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3 years ago

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A one-dimensional plane wall of thickness 2l= 100 mm experiences uniform thermal energy generation of q˙= 800 w/m3 and is convec

tively cooled at x= ±50 mm by an ambient fluid characterized by [infinity] t[infinity]= 26.0°c. if the steady-state temperature distribution within the wall is t(x)=a(l2-x2)+b where a= 10°c/m2 and b= 30°c, what is the thermal conductivity of the wall? what is the value of the convection heat transfer coefficient, h?

1 answer:

slega [8]3 years ago

4 0

Answer:

The thermal conductivity of the wall = 40W/m.C

h = 10 W/m^2.C

Explanation:

The heat conduction equation is given by:

d^2T/ dx^2 + egen/ K = 0

The thermal conductivity of the wall can be calculated using:

K = egen/ 2a = 800/2×10

K = 800/20 = 40W/m.C

Applying energy balance at the wall surface

"qL = "qconv

-K = (dT/dx)L = h (TL - Tinfinity)

The convention heat transfer coefficient will be:

h = -k × (-2aL)/ (TL - Tinfinty)

h = ( 2× 40 × 10 × 0.05) / (30-26)

h = 40/4 = 10W/m^2.C

From the given temperature distribution

t(x) = 10 (L^2-X^2) + 30 = 30°

T(L) = ( L^2- L^2) + 30 = 30°

dT/ dx = -2aL

d^2T/ dx^2 = - 2a

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The diagram for this question is shown on the first uploaded image

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b E = $36.69*10^3 N/C$

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d $V = 6.7*10^3 V$

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And $Potential \ Difference = \frac{1}{4\pi \epsilon_0} [\frac{q_1 }{r}+\frac{q_2}{R_2} ]$

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$r = 0.700 m \ , R_2 > r > R_1$

This implies that the electric field would be

$E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}$

This because it the electric filed of the charge which is below it in distance that it would feel

$E = 8*99*10^9 \frac{2*10^{-6}}{0.4900}$

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Considering c

r = 0.200 m

=> $r$

The electric field = 0

This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

Considering d

r = 4.00 m

=> $r > R_1 >r>R_2$

Now the potential difference is

$V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V$

This so because the distance between the charge we are considering is further than the two charges given

Considering e

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$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V$

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$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V$

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