Anyone any good at physics?

[Ross]

Hi everyone,

I'm doing my apprenticeship in engineering at the minute as a machinist. The course I'm studying at college has a therory part which is mostly irrelevant to anything ill be doing but that's besides the point.

 there's this one example question which has been playing me up for days. And I'm at my wits end, if somone can make some sort of sense of it I'd be thankful I do not have a correct answer yet, becuase no one else could solve it! I have only got near the end of the first part still none the wiser.

But if anyone has any pointers of even where to start as it would be helpful to at least try and et to the bottom of it.

Anyhow:

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.


Good luck!

[mklotz]

You may find this helpful...

http://www.engineeringtoolbox.com/orifice-nozzle-venturi-d_590.html

[David Jupp]

On the basis that what goes in must come out, the flow rate is identical regardless of where you are along the nozzle.

There is information missing if you want to put a figure to the flow rate.

[Lew_Merrick_PE]

I'm doing my apprenticeship in engineering at the minute as a machinist. The course I'm studying at college has a therory part which is mostly irrelevant to anything ill be doing but that's besides the point.

 there's this one example question which has been playing me up for days. And I'm at my wits end, if somone can make some sort of sense of it I'd be thankful I do not have a correct answer yet, becuase no one else could solve it! I have only got near the end of the first part still none the wiser.

But if anyone has any pointers of even where to start as it would be helpful to at least try and et to the bottom of it.

Anyhow:

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.

You are missing a key part of the question, What is the gradient across the nozzle  This is most commonly provided as a delta-Pressure value.

[Ross]

The Gradient isn't given.

But it can be worked out by dividing the difference in the diameters by the length:

1.5/6=0.25

So every meter the cone expands by 0.25m

Not sure where to go next though...

[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]

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.

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.

[Lew_Merrick_PE]

Come on people, think this through.  (1) The cited specific gravity is 0.866.  Water has a specific gravity of 1.000, so this fluid is likely to be an oil of some kind.   Gasses would have a specific gravity fairly close to 0.  (2) Whereas the density of the fluid may change is it is being forced through a (reducing or expanding) nozzle, the mass flow rate through the nozzle will remain constant (Green's Theorem).  (3) If there is no energy gradient across the nozzle then there is no flow through the nozzle (First Law of Thermodynamics)!

I do a moderate amount of fluid flow work.  I don't do enough to know all this noise of the top of what's left of my mind (though I have friends who do).  Most normally, you are given: either a volumetric or mass flow requirement, required nozzle exit fluid speed, or a specific pressure delta to achieve.  You then work out a pressure/thermal head needed to fit that requirement and size the pump or "drop" (etc.) to suit.

Water clocks are the classical example of this type of flow.  You use a very large area, but otherwise shallow, collector at each stage of a water clock to minimum variations in pressure head.  You use a (relatively) small orifice to regulate the outflow into your collection cylinders that, when you measure the height of the fluid in the cylinder, gives you the amount of that gradient of time that has passed.  Such systems were common by 500 BC and were probably in existence prior to that.

[andyf]

... Water clocks are the classical example of this type of flow.  You use a very large area, but otherwise shallow, collector at each stage of a water clock to minimum variations in pressure head.  You use a (relatively) small orifice to regulate the outflow into your collection cylinders that, when you measure the height of the fluid in the cylinder, gives you the amount of that gradient of time that has passed.  Such systems were common by 500 BC and were probably in existence prior to that.

Ah, the clypsedra! The Babylonians and Egyptians used those around 1600 BC.

Having now exhibited my knowledge of useless trivia, all I can say about Ross's problem is that it looks like a trick question, incorporating data which is intended to confuse. Other more essential data is omitted, like the rate at which fluid is being fed into the nozzle, and its nature ("fluids" comprised both liquids and gases when I was at school).

Given this dearth of real information, I think the best that can be said is that (ignoring gases,  and assuming liquids to be incompressible) the flow rate in both volume and weight terms will be the same as the supply flow rate and will be uniform all the way along the nozzle. This will hold true whatever the relative density of the fluid, though if one fluid has a relative density of 1 and the other 0.866 and the volumetric flow rate at the input is the same for both, then (ignoring any difference in friction) the mass flow rate of the latter fluid will be 0.866 that of the former.

Andy

[Ross]

Thanks for your help guys, ill have a look in the morning and see what I can do.

I forgot to mention ism working this out and assuming the liquid is not compressible and the only forces that's needed to worry about is just the flow of the liquid. 

[Ross]

Hi guys,

I'm doing the first part (a) it seems I NEED the velocity to do anything and seeming as pressure isn't given or anything it makes it a lot harder.

I understand the flow rate will remain constant but the question asks for the midpoint which Is what I shall do. But having so little info is proving the most tricky 

[Lew_Merrick_PE]

I'm doing the first part (a) it seems I NEED the velocity to do anything and seeming as pressure isn't given or anything it makes it a lot harder.

I understand the flow rate will remain constant but the question asks for the midpoint which Is what I shall do. But having so little info is proving the most tricky

Ross,

The velocity of flow will be inversely-proportional to the cross-sectional area.  For an incompressible fluid, this means that, the larger the area, the lower the velocity.  Given the ø0.50 m to ø2 m ratio, the velocity of flow will be (1²/0.25²) 16X the velocity of the large end at the small end.  The velocity of flow will vary along the length of the taper in proportion to the square of the change in radius (Delta-r).  Assuming we go from small to large (i.e. ø0.50 m to ø2 m), then the increase in radius from X=0 (i.e. R0.25 m) to X=6 (i.e. R1 m) is given by: r(x) = 0.25 + x(0.75/6) = 0.25 + 0.125x.  r²(x) = (0.25 + 0.125x)² = 0.0625 + 0.0625x + 0.015625x².  The latter (r²(x)) gives you the r² at any point between x=0 and x=6.  r²(x)/0.0625 = 16r²(x) will give you a plot of the local (x-based) velocity with respect to the velocity at the small (i.e. ø0.5 m) end of the taper.  Plotting this out might be helpful to your understanding.

[Ross]

Though not having the original velocity at one end wouldn't enable me to be able to do any of the above right?

[andyf]

That's right, Ross. Anyway, the question you quoted doesn't ask anything about velocity, only about "flow rate" and "mass flow rate". The latter being weight per second (or minute, or hour)  I think you can assume that the former must refer to volume per sec, min or hour.

All you can say is that both flow rates will, midway along the nozzle, be the same as the input rate, though it might impress if you conclude with a comment on the velocity with a recital of Lew's mathematics.

[Lew_Merrick_PE]

Ross,

Did you get this straightened out?

Irrational minds want to know....  ;-)

[Ross]

Getting there,

I got the "flow rate". Which is 10 m/s this is actually the velocity of flow. though no info of where the point of  measurement was. 


So I've worked out the surface area of the midpoint of the tube to be 1.227.

And the velocity of the flow rate would be 8.1499. Flow rate/surface area.

Then 8.1499x1.227=9.999 which is the flow rate through that point.
 

I'm assuming that mu tutor mixed up the difference between flow rate and velocity of flow. This is probably why this has all been so hard to understand 


Make any sense?