# Smaller diameter pressure outlet to increase force?

1. Aug 5, 2013

### Pharrahnox

I have a hose that temporarily stores pressure in it, and it is about 19.05mm (3/4 inch). The outlet diameter is 8mm, so the air is squeezing into an area ~0.4 times the size. Does this increase the air flow speed, or the pressure?

Does it follow a rule similar to:

P1*V1 = P2*V2

except with area instead of volume?

2. Aug 5, 2013

### jfizzix

For an incompressible fluid, this can be worked out with Bernoulli's principle (which is just the conservation of energy for fluids).
$\frac{1}{2}\rho v^{2} +\rho g h + PV = \text{const}$

If you assume the fluid flows at a constant rate (that it is incompressible), it must be flowing faster (the velocity of the bits of water, not the flow rate) as the diameter of the hose goes down (so that the same number of liters per second is coming out the nozzle as comes into it). If the height doesn't change, then the pressure must go down.

Of course, talking about air means that the density can change too, but there is a similar equation governing compressible fluids.
http://en.wikipedia.org/wiki/Bernoulli's_principle

Hope this helps:)

3. Aug 5, 2013

### mark.watson

The velocity of the incompressible fluid increases. Due to this, the dynamic pressure increases as the static pressure decreases leaving no net change in pressure, overall (neglecting losses due to friction).

4. Aug 5, 2013

### boneh3ad

Incompressible flow and fluid flowing at a constant rate are not the same thing.

The density is not necessarily changing in air either. For air moving less than roughly Mach 0.3, the change in density is effectively zero in a flow and it can be treated as incompressible.

You also have to be very careful applying Bernoulli's equation with corrections for compressibility. Especially in compressible flows you don't necessarily have some neatly packaged equation that you can just apply willy-nilly. In particular, just blindly applying a Bernoulli-like relation tells you nothing about whether the flow is choked or if there are shocks anywhere in the system.

Anyway, assuming an incompressible, steady flow, the velocity will increase when the hose constricts since the mass flowing trough the tune must go somewhere (it remains constant). To pass it through a smaller tube the flow must speed up. Bernoulli's equation is the energy conservation relation in this situation for an inviscid flow, and shows that the static pressure drops as the velocity increases.

5. Aug 5, 2013

### jfizzix

Incompressible flow and constant flow rate are indeed not the same thing.

If the fluid is incompressible, the flow rate through any cross sectional area over the length of the nozzle must be the same (as seen from the continuity equation), though it need not be constant in time (say if the air hose is connected to a tank which depletes over time).

It could be that the flow rate is constant throughout the extent of the system, but changing in time.

It would have been better to say that the flow rate is not changing

6. Aug 5, 2013

### Pharrahnox

Ok, so the flow rate stays the same, and in order for that to be the case, each fluid particle must speed up through the smaller tube?

So if I were to measure the new force of that, from it being 800000pc in the 2.865x10-4m2 hose, it would now be ?pc in the 5.027x10-5m2 tube?

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook