Chapter 6, Problem 6.126P

To determine

The proper valve opening angle in degrees for pipe 1.

Answer to Problem 6.126P

$Openingangle=20°$

Explanation of Solution

Given information:

The fluid is water at $20°C$

The pipe 1 is fitted with a butterfly valve, since the flow rate is equal to $1.5f{t}^3/s$

At three reservoir junction, if all flows are considered positive towards the junction, then

$Q_1+Q_2+Q_3=0\to \left(1\right)$

Assuming the gauge pressure $p_1=p_2=p_3=0$, the head loss $\Delta h$ across each pipe is equal to,

$\Delta h=f\left(\frac{L}{D}\right)\left(\frac{V^2}{2g}\right)=z-h_j$

$f$ - Friction factor

$L$ - Length of the pipe

$D$ - Diameter of the pipe

$V$ -Velocity of the flow

$z$ - Elevation

To solve this type of problems, we should guess the position of $h_j$ and solve the above mentioned equation to find the relevant velocities and flow rates iterating until the flow rates satisfies equation 1.

If the $h_j$ guessed was too high, then the sum of equation (1) will be negative, therefore we have to reduce $h_j$ and vice versa.

Since pipe 1 is fitted with a butterfly valve, the above equation can be written as,

$\Delta h=\left(\frac{V^2}{2g}\right)\left(f\left(\frac{L}{D}\right)+K_{butterflyvalve}\right)=z-h_j$

Calculation:

Assume, the water at $20°C$ will have,

$\rho =1.94slug/f{t}^3$

$\mu =2.09\times {10}^{-5}slug/fts$

The roughness of galvanized iron is equal to, $\epsilon =0.0005ft$

Assume $f=0.02$ for all pipes,

$h_j=51ft$

For pipe 2,

Calculate the velocity,

$\begin{array}{l}f\left(\frac{L}{D}\right)\left(\frac{V^2}{2g}\right)=z-h_j\\ \left(0.02\right)\left(\frac{1200ft}{\frac{6}{12}ft}\right)\left(\frac{V_2{}^2}{2\left(32.2ft/s^2\right)}\right)=49ft\\ V_2=8.11ft/s\end{array}$

Calculate the flow rate,

$Q_2=V_{2}A=\left(8.11ft/s\right)\left(\pi {\left(\frac{3}{12}\right)}^2\right)=1.59f{t}^3/s$

For pipe 3,

Calculate the velocity,

$\begin{array}{l}f\left(\frac{L}{D}\right)\left(\frac{V^2}{2g}\right)=z-h_j\\ \left(0.02\right)\left(\frac{1600ft}{\frac{9}{12}ft}\right)\left(\frac{V_3{}^2}{2\left(32.2ft/s^2\right)}\right)=-1ft\\ V_3=-1.23ft/s\end{array}$

Calculate the flow rate,

$Q_3=V_{3}A=\left(-1.23ft/s\right)\left(\pi {\left(\frac{4.5}{12}\right)}^2\right)=-0.54f{t}^3/s$

But we know that,

$Q_1+Q_2+Q_3=0\to \left(1\right)$

Therefore,

$Q_1=-\left(Q_2+Q_3\right)=-\left(1.59f{t}^3/s+\left(-0.54f{t}^3/s\right)\right)=-1.05f{t}^3/s$

The velocity of the flow in pipe 1 is equal to,

$V_1=\frac{Q_1}{A}=\frac{1.05f{t}^3/s}{\left(\pi {\left(\frac{4}{12}\right)}^2\right)}=3ft/s$

To calculate the value of the loss co efficient,

$\begin{array}{l}\left(\frac{V^2}{2g}\right)\left(f\left(\frac{L}{D}\right)+K_{butterflyvalve}\right)=z-h_j\\ \left(\frac{{\left(3ft/s\right)}^2}{2\left(32.2ft/s^2\right)}\right)\left(\left(0.02\right)\left(\frac{1800ft}{\frac{8}{12}}\right)+K_{butterflyvalve}\right)=31\\ K_{butterflyvalve}=168\end{array}$

Therefore, according to the graph 6.19, we can say that,

The opening angle of the butterfly valve is equal to almost $20°$

By further iteration we can get more accurate answers.

Conclusion:

The opening angle of the butterfly valve is equal to almost $20°$.