Chapter 5, Problem 115P

In some applications, elbow-type flow meters like the one shown in Fig. P5−115 are used to measure flow rates. The pipe radius is R. the radius of curvature of the elbow is A, and the pressure difference AP across the curvature inside the pipe is measured. From the potential flow theory, it is known that $V_r=C$ , where V is the fluid velocity at a distance r from the center of curvature O. and C is a constant. Assuming frictionless steady-state flow and thus the Bernoulli equation across streamlines to be applicable, obtain a relation for the flow rate as a function of $\rho ,g,\Delta ,\lambda$ , and R.

To determine

A relation for the flow rate as a function of $\rho$, $g$, $\Delta P$, $I$ and $R$.

Answer to Problem 115P

A relation for the flow rate is $\pi {\left(\lambda -R\right)}^2{\left(\lambda +R\right)}^2\sqrt{\frac{g}{4\lambda R\left({\lambda }^2+R^2\right)}\left[\frac{\Delta P}{\rho g}+\lambda \right]}$

Explanation of Solution

Given information:

The pipe radius is $R$, the radius of the curvature of the elbow is $I$, the pressure difference across the curvature inside the pipe is $\Delta P$

The Free body diagram of the elbow pipe is shown below.

  Figure-(1)

Write the expression between the velocity and volume flow rate for inner pipe.

   $V_1=\frac{\dot{V}}{A_1}$    ...... (I)

Here, area of the pipe at point $1$ is $A_1$.

Write the expression between the velocity and volume flow rate for outer pipe.

   $V_2=\frac{\dot{V}}{A_2}$    ...... (II)

Here, area of the pipe at point $2$ is $A_2$.

Write the Bernoulli Equation

   $\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+z_2$    ...... (III)

Here, pressure at point $1$ of the pipe is $P_1$, pressure at point $2$ of the pipe is $P_2$, velocity at point $1$ of the pipe is $V_1$, velocity at point $2$ of the pipe is $V_2$, datum height at point $1$ of the pipe is $z_1$, datum height at point $2$ of pipe is $z_2$, acceleration due to gravity is $g$, and density of water is $\rho$.

Write the relation for the inner radius of the pipe.

   $r_1=\lambda -R$    ...... (IV)

Here, $\lambda$ is the mean radius of the pipe from the centre of the curvature $O$ and $R$ is the radius of the pipe.

Write the relation for the inner radius of the pipe.

   $r_2=\lambda +R$    ...... (V)

Write the expression for area of the pipe at point $1$.

   $A_1=\pi r_1^2$    ...... (VI)

Write the expression for area of the pipe at point $2$.

   $A_2=\pi r_2^2$    ...... (VII)

Substitute $\pi r_1^2$ for $A_1$ in Equation (I).

   $V_1=\frac{\dot{V}}{\pi r_1^2}$

Substitute $\pi r_1^2$ for $A_2$ in Equation (II).

   $V_2=\frac{\dot{V}}{\pi r_2^2}$

Substitute $\frac{\dot{V}}{\pi r_1^2}$ for $V_1$, $\frac{\dot{V}}{\pi r_2^2}$ for $V_2$ in Equation (III).

   $\begin{array}{l}\frac{P_1}{\rho g}+\frac{{\dot{V}}^2}{2{\pi }^{2}g{r}_1^4}+z_1=\frac{P_2}{\rho g}+\frac{{\dot{V}}^2}{2{\pi }^{2}g{r}_2^4}+z_2\\ \frac{P_1-P_2}{\rho g}+\left(z_2-z_1\right)=\frac{{\dot{V}}^2}{2{\pi }^{2}g}\left(\frac{r_2^4-r_1^4}{r_1^4{r}_2^4}\right)\\ {\dot{V}}^2=2{\pi }^{2}g\left(\frac{r_1^4{r}_2^4}{r_2^4-r_1^4}\right)\left[\frac{P_1-P_2}{\rho g}+\left(z_2-z_1\right)\right]\\ \dot{V}=r_1^2{r}_2^2\sqrt{\frac{2{\pi }^{2}g}{r_2^4-r_1^4}\left[\frac{P_1-P_2}{\rho g}+\left(z_2-z_1\right)\right]}\end{array}$    ...... (VIII)

Substitute $\Delta P$ for $P_2-P_1$ in Equation (VIII).

   $\dot{V}=r_1^2{r}_2^2\sqrt{\frac{2{\pi }^{2}g}{r_2^4-r_1^4}\left[\frac{\Delta P}{\rho g}+\left(z_2-z_1\right)\right]}$    ...... (IX)

Substitute $0$ for $z_1$ and $\lambda$ for $z_2$ in Equation (IX)

   $\dot{V}=r_1^2{r}_2^2\sqrt{\frac{2{\pi }^{2}g}{r_2^4-r_1^4}\left[\frac{\Delta P}{\rho g}+\left(\lambda -0\right)\right]}$    ...... (X)

Substitute $\lambda +R$ for $r_2$ and $\lambda +R$ for $r_2$ in Equation (X).

   $\begin{array}{c}\dot{V}={\left(\lambda -R\right)}^2{\left(\lambda +R\right)}^2\sqrt{\frac{2{\pi }^{2}g}{{\left(\lambda +R\right)}^4-{\left(\lambda -R\right)}^4}\left[\frac{\Delta P}{\rho g}+\lambda \right]}\\ ={\left(\lambda -R\right)}^2{\left(\lambda +R\right)}^2\sqrt{\frac{2{\pi }^{2}g}{\left({\left(\lambda +R\right)}^2+{\left(\lambda -R\right)}^2\right)\left({\left(\lambda +R\right)}^2-{\left(\lambda -R\right)}^2\right)}\left[\frac{\Delta P}{\rho g}+\lambda \right]}\\ ={\left(\lambda -R\right)}^2{\left(\lambda +R\right)}^2\sqrt{\frac{2{\pi }^{2}g}{2\left({\lambda }^2+R^2\right)\left(4\lambda R\right)}\left[\frac{\Delta P}{\rho g}+\lambda \right]}\\ =\pi {\left(\lambda -R\right)}^2{\left(\lambda +R\right)}^2\sqrt{\frac{g}{4\lambda R\left({\lambda }^2+R^2\right)}\left[\frac{\Delta P}{\rho g}+\lambda \right]}\end{array}$

Conclusion:

A relation for the flow rate is $\pi {\left(\lambda -R\right)}^2{\left(\lambda +R\right)}^2\sqrt{\frac{g}{4\lambda R\left({\lambda }^2+R^2\right)}\left[\frac{\Delta P}{\rho g}+\lambda \right]}$