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.

The rate of flow of water through a horizontal pipe is 0.25m3/s$0.25m^3/s$. 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$11.772N/c{m}^2$.

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$Q=0.25m^3/s$

Diameter of smaller pipe

D1=200mm=2001000=0.20m$D_1=200mm={\displaystyle \frac{200}{1000}}=0.20m$

Area, A1=π4×D21$A_1={\displaystyle \frac{\pi }{4}}\times D_1^2$

A1=π4×(0.20)2=0.03141m2$A_1={\displaystyle \frac{\pi }{4}}\times (0.20{)}^2=0.03141m^2$

Diameter of larger pipe

D2=400mm=4001000=0.40m$D_2=400mm={\displaystyle \frac{400}{1000}}=0.40m$

Area, A2=π4×D22$A_2={\displaystyle \frac{\pi }{4}}\times D_2^2$

A2=π4×(0.40)2=0.12566m2$A_2={\displaystyle \frac{\pi }{4}}\times (0.40{)}^2=0.12566m^2$

Pressure in smaller pipe,

P1=11.772N/cm2=11.772×104N/m2$P_1=11.772N/c{m}^2=11.772\times {10}^{4}N/m^2$

Velocity, V1=QA1=0.250.03141=7.96m/s$V_1={\displaystyle \frac{Q}{A_1}}={\displaystyle \frac{0.25}{0.03141}}=7.96m/s$

Velocity, V2=QA2=0.250.12566=1.996m/s$V_2={\displaystyle \frac{Q}{A_2}}={\displaystyle \frac{0.25}{0.12566}}=1.996m/s$

(i) Loss of head due to sudden enlargement

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

he=(7.96−1.99)22×9.81$h_e={\displaystyle \frac{(7.96-1.99{)}^2}{2\times 9.81}}$

he=1.816m$h_e=1.816m$ of water

(ii) Let pressure intensity in large pipe = P2$P_2$

Apply Bernoulli's equation,

P1ρg+V212×g+z1=P2ρ×g+V222×g+z2+he${\displaystyle \frac{P_1}{\rho g}}+{\displaystyle \frac{V_1^2}{2\times g}}+z_1={\displaystyle \frac{P_2}{\rho \times g}}+{\displaystyle \frac{V_2^2}{2\times g}}+z_2+h_e$

Since z1=z2$z_1=z_2$

P1ρg+V212×g=P2ρ×g+V222×g+he....................(1)${\displaystyle \frac{P_1}{\rho g}}+{\displaystyle \frac{V_1^2}{2\times g}}={\displaystyle \frac{P_2}{\rho \times g}}+{\displaystyle \frac{V_2^2}{2\times g}}+h_e....................(1)$

Rewrite as

P2ρg=P1ρg+V212g−V222g−he........................(2)${\displaystyle \frac{P_2}{\rho g}}={\displaystyle \frac{P_1}{\rho g}}+{\displaystyle \frac{V_1^2}{2g}}-{\displaystyle \frac{V_2^2}{2g}}-h_e........................(2)$

Putting values in equation (2), we get

P2ρg=11.772×104100×9.81+7.9622×9.81−1.9922×9.81−1.816${\displaystyle \frac{P_2}{\rho g}}={\displaystyle \frac{11.772\times {10}^4}{100\times 9.81}}+{\displaystyle \frac{{7.96}^2}{2\times 9.81}}-{\displaystyle \frac{{1.99}^2}{2\times 9.81}}-1.816$

P2ρg=12.0+3.229−0.2018−1.8160${\displaystyle \frac{P_2}{\rho g}}=12.0+3.229-0.2018-1.8160$

P2ρg=15.229−2.0178=13.21m${\displaystyle \frac{P_2}{\rho g}}=15.229-2.0178=13.21m$ of water

∴P2=13.21×ρg$\therefore P_2=13.21\times \rho g$

P2=13.21×1000×9.81$P_2=13.21\times 1000\times 9.81$

P2=12.96N/cm2$P_2=12.96N/c{m}^2$

(iii) Power lost due to sudden enlargement

P=ρ.g.Q.he1000$P={\displaystyle \frac{\rho .g.Q.h_e}{1000}}$

P=1000×9.81×0.25×1.8161000$P={\displaystyle \frac{1000\times 9.81\times 0.25\times 1.816}{1000}}$

P=4.453kW