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
![$P_1=P_{atm} \text{ and} \ V_1=0$](https://tex.z-dn.net/?f=%24P_1%3DP_%7Batm%7D%20%5Ctext%7B%20and%7D%20%5C%20V_1%3D0%24)
![$\frac{P_{atm}}{\rho g}+\frac{v_1^2}{2g} +h = \frac{P_2}{\rho g}$](https://tex.z-dn.net/?f=%24%5Cfrac%7BP_%7Batm%7D%7D%7B%5Crho%20g%7D%2B%5Cfrac%7Bv_1%5E2%7D%7B2g%7D%20%2Bh%20%3D%20%5Cfrac%7BP_2%7D%7B%5Crho%20g%7D%24)
![$\frac{101.325}{1000 \times 9.81}+0.02 =\frac{P_2}{\rho g}$](https://tex.z-dn.net/?f=%24%5Cfrac%7B101.325%7D%7B1000%20%5Ctimes%209.81%7D%2B0.02%20%3D%5Cfrac%7BP_2%7D%7B%5Crho%20g%7D%24)
![$P_2 = 111.35 \ kPa$](https://tex.z-dn.net/?f=%24P_2%20%3D%20111.35%20%5C%20kPa%24)
Therefore, 111.325 kPa is the gas supply pressure required to keep the water from leaking back into the tube.
Velocity at point 2,
![$V_2=\sqrt{\left(\frac{111.135}{\rho g}+0.02}\right)\times 2g$](https://tex.z-dn.net/?f=%24V_2%3D%5Csqrt%7B%5Cleft%28%5Cfrac%7B111.135%7D%7B%5Crho%20g%7D%2B0.02%7D%5Cright%29%5Ctimes%202g%24)
= 1.617 m/s
Flow of water, ![$Q_2 = A_{tube} \times V_2$](https://tex.z-dn.net/?f=%24Q_2%20%3D%20A_%7Btube%7D%20%5Ctimes%20V_2%24)
![$=\frac{\pi}{4} \times (10^{-3})^2 \times 1.617 $](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%20%5Ctimes%20%2810%5E%7B-3%7D%29%5E2%20%5Ctimes%201.617%20%24)
![$1.2695 \times 10^{-6} \ m^3/s$](https://tex.z-dn.net/?f=%241.2695%20%5Ctimes%2010%5E%7B-6%7D%20%5C%20m%5E3%2Fs%24)
Minimum air flow rate,
![$Q_2 = Q_3 = A_{hose} \times V_3$](https://tex.z-dn.net/?f=%24Q_2%20%3D%20Q_3%20%3D%20A_%7Bhose%7D%20%5Ctimes%20V_3%24)
![$V_3 = \frac{Q_2}{\frac{\pi}{4}D^2}$](https://tex.z-dn.net/?f=%24V_3%20%3D%20%5Cfrac%7BQ_2%7D%7B%5Cfrac%7B%5Cpi%7D%7B4%7DD%5E2%7D%24)
![$V_3 = \frac{1.2695 \times10^{-6}}{\pi\times 0.25 \times 36 \times 10^{-6}}$](https://tex.z-dn.net/?f=%24V_3%20%3D%20%5Cfrac%7B1.2695%20%5Ctimes10%5E%7B-6%7D%7D%7B%5Cpi%5Ctimes%200.25%20%5Ctimes%2036%20%5Ctimes%2010%5E%7B-6%7D%7D%24)
= 0.0449 m/s
b). Reynolds number in hose,
![$Re = \frac{\rho V_3 D}{\mu} = \frac{V_3 D}{\nu}$](https://tex.z-dn.net/?f=%24Re%20%3D%20%5Cfrac%7B%5Crho%20V_3%20D%7D%7B%5Cmu%7D%20%3D%20%5Cfrac%7BV_3%20D%7D%7B%5Cnu%7D%24)
υ for water at 25 degree Celsius is ![$8.9 \times 10^{-7} \ m^2/s$](https://tex.z-dn.net/?f=%248.9%20%5Ctimes%2010%5E%7B-7%7D%20%5C%20m%5E2%2Fs%24)
υ for air at 25 degree Celsius is ![$1.562 \times 10^{-5} \ m^2/s$](https://tex.z-dn.net/?f=%241.562%20%5Ctimes%2010%5E%7B-5%7D%20%5C%20m%5E2%2Fs%24)
![$Re_{hose}=\frac{0.0449 \times 6 \times 10^{-3}}{1.562 \times 10^{-5}}$](https://tex.z-dn.net/?f=%24Re_%7Bhose%7D%3D%5Cfrac%7B0.0449%20%5Ctimes%206%20%5Ctimes%2010%5E%7B-3%7D%7D%7B1.562%20%5Ctimes%2010%5E%7B-5%7D%7D%24)
= 17.25
Therefore the flow is laminar.
Reynolds number in the pipe
![$Re = \frac{V_2 d}{\nu} = \frac{1.617 \times 10^{-3}}{8.9 \times 10^{-7}}$](https://tex.z-dn.net/?f=%24Re%20%3D%20%5Cfrac%7BV_2%20d%7D%7B%5Cnu%7D%20%3D%20%5Cfrac%7B1.617%20%5Ctimes%2010%5E%7B-3%7D%7D%7B8.9%20%5Ctimes%2010%5E%7B-7%7D%7D%24)
= 1816.85, which is less than 2000.
So the flow is laminar inside the tube.