Question

Water flows though a vertical contraction from a pipe of diameter d to another of diameter d/2 (see fig.). The flow velocity at the inlet to the contraction is 2 m/s and pressure 200 kN/$m^2$. If the height of the contraction measures 2 m,then pressure at the exit of the contraction will be very nearly

• A. 168 kN/$m^2$
• B. 192 kN/$m^2$
  
• C. 150 kN/$m^2$
  
• D. 174 kN/$m^2$
  

Answer 1

The correct option is C 150 kN/$m^2$




From equation of continuity,we get
$A_1{V}_1=A_2{V}_2$
$\frac{\pi }{4}{d}^2\times 2=\frac{\pi }{4}\times {\left(\frac{d}{2}\right)}^2\times V_2$
or $V_2=8m/s$
Using Bernoulli's equations,we get
$\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2$
$\frac{200\times {10}^3}{{10}^3\times 9.81}+\frac{(2{)}^2}{2\times 9.81}+0$
$=\frac{p_2}{{10}^3\times 9.81}+\frac{(8{)}^2}{2\times 9.81}+2$
$20.38+0.20=\frac{p_2}{9810}+3.26+2$
$or\frac{p_2}{9810}=15.32$
$or{p}_2=15.32\times 9810=150289.2N/m^2$
$=150.289kN/m^2=150kN/m^2$