Solving Old Test Question - Flow Speed, Flow Rate & Pressure

[edge]

In studying for my final, I'm working through the problems I don't understand from my old tests. The only one I can't go back and figure out is the second part of a 2 part question. I'll write it all so you can read it as I would, just ignore part c (it was discussed in another post.

A liquid of density equal to 1.85 g/cm^3 flows through two horizontal sections of tubing joined end to end. In the first section (large section) the cross-sectional area is 10 cm^2, the flow speed is 275 cm/s, and the pressure is 1.2E5 Pa. In the second section (small section), the diameter of the tube is 2.5 cm.

c) Find the flow speed, flow rate, and the pressure in the small tube section.

d) The system is modified adding a vertical section between the 2 horizontal tubes. The first section is now 1.5m below the 2nd section. In the first section, the flow and the pressure remains the same as in part c). Find the pressure and the flow rate in the small tube section.

Now, I'm also assuming that the cross-sectional areas are the same as in the other part (.1 m^2 for the big section and .049 m^2 in the small section). I then use bernoulli's equation and get:

(1.2E5) + (.5)(1850)(2.75^2) = P2 + (1850)(9.8)(1.5) + (.5)(1850)(v2^2)

This leaves me with 2 unknowns. Can I use A1v1 = A2v2 to get v2? That just make the flow rate the same in part d) as in part c) but the pressure would still be different. I just didn't think I could do that. If not, what should I do to get the second equation?

Thank you for your time!

-edge

 

[member 553083]

Yes, you can use the equation A1v1 = A2v2 to get v2. This will allow you to solve for the pressure and flow rate in the small tube section. However, the pressure in the small tube section will be different than the pressure in the first section (due to the change in elevation). Therefore, you will need to use Bernoulli's equation to solve for the pressure in the small tube section. The equation should look something like this:P1 + .5*ρ*v1^2 = P2 + .5*ρ*v2^2 + ρ*g*h Where P1 is the pressure in the first section, ρ is the density of the liquid, v1 is the velocity in the first section, P2 is the pressure in the second section, v2 is the velocity in the second section, g is the acceleration due to gravity, and h is the difference in elevation between the two sections.

 

[M98Ranger]

Hi Edge,

Thank you for sharing your question and thought process. It seems like you have a good understanding of the concepts involved in this problem.

To solve part c), you can use the equation of continuity, which states that the flow rate (Q) is equal at all points in a continuous flow system. This means that the flow rate in the large section (Q1) is equal to the flow rate in the small section (Q2). So, using the formula Q = Av, where A is the cross-sectional area and v is the flow speed, we can set up the following equation:

Q1 = Q2
A1v1 = A2v2

Solving for v2, we get:

v2 = (A1v1)/A2

Substituting in the values given, we get:

v2 = (10 cm^2)(275 cm/s)/(2.5 cm^2) = 1100 cm/s

To find the flow rate (Q2), we can use the formula Q = Av again:

Q2 = A2v2 = (2.5 cm^2)(1100 cm/s) = 2750 cm^3/s

To find the pressure (P2) in the small section, we can use the Bernoulli's equation that you mentioned:

P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2

Since the flow speed and height are the same in both sections, we can simplify the equation to:

P1 = P2 + (1/2)ρv2^2

Substituting in the values given, we get:

P2 = 1.2E5 Pa + (1/2)(1850 g/cm^3)(1100 cm/s)^2 = 1.35E5 Pa

For part d), we can use the same approach. The pressure in the large section (P1) remains the same, so we can use the same equation as before:

P1 = P2 + (1/2)ρv2^2

To find the pressure in the small section (P2), we need to take into account the change in height. Using the equation P1 + ρgh1 = P2 + ρgh2, we get:

P2 = P1 -