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Tpy6a [65]
3 years ago
9

A one-dimensional plane wall of thickness 2L=80 mm experiences uniform thermal generation of q= 1000 W/m^3 and is convectively c

ooled at x=±40 mm by an ambient fluid characterized byT infinity=20degreesC. If the steady-state temperature distribution within the wall is T(x) = a(L2-x2)+b where a=15oC/m2and b=40oC, what is the thermal conductivityof the wall? What is the value of the convection heat transfer coefficient?
Engineering
1 answer:
Eduardwww [97]3 years ago
7 0

Answer:

h=1.99998\ W/m^2.C

k=33.333\ W/m.C

Explanation:

Considering the one dimensional and steady state:

From Heat Conduction equation considering the above assumption:

\frac{\partial^2T}{\partial x^2}+\frac{\dot e_{gen}}{k}=0       Eq (1)

Where:

k is thermal Conductivity

\dot e_{gen} is uniform thermal generation

T(x) = a(L^2-x^2)+b

\frac{\partial\ T(x)}{\partial x}=\frac{\partial\ a(L^2-x^2)+b}{\partial x}=-2ax\\\frac{\partial^2\ T(x)}{\partial x^2}=\frac{\partial^2\ -2ax}{\partial x^2}=-2a

Putt in Eq (1):

-2a+\frac{\dot e_{gen}}{k}=0\\ k=\frac{\dot e_{gen}}{2a}\\ k=\frac{1000}{2*15}\\ k=33.333\ W/m.C

Energy balance is given by:

Q_{convection}=Q_{conduction}

h(T_L-T_{inf})=-k(\frac{dT}{dx}) _L     Eq  (2)

T(x) = a(L^2-x^2)+b

Putting x=L

T(L) = a(L^2-L^2)+b\\T(L)=b\\T(L)=40^oC

\frac{dT}{dx}=\frac{d(a(L^2-x^2)+b}{dx}=-2ax\\Put\ x\ =\ L\\\frac{dT}{dx}=-2aL\\(\frac{dT}{dx})_L=-2*15*0.04=-1.2

From Eq (2)

h=\frac{-k*-1.2}{(40-20)} \\h=\frac{-33.333*-1.2}{(40-20)}\\h=1.99998\ W/m^2.C

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