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2 years ago

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Consider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 50 W/m · K and

a thickness L = 0.35 m, with no internal heat generation. Determine the heat flux and the unknown quantity for each case and sketch the temperature distribution, indicating the direction of the heat flux.

1 answer:

tatuchka [14]2 years ago

6 0

Answer:

solution:

dT/dx =T2-T1/L

&

q_x = -k*(dT/dx)

<u>Case (1) </u>

dT/dx= (-20-50)/0.35==> -280 K/m

q_x =-50*(-280)*10^3==>14 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

q_x =-50*(80)*10^3==>-4 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

q_x =-50*(80)*10^3==>-4 kW

Case (3)

q_x =-50*(160)*10^3==>-8 kW

T2=T1+dT/dx*L=70+160*0.25==> 110° C

Case (4)

q_x =-50*(-80)*10^3==>4 kW

T1=T2-dT/dx*L=40+80*0.25==> 60° C

Case (5)

q_x =-50*(200)*10^3==>-10 kW

T1=T2-dT/dx*L=30-200*0.25==> -20° C

note:

all graph are attached

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If you take the relation E2/E1 = 4.

B.

We have the next facts:

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Using the energy equation for a spring:

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