Chapter 13, Problem 85A

To determine

The quantity of methanol that will flow out of the tank.

Answer to Problem 85A

  $\text{500}\text{L}$

Explanation of Solution

Given:

Diameter of the tank, $d=2.0000\text{ m}$

Height of the tank, $h=5.0000\text{ m}$

Initial temperature, $T_1={10}^0\text{C}$

Final temperature, $T_2={40}^0\text{C}$

Formula used:

Change in volume, $\Delta V=\beta V_1\Delta T$

Here, $\beta$ is the coefficient of thermal expansion.

Calculation:

Change in temperature

  $\begin{array}{l}\Delta T=({40}^0{\text{ C-10}}^0\text{ C)}\\ {\text{ = 30}}^0\text{C}\end{array}$

Coefficient of thermal expansion of methanol and the steel is,

  $\begin{array}{l}{\beta }_{methanol}=1100\times {10}^{-6}{/}^0\text{C}\\ {\beta }_{steel}=35\times {10}^{-6}{/}^0\text{C}\end{array}$

Volume of the steel tank,

  $V_1=\frac{\pi }{4}{d}^{2}h$

  $V_1=\frac{\pi }{4}{(2.0000)}^2\times \text{5}{\text{.0000 m}}^3$

  $V_1=15.7{\text{ m}}^3$

Now substitute all the values in the formula, $\Delta V$

  $\Delta V=({\beta }_{methanol}-{\beta }_{steel})V_1\Delta T{\text{ m}}^3$

  $\Delta V=(1100\times {10}^{-6}{/}^0\text{C - 35}\times {10}^{-6}{/}^0\text{C})\times 15.7{\text{ m}}^3\times {30}^0\text{C }$

  $\begin{array}{l}\Delta V=0.5{\text{ m}}^3=0.5\times {10}^3\text{L}\\ \text{ = 500}\text{L}\end{array}$

Conclusion:  The volume of methanol that will flow out from tank is $\text{500}\text{L}$ .