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A cylindrical tube has an inner radius of r1 and an outer radius of r2. Using the heat equation, derive an expression for the steady-state temperature distribution across the thickness of the tube wall, if the heat flux and temperature on the inner surface of the tube are q1" and Ti, respectively. There is no heat generation within the tube wall.

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Answer: q1 = (-k¶l(r2+r1)(TI-To))/(r2-r1)

Explanation: consider a tube of inner radius r1 and outer radius r2, let the temperature of the outer wall be To and temperature of inner wall be TI and length of tube to be l. Let heat flow rate be q1.

From heat flow rate equation,

q1 = -kAdT/dx,

dT = Ti-To,

dx = r2-r1 (wall thickness)

q1 = -kA(Ti-To) /(r2-r1)

Where A is the mean surface area of the inner and outer walls of the tube. If wall of tube is not too thick we can take A as,

A = 2¶l(r2+r1)/2 = ¶l(r2+r1) for a length of l

Therefore,

q1 = (-k¶l(r2+r1)(TI-To))/(r2-r1)

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