Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne Pressure rise in pipe expansion
Name: ericschotz
Status: N/A
Age: N/A
Location: N/A
Country: N/A
Date: Around 1993


Question:
I have learned that while fluid is flowing through a pipe and the pipe expands the pressure is higher in the segment of the tube with greater diameter. I have seen the equations that prove this but I am having a hard time grasping in physically. I have this idea that faster flow implies high pressure but in this case the high pressure segment is slower. If anyone has a good explanation, I would appreciate it.
Replies:
Faster flow does not necessarily mean higher pressure. In the case of pipe expansion: In order to maintain a constant flow rate (flow rate = Q), an increase in cross-sectional area (pipe expansion) means a lower velocity because Q = A1 * V1 = A2 * V2. An increase in A2 (pipe expansion) means that V2 must be lower to maintain the same flow rate. When V1 > V2, then the change in pressure (Delta P) is positive due to P2 - P1 = Delta P = (V1^2 - V2^2)/2. I know you have seen the equations to prove it, but I think your idea of faster flow is always high pressure is holding you back. With a little manipulation of Bernoulli's equation, one can show that pressure change in a straight pipe is related to the density of the liquid, the flow rate of the liquid and the diameters of the pipes by (I hope this makes sense here) P2 - P1 = (8 * rho * Q^2/pi^2) * (1/D1^4 - 1/D2^4). Here it is easy to see that with a constant flow rate, the change in pipe diameters is the driving force to the pressure change. As for physically explaining this, I do not think I did that, but I hope this helps.

cdmurphy


The thing to focus on here is the kinetic energy of a given volume of fluid as it passes from the large-diameter pipe to the small diameter pipe. In the large pipe, the volume is moving more slowly and has less kinetic energy than it will have when it gets into the small pipe. So, the volume must be ACCELERATED as it moves from one pipe to the other, and this requires a force. In fact, it requires a larger force from the fluid still in the large pipe that will more than balance the force acting backwards from the fluid in the small pipe, so then net force on the volume of fluid will be sufficient to accelerate it to its higher velocity. The only place that force can come from is the pressure (pressure, as you probably know, is just force per unit area). What we need here is actually, a larger force per unit VOLUME, since this is equivalent to an acceleration (for an incompressible fluid, volume and mass are proportional.)

mooney


Both of the above answers are very good. I would just like to add this example for you to picture. To get honey from a squeeze bottle, the bigger the hole, the less you have to squeeze (and the more you can eat).

dipper



Click here to return to the Engineering Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (help@newton.dep.anl.gov), or at Argonne's Educational Programs

NEWTON AND ASK A SCIENTIST
Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory