Question

The figure shows a crude type of atomizer. When bulb A is compressed, air flows swiftly through tube BC causing reduced pressure in the particles of the vertical tube. The liquid rises in the tube, enters

BC and is sprayed out. If the pressure in the bulb is $P_a+P$, where $P$ is the gauge pressure and

$P_a$ is the atmospheric pressure, $v$ is the speed of air in BC, find how large would $v$ need to be to cause the liquid to rise to BC.

(Density of air $=1.3kg/m^3$.)

• A. $\sqrt{\frac{P+\rho gh}{0.65}}$
• B. $\frac{Ph\rho g}{\sqrt{0.65}}$
• C. $\frac{P\sqrt{h\rho g}}{0.65}$
• D. $\frac{P}{h\rho g\sqrt{0.65}}$

Answer 1

The correct option is A $\sqrt{\frac{P+\rho gh}{0.65}}$

Absolute pressure in bulb $=$ Gauge pressure + Atmospheric pressure $=P+P_a$
Using Bernoulli's equation,
$P_a+P=P_{BC}+\frac{{\rho }_a{v}^2}{2}$
where ${\rho }_a$ is the density of air.
$\therefore P_{BC}=P_a+P=\left(\frac{1.3}{2}\right)v^2$
$P_{BC}=P_a-\rho gh$

So equating these two values for $P_{BC}$, we get:
$v=\sqrt{(\frac{P+\rho gh}{0.65})}$