Question

Consider steady, incompressible and irrotational flow through a reducer in a horizontal pipe where the diameter is reduced from 20 cm to 10 cm. The pressure in the 20 cm pipe just upstream of the reducer is 150 kPa. The fluid has a vapour pressure of 50 kPa and a specific weight of 5 kN/$m^3$. Neglecting frictional effects, the maximum discharge $(in{m}^3/s$) that can pass through the reducer without causing cavitation is

• A. 0.05
• B. 0.16
• C. 0.27
• D. 0.38

Answer 1

The correct option is B 0.16

Given data:
$d_1=20cm=0.2m$
$d_2=10cm=0.1m$
$p_1=150kPa=150000Pa$
Vapour pressure,
$p_v=50kPa$
Specific weight,
w = $5kN/m^3=5000N/m^3=\rho g$

Applying continuity equation at sections(1)-(1) and (2)-(2). we get
$A_1{V}_1=A_2{V}_2$ for incompressible flow
$\frac{\pi }{4}{d}_1^2{V}_1=\frac{\pi }{4}{d}_2^2{V}_2$
$d_1^2{V}_1=d_2^2{V}_2$
$or{V}_1={\left(\frac{d_2}{d_1}\right)}^2{V}_2{\left(\frac{10}{20}\right)}^2{V}_2=0.25V_2$
The maximum pressure in downstream of reducer should be greater or equal to vapour pressure$\left(p_v\right)$ to avoid cavitation. Therefore for maximum discharge,
$p_2=p_v=50kPa=50000Pa$
Applying Bernoulli's equation at sections(1)-(1) and (2)-(2), we get
$\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2$
$\frac{p_1}{\rho g}+\frac{V_1^2}{2g}=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}$
$\because z_1=z_2$
$\frac{150000}{5000}+\frac{(0.25{)}^2{V}_2^2}{2\times 9.81}=\frac{50000}{5000}+\frac{V_2^2}{2\times 9.81}$
$30+0.00318V_2^2=10+0.0509V_2^2$
$V_2^2=418.51$
$or{V}_2=20.45m/s$
$\therefore Q=A_2{V}_2=\frac{\pi }{4}{d}_2^2{V}_2$
$=\frac{3.14}{4}\times (0.1{)}^2\times 20.45=0.160m^3/s$