The difference in water surface levels in the two tanks is 18 m. Determine the rate of flow of water if coefficients of friction are 0.0075, 0.0078 & 0.0072 respectively (consider minor losses).

Given:

For the three pipes in series connected between two tanks

Now, Applying Bernoulli’s Principle between the liquid surfaces in the two tanks, $\frac{P_1}{ρg}+\frac{v_1^2}{2g}+Z_1=\frac{P_2}{ρg}+\frac{v_2^2}{2g}+Z_2+h_l$

Now,

Assuming both the tanks are open to atmosphere ⇒$P_1=P_2$

Assuming the tanks are infinitely large such that the water level does not change as flow occurs

⇒$Z_1-Z_2$=Constant=h=18 m and $v_1=v_2$=0

Substituting in the above equation, ∴$Z_1-Z_2=h_l$=18

Now, $h_l$ is the sum of the major and minor losses between the two tanks,

∴$h_l$=Major losses+Minor losses

Now, Major losses=Friction losses through $(Pipe_a+〖Pipe〗_b+Pipe_c)$ Friction loss through a pipe is given by,

$h_f=\frac{flv^2}{2gd}$

(Assuming the Coefficient of friction values given are the Darcy Weisbach friction factor)

∴Major losses=$(\frac{flv^2}{2gd})_a+(\frac{flv^2}{2gd})_b+(\frac{flv^2}{2gd})_c$

Now, For pipes in series

Q ̇=$(vA)_a=(vA)_b=(vA)_c$

∴Major losses=$(\frac{flQ ̇^{2}}{2gdA^{2}})_a+(\frac{flQ ̇^{2}}{2gdA^{2}})_b+(\frac{flQ ̇^{2}}{2gdA^{2}})_c$

∴Major losses=$\frac{Q ̇^2}{2g} \{\frac{fl}{dA^2})_a+(\frac{fl}{dA^2})_b+(\frac{fl}{dA^2})_c\}$

Substituting in the above equation,

∴Major losses=$\frac{12687.03Q ̇^2}{2g}$

Now, Minor losses=Entrance loss+Contraction loss+Enlargement loss+Exit loss

We know that,

Entrance loss=$\frac{0.5 v_a^{2}}{2g}$

Contraction loss=$\frac{0.5v_b^{2}}{2g}$

Enlargement loss=$(1-\frac{A_b}{A_c})^2 \frac{v_b^{2}}{2g}$

Exit loss=$\frac{v_c^2}{2g}$

∴Minor losses=$\frac{0.5v_a^{2}}{2g}+\frac{0.5v_b^{2}}{2g}+(1-\frac{A_b}{A_c})^2 \frac{v_b^2}{2g}+\frac{v_c^2}{2g}$

∴Minor losses=$\frac{0.5Q ̇^2}{2gA_a^2}+\frac{0.5Q ̇^2}{2gA_b^2}+(1-\frac{A_b}{A_c})^2 \frac{Q ̇^2}{2gA_b^2 }+\frac{Q ̇^2}{(2gA_c^2}$

∴Minor losses=$\frac{Q ̇^2}{2g} \{\frac{0.5}{A_a^2}+\frac{0.5}{A_b^2}+(1-\frac{A_b}{A_c})^2 \frac{1}{A_b^2}+\frac{1}{(A_c^2}\}$

Substituting in the above equation,

∴Minor losses=$\frac{1239.934Q ̇^2}{2g}$

∴$h_l=\frac{f12687.03Q ̇^2}{2g}+\frac{1239.934Q ̇^2}{2g}$

∴$h_l=\frac{13926.96Q ̇^2}{2g}$

∴18=$\frac{13926.96Q ̇^2}{2g}$

∴$Q ̇^2$=0.02536

∴Q ̇=$0.1592 m^3/s$

…this is the required rate of flow of water.