Time taken for pressure to equalise?

  • Context: Graduate 
  • Thread starter Thread starter serverxeon
  • Start date Start date
  • Tags Tags
    Pressure Time
Click For Summary
SUMMARY

The discussion centers on calculating the time taken for pressure to equalize in a hemisphere dome after puncturing a hole. The internal pressure is 1 atm (1.05 kPa) and external pressure is 0.01 atm (10.5 Pa). Participants suggest using the differential form of Bernoulli's equation, alongside the ideal gas law and the rate of change of mass equation, to model the non-linear dynamics of the system. Key considerations include the changing density of air and the assumption of non-turbulent flow.

PREREQUISITES
  • Understanding of Bernoulli's equation and its differential form
  • Familiarity with the ideal gas law
  • Basic knowledge of differential equations
  • Concept of mass flow rate and its relation to area and velocity
NEXT STEPS
  • Study the differential form of Bernoulli's equation in fluid dynamics
  • Explore the ideal gas law and its applications in varying pressure scenarios
  • Learn about solving non-linear differential equations
  • Investigate the effects of compressibility in gas flow
USEFUL FOR

Students and professionals in physics, engineering, and fluid dynamics who are interested in pressure dynamics and gas behavior in confined spaces.

serverxeon
Messages
100
Reaction score
0
I have a hemisphere dome, which volume = \frac{2}{3}∏r3
Inside the dome, pressure is 1atm (assume 1.05 kPa)
Outside the dome, pressure is 0.01atm (10.5 Pa)
I then puncture a 0.2 cm2 (2 x 10-5 m2) hole on the flat side of the dome.

Question: Find time taken for pressure to equalise.

--------------

I have searched up the internet, and seems that I have to use bernoulli. But I have a few problems.
8fe8d95e8b31880e495d141a3256e3db.png

1) I have no idea what to substitute for the variables.
2) Are the density of air inside and outside the dome different?
3) How should I use calculus (or differential equations, or anything else) to find the time taken, given the process is non-linear in relation.

Thanks
 
Last edited:
Physics news on Phys.org
I don't think you can use Bernoulli's equation to solve, as most requirements are not met (dm/dt is not constant, air is not incompressible. You may use its differential form, though:

\frac{dp}{\rho} + d(\frac{v^2}{2}) + g dz = 0

Now you'll have to make a few hypothesis, such as the processes is not turbulent (which it is, however, though it would be almost impossible to accurately describe this phenomenon without this consideration, so your answer will be physically wrong), the air is an ideal gas (or find a good equation of state, though ideal gas law is good).

The specific mass of air in both situations are different, but remember that they must equal when in equilibrium.

The differential equation you will have to solve will be a combination of that differential form of Bernoulli's law and the fact that

\frac{dm}{dt} = V \frac{d\rho}{dt} = Av

Interesting problem, by the way.
 
Thanks for the reply,

But I'll like to clarify, how is rate of change of mass = area * velocity?
\frac{dm}{dt} = V \frac{d\rho}{dt} = Av
 
I'm sorry, that's wrong of course, must've been my distraction. It should be

\frac{dm}{dt} = V\frac{d\rho}{dt} = \rho A v
 
thanks pal,

but I am looking at the equations, and am thinking...
Pressure will change as the process equalise, and density will change as pressure change, so mass in the dome will change too. At the same time velocity changes too... So I have multiple interrelated quantities.. and er...

Can someone please guide me on how should I start?
I'm like just stuck!
 
Last edited:
Yes, everything changes and that's why you use a differential equation to solve this problem.

You will have to work with three equations:

1) Differential form of Bernoulli's
2) Rate of change of mass
3) Ideal gas law

Can you find a equation for velocity using 1) and 3)? Tip: make a few assumptions, one of them is to ignore gravity effects (the dome is too small).
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
10K
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 8 ·
Replies
8
Views
17K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 30 ·
2
Replies
30
Views
4K
Replies
2
Views
1K
Replies
16
Views
3K