Anyone any good at physics?

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andyf:
I am no physicist, so take the following with a large pinch of salt.

If fluid is entering the wide end at a given flow rate (in units of volume per unit of time) it must be exiting the narrow end at the same flow rate, or the nozzle would explode. So the flow rate will be the same at the entry and the exit, and indeed halfway along, or at any point along, the nozzle. Of course, the speed will increase between entry and exit.

The same will work in reverse, with fluid exiting the wide end at the same flow rate as it enters the narrow end, though in this case the fluid will slow down between entry and exit.

Assuming a fluid which is for practical purposes incompressible (a liquid rather than a gas), its mass will be directly proportional to its volume, so the mass flow rate will be the same all the way along the nozzle.

If (a big if) I am right, then the dimensions of the nozzle are immaterial.

Andy

DavidA:
There seems to be no mention of presure or time..

Looking at the equations linked to by Ross, it appears to me you need these to proceed.

Also not a physicist.

Dave.

tomrux:
Seems to me the flow rate has change/increase by the ratio of inlet to oulet eg:.5 to 2 meter or 4 to 1: therfore the flow rate would be twice the inlet flow rate at the half way point. the flow rate must increase to get the same volume through in the time (the volume in must equal volume out). without some idea of pressure or head height there is no way of telling absolute flow rates.

short answer, given the information at hand, is going to be: twice the inlet flow or half the outlet

Tom

Lew_Merrick_PE:

--- Quote from: Ross on December 05, 2012, 12:34:25 PM ---A conical Nozzle has a diameter on one end of 0.5m and 2m on the other. The nozzle is 6m long.

-Calculate the flow rate at midpoint along the nozzle.

-if the Fluid has a relative density of 0.866 calculate the mass flow rate.
--- End quote ---

What is the orientation of the nozzle? If it is vertical (2m on "top" & 0.5m on "bottom), then, assuming that the fluid is appearing magically at the input, your gradient is the 6m of height (density * gravity * height). Otherwise, it is insolvable as if there is no energy gradient there is (by definition) no flow! Now, if my assumption as to the orientation of the nozzle is correct, then the velocity at which the fluid is leaving the nozzle is V = (9.8066 m/s² * 6 m)0.5 m/s (assuming no friction of flow losses). Then V (m/s) * pi (m²) will be your volumetric flow rate (m³/s). Assuming the trivial density of water (1000 kg/m³) is accurate enough, then you fluid has a density of 866 kg/m³ and, if you multiply that times your volumetric flow rate, you get your mass flow rate.

However, this requires a specific orientation and a it is miraculously fed with fluid assumption...

GerryB:
G Day Ross,
You certainly have set one teaser.
Now i am not your intellectual but reading the problem it would seem that the answer should be simple.
As they do not give you as to what direction the nozzle is pointing, gravity does not come into the equation which would account for pressure differences.
I think that you will find that the output diameter dictates the flow rate so finding this should give the flow rate for any part of the length.
Good luck,
GerryB

Also if the nozzle had a smaller diameter at the end then the flow rate would be lower,but the pressure would be higher so i seems that the area of the output dictates the flow rate along the whole length of the unit
So the flow rate is going to be the same no matter at what point along its length it is measured.

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