Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter

    Why is the pressure at the surface less than the atmospheric pressure.

    The effect makes sense if the pressure is less near the surface, but mathematically, I don't understand how is that the case.

    I think this equation is relevant to this:

    \frac{\partial p}{\partial n} + \rho g \frac{\partial z}{\partial n} = \rho \frac{V^2}{r_s}

    where V is the velocity, rs is the radius of curvature, and n is the distance from surface.
Submit reply
Turn on thread page Beta
Updated: May 17, 2013
Do I go to The Streets tomorrow night?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.