Question

A temperature difference of 15°C is impressed across a brick wall of a house which is 15 cm in thickness. The thermal conductivity of the brick is 1.0 W/m °C. The face of the wall is 6 meters high and 12 meters long. Compute both the heat flux and the heat transfer rate through the wall. Why aren't these numerical values the same?

Answer 1

Q=7200 W

q=7200/72=100
`W/m^2`

Step-by-step explanation:

Given that

ΔT=15° C

Thickness ,t=15 cm

Thermal conductivity ,K=1 W/m.°C

Height,h=6 m

Length ,L=12 m

As we know that heat conduction through wall given as

`Q=(KA)/(t)\Delta T`

Now by putting the values

A= 6 x 12 =72
`m^2`

`Q=(KA)/(t)\Delta T`

`Q=(1* 72)/(0.15)* 15\ W`

Q=7200 W

Q is the total heat transfer.

Heat flux q

q=Q/A
`W/m^2`

q=7200/72=100
`W/m^2`

q is the heat flux.

As w know that heat flux(q) is the heat transfer rate from per unit area and on the other hand heat transfer(Q) is the total heat transfer from the surface.

Heat flux q=Q/A

That is why these both are different.