Bernoulli’s Equation and External Forces

AI Thread Summary
The discussion focuses on calculating the external force needed to stabilize a spray nozzle with given dimensions and pressures using Bernoulli's Equation. A participant is confused about determining velocities and forces, having attempted to calculate the forces at the nozzle's entrance and exit but struggling with the momentum change and time interval. They are encouraged to clarify their calculations and consider extreme cases to better understand the force dynamics. Key questions raised include the consistency of pressures with Bernoulli's Equation and the application of a macroscopic momentum balance. The conversation aims to resolve the confusion around these calculations and concepts.
Khalid Qasim
Messages
2
Reaction score
0
Thread moved from the technical forums
Summary:: Bernoulli’s Equation and External Forces

Hello all, I am a bit confused on this question:

Water flows through a spray nozzle (as shown in the figure below) at a rate of 50 cm3 h -1 .
The internal inlet diameter of the nozzle entrance is 5 cm and the internal exit diameter of the nozzle is 1 cm.
The gauge pressure is 2.75 bar at the entrance of the nozzle and the pressure outside the nozzle exit is atmospheric.
Calculate the external force required to keep the spray nozzle in place, and the direction of this force.
State any assumptions you make and give all the steps in your calculation.

I have tried using bernoulli's eq to find different velocities to calculate momentums to hopefully find a thrust force, but to no avail. Can someone help?
 

Attachments

  • thrust.png
    thrust.png
    2.6 KB · Views: 145
Physics news on Phys.org
Please show us what you tried. Then perhaps we can diagnose why "to no avail."
 
kuruman said:
Please show us what you tried. Then perhaps we can diagnose why "to no avail."
I've crossed it out now, but what I had done:
Found force at the entrance by multiplying 3.76 bar and area at start.
Found force at the exit by multipying 1.01bar and area at exit.

Now for the next part:
Finding the velocities I did: Start: (50x(10^-2)^3)/(3600x(area at start)) = 0.0000708 (which already seems wrong)
Now subbing into bernoullis eq i got the exit velocity as 23.45
so i tried to do a momentum change, but was unsure of the time interval to even use, and I got a bit too confused to continue
 
What is your strategy for finding the external force required to keep the spray nozzle in place? To guide your thinking, consider two extreme situations in which the pressure at the entrance is 2.75 bar: (a) the exit diameter is plugged and there is no water flow; (b) the exit diameter is the same as the entrance diameter and the fluid flows freely. How would you calculate the required force in each case?

When you respond, please use equations in symbolic form, not numbers and especially not numbers without units attached to them.
 
My first question for you is "Is 2.75 bars at the entrance to the nozzle together with the inlet and exit velocities and the exit pressure of 1 bar consistent with the Bernoulli equation?"

My second question is whether you are familiar with the use of a macroscopic momentum balance to determine the force exerted by the nozzle on the fluid within the nozzle?
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top