Chapter 10, Problem 41P

To determine

The rate of flow of volume is to be calculated on the basis of given information.

Answer to Problem 41P

The volume rate of flow is $5.6\times {10}^{-3}{m}^3/\sec$ .

Explanation of Solution

Given info:

Diameter, $d=6.0cm=0.06m\Rightarrow r=0.03m$

Reduced diameter, $d^{\text{'}}=4.0cm=0.04m\Rightarrow r=0.02m$

Gauge pressures in two sections are, $P_1=32.0kPaand{P}_2=24.0kPa$

Density of water, $\rho =1000kg/m^3$

Formula used:

• $A_1{v}_1=A_2{v}_2$, the equation of continuity
• $P_1+\rho g{h}_1+\frac{1}{2}\rho v_1{}^2=P_2+\rho g{h}_2+\frac{1}{2}\rho v_2{}^2$

Calculation:

When there is no change in height, then the Bernoulli’s equation becomes,

  $\begin{array}{l}{P}_1-P_2+\rho g\left(h_2-h_1\right)+\frac{1}{2}\rho \left(v_1{}^2-v_2{}^2\right)=0\\ \Delta P+0+\frac{1}{2}\rho {\left(\Delta v\right)}^2=0\left(\because h_2=h_1\right)\\ \Delta P+\frac{1}{2}\rho {\left(\Delta v\right)}^2=0\end{array}$

Now, from the continuity equations,

  $\begin{array}{l}{A}_1{v}_1=A_2{v}_2\\ v_2=\frac{A_1{v}_1}{A_2}\\ v_2=\frac{A_1{v}_1}{A_2}=\frac{\pi r_1^2}{\pi r_2^2}{v}_1=\frac{\pi r_1^2}{\pi r_2^2}{v}_1=\frac{{\left(r_1\right)}^2}{{\left(r_2\right)}^2}{v}_1\end{array}$

Now use these values in reduces form of Bernoulli’s equation

  $\begin{array}{l}\Delta P+\frac{1}{2}\rho {\left(v_2-v_1\right)}^2=0\\ \Delta P+\frac{1}{2}\rho {\left(\frac{{\left(r_1\right)}^2}{{\left(r_2\right)}^2}{v}_1-v_1\right)}^2=0\\ v_1=\sqrt{\frac{2\frac{\Delta P}{\rho }}{1-\frac{{\left(r_1\right)}^4}{{\left(r_2\right)}^4}}}\end{array}$

Now substitute the values in the s relation then the resulted outcomes is,

  $v_1=\sqrt{\frac{2\frac{\Delta P}{\rho }}{1-\frac{{\left(r_1\right)}^4}{{\left(r_2\right)}^4}}}=\sqrt{\frac{2\times \frac{\left(32-24\right)\times 1000}{1000}}{\left(1-{\left(\frac{0.03}{0.02}\right)}^4\right)}}=3.93846m^3$

Now calculate the volume rate flow such that

  $A_1{v}_1=\pi r_1{}^{2}3.93846m^3=3.93846m^3\times 3.14\times {\left(0.03\right)}^2=5.6\times {10}^{-3}{m}^3/\sec$

Conclusion:

The flow rate of volume is $5.6\times {10}^{-3}{m}^3/\sec$ .