[Fluids] Determine pressure change from industrial jet pump

[Bluestribute]

Homework Statement

The apparatus or "jet pump" used in an industrial plant is constructed by placing the tube within the pipe. The velocity of the flow within the 200-mm-diameter pipe is 2 m/s, and the velocity of the flow through the 20-mm-diameter tube is V = 37 m/s . The fluid is ethyl alcohol having a density of ρea = 790 kg/m3. Assume the pressure at each cross section of the pipe is uniform. Assume ethyl alcohol is ideal fluid, that is, incompressible and frictionless.

Determine the increase in pressure PB−PA that occurs between the back A and front B of the pipe.

Homework Equations

I'm assuming CoE since it has a pressure term explicitly

The Attempt at a Solution

So I don't really know where to begin. What I have:

V_in = 37 m/s
V_out = 2 m/s
ρ = 790 kg/m³
A_in = π(0.01m)²
A_out = π(0.1m)²

I'm using "in" and "out", but in the end, it's just deltas I'm looking for. But, like . . . I don't know where to begin with these numbers? Do I find kinetic energy? Set that equal to . . . something? Or find a force first. And what to do after that (I'm going to have a lot of questions with CoM and CoE)?

 

[KataKoniK]

Based on the given information, we can use the Bernoulli's equation to solve for the pressure difference between points A and B. The Bernoulli's equation states that the total energy at any point in a fluid flow system is constant. This means that the sum of kinetic energy, potential energy, and pressure energy at point A is equal to the sum of kinetic energy, potential energy, and pressure energy at point B.

In this case, we can ignore the potential energy term since the height of the fluid is not changing. Therefore, we can write the Bernoulli's equation as:

P1 + 1/2ρV1^2 = P2 + 1/2ρV2^2

where P1 and V1 are the pressure and velocity at point A, and P2 and V2 are the pressure and velocity at point B.

Substituting the given values, we get:

P1 + 1/2(790)(37)^2 = P2 + 1/2(790)(2)^2

Solving for P2, we get:

P2 = P1 + 1/2(790)(37)^2 - 1/2(790)(2)^2

P2 = P1 + 692,050 Pa

Therefore, the pressure difference between points A and B is 692,050 Pa or 692.05 kPa.