Saturday, August 14, 2021

The rate of flow of water through a horizontal pipe is 0.25m3/s0.25m3/s. The diameter of the pipe which is 200 mm is suddenly enlarged to 400 mm. The pressure intensity in the smaller pipe is 11.772N/cm211.772N/cm2. Find:- 1) loss of head due to sudden enlargement 2) pressure intensity in large pipe 3) power lost due to enlargement

 The rate of flow of water through a horizontal pipe is 

0.25m3/s. The diameter of the pipe which is 200 mm is suddenly enlarged to 400 mm. The pressure intensity in the smaller pipe is 11.772N/cm2.

Find:-

1) loss of head due to sudden enlargement

2) pressure intensity in large pipe

3) power lost due to enlargement


Solution:

 Given:

Discharge, Q=0.25m3/s

Diameter of smaller pipe

D1=200mm=2001000=0.20m

Area, A1=π4×D12

A1=π4×(0.20)2=0.03141m2


Diameter of larger pipe

D2=400mm=4001000=0.40m

Area, A2=π4×D22

A2=π4×(0.40)2=0.12566m2

Pressure in smaller pipe,

P1=11.772N/cm2=11.772×104N/m2

Velocity, V1=QA1=0.250.03141=7.96m/s

Velocity, V2=QA2=0.250.12566=1.996m/s


(i) Loss of head due to sudden enlargement

he=(V1V2)22×g

he=(7.961.99)22×9.81

he=1.816m of water


(ii) Let pressure intensity in large pipe = P2

Apply Bernoulli's equation,

P1ρg+V122×g+z1=P2ρ×g+V222×g+z2+he

Since z1=z2

P1ρg+V122×g=P2ρ×g+V222×g+he....................(1)

Rewrite as

P2ρg=P1ρg+V122gV222ghe........................(2)

Putting values in equation (2), we get

P2ρg=11.772×104100×9.81+7.9622×9.811.9922×9.811.816

P2ρg=12.0+3.2290.20181.8160a

P2ρg=15.2292.0178=13.21m of water

P2=13.21×ρg

P2=13.21×1000×9.81

P2=12.96N/cm2


(iii) Power lost due to sudden enlargement

P=ρ.g.Q.he1000

P=1000×9.81×0.25×1.8161000

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