Chapter 6, Problem 6.103P

To determine

Estimate the flow of water at 20° C.

Answer to Problem 6.103P

$Q=0.11f{t}^3/s$

Explanation of Solution

Given information:

Both reservoirs are connected by cast iron pipes with sharp edged entrance and exit. Reservoir 1 is $45ft$ higher than reservoir 2. The fluid is water at $20°C$

The volumetric flow rate Q can be defined as below,

$Q=VA$

In above equation,

$V$ - Velocity of the flow

$A$ -Area

The steady flow equation between the reservoir 1 & 2 will be,

$\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2+h_f+{\displaystyle \sum h_m-h_p}$

We know that, $p_1=p_2$ and $V_1=V_2\approx 0$, therefore the above equation can be rewritten as,

$h_p=z_2-z_1+h_f+{\displaystyle \sum h_m}$

The Reynolds’s number is defined as,

$R_e=\frac{DV}{\nu }$

$D$ - Diameter of the pipe

$V$ - Velocity of the flow

$\nu$ - Kinematic viscosity

The friction factor can be determined by,

$\frac{1}{f^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)$

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

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

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

Assume that the wrought iron will have a roughness of,

$\epsilon =0.00015ft$

Calculation:

Assume small pipe as (a) and the large pipe as (b).

First of all, the roughness ratio will be,

For pipe (a)

$\frac{\epsilon }{d_a}=\frac{0.00015ft}{\frac{1}{12}ft}=1.8\times {10}^{-3}$

For pipe (b)

$\frac{\epsilon }{d_b}=\frac{0.00015ft}{\frac{2}{12}ft}=9\times {10}^{-4}$

Loss co-efficient for pipe (a) will be,

$\begin{array}{l}{K}_{sharpentrance}=0.5\\ K_{sudden\exp ansion}=0.56\end{array}$

Similarly for pipe (b),

$K_{submergedexit}=1.0$

By equating the volumetric flow for both pipes,

$\begin{array}{l}{Q}_a=Q_b\\ V_a\left(\frac{\pi d_a{}^2}{4}\right)=V_b\left(\frac{\pi d_b{}^2}{4}\right)\\ V_a\left(\frac{\pi d_a{}^2}{4}\right)=V_b\left(\frac{\pi {\left(2d_a\right)}^2}{4}\right)\\ V_a=4V_b\end{array}$

According to the steady flow equation between reservoir (1) & (2),

$45ft=\frac{V_a{}^2}{2\left(32.2ft/s^2\right)}\left(f_a\frac{\left(20ft\right)}{\frac{1}{12}ft}+\left(0.5+0.56\right)\right)+\frac{V_b{}^2}{2\left(32.2ft/s^2\right)}\left(f_b\frac{\left(20ft\right)}{\frac{2}{12}ft}+\left(1.0\right)\right)$

We know that,

$V_b=\frac{V_a}{4}$

Therefore,

$\begin{array}{l}45ft=\frac{V_a{}^2}{2\left(32.2ft/s^2\right)}\left(240f_a+1.06\right)+\frac{{\left(\frac{V_a}{4}\right)}^2}{2\left(32.2ft/s^2\right)}\left(120f_b+\left(1.0\right)\right)\\ =\left[\frac{V_a{}^2}{2\left(32.2ft/s^2\right)}\right]\left(\left(240f_a+1.06\right)+\left(\frac{120f_b+\left(1.0\right)}{16}\right)\right)\end{array}$

To find $V_a$ the friction factor needs to be known, therefore assume $f_a=f_b=0.02$

$\begin{array}{l}45ft=\left[\frac{V_a{}^2}{2\left(32.2ft/s^2\right)}\right]\left(\left(240(0.02)+1.06\right)+\left(\frac{120(0.02)+\left(1.0\right)}{16}\right)\right)\\ V_a=21.845ft/s\end{array}$

To find Reynolds’s number,

$\begin{array}{l}{R}_e=\frac{DV}{\nu }\\ =\frac{\frac{1}{12}ft(21.845ft/s)}{1.09\times {10}^{-5}}\\ =167010.70\end{array}$

To find friction factor,

$\begin{array}{l}\frac{1}{f^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\\ \approx -1.8\log \left(\frac{6.9}{167010.7}+{\left(\frac{1.8\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\\ \approx 0.0238\end{array}$

Similarly for pipe (b),

$V_b=\frac{V_a}{4}=\frac{21.845ft/s}{4}=5.46125ft/s$

To find Reynolds’s number,

$\begin{array}{l}{R}_e=\frac{DV}{\nu }\\ =\frac{\frac{2}{12}ft(5.46125ft/s)}{1.09\times {10}^{-5}}\\ =83505.35\end{array}$

To find the friction factor,

$\begin{array}{l}\frac{1}{f^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\\ \approx -1.8\log \left(\frac{6.9}{83505.35}+{\left(\frac{9\times {10}^{-4}}{3.7}\right)}^{1.11}\right)\\ \approx 0.022\end{array}$

Therefore, to find the absolute value of $V_a$

$\begin{array}{l}45ft=\left[\frac{V_a{}^2}{2\left(32.2ft/s^2\right)}\right]\left(\left(240(0.0238)+1.06\right)+\left(\frac{120(0.022)+\left(1.0\right)}{16}\right)\right)\\ V_a=20.347ft/s\end{array}$

To calculate the flow of water,

$\begin{array}{l}Q=VA\\ =\left(20.347ft/s\right)\left(\pi {\left(\frac{0.5}{12}\right)}^2\right)\\ =0.11f{t}^3/s\end{array}$

Conclusion:

The flow of water is equal to $Q=0.11f{t}^3/s$.