How Does Air Flow Affect Pressure and Velocity in a Venturi Tube?

[SNOOTCHIEBOOCHEE]

Air flows from left to right though a horizontal pipe that has different diameters in two different sections. the narrow section has a radius of 5 cm and the wider sectio nhas radius of 20 cm. At point 1 (r=5) the air is flowing with a velocity of 10 m/s and has an absolute pressure of 150000 Pa. A narrow u shaped tup has been attached to the air tube and this u tube is filled with water. note that no air actually flows into or out of the u shaped tube, but instead flows right across its openings. The pressures at points 1 and 2 are different so the water in the u shaped tube is not leve. Assume that the transitions between the sections have a negligible effect on the frictionless flow of air. and that the density of air is constant throughtout the pipe. Find the speed of the air and its pressure at point two and the differenece in heights of the water in the u shaped tube.

I have no clue where to start

 

[MexChemE]

A device like this is called a Venturi tube. Since air is being treated as incompressible and inviscid, we can easily model this situation with a simple continuity equation and Bernoulli's equation. We will use the continuity equation to find the velocity of air at point two.
$$A_1 v_1 = A_2 v_2$$
$$v_2 = \frac{A_1}{A_2} v_1 = 10 \frac{m}{s} \left( \frac{5 \ cm}{20 \ cm} \right)^2 = 0.625 \frac{m}{s}$$
This is a logical result, in order for mass to be conserved, the section with the smaller cross-section area must have a higher velocity than the section with greater cross-section area. Now, we will use Bernoulli's equation in order to find the pressure at point two. When we analyze a Venturi tube we normally neglect the change in potential energy of the fluid, so we have
$$P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$$
Notice we need the density of air, so we'll just use an average value of 1.2 kg/m³.
$$P_2 = P_1 + \frac{1}{2} \rho \left(v_1^2 - v_2^2 \right) = 150000 \ Pa + 0.5 \left( 1.2 \frac{kg}{m^3} \right) \left( \left(10 \frac{m}{s} \right)^2 - \left(0.625 \frac{m}{s} \right)^2 \right) = 150060 \ Pa$$
According to the Bernoulli Effect, the section with the higher velocity will have a smaller pressure, so the results are coherent. Finally, in order to calculate the height difference in the manometer, we use the following equation
$$\Delta P = \rho g \Delta h$$
This time, we need the density of water, so we'll just use ρ = 1000 kg/m³.
$$\Delta h = \frac{\Delta P}{\rho g} = \frac{60 \ Pa}{\left( 1000 \frac{kg}{m^3} \right) \left(9.8 \frac{m}{s^2} \right)} = 0.006 \ m = 0.6 \ cm$$