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.

 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.

Solution:

Given:

Total length of pipe, L=40m

Length of first pipe, L1=25m$L_1=25m$

Diameter of first pipe, d1=150mm=0.15m$d_1=150mm=0.15m$

Length of second pipe, L2=40−25=15m$L_2=40-25=15m$

Diameter of second pipe, d2=300mm=0.30m$d_2=300mm=0.30m$

Height of water, H=8m$H=8m$

Coefficient of friction, f=0.01

Applying Bernoulli's Theorem,

0+0+8=P2ρg+V222g+0+all losses$0+0+8={\displaystyle \frac{P_2}{\rho g}}+{\displaystyle \frac{V_2^2}{2g}}+0+\text{all losses}$

8=0+V222g+hi+hf1+he+hf2................(A)$8=0+{\displaystyle \frac{V_2^2}{2g}}+h_i+h{f}_1+h_e+h{f}_2................(A)$

where,

loss at entrance, hi=0.5V212g$h_i=0.5{\displaystyle \frac{V_1^2}{2g}}$

Head lost due to friction in pipe 1, hf1=4×f×L1×V21d1×2g$h{f}_1={\displaystyle \frac{4\times f\times L_1\times V_1^2}{d_1\times 2g}}$

Loss due to sudden enlargement, he=(V1−V2)22g$h_e={\displaystyle \frac{(V_1-V_2{)}^2}{2g}}$

Head lost due to friction in pipe 2, hf2=4×f×L2×V22d2×2g$h{f}_2={\displaystyle \frac{4\times f\times L_2\times V_2^2}{d_2\times 2g}}$

But continuity equation,

A1V1=A2V2$A_1{V}_1=A_2{V}_2$

∴,V1=A2V2A1$\therefore ,V_1={\displaystyle \frac{A_2{V}_2}{A_1}}$

=π4(d2)2×V2π4(d1)2$={\displaystyle \frac{{\displaystyle \frac{\pi }{4}}(d_2{)}^2\times V_2}{{\displaystyle \frac{\pi }{4}}(d_1{)}^2}}$

=(d2d1)2×V2$={\left({\displaystyle \frac{d_2}{d_1}}\right)}^2\times V_2$

=(0.30.15)2×V2$={\left({\displaystyle \frac{0.3}{0.15}}\right)}^2\times V_2$

V1=4V2......................(1)$V_1=4V_2......................(1)$

Substituting the values of V1$V_1$ in different losses, we get

hi=0.5V212g=0.5(4V2)22g=8V222×9.81$h_i=0.5{\displaystyle \frac{V_1^2}{2g}}=0.5{\displaystyle \frac{(4V_2{)}^2}{2g}}={\displaystyle \frac{8V_2^2}{2\times 9.81}}$

hf1=4×f×L1×V21d1×2g=4×0.01×25×(4V2)20.15×2×9.81=106.67V222×9.81$h{f}_1={\displaystyle \frac{4\times f\times L_1\times V_1^2}{d_1\times 2g}}={\displaystyle \frac{4\times 0.01\times 25\times (4V_2{)}^2}{0.15\times 2\times 9.81}}=106.67{\displaystyle \frac{V_2^2}{2\times 9.81}}$

he=(V1−V2)22g=(4V2−V2)22g=9V222×9.81$h_e={\displaystyle \frac{(V_1-V_2{)}^2}{2g}}={\displaystyle \frac{(4V_2-V_2{)}^2}{2g}}={\displaystyle \frac{9V_2^2}{2\times 9.81}}$

hf2=4×f×L2×V22d2×2g=4×0.01×15×V220.3×2×9.81=2.0V222×9.81$h{f}_2={\displaystyle \frac{4\times f\times L_2\times V_2^2}{d_2\times 2g}}={\displaystyle \frac{4\times 0.01\times 15\times V_2^2}{0.3\times 2\times 9.81}}=2.0{\displaystyle \frac{V_2^2}{2\times 9.81}}$

Put the values in equation (A), we get

8=0+V222g+8V222×9.81+106.67V222×9.81+9V222×9.81+2.0V222×9.81$8=0+{\displaystyle \frac{V_2^2}{2g}}+{\displaystyle \frac{8V_2^2}{2\times 9.81}}+106.67{\displaystyle \frac{V_2^2}{2\times 9.81}}+{\displaystyle \frac{9V_2^2}{2\times 9.81}}+2.0{\displaystyle \frac{V_2^2}{2\times 9.81}}$

8=V222×9.81[1+8+106.67+9+2]$8={\displaystyle \frac{V_2^2}{2\times 9.81}}[1+8+106.67+9+2]$

∴8.0=126.67V222×9.81$\therefore 8.0=126.67{\displaystyle \frac{V_2^2}{2\times 9.81}}$

∴V2=8.0×2×2.91126.67−−−−−−−−−−−−√$\therefore V_2=\sqrt{{\displaystyle \frac{8.0\times 2\times 2.91}{126.67}}}$

∴V2=1.2391−−−−−√$\therefore V_2=\sqrt{1.2391}$

V2=1.113m/s$V_2=1.113m/s$

Therefore the rate of flow,

Q=A2×V2$Q=A_2\times V_2$

Q=π4×(0.3)2×1.113$Q={\displaystyle \frac{\pi }{4}}\times (0.3{)}^2\times 1.113$

Q=0.07867m3/s$Q=0.07867m^3/s$

or, Q=78.67$Q=78.67$ litres/sec