0
11kviews
Three pipes of diameter 300 mm, 200 mm & 400 mm and lengths 450 m, 255 m & 315 m respectively are connected in series

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).

1 Answer
0
851views

Given:

For the three pipes in series connected between two tanks

enter image description here

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,

enter image description here

∴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,

enter image description here

∴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.

Please log in to add an answer.

Continue reading...

The best way to discover useful content is by searching it.

Search