Question

The outlet at the bottom of a tank is so formed that the velocity of water at a point $A$ as shown in the figure below is $2.0$ times the mean velocity within the outlet pipe (atmospheric pressure $=95.48kPa$(absolute) and vapour pressure $=4.00kPa$ (absolute) and neglect all other losses). The greatest length of pipe $L$
Which may be used without cavitation is ___ $m$

• A. $1.21$
• B. $1.21$
• C. $1.21$
• D. $1.21$

Answer 1

The correct option is C $1.21$

$P_A=$ Vapour pressure $=4.00kPa$ (abs)
$P_1=P_2=$ atmospheric pressure $=95.47kPa$(abs)
By applying Bernoulli's equation to point $1$ and $A$.
$\frac{P_1}{\rho g}+\frac{v_1^2}{2g}+z_1=\frac{P_A}{\rho g}+\frac{v_A^2}{2g}+z_A$
$\left(\frac{95.48}{9.81}\right)+0+1.5=\left(\frac{4.00}{9.81}\right)+\frac{v_A^2}{2g}+0$
$\frac{v_A^2}{2g}=\frac{95.48-4.00}{9.81}+1.5=10.825m$
$V_A=(2\times 9.81\times 10.825{)}^{1/2}=14.57m/sec$
$v_2=\frac{V_A}{2}=\frac{14.57}{2}=7.28m/sec$
By applying Bernoulli's equation to pioint $1$ and $2$, with datum at point $2$.
$\frac{P_1}{\rho g}+\frac{v_1^2}{2g}+z_1=\frac{P_2}{\rho g}+\frac{v_2^3}{2g}+z_2$
$\frac{95.48}{9.81}+0+(L+1.50)=\left(\frac{95.48}{9.81}\right)+\frac{(7.28{)}^2}{2\times 9.81}+0$
$L=1.21m$