In the figure-II temperature of inner surface of cylinder is 0 ∘ C and temperature of outer surface is 10 0 ∘ C . In figure-I temperature of one end of cylinder is 10 0 ∘ C while another end is at 0 ∘

Question

In the figure-II temperature of inner surface of cylinder is $0^{\circ} C$ and temperature of outer surface is $10 0^{\circ} C$ . In figure-I temperature of one end of cylinder is $10 0^{\circ} C$ while another end is at $0^{\circ} C$ .

The temperature gradients dT/dr and dT/dℓ are measured radially in Figure II and along the cylinder in Figure I, respectively. Both cylinders are made of uniform material.

TestHub · Accepted Answer

(I) Figure (I) dt dQ ​ = − kA d ℓ dT ​ R th ​ = kA ℓ ​ in figure (I) heat flow from A side to side B. In this case k = constant and area of cross section constant. So in steady state d ℓ dT ​ is constant. Figure (II) heat flows in radial direction and in this case area is different for every radius so A is not constant so in steady state dr dT ​  = constant

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