Question:

A pipe with an outer diameter of \(0.02\ \mathrm{m}\) carries hot water. The temperature at the outer surface of the pipe is \(70^\circ\mathrm{C}\). This pipe is covered with two concentric layers of insulation, each having a thickness of \(0.01\ \mathrm{m}\). The first layer (adjacent to the pipe) is made up of felt material (thermal conductivity \(k_f = 0.12\ \mathrm{W\,m^{-1}\,^\circ C^{-1}}\)), and the next layer is made up of asbestos (thermal conductivity \(k_{as} = 0.15\ \mathrm{W\,m^{-1}\,^\circ C^{-1}}\)). The insulated pipe is exposed to ambient air at \(20^\circ\mathrm{C}\). The convection heat transfer coefficient at the outer insulation layer is \(3\ \mathrm{W\,m^{-2}\,^\circ C^{-1}}\). At steady state, neglecting heat transfer due to radiation, the heat loss at the outer insulation layer (in \(\mathrm{W\,m^{-1}}\)) is ______ (rounded off to two decimal places).

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Model the felt layer, asbestos layer, and outer convection film as three series resistances per unit pipe length, then divide the overall temperature drop by their sum.
Updated On: Jul 17, 2026
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Correct Answer: 16.04

Solution and Explanation

Step 1: Radii.

\[ r_1 = 0.01,\ r_2 = 0.02,\ r_3 = 0.03\ \mathrm{m} \]

Step 2-5: Series resistances per unit length.

\[ R_{felt}' = 0.91932,\quad R_{as}' = 0.43025,\quad R_{conv}' = 1.76839\ \mathrm{m\,^\circ C/W} \]

Step 6: Total and heat loss.

\[ R_{total}' = 3.11796,\quad \Delta T=50,\quad q' = 50/3.11796 = 16.037\ \mathrm{W/m} \]
\[ \boxed{q' \approx 16.04\ \mathrm{W/m}} \]
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