Respuesta :
Answer:
T = 167 ° C
Explanation:
To solve the question we have the following known variables
Type of surface = plane wall ,
Thermal conductivity k = 25.0 W/m·K,
Thickness L = 0.1 m,
Heat generation rate q' = 0.300 MW/m³,
Heat transfer coefficient hc = 400 W/m² ·K,
Ambient temperature T∞ = 32.0 °C
We are to determine the maximum temperature in the wall
Assumptions for the calculation are as follows
- Negligible heat loss through the insulation
- Steady state system
- One dimensional conduction across the wall
Therefore by the one dimensional conduction equation we have
[tex]k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}[/tex]
During steady state
[tex]\frac{dT}{dt}[/tex] = 0 which gives [tex]k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0[/tex]
From which we have [tex]\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}[/tex]
Considering the boundary condition at x =0 where there is no heat loss
[tex]\frac{dT}{dt}[/tex] = 0 also at the other end of the plane wall we have
[tex]-k\frac{dT }{dx } =[/tex] hc (T - T∞) at point x = L
Integrating the equation we have
[tex]\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}[/tex] from which C₁ is evaluated from the first boundary condition thus
0 = [tex]\frac{q'_{G}}{k} (0)+ C_{1}[/tex] from which C₁ = 0
From the second integration we have
[tex]T = -\frac{q'_{G}}{2k} x^{2} + C_{2}[/tex]
From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows
[tex]-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)[/tex] → C₂ = [tex]q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞[/tex]
T(x) = [tex]\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞[/tex] and T(x) = T∞ + [tex]\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )[/tex]
∴ Tmax → when x = 0 = T∞ + [tex]\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))[/tex]
Substituting the values we get
T = 167 ° C
