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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.

Respuesta :

Answer:

Heat generation per unit volume is 45 KW.

Explanation:

Given that

Thickness of wall = 6 cm

Temperature distribution

[tex]T(z)=145+3000z-1500z^2[/tex]    -----1

K= 15 W/m.k

As we know that at steady state condition

[tex]\dfrac{d^2T}{dz^2}+\dfrac{q}{K}=0[/tex]   -----2

Where q is the heat generation per unit volume.

So from equation 1

[tex]\dfrac{dT}{dz}=3000-3000z[/tex]

[tex]\dfrac{d^2T}{dz^2}=-3000[/tex]

Now from equation 2

[tex]\dfrac{d^2T}{dz^2}+\dfrac{q}{K}=0[/tex]  

[tex]-3000+\dfrac{q}{15}=0[/tex]  

So q= 45 KW

So heat generation per unit volume is 45 KW.

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