Question

The heat flux for a given wall is in the x-direction and given as q^n = 11 W/m^2, the walls thermal conductivity is 1.7 W/mK and the walls thickness is 1.3 cm. Assume steady state conditions exist. Determine the temperature gradient in units of K/m and degree C/m. If the temperature gradient were larger what happens to the heat flux and why? Comment on the direction of heat flux given a negative temperature gradient and again for the case of a positive temperature gradient.

Answer 1

`(dT)/(dx) = 6.47 ^oC/m`

Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.

For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards

Step-by-step explanation:

As we know that heat flux is given by the formula

`q^n = K(dT)/(dx)`

here we know that

K = thermal conductivity

`(dT)/(dx)` = temperature gradient

now we know that

`q^n = 11 W/m^2`

also we know that

K = 1.7 W/mK

now we have

`11 = 1.7 (dT)/(dx)`

so temperature gradient is given as

`(dT)/(dx) = (11)/(1.7) = 6.47 K/m`

also in other unit it will be same

`(dT)/(dx) = 6.47 ^oC/m`

Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.

For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards