Force resulting from a pressurized departing gas

  • Context: Undergrad 
  • Thread starter Thread starter tjhj
  • Start date Start date
  • Tags Tags
    Force Gas
Click For Summary
SUMMARY

The discussion focuses on calculating the force exerted by gas escaping from a cylinder with a hole. The key parameters provided are the gas density of 2.85 lb/ft³ and a mass flow rate of 69.33 lb/s. To determine the resulting force, the principle of momentum change is applied, emphasizing the need to convert to SI units for accurate calculations. The velocity of the gas, approximately 4000 ft/s, is also a critical factor in the force calculation.

PREREQUISITES
  • Understanding of fluid dynamics principles
  • Knowledge of momentum and force equations
  • Familiarity with unit conversion, particularly to SI units
  • Basic grasp of gas properties, including density and flow rate
NEXT STEPS
  • Study the concept of momentum conservation in fluid dynamics
  • Learn about the application of the continuity equation in gas flow
  • Explore the relationship between mass flow rate and force in compressible fluids
  • Investigate the effects of gas velocity on force calculations
USEFUL FOR

Engineers, physicists, and students in fluid mechanics or thermodynamics who are interested in understanding the dynamics of gas flow and force calculations in pressurized systems.

tjhj
Messages
2
Reaction score
0
Hello,
Thank you for your help.

So if I have a cylinder with a hole in it, and gas escaping from it. How can I find the resulting force that gas is exerting on the cylinder. So I can then find the acceleration it is imparting.

I know the density of the gas 2.85 lb/ft^3, and the mass flow rate of the escaping gas is 69.33 lb/s.

In a very small increment of time .001 milliseconds. There has been 6.933*10-5 lbs discharged.

What is the resulting force on the cylinder? Is it just the lbs of Gas discharged? I know the velocity of the gas is around 4000 ft/s, do I need this to find this out? or is the mass flow rate accounting for it?

Thank you,
 
Physics news on Phys.org
I would say that you need to use force = rate of change of momentum
You should really be using SI units...m, kg, s
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
7
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K