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Radial heat flow through the sides of a tube

In most of our homes hot water flows through pipes to taps or in hot water systems. If these pipes are not lagged a great deal of heat will be lost. It is no use turning on the hot tap to find that so much heat has been lost that the water is cold before it reaches you from the hot water tank!

Consider a tube of inner radius r and outer radius R. Let the temperature on the outer surface be θ₁ and that inside the tube θ₂. The rate of flow of heat through the walls is then given approximately by:

dH/dt = -kAdθ/dx = - kA(θ₂ - θ1)/(R – r)
where A is the mean surface area of the inner and outer walls of the tube. However, if the walls of the tube are not too thick we can take A as [2π(R + r)L]/2 = pL(R + r), for a length of tube L.

Therefore:

Rate of loss of heat = dH/dt = - kπL[(R + r)/(R - r)](θ₂ – θ₁)

The loss of heat from unlagged central heating pipes can be found by this method. As with the loss of heat through a window, we start by assuming that the temperature outside the pipe is the air temperature but in reality this will not be so.

Example problem

Consider a copper hot water pipe that delivers a flow of water at 0.2 kg s^-1 (m). The pipe has a length of 20 m and the inlet temperature is 60 ^oC.
Calculate the outlet temperature if the exterior of the pipe is at a temperature of 50 ^oC.
(Thermal conductivity of copper (k) = 385 W m^-1K^-1 and specific heat capacity of water (c) = 4200 J kg^-1K^-1)
Let the pipe have an internal diameter of 2.0 cm and an exterior diameter of 3.5 cm.
Let the outlet temperature be q.
Therefore:
mc(60 - q) = kAdq/dx = kA(50 – ½(q + 60))
This gives q = 39.4 ^oC

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