Derive an expression for equivalent size of pipe in series. A piping system consists of three pipes arranged in series. The lengths of the pipes are 1000m, 800m, 300m and the diameters are 500mm, 400mm and 300mm respectively when they are connected in series. These pipes are to be replaced by a single pipe of length 2100m. Find the diameter of single pipe

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.

Given :

Length of pipe 1 : $L_1$ = 1000 m and $d_1$ = 500 mm = 0.5 m

Length of pipe 2 : $L_2$ = 800 m and $d_2$ = 400 mm = 0.4 m

Length of pipe 3 : $L_3$ = 300 m and $d_3$ = 300 mm = 0.3 m

Length of single pipe L = 2100 m

Diameter of single pipe d = ?

Dupit Equation is,

$\frac{L}{d^5} = \frac{L_1}{d_1^5} + \frac{L_2}{d_2^5} + \frac{L_3}{d_3^5}$

$\therefore$ $\frac{2100}{d^5} = \frac{1000}{0/5^5} + \frac{800}{0.4^5} + \frac{300}{0.3^5}$

$\therefore 1/d^5 = \frac{233581.7901}{2100}$

$\therefore$ d = 0.3897 m

i.e. d = 389.7 mm

$\therefore$ The diameter of single pipe is 389.7 mm.