Saturday, August 14, 2021

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:

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Given:

Total length of pipe, L=40m

Length of first pipe, L1=25m

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

Length of second pipe, L2=4025=15m

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

Height of water, H=8m

Coefficient of friction, f=0.01

Applying Bernoulli's Theorem,

0+0+8=P2ρg+V222g+0+all losses

8=0+V222g+hi+hf1+he+hf2................(A)

where,

loss at entrance, hi=0.5V122g

Head lost due to friction in pipe 1, hf1=4×f×L1×V12d1×2g

Loss due to sudden enlargement, he=(V1V2)22g

Head lost due to friction in pipe 2, hf2=4×f×L2×V22d2×2g

But continuity equation,

A1V1=A2V2

,V1=A2V2A1

=π4(d2)2×V2π4(d1)2

=(d2d1)2×V2

=(0.30.15)2×V2

V1=4V2......................(1)

Substituting the values of V1 in different losses, we get

hi=0.5V122g=0.5(4V2)22g=8V222×9.81

hf1=4×f×L1×V12d1×2g=4×0.01×25×(4V2)20.15×2×9.81=106.67V222×9.81

he=(V1V2)22g=(4V2V2)22g=9V222×9.81

hf2=4×f×L2×V22d2×2g=4×0.01×15×V220.3×2×9.81=2.0V222×9.81

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=V222×9.81[1+8+106.67+9+2]

8.0=126.67V222×9.81

V2=8.0×2×2.91126.67

V2=1.2391

V2=1.113m/s

Therefore the rate of flow,

Q=A2×V2

Q=π4×(0.3)2×1.113

Q=0.07867m3/s

or, Q=78.67 litres/sec

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