (-200+60x)_{x=0}[/tex]
[tex]\mathbf{q__{in}} \mathbf{ = 200 W/m^2}[/tex]
Thus , the rate of heat transfer into the wall is [tex]\mathbf{q__{in}} \mathbf{ = 200 W/m^2}[/tex]
The rate of heat output is:
[tex]q_{out} = q_{x=L}[/tex]; where x = 0.3
[tex]q_{out} = -k (\dfrac{dT}{dx})_{x=0.3}[/tex]
replacing T with [tex]200x-200x+30x^2[/tex] and k with 1 W/m.K
[tex]q_{out} = -1 (\dfrac{d(200x-200x+30x^2)}{dx})_{x=0.3}[/tex]
[tex]q_{out} = -1 (-200+60x)_{x=0.3}[/tex]
[tex]q_{out} = 200-60*0.3[/tex]
[tex]\mathbf{q_{out} =182 \ W/m^2}[/tex]
Therefore , the rate of heat output is [tex]\mathbf{q_{out} =182 \ W/m^2}[/tex]
Using energy balance to determine the change of energy(internal energy) stored by the wall.
[tex]\Delta E_{stored} = E_{in}-E_{out} \\ \\ \Delta E_{stored} = q_{in}- q_{out} \\ \\ \Delta E_{stored} = (200 - 182 ) W/m^2 \\ \\[/tex]
[tex]\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }[/tex]
Thus; the rate of change of energy stored by the wall is [tex]\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }[/tex]
We all know that for a steady state, the heat conducted to the end of the plate must be convected to the surrounding fluid.
So:
[tex]q_{x=L} = q_{convected}[/tex]
[tex]q_{x=L} = h(T(L) - T _ \infty)[/tex]
where;
h is the convective heat transfer coefficient.
Then:
[tex]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[/tex] We have:
182 = h(200-200×0.3 + 30 ×0.3² - 100 )
182 = h (42.7)
h = 4.26 W/m².K
Thus, the convection coefficient is h = 4.26 W/m².K