Chapter 13, Problem 25P

(a)

To determine

To Calculate: The temperature that required changing the diameter of the barrel.

Answer to Problem 25P

The temperature that required changing the diameter of the barrel is $27°\text{C}$ .

Explanation of Solution

Given:

Final length $L=134.122\text{cm}$

Initial length $L_0=134.110\text{cm}$

Coefficient of thermal expansion $\alpha =12\times {10}^{-6}/°\text{C}$

Initial temperature $T_0=20°\text{C}$

Formula used:

  $\begin{array}{l}\Delta L=\alpha L_0\Delta T\\ \Delta L=L-L_0\\ \Delta L=\alpha L_0(T-T_0)\end{array}$

This expression can be written as to find out final temperature

  $T=T_0+\frac{L-L_0}{\alpha L_0}$

Here, $\Delta L$ is the change in length,

  $\alpha$ is the coefficient of thermal expansion of iron,

  $\Delta T$ is the change in temperature.

Calculation:

Substitute the values in the above expression.

  $\begin{array}{c}T=20\text{°C}+\frac{134.122\text{cm}-134.110\text{cm}}{(12\times {10}^{-6}\text{/°C})(134.110\text{cm})}\\ =27.457\text{°C}\approx 27\text{°C}\end{array}$

(b)

To determine

To calculate: The tension of the band when the temperature is decreased.

Answer to Problem 25P

The tension of the band when the temperature is decreased by $20°C$ is $4.3\times {10}^3\text{N}$ .

Explanation of Solution

Given:

Width if the barrel $b=7.4\text{cm}$ .

The Thickness of the barrel $t=0.65cm$ .

Change in temperature $\Delta T=27.457°\text{C}-20°\text{C}=7.457°\text{C}$ .

Formula:

Write the expression for the stress:

  $Stress\left(\sigma \right)=\frac{F}{A}=\alpha E\Delta T$

This can be written as:

  $F=AE\frac{\Delta L}{L_0}=AE\alpha \Delta T$

Here, $A$ is the area of the barrel,

  $E$ is the young modulus,

  $\Delta T$ is the change in the temperature,

  $\alpha$ is the coefficient of thermal conductivity.

Calculation:

Substitute the values in the above expression.

  $\begin{array}{l}=(7.4\times {10}^{-2}\text{m})(6.5\times {10}^{-3}\text{m})(100\times {10}^9{\text{N/m}}^{\text{2}})(12\times {10}^{-6}\text{/°C})(7.457°\text{C})\\ =4.3\times {10}^3\text{N}\end{array}$