Derive an expression for the equivalent size of the pipe to replace the pipes in series.

Equivalent pipe : The pipe of uniform diameter having loss of head and discharge equal to the loss of head and discharge of a compound pipe, consisting of several pipe of different length and diameters.

Now, Let $L_1$ = Length of pipe 1 and $d_1$ = dia.

$L_2$ = Length of pipe 2 and $d_2$ = dia.

$L_3$ Length of pipe 3 and $d_3$ = dia.

H = total head loss

L = length of equivalent pipe

d = dia. of equivalent pipe.

$\therefore L = L_1 + L_2 + L_3$

neglect minor losses, the head loss in compound pipe,

$H = \frac{4f_1L_1v_1^2}{d\times2g} + \frac{4f_2L_2V_2^2}{d\times2g} + \frac{4f_3L_3V_3^2}{d\times2g} ----- (1)$

(Assume $(f_1 = f_2 = f_3 = f)$

Discharge $Q = Av_1 = A_2 V_2 = A_3 V_3 = \frac{\pi}{4} d_1^2v_1 = \frac{\pi}{4} d_2^2v_2 = \frac{\pi}{4} d_3^2 v_3^2$

Substituting these values in the form of

$v = (\frac{4Q}{\pi d^2})$ in (1)

$H = 4fL_1 \times \frac{ \bigg( \frac{4Q}{\pi d_1^2 } \bigg)^2 }{d_1 \times 2g} + 4fL_2 \times \frac{ \bigg( \frac{4Q}{\pi d_2^2 } \bigg)^2 }{d_2 \times 2g} + 4fL_3 \times \frac{ \bigg( \frac{4Q}{\pi d_3^2 } \bigg)^2 }{d_3 \times 2g} $

$H = \frac{4\times1bfQ}{\pi ^2 \times 2g} (\frac{L_1}{d1^5} + \frac{L_2}{d2^5} + \frac{L_3}{d3^5}) $

$\therefore \text{Head loss in equivalent pipe}, H = \frac{4fLv^2}{d\times2g}$

where, v = $\frac{4Q}{\pi d^2}$

$\therefore$ H = 4fL x $(\frac{4Q}{\pi d^2})^2$ = $\frac{4\times16Q^2f}{\pi ^2 \times 2g}$ $(\frac{L}{d^5})$ ----- (2)

Head loss in compound pipe and in equivalent pipe is same hence equating equation (2) and (1)

$\therefore \frac{4\times16fQ^2}{\pi ^2\times2g}$ $(\frac{L_1}{d1^5} + \frac{L_2}{d2^5} + \frac{L_3}{d3^5}) = \frac{4\times16Q^2}{\pi ^2 \times 2g} (\frac{L}{d^5})$

$\therefore$ $\frac{L_1}{d1^5} + \frac{L_2}{d2^5} + \frac{L_3}{d3^5} = \frac{L}{d^5}$

Equation (3) is known as Dupuit's Equation where L = $L_1 + L_2 + L_3$ and $d_1, d_2, d_3$ are known. Hence equivalent size of pipe (d) can be found out.