Chapter 10, Problem 42P

To determine

To Find: The volume rate of flow of water.

Answer to Problem 42P

The volume rate of flow of water is $4.6\times {10}^{-3}\frac{{\text{m}}^{\text{3}}}{\text{s}}$ .

Explanation of Solution

Given info:

Diameter:

  $d=1.85\text{cm}=0.0185\text{m}$

Height, $h=15.0\text{m}$

Formula used:

Bernoulli’s equation:

  $P_{faucet}+\frac{1}{2}\rho v_{faucet}{}^2+\rho g{h}_{faucet}=P_{head}+\frac{1}{2}\rho v_{head}{}^2+\rho g{h}_{head}$

The velocity of water at the faucet is,

  $\begin{array}{l}{v}^2=2gh\\ v=\sqrt{2gh}\end{array}$

The volume flow rate is given as

  $\text{volume}\text{of}\text{flow}\text{rate}\text{=}\text{area×velocity}$

Calculation:

The pressure head is considered as the initial height for the water with zero initial speed and atmospheric pressure. So, for the calculation for volume rate, Bernoulli’s equation is used.

  $\begin{array}{l}{P}_{faucet}+\frac{1}{2}\rho v_{faucet}{}^2+\rho g{h}_{faucet}=P_{head}+\frac{1}{2}\rho v_{head}{}^2+\rho g{h}_{head}\\ v_{faucet}{}^2=\frac{2}{\rho }\left(P_{head}-P_{faucet}\right)+v_{head}{}^2+2g\left(h_{head}-h_{faucet}\right)\\ v_{faucet}{}^2=2g{h}_{head}\\ v_{faucet}=\sqrt{2g{h}_{head}}\end{array}$

The volume flow rate is given by:

Volume of flow rate= $Av=\pi r^2\times \sqrt{2g{h}_{head}}$

By putting all the values in the above equation

  $\begin{array}{l}=3.14\times {\left(\frac{1.85\times {10}^{-2}}{2}\right)}^2\times \sqrt{2\times 9.8\times 15.0}\\ =4.6\times {10}^{-3}\frac{{\text{m}}^{\text{3}}}{\text{s}}\end{array}$

Conclusion:

The flow rate of the volume is $4.6\times {10}^{-3}\frac{{\text{m}}^{\text{3}}}{\text{s}}$ .