17 years ago
Pi + 1/2(density)(Vi)^2 = Pe + 1/2(density)(Ve)^2 , where P and V are
the pressure and velocity and the i and e subscripts indicate initial
and exit values. There is no height term since the pipe is horizontal.
Let's say the inital pressure Pi is very high say thousands of bars. I
want to find the velocity on exit from the pipe when it empties to the
ambient pressure, Pe = 1 bar.
In addition to the Bernoulli principle an equation used to calculate
the velocity of a fluid stream is the continuity equation AeVe = AiVi,
where A is the cross-sectional area of the pipe, with e and i again
meaning initial and exit values. This equation holds since assuming an
incompressible liquid the total volume flowing in must equal the volume
Therefore if the pipe is at constant diameter the velocity should stay
the same. But that means in the Bernoulli equation above the velocity
cancels out so the equation becomes Pi = Pe = 1 bar, which is
incorrect. What's the resolution of this?
Secondly let's say the pipe is of smaller diameter on exit. Then Vi =
(Ae/Ai)Ve. Substituting this into Bernoulli gives:
Pi + 1/2(density)(AeVe/Ai)^2 = Pe + 1/2(density)(Ve)^2.
Then Pi-Pe = 1/2(density)(1-(Ae/Ai)^2)Ve^2. And solving for Ve this
Ve = [2(Pi-Pe)/(density)(1-(Ae/Ai)^2)]^.5
The curious thing about this equation is if we make the exit
cross-section arbitrarily close to the initial cross-section then we
can make the exit velocity arbitrarily large.
Is this correct?