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Q115CP

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Found in: Page 582

### University Physics with Modern Physics

Book edition 14th edition
Author(s) Hugh D. Young, Roger A. Freedman
Pages 1596 pages
ISBN 9780321973610

# A hollow cylinder has length $$L$$, inner radius $$a$$, and outer radius $$b$$, and the temperatures at the inner and outer surfaces are $${T_2}$$ and $${T_1}$$. (The cylinder could represent an insulated hot-water pipe.) The thermal conductivity of the material of which the cylinder is made is $$k$$. Derive an equation for (a) the total heat current through the walls of the cylinder; (b) the temperature variation inside the cylinder walls. (c) Show that the equation for the total heat current reduces to Eq. (17.21) for linear heat flow when the cylinder wall is very thin. (d) A steam pipe with a radius of $$2.00\;cm$$, carrying steam at $$140^\circ C$$, is surrounded by a cylindrical jacket with inner and outer radii $$2.00\;cm$$ and $$4.00\;cm$$ and made of a type of cork with thermal conductivity $$4 \times 1{0^{ - 2}}\;{W \mathord{\left/{\vphantom {W {m \cdot K}}} \right.\\} {m \cdot K}}$$ . This in turn is surrounded by a cylindrical jacket made of a brand of Styrofoam with thermal conductivity $$2.70 \times 1{0^{ - 2}}\;{W \mathord{\left/{\vphantom {W {m \cdot K}}} \right.\\} {m \cdot K}}$$ and having inner and outer radii $$4.00\;cm$$ and $$6.00\;cm$$ (Fig. P17.115). The outer surface of the Styrofoam has a temperature of $$15^\circ C$$. What is the temperature at a radius of $$4.00\;cm$$, where the two insulating layers meet? (e) What is the total rate of transfer of heat out of a $$2.00\;m$$ length of pipe?

The total heat current through the walls of the cylinder is $$\frac{{2\pi kL\left( {{T_2} - {T_1}} \right)}}{{\ln \left( {{b \mathord{\left/{\vphantom {b a}} \right.\\} a}} \right)}}$$.

See the step by step solution

## Step 1: Write the given data from the question.

The length of the cylinder is $$L$$.

Inner radius of cylinder is $$a$$.

Outer radius of cylinder is $$b$$.

The temperature of inner surface is $${T_2}$$ and outer surface is $${T_1}$$.

Thermal conductivity of material is $$k$$.

## Step 2: Determine the formulas to derive the equation for total heat current through the walls of the cylinder.

The expression to calculate the heat current through the walls of cylinder is given as follows.

$$H = kA\frac{{\left( {{T_2} - {T_1}} \right)}}{L}$$ …… (i)

Here, $$A$$ is the area of the cylinder.

## Step 3: Derive the equation for total heat current through the walls of the cylinder.

Let the change in the temperature of inside and outer walls of the cylinder is $$dT$$ and thickness of the shells is $$dr$$.

Substitute $$dT$$ for $${T_2} - {T_1}$$ and $$dr$$ for $$L$$ into equation (i).

$$H = - kA\frac{{dT}}{{dr}}$$

Substitute $$2\pi rL$$ for $$A$$ into above equation.

$$\begin{array}{c}H = - k\left( {2\pi rL} \right)\frac{{dT}}{{dr}}\\\frac{H}{{2\pi r}}dr = - \left( {kL} \right)dT\end{array}$$

Integrate the left side of the above equation from $$a$$ for $$b$$ and right of above equation from $${T_2}$$ to $${T_1}$$.

$$\begin{array}{c}\int_a^b {\frac{H}{{2\pi r}}} dr = - \int_{{T_2}}^{{T_1}} {\left( {kL} \right)dT} \\\frac{H}{{2\pi }}\int_a^b {\frac{1}{r}dr} = - \left( {kL} \right)\int_{{T_2}}^{{T_1}} {dT} \\\frac{H}{{2\pi }}\left( {\ln } \right)_a^b = - \left( {kL} \right)\left( T \right)_{{T_2}}^{{T_1}}\\\frac{H}{{2\pi }}\left( {{\mathop{\rm lnb}\nolimits} - {\mathop{\rm lna}\nolimits} } \right) = - kL\left( {{T_1} - {T_2}} \right)\end{array}$$

Solve further as,

$$\begin{array}{c}\frac{H}{{2\pi }}\ln \left( {\frac{b}{a}} \right) = kL\left( {{T_2} - {T_1}} \right)\\H = \frac{{2\pi kL\left( {{T_2} - {T_1}} \right)}}{{\ln \left( {{b \mathord{\left/{\vphantom {b a}} \right.\\} a}} \right)}}\end{array}$$

Hence the total heat current through the walls of the cylinder is $$\frac{{2\pi kL\left( {{T_2} - {T_1}} \right)}}{{\ln \left( {{b \mathord{\left/{\vphantom {b a}} \right.\\} a}} \right)}}$$.