Chapter 5, Problem 57EP

Air is flowing through a venturi meter whose diameter is 2.6 in at the entrance part (location 1) and 1.8 in at the throat (location 2). The gage pressure is measured 239 to be 12.2 psia at the entrance and 11.8 psia at the throat. Neglecting frictional effects, show that the volume how rate can be expressed as $V=A_2\sqrt{\frac{2\left(P_1-P_2\right)}{\upsilon \left(1-A_2^2/A_1^2\right)}}$and determine the flow rate of air. Take the air density to be 0.075 Ibm/ft³.

To determine

The volume flow rate.

The flow rate of the air.

Answer to Problem 57EP

The volume flow rate can be expressed as $A_2\sqrt{\frac{2\left(P_1-P_2\right)}{\rho \left(1-A_2{}^2/A_1{}^2\right)}}$.

The flow rate of the air is $175.96{\text{ft}}^{\text{3}}/\text{s}$.

Explanation of Solution

Given information:

The inlet diameter of the venturimeter is $2.6\text{in}$, the diameter of the throat is $1.8\text{in}$, the inlet pressure of the venturimeter is $12.2\text{psia}$, the throat pressure is $11.8\text{psia}$ and the density of the air is $0.075\text{lbm}/{\text{ft}}^3$.

The figure below shows the different points of venturimeter.

  

  Figure-(1)

Write the expression for Bernoulli's equation between the points $1$ and $2$.

  $\frac{P_1}{\rho g}+\frac{V_1{}^2}{2g}+Z_1=\frac{P_2}{\rho g}+\frac{V_2{}^2}{2g}+Z_2$   .......(I)

Here, the pressure at point $1$ is $P_1$, the pressure at point $2$ is $P_2$, the density of air is $\rho$, acceleration due to gravity is $g$, the velocity at point $1$ is $V_1$, the velocity at point $2$ is $V_2$, the datum head at point $1$ is $Z_1,$ and the datum head at point $2$ is $Z_2$.

The point $1$ and $2$ is taken as datum line, the datum head at point $1$ and $2$ becomes zero.

Substitute $0$ for $Z_1$ and $0$ for $Z_2$ in Equation (I).

  $\begin{array}{l}\frac{P_1}{\rho g}+\frac{V_1{}^2}{2g}+0=\frac{P_2}{\rho g}+\frac{V_2{}^2}{2g}+0\\ P_1-P_2=\frac{\rho }{2}\left(V_2{}^2-V_1{}^2\right)\end{array}$  .......(II)

Write the expression for continuity equation at point $1$.

  $V_1=\frac{\dot{V}}{A_1}$

Here, the flow rate is $\dot{V}$ and the area of the section $1$ is $A_1$.

Write the expression for continuity equation at point $2$.

  $V_2=\frac{\dot{V}}{A_2}$

Here, the flow rate is $\dot{V}$ and the area of the section $2$ is $A_2$.

Write the expression for area of the inlet.

  $A_1=\frac{\pi }{4}{d}_1{}^2$   .......(IV)

Here, the inlet diameter of the venturimeter is $d_1$.

Write the expression for area of the throat.

  $A_2=\frac{\pi }{4}{d}_2{}^2$   .......(V)

Here, the throat diameter of the venturimeter is $d_2$.

Calculation:

Substitute $\dot{V}/A_1$ for $V_1$ and $\dot{V}/A_2$ for $V_2$ in Equation (III).

  $\begin{array}{l}{P}_1-P_2=\frac{{\rho }_{air}}{2}\left({\left(\frac{\dot{V}}{A_2}\right)}^2-{\left(\frac{\dot{V}}{A_1}\right)}^2\right)\\ {\dot{V}}^2=\frac{2\left(P_1-P_2\right){\left(A_2{A}_1\right)}^2}{\rho \left(A_1{}^2-A_2{}^2\right)}\\ \dot{V}=A_2\sqrt{\frac{2\left(P_1-P_2\right)}{\rho \left(1-\frac{A_2{}^2}{A_1{}^2}\right)}}\end{array}$  .......(VI)

Substitute $2.6\text{in}$ for $d_1$ in equation (IV).

  $\begin{array}{c}{A}_1=\frac{\pi }{4}{\left(2.6\text{in}\right)}^2\\ =0.785{\left(2.6\text{in}\right)}^2\\ =5.30{\text{in}}^2\end{array}$

Substitute $1.8\text{in}$ for $d_2$ in equation (V).

  $\begin{array}{c}{A}_2=\frac{\pi }{4}{\left(1.8\text{in}\right)}^2\\ =0.785{\left(1.8\text{in}\right)}^2\\ =2.54{\text{in}}^2\end{array}$

Substitute $5.30{\text{in}}^2$ for $A_1$, $2.54{\text{in}}^2$ for $A_2$, $0.075\text{lbm}/{\text{ft}}^{\text{3}}$ for $\rho$, $12.2\text{psia}$ for $P_1,$ and $11.8\text{psia}$ for $P_2$ in Equation (VI).

  $\begin{array}{c}\dot{V}=2.54{\text{in}}^2\sqrt{\frac{2\left(12.2\text{psia}-11.8\text{psia}\right)}{0.075\text{lbm}/{\text{ft}}^{\text{3}}\left(1-\frac{{\left(2.54{\text{in}}^2\right)}^2}{{\left(5.30{\text{in}}^2\right)}^2}\right)}}\\ =2.54{\text{in}}^2\sqrt{\frac{2\left(0.4\text{psia}\left(\frac{6894.75728\text{N}/{\text{m}}^{\text{2}}}{1\text{psia}}\right)\right)}{0.075\text{lbm}/{\text{ft}}^{\text{3}}\left(1-\frac{{\left(2.54{\text{in}}^2\right)}^2}{{\left(5.30{\text{in}}^2\right)}^2}\right)}}\\ =2.54{\text{in}}^2\left(\frac{6.94\times {10}^{-3}{\text{ft}}^2}{1{\text{in}}^2}\right)\sqrt{\frac{5515.8\text{N}/{\text{m}}^{\text{2}}}{0.075\text{lbm}/{\text{ft}}^{\text{3}}\left(0.77{\text{in}}^2\right)\left(\frac{6.94\times {10}^{-3}{\text{ft}}^2}{1{\text{in}}^2}\right)}}\\ =0.0176{\text{ft}}^2\sqrt{\frac{5515.8\text{N}/{\text{m}}^{\text{2}}}{4.01\times {10}^{-4}\text{lbm}/\text{ft}}}\end{array}$

  $\begin{array}{c}\dot{V}=0.0176{\text{ft}}^2\sqrt{\frac{5515.8\text{N}/{\text{m}}^{\text{2}}\left(\frac{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{N}}\right)}{4.01\times {10}^{-4}\text{lbm}/\text{ft}\left(\frac{0.453592\text{kg}}{1\text{lbm}}\right)}}\\ =0.0176{\text{ft}}^2\sqrt{\frac{5515.8\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\left(\frac{3.28\text{ft}}{1\text{m}}\right)}{1.81\times {10}^{-4}\text{kg}/\text{ft}}}\\ =175.96{\text{ft}}^{\text{3}}/\text{s}\end{array}$

Conclusion:

The volume flow rate can be expressed as $A_2\sqrt{\frac{2\left(P_1-P_2\right)}{\rho \left(1-A_2{}^2/A_1{}^2\right)}}$.

The flow rate of the air is $175.96{\text{ft}}^{\text{3}}/\text{s}$.