Change in pressure when merging tubes

  • Thread starter Thread starter Gerald Funk
  • Start date Start date
  • Tags Tags
    Change Pressure
AI Thread Summary
When merging two tubes with the same pressure and diameter into a single tube, the pressure does not simply double. Instead, the Bernoulli equation must be applied to analyze the flow dynamics, considering the velocities of air in each section. Since both tubes have identical conditions, the combined flow rate into the exit tube can be determined, which affects the velocity and pressure in the merged section. The discussion emphasizes the importance of understanding laminar flow and how to integrate the Bernoulli equation for this specific scenario. Ultimately, the relationship between pressure and velocity in the merged tube needs careful consideration to arrive at the correct conclusion.
Gerald Funk
Messages
6
Reaction score
0

Homework Statement


hello, I have a problem that I couldn't find a clear answer to. Attached is a diagram of a related problem that explains the problem

Air traveling through two tubes with the same direction have the same PSI (say 75). If you merge the two tubes into a tube with the same diameter as the previous two, what happens to the pressure of the moving air? Does it double?

20z7n6s.jpg


Homework Equations

The Attempt at a Solution


I think the pressure would double, but I am unsure
 
Physics news on Phys.org
Have you investigated the Bernoulli equation for this setup? If the tube diameters are all the same, how are the velocities of the air in each section related?
 
  • Like
Likes Gerald Funk
I don't know if the original problem is that complicated. Let's say the source of air is two identical air compressors set to the same settings, so the PSI, the velocities, and the volume is the exact same for both tubes.
 
Apply Bernoulli. Really :smile:

You'll have to put a small bit of thought into deducing the velocity in the merged tube. Assume that the tubes are lying on the ground (you're looking down on them) so there's no change in height involved.
 
I am still having troubles getting my head around this. The problem is that any time I've used the Bernoulli equation, its only looking at one tube (like a Laval nozzle where one tube gets smaller), but this question has two tubes feeding into one. So this equation:
a71608f0f41473679c7dce52f525d232.png


would need a p3 and v3 somewhere in there, but I can't figure out how to integrate it
 
You said that the pressure and flow rate in the feeds were identical. Thus each feed moves the same volume of air into the exit tube per unit time. Knowing the dimensions of the tubes, that should allow you to determine what the flow rate (hence velocity) must be in the exit tube. A flow streamline can be chosen in one feed tube and followed to the exit tube. We assume no turbulence, hence laminar flow. Apply Bernoulli to that streamline. Two velocities, two pressures. You know the two velocities and one of the pressures...
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top