Fluid mechanics problem involving a hose and water coming out.

Click For Summary
SUMMARY

The discussion focuses on solving a fluid mechanics problem involving a hose and water flow, specifically applying the conservation of mass principle. The participants emphasize the equation V1*A1 = V2*A2 for steady-state incompressible flow, while clarifying that the flow rate must account for varying velocity distributions across the cross-section. The correct approach involves calculating the total volumetric flow rate and using Bernoulli's equation to determine the uniform exit velocity at the nozzle, leading to the diameter ratio calculations.

PREREQUISITES
  • Understanding of fluid mechanics principles, particularly conservation of mass.
  • Familiarity with Bernoulli's equation and its applications.
  • Knowledge of volumetric flow rate calculations.
  • Ability to perform integral calculus for velocity distributions.
NEXT STEPS
  • Study the application of Bernoulli's equation in fluid dynamics.
  • Learn about calculating volumetric flow rates in varying cross-sections.
  • Explore the concept of velocity profiles in fluid flow.
  • Investigate the relationship between diameter ratios and flow rates in hoses.
USEFUL FOR

Students and professionals in engineering, particularly those specializing in fluid mechanics, as well as anyone involved in designing or analyzing fluid flow systems.

guitar24
Messages
9
Reaction score
0

Homework Statement



2jw6rq.png


Homework Equations



I am having some trouble starting this problem. At first look I would say that a conservation of mass will give V1*A1=V2*A2 for steady state incompressible fluid. Then I would solve for A2 and relate the two diameters but I don't have the velocity distribution of the water stream. Any hints??
 
Physics news on Phys.org
You're spot on to include conservation of volume, but V1*A1 = V2*A2 can't be right since the flow of water thru A1 varies with r. So think of the total volume of water flowing in 1 sec. thru A1 and equate that to the total volume of water flowing in 1 sec. thru A2.
 
Thank you for your reply! could you elaborate a bit more on how I would do that?
The total volumetric flow rate in the tube is Vavg*Area of the tube. Multiply by one and you get the volume of water entering the cross section in one second. Then i would just divide that by the area of the water stream to get the volume of water flowing in one sec through the area of the water stream. Then just rearrange the equation to get Dout/Dhose. Is that correct?

Thank you!
 
guitar24 said:
Thank you for your reply! could you elaborate a bit more on how I would do that?
The total volumetric flow rate in the tube is Vavg*Area of the tube. Multiply by one and you get the volume of water entering the cross section in one second. Then i would just divide that by the area of the water stream to get the volume of water flowing in one sec through the area of the water stream. Then just rearrange the equation to get Dout/Dhose. Is that correct?

Thank you!

I computed the average flow per unit time and found it to be not Vavg but (1/R)∫v(r)dr with limits 0 and R, = (4/3)Vavg which is somewhat weird.

Anyway, whatever average V is, you are right in using mass conservation flow to get the answer. But you still need to get v, the (uniform) velocity of the stream exiting at Dout. For that, go to Bernoulli.
 
Given the assumption that there is no interaction with the surrounding air and thus no momentum loss, that suggests momentum is constant. Write integrals representing the momentum at the nozzle using the given velocity profile and one for the region downstream (utilizing conservation of mass flow) where the velocity is uniform. Equate them and solve for diameter ratios.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
Replies
31
Views
4K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
Replies
15
Views
3K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
22
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K