A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heate
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Answer:
Attached below are the sketches
answer :
c) G(s) = 100 / ( s + 100 )
d) y'(t) + 100Y(s) = 100 X(s)
e) g(t) = e^-100t u(t)
Explanation:
a) Sketch the bode plot
The filter here is a low pass filter
b) Sketch the s-plane
attached below. pole ( s ) is at 100
c) write the transfer function of the filter
Transfer function ; G(s) = 100 / ( s + 100 )
d) write the differential equation
Y(s) / X(s) = 100 / s + 100
Y(s) [ s + 100 ] = 100 X(s)
= sY(s) + 100Y = 100 X(s)
∴ differential equation = y'(t) + 100Y(s) = 100 X(s)
e) write out the unforced transient response
g(t) = e^-100t u(t)
f) write out the frequency response
attached below
Answer:
Now find the temperature of each surface, we have that the the temperature on the left side of the wall is T∞₁ - Q/h₁A and the temperature on the right side of the wall is T∞₂ + Q/h₂A.
Note: kindly find an attached diagram to the complete question given below.
Sources: The diagram/image was researched and taken from Slader website.
Explanation:
Solution
Let us consider the rate of heat transfer through the plane wall which can be obtained from the relations given below:
Q = T∞₁ -T₁/1/h₁A = T₁ -T₂/L/kA =T₂ -T∞₂/1/h₂A
= T∞₁ - T∞₂/1/h₁A + L/kA + 1/h₂A
Here
The convective heat transfer coefficient on the left side of the wall is h₁, while the convective heat transfer coefficient on the right side of the wall is h₂. the thickness of the wall is L, the thermal conductivity of the wall material is k, and the heat transfer area on one side of the wall is A. Q is refereed to as heat transfer.
Thus
Let us consider the convection heat transfer on the left side of the wall which is given below:
Q = T∞₁ -T₁/1/h₁A
T₁ = T∞₁ - Q/h₁A
Therefore the temperature on the left side of the wall is T∞₁ - Q/h₁A
Now
Let us consider the convection heat transfer on the left side of the wall which is given below:
Q= T₂ -T∞₂/1/h₂A
T₂ = T∞₂ + Q/h₂A
Therefore the temperature on the right side of the wall is T∞₂ + Q/h₂A
