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In the above example, if temperature of ...

In the above example, if temperature of inner surface P is kept constant at `theta_1` and of the outer surface Q at `theta_2(lttheta_1)`. Then,
Find.
(a) rate of heat flow or heat current from inner surface to outer surface.
(b) temperature `theta` at a distance `r (altrltb)` from centre.

Text Solution

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The correct Answer is:
A, B, D

(a)
`H or (dQ)/(dt) = ("Temperature difference")/("Thermal resistance")`
`= ((theta_1 - theta_2)(4piK))/(1/a-1/b)`
`= ((theta_1-theta_2)(ab)(4piK))/(b-a)`
(b)
In the figure, we can see that
Heat current `H_1` = Heat current `H_2`
`:. ((TD)_(PM)/R_(PM)) = ((TD)_(MQ)/R_(MQ))` ..........(i)
Using the result obtained in Example-3 of thermal resistance, we can find
` R_(PM) = 1/(4piK) (1/a-1/r)`
and `R_(MQ) = 1/(4piK) (1/r - 1/b)`
Substituting the values in Eq. (i), we have
`((theta_1-theta)/((1/(4piK)) (1/a-1/r))) = (theta - theta_2)/((1/4piK)(1/r-1/b))`
Solving this equation we can find `theta`.
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