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gfd43tg
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Homework Statement
Consider a large plane wall with thickness L whose thermal conductivity varies in a specified temperature range as
k(T) = k_{0}(1 + \beta T)
where ##k_{0}## and ##\beta## are two specified constants. The wall surface at ##x=0## is maintained at a constant temperature of ##T_{1}##, while the surface at ##x = L## is maintained at ##T_{2}##. Assuming steady one dimensional heat transfer by conduction through the wall only, obtain a relation for the heat transfer rate through the wall.
Sketch temperature and heat flux profiles for zero, positive and negative coefficients. Explain briefly why gases usually have a positive coefficient a, but liquids and solids show a negative coefficient. Water is an exceptional liquid that shows a positive coefficient β, explain.
Homework Equations
The Attempt at a Solution
I tried this two different ways, and am stuck a little bit. The first approach I did a general energy balance
FIRST METHOD
\frac {dE}{dt} = \dot Q_{x} - \dot Q_{x + \Delta x} + \dot e_{gen}A \Delta x
\frac {d(\rho \hat c_{p} A \Delta x T)}{dt} = \dot Q_{x} - \dot Q_{x + \Delta x} + \dot e_{gen}A \Delta x
\rho \hat c_{p} A \Delta x \frac {dT}{dt} = \dot Q_{x} - \dot Q_{x + \Delta x} + \dot e_{gen}A \Delta x
divide through by ##A \Delta x##
\lim_{\Delta x \to 0} \rho \hat c_{p} \frac {dT}{dt} = \frac {1}{A \Delta x} (\dot Q_{x} - \dot Q_{x + \Delta x}) + \dot e_{gen}
Assuming that no heat is generated and using Fourier's law, ##\dot Q = -kA \frac {dT}{dx}##, this reduces to
\rho \hat c_{p} \frac {dT}{dt} = \frac {-1}{A} \frac {d \dot Q}{dx}
\rho \hat c_{p} \frac {dT}{dt} = \frac {-1}{A} \frac {d}{dx} (-kA \frac {dT}{dx})
The Area terms cancel out as well as the negative terms on the RHS,
\rho \hat c_{p} \frac {dT}{dt} = \frac {d}{dx} (k \frac {dT}{dx})
With the assuming of steady state and the problem statement stating that ##k = k_{0}(1 + \beta T)##, we get
##\frac {d}{dx}[k_{0}(1 + \beta T) \frac {dT}{dx}] = 0##
Now I am hesitant to try to integrate this thing, with the ##\frac {d}{dx}## not combining to make ##k_{0}(1+ \beta T) \frac {d^{2}T}{dx^2}##
SECOND METHOD
For this, I just went straight away with Fourier's Law
\dot Q = -kA \frac {dT}{dx}
\dot Q = -k_{0}(1 + \beta T)A \frac {dT}{dx}
\frac {\dot Q}{-k_{0}A} dx = (1 + \beta T) dT
\frac {\dot Q}{-k_{0}A} \int dx = \int (1 + \beta T) dT
This integrates to be
\frac {\dot Q}{-k_{0}A}x = T + \frac {\beta}{2}T^{2} + C
Using the boundary condition of ##T = T_{1}## at ##x = 0##, this will be
0 = T_{1} + \frac {\beta}{2}T_{1}^{2} + C
C = -T_{1} - \frac {\beta}{2}T_{1}^{2}
Putting back into the original equation
\frac {\dot Q}{-k_{0}A}x = T + \frac {\beta}{2}T^{2} -T_{1} - \frac {\beta}{2}T_{1}^{2}
\dot Q = \frac {-k_{0}A}{x} [T + \frac {\beta}{2} T^{2} - T_{1} - \frac {\beta}{2}T_{1}^{2}]
So I do get ##\dot Q = f(T)##, but it seems like I completely ignored the second boundary condition. It seems like this should have two constants of integration, which would be closer to what I saw in the first method. So I am both not sure how to integrate the first method, and here unsure of my boundary conditions for the second method.
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Dunno where my brain went to for that one. Your original interpretation is correct.
