Question

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 Celsius 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

Answer 1

the rate of heat transfer into the wall is
`\mathbf{q__(in)} \mathbf{ = 200 W/m^2}`

the rate of heat output is
`\mathbf{q_(out) =182 \ W/m^2}`

the rate of change of energy stored by the wall is
`\mathbf{ \Delta E_(stored) = 18 \ W/m^2 }`

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

Step-by-step explanation:

From the question:

The temperature distribution across the wall is given by :

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

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 :

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

According to the application of Fourier's Law of heat conduction.

`q_x = -k (dT)/(dx)`

where the rate of heat input
`q_(in) = q_k` ; Then x= 0

So:

`q_(in)= -k ((d( 200x-200x+30x^2))/(dx))_(x=0)`

`q_(in)= -1 (-200+60x)_(x=0)`

`\mathbf{q__(in)} \mathbf{ = 200 W/m^2}`

Thus , the rate of heat transfer into the wall is
`\mathbf{q__(in)} \mathbf{ = 200 W/m^2}`

The rate of heat output is:

`q_(out) = q_(x=L)`; where x = 0.3

`q_(out) = -k ((dT)/(dx))_(x=0.3)`

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

`q_(out) = -1 ((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}`

Therefore , the rate of heat output is
`\mathbf{q_(out) =182 \ W/m^2}`

Using energy balance to determine the change of energy(internal energy) stored by 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 rate of change of energy stored by the wall is
`\mathbf{ \Delta E_(stored) = 18 \ W/m^2 }`

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:

`q_(x=L) = q_(convected)`

`q_(x=L) = h(T(L) - T _ \infty)`

where;

h is the convective heat transfer coefficient.

Then:

`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 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