A large plane wall has a thickness L = 160 cm and thermal conductivity k = 25 W/m∙K. On the left surface (x = 0), it is subjected to a uniform heat flux q(dot)0. Picture while the surface temperature T0 is constant. On the right surface, it experiences convection and radiation heat transfer while the surface temperature is TL = 225°C and the surrounding temperature is 25°C. The emissivity and the convection heat transfer coefficient on the right surface are 0.7 and 15 W/m2∙K, respectively.

Determine the temperature of the left surface of the wall at x = 0 (in °C)

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dfbustos dfbustos · Answer 1 · 2020-02-20T11:53:43+01:00

Answer:

[tex] T(x)= \frac{q*0}{k} (L-x) +T_L [/tex]

And if we find the temperature for x=0 we got:

[tex] T(0) = \frac{q*o}{25 W/ mK} (1.6 m -0) +225C[/tex]

[tex] T(0) = 0.064 q*0 +225[/tex]

Figure attached.

Explanation:

For this case we can use the Heat equation given by:

[tex] \frac{d^2 T}{dx^2}=0[/tex] (1)

Since we have all the hest in one dimension. The boundary conditions for this case are:

[tex] q* (x=0) = -k \frac{dT(0)}{dx}= q*0[/tex]

[tex] T(L) = T_L[/tex]

If we integrate the equation (1) we got:

[tex] \frac{dT}{dx} = a_1 [/tex]

And if we integreate again we got:

[tex] dT= a_1 dx[/tex]

[tex] T= a_1 x + a_2[/tex]

Where [tex] a_1 , a_2[/tex] are constants. Now we can apply the boundary conditions:

[tex] -k a_1 = q*0[/tex] so then the constant [tex] a_1[/tex] would be:

[tex] a_1 = -\frac{q*0}{k}[/tex]

Using the second boundary condition we have:

[tex] T_L = -\frac{q*0}{k} L +a_2[/tex]

[tex] a_2 = T_L +\frac{q*0}{k}L[/tex]

And then our general solution would be given by:

[tex] T(x) = -\frac{q*0}{k} x + T_L + \frac{q*0}{k} L[/tex]

Taking common factor we got:

[tex] T(x)= \frac{q*0}{k} (L-x) +T_L [/tex]

And if we find the temperature for x=0 we got:

[tex] T(0) = \frac{q*o}{25 W/ mK} (1.6 m -0) +225C[/tex]

[tex] T(0) = 0.064 q*0 +225[/tex]

And the figure is on the plot attached.