Energy is generated uniformly in a 6 cm thick wall. the steady-state temperature distribution
t(z) = 145 + 3000z - 1500z^2
where t is temperature measured in c, and z is the distance measured from one side of the walls in meters. determine the rate of heat generation per unit volume if the thermal conductivity of the wall is 15 w/m. k.

Heat generation per unit volume is 45 KW.

Explanation:

Given that

Thickness of wall = 6 cm

Temperature distribution

   -----1

K= 15 W/m.k

As we know that at steady state condition

  -----2

Where q is the heat generation per unit volume.

So from equation 1

Now from equation 2

So q= 45 KW

So heat generation per unit volume is 45 KW.

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