Question

(20 points) A 1 mm diameter tube is connected to the bottom of a container filled with water to a height of 2 cm from the bottom. Air flows from the tube into the liquid and creates spherical bubbles with diameter about the diameter of the tube (1 mm). Everything is at 298 K. The tube is short but is connected to a much longer 2 m long hose that is 6 mm in diameter. The hose is connected to the gas supply. If there is no gas flow the water will leak into the tube and into the supply hose. When gas flows the water is blocked from entering the tube and bubbling starts. State all assumptions in answering the following questions. (a) What should be the minimum air flow rate and the gas supply pressure to keep the water from leaking back into the tube? (b) Is the flow in the hose laminar or turbulent? Is the flow in the tube laminar or turbulent?

Answer 1

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

$P_1=P_{atm} \text{ and} \ V_1=0$

$\frac{P_{atm}}{\rho g}+\frac{v_1^2}{2g} +h = \frac{P_2}{\rho g}$

$\frac{101.325}{1000 \times 9.81}+0.02 =\frac{P_2}{\rho g}$

$P_2 = 111.35 \ kPa$

Therefore, 111.325 kPa is the gas supply pressure required to keep the water from leaking back into the tube.

Velocity at point 2,

$V_2=\sqrt{\left(\frac{111.135}{\rho g}+0.02}\right)\times 2g$

   = 1.617 m/s

Flow of water,  $Q_2 = A_{tube} \times V_2$

                               $=\frac{\pi}{4} \times (10^{-3})^2 \times 1.617$

                               $1.2695 \times 10^{-6} \ m^3/s$

Minimum air flow rate,

$Q_2 = Q_3 = A_{hose} \times V_3$

$V_3 = \frac{Q_2}{\frac{\pi}{4}D^2}$

$V_3 = \frac{1.2695 \times10^{-6}}{\pi\times 0.25 \times 36 \times 10^{-6}}$

    = 0.0449 m/s

b). Reynolds number in hose,

$Re = \frac{\rho V_3 D}{\mu} = \frac{V_3 D}{\nu}$

υ for water at 25 degree Celsius is $8.9 \times 10^{-7} \ m^2/s$

υ for air at 25 degree Celsius is $1.562 \times 10^{-5} \ m^2/s$

$Re_{hose}=\frac{0.0449 \times 6 \times 10^{-3}}{1.562 \times 10^{-5}}$

           = 17.25

Therefore the flow is laminar.

Reynolds number in the pipe

$Re = \frac{V_2 d}{\nu} = \frac{1.617 \times 10^{-3}}{8.9 \times 10^{-7}}$

                = 1816.85, which is less than 2000.

So the flow is laminar inside the tube.