Two pipes of diameter D and d and of equal length L are considered. If the pipes are arranged in parallel, the loss of head for either pipe when a total quantity of water flows through them is h. If the pipes are arranged in series and the same quantity Q flows through them, the loss of head is H. If d= 0.5D, find the percentage of the total flow through each pipe when placed in parallel and the ratio of H to h. Neglect minor losses and assume friction coefficients to be constants.

Solution :

Given :

Length of pipe 1,

$ L_{1}=L$ and its diameter $d_{1}=D$

Length of pipe 2,

$ L_{2}=L$ and its diameter $d_{2}=d$

Total discharge $=Q$

Head loss when pipes are in parallel $=h$

Head loss when pipes are in series $=H$

$d=\frac{D}{2}$ and f is constant

Case 1:

When pipes are connected to parallel

$Q=Q_{1}+Q_{2}................(i)$

Loss of head in each pipe $=h$

For pipe $A B$,$\quad \frac{4 f L_{1} V_{1}^{2}}{d_{1} \times 2 g}=h$,

where

$$\begin {aligned} V_{1}=\frac{Q_{1}}{A_{1}}=\frac{Q_{1}}{\frac{\pi}{4} D^{2}}=\frac{4 Q_{1}}{\pi D^{2}} \end {aligned}$$

$$ d_{1}=D $$

For pipe A C,

$\begin{aligned}h= \frac{32 f L Q_{2}^{2}}{\pi^{2} d^{5}\times g} \end{aligned} ...............(ii)$

$$ \begin {aligned} \therefore \frac{32 f L Q_{1}^{2}}{\pi^{2} D^{5} g} &=\frac{32 f L Q_{2}^{2}}{\pi^{2} d^{5} g} \text { or } \\ \frac{Q_{1}^{2}}{D^{5}} &=\frac{Q_{2}^{2}}{d^{5}} \end {aligned} $$

or

$ \begin{aligned} \left(\frac{Q_{1}}{Q_{2}}\right)^{2} &=\frac{D^{5}}{d^{5}}=\frac{(2 d)^{5}}{d^{5}} \\ &=2^{5}=32 \\ \frac{Q_{1}}{Q_{2}} &=\sqrt{32}=5.657 or \\ Q_{1} &=5.657 Q_{2} \end{aligned} $

Substituting the values of $Q_{1}$ in equation (i), we get

$\begin{aligned} Q &=5.657 Q_{2}+Q_{2} =6.657 Q_{2}\\ \therefore Q_{2} &=\frac{Q}{6.657}\\ &=0.15 Q \\ \text{From (i)} \\ \therefore Q_{1} &=Q-Q_{2} ...........(iii)\\ &=Q-0.15 Q \\ &=0.85 Q.............(iv) \end{aligned}$

Case 2 :

When the pipes are connected in series.

Total loss = Sum of head losses in the two pipes

$$ \therefore H=\frac{4 f L V_{1}^{2}}{d_{1} \times 2 g}+\frac{4 f L V_{2}^{2}}{d_{2} \times 2 g} $$

where

$\begin {aligned} V_{1} &=\frac{Q}{\frac{\pi}{4} D^{2}}=\frac{4 Q}{\pi D^{2}}, \\ V_{2} &=\frac{Q}{\frac{\pi}{4} d^{2}}=\frac{4 Q}{\pi d^{2}} \end {aligned}$

$ \therefore H=\frac{4 fL \left(\frac{4Q}{\pi D^{2}}\right)^2}{D \times 2 g}+\frac{4 f L \left( \frac{4Q}{\pi d^{2}}\right)^2}{d \times 2 g} $

or

$$\begin {aligned} H=\frac{32 f L Q^{2}}{D^{5} \pi^{2} \times g}+\frac{32 f L Q^{2}}{d^{5} \pi^{2} \times g}....(v) \end{aligned} $$

From equation (ii),

$$\begin {aligned}\frac{32 f L}{\pi^{2} D^{5} \times g}=\frac{h}{Q_{1}^{2}}\end{aligned}$$

and from equation (ii),

$$ \begin {aligned} \frac{32 f L}{\pi^{2} d^{5} \times g}=\frac{h}{Q_{2}^{2}} \end{aligned}$$

Substituting these values in equation (v), we have

$\begin {aligned} H &=Q^{2} \times \frac{h}{Q_{1}^{2}}+Q^{2} \times \frac{h}{Q_{2}^{2}}\\ &=\frac{Q^{2}}{Q_{1}^{2}} h+\frac{Q^{2}}{Q_{2}^{2}} h\\ &=h\left[\frac{Q^{2}}{Q_{1}^{2}}+\frac{Q^{2}}{Q_{2}^{2}}\right] \\ H &=Q^{2} \times \frac{h}{Q_{1}^{2}}+Q^{2} \times \frac{h}{Q_{2}^{2}}\\ \therefore \frac{H}{h} &=\frac{Q^{2}}{Q_{1}^{2}}+\frac{Q^{2}}{Q_{2}^{2}} \end{aligned} $

But from equations (iii) and (iv),

$ Q_{1}=0.85 Q$ and $Q_{2}=0.15 Q$

$\begin {aligned} \therefore \frac{H}{h} &=\frac{Q^{2}}{0.85^{2} Q^{2}}+\frac{Q^{2}}{0.15^{2} Q^{2}} \\ &=\frac{1}{0.85^{2}}+\frac{1}{0.15^{2}}\\ &=1.384+44.444\\ &=45.828 \end {aligned}$