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Q115CP

Expert-verifiedFound in: Page 582

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 \(\frac{{2\pi kL\left( {{T_2} - {T_1}} \right)}}{{\ln \left( {{b \mathord{\left/{\vphantom {b a}} \right.\\} a}} \right)}}\).

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 \(k\).

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.**

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 \(\frac{{2\pi kL\left( {{T_2} - {T_1}} \right)}}{{\ln \left( {{b \mathord{\left/{\vphantom {b a}} \right.\\} a}} \right)}}\).

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