A horizontal pipe of diameter 500 mm is suddenly contracted to a diameter of 250 mm. The pressure intensities in large and smaller pipe are given as 13.734N/cm213.734N/cm2 and 11.772N/cm211.772N/cm2. Find loss of head due to contraction if Cc=0.62Cc=0.62. Also determine the rate of flow of water.

  A horizontal pipe of diameter 500 mm is suddenly contracted to a diameter of 250 mm. The pressure intensities in large and smaller pipe are given as 

13.734N/cm2$13.734N/c{m}^2$ and 11.772N/cm2$11.772N/c{m}^2$. Find loss of head due to contraction if Cc=0.62$C_c=0.62$. Also determine the rate of flow of water.

Solution:

 Given:-

Diameter of large pipe

D1=500mm=5001000=0.50m$D_1=500mm={\displaystyle \frac{500}{1000}}=0.50m$

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

A1=π4×(0.50)2=0.1963m2$A_1={\displaystyle \frac{\pi }{4}}\times (0.50{)}^2=0.1963m^2$

Diameter of small pipe

D2=250mm=2501000=0.25m$D_2=250mm={\displaystyle \frac{250}{1000}}=0.25m$

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

A2=π4×(0.25)2=0.04908m2$A_2={\displaystyle \frac{\pi }{4}}\times (0.25{)}^2=0.04908m^2$

Pressure in large pipe,

P1=13.734N/cm2=13.734×104N/m2$P_1=13.734N/c{m}^2=13.734\times {10}^{4}N/m^2$

Pressure in smaller pipe,

P2=11.772N/cm2=11.772×104N/m2$P_2=11.772N/c{m}^2=11.772\times {10}^{4}N/m^2$

To find the head loss due to contraction,

=V222g[1Cc−1.0]2$={\displaystyle \frac{V_2^2}{2g}}{\left[{\displaystyle \frac{1}{C_c}}-1.0\right]}^2$

=V222g[10.62−1.0]2$={\displaystyle \frac{V_2^2}{2g}}{\left[{\displaystyle \frac{1}{0.62}}-1.0\right]}^2$

=0.375V222g$=0.375{\displaystyle \frac{V_2^2}{2g}}$

From continuity equation,

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

∴,V1=A2V2A1=π4×(D2)2×V2π4(D1)2$\therefore ,V_1={\displaystyle \frac{A_2{V}_2}{A_1}}={\displaystyle \frac{{\displaystyle \frac{\pi }{4}}\times (D_2{)}^2\times V_2}{{\displaystyle \frac{\pi }{4}}(D_1{)}^2}}$

V1=[D2D1]2×V2$V_1={\left[{\displaystyle \frac{D_2}{D_1}}\right]}^2\times V_2$

V1=(0.250.50)2×V2$V_1={\left({\displaystyle \frac{0.25}{0.50}}\right)}^2\times V_2$

V1=V24$V_1={\displaystyle \frac{V_2}{4}}$

Apply Bernoulli's equation, (Z1=z2$Z_1=z_2$)

P2ρg=P1ρg+V212g−V222g−hc${\displaystyle \frac{P_2}{\rho g}}={\displaystyle \frac{P_1}{\rho g}}+{\displaystyle \frac{V_1^2}{2g}}-{\displaystyle \frac{V_2^2}{2g}}-h_c$

But hc=0.375V222g$h_c=0.375{\displaystyle \frac{V_2^2}{2g}}$ and V1=V24$V_1={\displaystyle \frac{V_2}{4}}$

Putting values in equation, we get

13.374×1049.81×1000+(V24)22×9.81=11.772×1041000×9.81+V222g+0.375V222g${\displaystyle \frac{13.374\times {10}^4}{9.81\times 1000}}+{\displaystyle \frac{{\left({\displaystyle \frac{V_2}{4}}\right)}^2}{2\times 9.81}}={\displaystyle \frac{11.772\times {10}^4}{1000\times 9.81}}+{\displaystyle \frac{V_2^2}{2g}}+0.375{\displaystyle \frac{V_2^2}{2g}}$

14.0+V2216×2×9.81=12.0+1.375V222×9.8$14.0+{\displaystyle \frac{V_2^2}{16\times 2\times 9.81}}=12.0+1.375{\displaystyle \frac{V_2^2}{2\times 9.8}}$

∴,14−12=1.375V222×9.81−116V222×9.81$\therefore ,14-12=1.375{\displaystyle \frac{V_2^2}{2\times 9.81}}-{\displaystyle \frac{1}{16}}{\displaystyle \frac{V_2^2}{2\times 9.81}}$

2=1.3125V222×9.81$2=1.3125{\displaystyle \frac{V_2^2}{2\times 9.81}}$

∴V2=2×2×9.811.3125−−−−−−−−−−√=5.467m/s$\therefore V_2=\sqrt{{\displaystyle \frac{2\times 2\times 9.81}{1.3125}}}=5.467m/s$

∴hc=0.375×V222g$\therefore h_c=0.375\times {\displaystyle \frac{V_2^2}{2g}}$

h=0.375×(5.467)22×9.81=0.571