Chapter 10, Problem 85GP

To determine

To Find: The rate of flow.

Answer to Problem 85GP

The rate of flow of the tube is $4\times {10}^{-4}{\text{m}}^{\text{3}}\text{/s}$ .

Explanation of Solution

Given info:

Diameter of the tube $d=1.2\text{ cm }=0.012\text{ m}$ .

Height $h=64\text{ cm }=0.64\text{ m}$ .

Formula used:

Expression of $\Delta P$ :

  $\begin{array}{l}\Delta P=\frac{1}{2}\times \rho \times v^2\\ \Delta P=\rho \times g\times h\\ Q=A\times v\\ A=\frac{\pi }{4}\times d^2\end{array}$

Here,

  $\Delta P$ is the pressure difference,

  $\rho$ is density of fluid,

  $v$ is velocity of fluid,

  $h$ is the height,

  $Q$ is the rate of fluid flow,

  $A$ is the area of flow tube,

  $d$ is the diameter of tube.

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}\rho \times g\times h=\frac{1}{2}\times \rho \times v^2\\ g\times h=\frac{1}{2}\times v^2\\ v^2=2\times g\times h=2\times 9.8\times 0.64=12.54\\ v=3.54\text{ m/s }\end{array}$

Substitute the value of velocity in the equation of discharge.

  $\begin{array}{l}Q=A\times v=\frac{\pi }{4}\times d^2\times v\\ Q=\frac{\pi }{4}\times {\left(0.012\right)}^2\times 3.54\\ Q=4\times {10}^{-4}{\text{ m}}^{\text{3}}\text{/s}\end{array}$

Conclusion:

Thus, the rate of flow of the tube is $4\times {10}^{-4}{\text{ m}}^{\text{3}}\text{/s}$ .