Hi - Can someone figure out how to calculate the air velocity escaping from a balloon through a pinhole, if you know the pressure in the balloon and the diameter of the pinhole. For small diameters, I think the velocity depends on both parameters. Also, how to calculate the friction in a very very small air pipe that I might insert through the pin hole to re-inflate the balloon. Thank you, Ed

Can someone figure out how to calculate the air velocity escaping from a balloon through a pinhole, if you know the pressure in the balloon and the diameter of the pinhole. For small diameters, I think the velocity depends on both parameters. No, It doesn't depend on eigther.  The velocity will be equal to the speed of sound because the pinhole is a choke point in the flow. George Marklin This e-mail address is being protected from spam bots, you need JavaScript enabled to view it

I observe that when I poke two pinholes of diferent diameter at the same height at the bottom of a milk carton water from the slightly larger hole projects further. How come?

Can someone figure out how to calculate the air velocity escaping from a balloon through a pinhole, if you know the pressure in the balloon and the diameter of the pinhole. For small diameters, I think the velocity depends on both parameters. No, It doesn't depend on eigther.  The velocity will be equal to the speed of sound because the pinhole is a choke point in the flow. That's only true if the ratio of upstream to downstream pressure is above a critical value, around 2 for air.

I observe that when I poke two pinholes of diferent diameter at the same height at the bottom of a milk carton water from the slightly larger hole projects further. How come? The physics of a liquid going through a pinhole is different from a gas.  Water will not go at the sound speed.  A crude estimate may be obtained from Bernoulli's law: 1/2 rho v^2 = rho g h giving v = sqrt(2gh) where h is the height of the water above the hole.  This does not depend on the hole size but the effects of viscosity and turbulence have been ignored. A very crude estimate of the effect of viscosity may be obtained by modifying Bernoulli's law to: 1/2 rho v^2 = rho g h + rho nu k^2 v l where nu is the coefficient of viscosity and k is a radial wavenumber measuring the shear in the flow and l is a coordinate along the flow-lines.  This gives: v = sqrt(2gh+(nu k^2 l)^2)-nu k^2 l A crude estimate of the flow geometry for the pinhole would be to take k=1/a where a is the radius of the hole, and l=t where t is the thickness of the wall.  Thus: v = sqrt(2gh+(nu t/a^2)^2)-nu t/a^2                         (1) which shows that the velocity can be increased by making the hole bigger or making the wall thinner.  But for very thin walls (t<a) you probably have to take l=a. Then you have: v = sqrt(2gh+(nu/a)^2)-nu/a                                 (2) which still shows that the velocity can be increased by making the hole bigger, but the dependence on a is different. Eq. (1) is for t a and eq. (2) is for t < a.  You can do some experiments to determine if these crude models give the correct dependence on a. George Marklin This e-mail address is being protected from spam bots, you need JavaScript enabled to view it

It still seemsl partly unresolved. A scaling factor perhaps ... From equation (2) by Marklin: v = sqrt (pressure / density - (viscosity / radius)^2) - viscosity / radius Try, for air through a pinhole from 2 atm: pressure drop = 100000   kg-m/s^2 density       = 1.5      kg/m^3 viscosity     = .000183  kg/m/sec radius        = .0001    m then p/d      = 60000.00    m^2/s^2      v/r      =     1.83    kg/sec v = sqrt(60000) = 250 m/s, from p/d only The p/d term seems about right and the v/r term does not. I assume I have lost something in the translation to do with the scaling and units of the viscosity / radius term. Ed