Solution :
Given :
h = 2 cm
Diameter of the tube , d = 1 mm
Diameter of the hose, D = 6 mm
Between 1 and 2, by applying Bernoulli's principle, we get
As point 1 is just below the free surface of liquid, so
[tex]$P_1=P_{atm} \text{ and} \ V_1=0$[/tex]
[tex]$\frac{P_{atm}}{\rho g}+\frac{v_1^2}{2g} +h = \frac{P_2}{\rho g}$[/tex]
[tex]$\frac{101.325}{1000 \times 9.81}+0.02 =\frac{P_2}{\rho g}$[/tex]
[tex]$P_2 = 111.35 \ kPa$[/tex]
Therefore, 111.325 kPa is the gas supply pressure required to keep the water from leaking back into the tube.
Velocity at point 2,
[tex]$V_2=\sqrt{\left(\frac{111.135}{\rho g}+0.02}\right)\times 2g$[/tex]
= 1.617 m/s
Flow of water, [tex]$Q_2 = A_{tube} \times V_2$[/tex]
[tex]$=\frac{\pi}{4} \times (10^{-3})^2 \times 1.617 $[/tex]
[tex]$1.2695 \times 10^{-6} \ m^3/s$[/tex]
Minimum air flow rate,
[tex]$Q_2 = Q_3 = A_{hose} \times V_3$[/tex]
[tex]$V_3 = \frac{Q_2}{\frac{\pi}{4}D^2}$[/tex]
[tex]$V_3 = \frac{1.2695 \times10^{-6}}{\pi\times 0.25 \times 36 \times 10^{-6}}$[/tex]
= 0.0449 m/s
b). Reynolds number in hose,
[tex]$Re = \frac{\rho V_3 D}{\mu} = \frac{V_3 D}{\nu}$[/tex]
υ for water at 25 degree Celsius is [tex]$8.9 \times 10^{-7} \ m^2/s$[/tex]
υ for air at 25 degree Celsius is [tex]$1.562 \times 10^{-5} \ m^2/s$[/tex]
[tex]$Re_{hose}=\frac{0.0449 \times 6 \times 10^{-3}}{1.562 \times 10^{-5}}$[/tex]
= 17.25
Therefore the flow is laminar.
Reynolds number in the pipe
[tex]$Re = \frac{V_2 d}{\nu} = \frac{1.617 \times 10^{-3}}{8.9 \times 10^{-7}}$[/tex]
= 1816.85, which is less than 2000.
So the flow is laminar inside the tube.
