First slide

Fluid dynamics

Question

Water is falling down at the rate of 1.8 Kg per minute from a vertical tube. The radius at the bottom of the tube is 0.004 m and pressure around it is 0.76m of Hg. The diameter of the tube at a height of 0.5 m from the bottom is 0.005 m. find the pressure at that point.

Moderate

A

0.724 m of Hg
B

0.824 m of Hg
C

0.924 m of Hg
D

0.624 m of Hg

Solution

p₂ = 0.76 m of Hg = 1.013 x 10⁵ N/m²
$\large \frac mt=Av\rho=\pi r^2\rho v$
$\large \therefore v_1=\frac {(1.8/60)}{(3.14)\left [ \frac 52\times10^{-3} \right ]^2\times 10^3}=1.5ms^{-1}$
$\large A_1v_1=A_2v_2\Rightarrow v_2=\left ( \frac {r_1}{r_2} \right )^2 v_1$
$\large p_1+\frac 12\rho v_1^2+\rho gh=p_2+\frac 12\rho v_2^2$
$\large p_1=p_2+\frac 12\rho(v_2^2-v_1^2)-\rho gh$
$\large =p_2+\frac 12\rho v_1^2\left [ \left (\frac {r_1}{r_2} \right )^2 -1\right ]-\rho gh$
$\large =[1.013\times 10^{5}]+\frac 12\times 10^3\times 2.25\left [ \left (\frac {0.005/2}{0.004} \right )^2 -1\right ]-10^3\times 9.8\times 0.5=0.72m\;of\; Hg$

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