The difference in water surface levels in two tanks, which are connected by 3 pipes in series of length 300 m, 170 m, and 210 m and of diameter 300 mm, 200 mm and 400 mm respectively is 12 m. Determine the rate of flow of water is coefficient of friction are 0.005,0.0052 and 0.0048 respectively, considering major losses only.

Data:-

$L_1=300 m$

$L_2=170 m$

$L_3=200 m$

$D_1=300 mm=0.3m$

$D_2=200 mm=0.2m$

$D_3=400 mm=0.4m$

$f_1=0.05,f_2=0.0052,f_3=0.0048$

$H=12m$

To find:- Q=?

Solution:-

Pipes are in series

$Q=A_1V_1=A_2V_2=A_3V_3$

$V_1=\frac{4Q}{\pi D_1^2},V_2=\frac{4Q}{\pi D_2^2},V_3=\frac{4Q}{\pi D_3^2}$

$H=\frac{4f_1L_1V_1^2}{2gD_1}+4\frac{4f_2L_2V_2^2}{2gD_2}+\frac{4f_3L_3V_3^2}{2gD_3}$

$=\frac{4f_1L_1}{2gD_1}(\frac{4Q}{\pi D_1^2})^2+4\frac{4f_2L_2}{2gD_2}(\frac{4Q}{\pi D_2^2})^2+\frac{4f_3L_3}{2gD_3}(\frac{4Q}{\pi D_3^2})^2$

$l^2=\frac{4\times 0.005\times 300}{2\times 9.81\times 0.3}\times(\frac{4Q}{\pi\times 0.3^2})^2+\frac{4\times 0.0052\times 140}{2\times 9.81\times 0.2}\times(\frac{4Q}{\pi\times 0.2^2})^2+\frac{4\times 0.0048\times 200}{2\times 9.81\times 0.4}\times(\frac{4Q}{\pi\times 0.4^2})^2$

$\therefore l^2=204.016Q^2+913.02Q^2+30.98Q^2$

$\therefore Q=0.1022m^3/s$