Page: Loss of head continued

Loss of head at the entrance of pipe:-

This loss is similar to loss due to sudden contraction. This loss occurs when a liquid enters A pipe which is connected to a large tank.

This loss is also dependent on the type of entrance.

$h_i=0.5\dfrac {V^2}{2g}$

Loss of head at the exit of pipe:-

This loss is mainly due to the velocity of liquid at the outlet of the pipe.

$h_o=\dfrac {V^2}{2g}$

Loss of head due to an obstruction in a pipe:-

This loss occurs when there is a reduction in the cross-section area of the pipe due to some obstruction. The area suddenly increases or enlarges after the obstruction.

$=\dfrac {(V_c-V)^2}{2g}=\dfrac {\dfrac {A\times V}{C_c(A-a)}-V}{2g}$

$\therefore$ Loss of head due to obstruction

$=\dfrac {V^2}{2g}\left(\dfrac A{C_c(A-a)}-1\right)^2$

Loss of head due to bend in a pipe:-

This loss occurs when there is a bend in a pipe which causes a change in velocity which leads to separation of the flow from the boundaries and it also forms eddies.

Thus the energy is lost.

$h_b=\dfrac {KV^2}{2g}$

Loss of head in various pipe fittings:-

$=\dfrac {KV^2}{2g}$

Q4) A horizontal pipeline 40m long is connected to a water tank at one end and discharges freely into the atmosphere at the other end. For the first 25 m of its length from the tank, the pipe is 150 mm diameter and its diameter is suddenly enlarged to 300 mm. The height of the water level in the tank is 8 m above the centre of the pipe. Considering all losses of head which occurs, determine the rate of flow. Take f=0.01 for both sections of the pipes.


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Total length of pipe, L=40m

Length of first pipe, $L_1=25m$

Diameter of first pipe, $d_1=150mm=0.15m$

Length of second pipe, $L_2=40-25=15m$

Diameter of second pipe, $d_2=300mm=0.30m$

Height of water, $H=8m$

Coefficient of friction, f=0.01

Applying Bernoulli's Theorem,

$0+0+8=\dfrac {P_2}{\rho g}+\dfrac {V_2^2}{2g}+0+\text{all losses}$

$8=0+\dfrac {V_2^2}{2g}+h_i+hf_1+h_e+hf_2................(A)$


loss at entrance, $h_i=0.5\dfrac {V_1^2}{2g}$

Head lost due to friction in pipe 1, $hf_1=\dfrac {4\times f\times L_1\times V_1^2}{d_1\times 2g}$

Loss due to sudden enlargement, $h_e=\dfrac {(V_1-V_2)^2}{2g}$

Head lost due to friction in pipe 2, $hf_2=\dfrac {4\times f\times L_2\times V_2^2}{d_2\times 2g}$

But continuity equation,


$\therefore, V_1=\dfrac {A_2V_2}{A_1}$

$=\dfrac {\dfrac \pi 4 (d_2)^2\times V_2}{\dfrac \pi 4(d_1)^2}$

$=\left(\dfrac {d_2}{d_1}\right)^2\times V_2$

$=\left(\dfrac {0.3}{0.15}\right)^2\times V_2$


Substituting the values of $V_1$ in different losses, we get

$h_i=0.5\dfrac {V_1^2}{2g}=0.5\dfrac {(4V_2)^2}{2g}=\dfrac {8V_2^2}{2\times9.81}$

$hf_1=\dfrac {4\times f\times L_1\times V_1^2}{d_1\times 2g}=\dfrac {4\times 0.01\times 25\times (4V_2)^2}{0.15\times 2\times 9.81}=106.67\dfrac {V^2_2}{2\times 9.81}$

$h_e=\dfrac {(V_1-V_2)^2}{2g}=\dfrac {(4V_2-V_2)^2}{2g}=\dfrac {9V_2^2}{2\times 9.81}$

$hf_2=\dfrac {4\times f\times L_2\times V_2^2}{d_2\times 2g}=\dfrac {4\times 0.01\times15\times V_2^2}{0.3\times 2\times 9.81}=2.0\dfrac {V^2_2}{2\times 9.81}$

Put the values in equation (A), we get

$8=0+\dfrac {V_2^2}{2g}+\dfrac {8V_2^2}{2\times9.81}+106.67\dfrac {V^2_2}{2\times 9.81}+\dfrac {9V_2^2}{2\times 9.81}+2.0\dfrac {V^2_2}{2\times 9.81}$

$8=\dfrac {V_2^2}{2\times9.81}[1+8+106.67+9+2]$

$\therefore 8.0=126.67\dfrac {V_2^2}{2\times9.81}$

$\therefore V_2=\sqrt{\dfrac {8.0\times 2\times 2.91}{126.67}}$

$\therefore V_2=\sqrt{1.2391}$


Therefore the rate of flow,

$Q=A_2\times V_2$

$Q=\dfrac \pi4\times (0.3)^2\times 1.113$


or, $Q=78.67$ litres/sec

page fluid mechanics 2 fm2 • 32 views
written 4 months ago by gravatar for Syedahina Mohi Syedahina Mohi0
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