Question

Water is coming out from a tap and a falls vertically downwards. At the tap opening, the stream diameter is 20 mm with uniform velocity of 2 m/s. Acceleration due to gravity is 9.81 $m/s^2$. Assuming steady, inviscid flow, constant atmospheric pressure everywhere and neglecting curvature and surface tension effects, the diameter in mm of the stream 0.5 m below the tap is approximately

• A. 10
• B. 15
• C. 20
• D. 25

Answer 1

The correct option is B 15


Given data:
At section 1-1,
Diameter: $d_1=20mm=0.02m$
Velocity: $V_1=2m/s$
Acceleration due to gravity,
$g=9.81m/s^2$
Applying Bernoulli's equation between sections 1 and 2, we get
$\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2$
$0+\frac{V_1^2}{2g}+0.5=0+\frac{V_2^2}{2g}+0$
$\frac{V_1^2}{2g}+0.5=\frac{V_2^2}{2g}$
$or{V}_2^2=V_1^2+0.5\times 2g$
$or{V}_2=\sqrt{V_1^2+0.5\times 2\times g}$
$=\sqrt{(2{)}^2+0.5\times 2\times 9.81}$
$=\sqrt{13.81}=3.716m/s$
Applying continuity,equation between sections 1 and 2, we get
$A_1{V}_1=A_2{V}_2$
$\frac{\pi }{4}{d}_1^2{V}_1=\frac{\pi }{4}{d}_2^2{V}_2$
$or{d}_1^2{V}_1=d_2^2{V}_2$
$(0.02{)}^2\times 2=d_2^2\times 3.716$
$or{d}_2^2=2.1528\times {10}^{-4}$
$or{d}_2=0.01467m$
=14.67 mm $\simeq$ 15 mm