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:
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
You might be interested in
Answer:
Explanation:
When the unpolarized light passes through the first polarizer, only the component of the light parallel to the axis of the polarizer passes through.
Therefore, after the first polarizer, the intensity of light passing through it is halved, so the intensity after the first polarizer is:
Then, the light passes through the second polarizer. In this case, the intensity of the light passing through the 2nd polarizer is given by Malus' law:
where
is the angle between the axes of the two polarizer
Here we have
So the intensity after the 2nd polarizer is
And substituting the expression for I1, we find: