Question

As shown in the figure, a liquid of density $\rho$ is filled in a sealed container to a height $h$. The container contains compressed air at a gauge pressure of $p$. The horizontal outlet pipe has a cross-sectional area $A$ at point $C$ and $A/2$ at $E$. Find the correct options.

• A. The velocity of the liquid at $C$ will be ${\left[\frac{p+\rho gh}{2\rho }\right]}^{1/2}$.
• B. The velocity of the liquid at $C$ will be ${\left[\frac{2(p+\rho gh)}{\rho }\right]}^{1/2}$.
• C. The discharge rate is given by $\frac{A}{2\rho }(p+\rho gh{)}^{1/2}$.
• D. The discharge rate is given by $\frac{A}{\sqrt{2\rho }}(p+\rho gh{)}^{1/2}$.

Answer 1

Answer: (A), (D)

The discharge rate is given by $\frac{A}{\sqrt{2\rho }}(p+\rho gh{)}^{1/2}$.

Pressure of the compressed air $=p+p_o$ where $p_0$ is the atmospheric pressure.
Then,
$p_A=p+p_0+\rho gh$ ---- (i)

Applying Bernoulli's theorem at $A$, $C$ and $E$ ,

$p_A+\frac{\rho v_A^2}{2}=p_C+\frac{\rho v_C^2}{2}=p_E+\frac{\rho v_E^2}{2}$ ----- (ii)

(because $A,C,E$ are at the same heights)

$v_A=0$ (because the liquid is at rest) and $p_E=0$ because the outlet is open to the atmosphere.

Therefore,

$p+p_0+\rho gh=p_C+\frac{\rho v_C^2}{2}=p_0+\frac{\rho v_E^2}{2}$

Also,
$p_C=p_0+\rho g{h}_1$ ($h_1$ is the level of water above point $C$)

$A_C{v}_C=A_E{v}_E=\frac{A_C}{2}{v}_E$

From equations (i) and (ii),

$p_C+\frac{\rho v_C^2}{2}=p_0+\rho g{h}_1+\frac{\rho v_C^2}{2}$
$=p_0+\frac{\rho v_E^2}{2}=p_0+\rho 2v_C^2$

$\Rightarrow \rho g{h}_1=2\rho v_C^2-\frac{\rho v_C^2}{2}=\frac{3}{2}\rho v_C^2$

$\Rightarrow h_1=\frac{3v_C^2}{2g}$

Also from equation (ii),
$p+p_0+\rho gh=p_C+\frac{\rho v_C^2}{2}=p_0+\rho g{h}_1+\frac{\rho v_C^2}{2}$
$\Rightarrow p+\rho gh=\rho g{h}_1+\frac{\rho v_C^2}{2}$
$=\rho g\frac{3v_C^2}{2g}+\frac{\rho v_C^2}{2}$
$=2\rho v_C^2$
$\Rightarrow v_C^2=\frac{p+\rho gh}{2\rho }$

Discharge rate $=A_C{v}_C=A{\left(\frac{p+\rho gh}{2\rho }\right)}^{1/2}$