Thermodynamics and Fluid Mechanics Thermodynamics and Fluid Mechanics Physics Help Forum 
Dec 9th 2012, 07:42 AM

#1  Junior Member
Join Date: Dec 2012
Posts: 6
 Flow rates and mass flow rates.
Hi everyone,
First off, hello and I hope this is the right place to post this.
I'm doing an apprenticeship in engineering at the minute as a machinist. The course I'm studying at college has a mechanical principle unit. I understand 75% of the things taught but things get pear shaped when different situations and problems occur.
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!
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.
I am working at Level three or A level, so other forces don't really apply and I am assuming that the fluid is non compressible.
Anyhow:
A conical Nozzle has a diameter on one end of 0.5m and 2m on the other. The nozzle is 6m long.
1. Calculate the flow rate at midpoint along the nozzle.
2. if the Fluid has a relative density of 0.866 calculate the mass flow rate.
All I have done so Far on part 1 is worked out the rate of expansion of the cone by doing:
The difference in diameters is 1.5
So 1.5/6=0.25
So the cone expands 0.25 every meter.
Then I times the rate of expansion by half the length:
0.25x3=0.75
Then the surface area of the midpoint
PI x 1.25^2/4 = 1.227
I don't know where to go next as there is no velocity or pressure given.
Thanks,
Ross
Last edited by Ross891; Dec 9th 2012 at 07:56 AM.

 
Dec 9th 2012, 11:49 AM

#2  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,354

So far so good. If you assume incompressible flow then the the flow rate is
Where rho is the density, v is velocity, and A is crosssection area. The mass flow is constant throughout the length of the nozzle, and if the flow is incompressible then the quantity vA is constant along the length of the nozzle. You've figured out A at the midpoint, but without knowing either the mass flow, or the velocity at some other point, you can't do much more. Are you sure that they don't give some other data, such as velocity at the nozzle entrance or exit?
Last edited by ChipB; Dec 10th 2012 at 12:56 PM.

 
Dec 9th 2012, 02:15 PM

#3  Junior Member
Join Date: Dec 2012
Posts: 6

Originally Posted by ChipB So far so good. If you assume incompressible flow then the the flow rate is
Where rho is the density, v is velocity, and A is crosssection area. The mass flow is constant throughout the length of the nozzle, and if the flow is incompressible then the quantity vA is constant along the length of the nozzle. You've figured out A at the midpoint, but without knowing either the mass flow, or the velocity at some other point, you can't do much more. Are you sure that they don't give some other data, such as velocity at the nozzle entrance or exit? 
Thanks for that, That is all the information I am given. I wouldn't be surprised if the tutor has made a mistake with that question.
So there is no way to work out any of these without the flow rate or velocity?

 
Dec 10th 2012, 07:41 AM

#4  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,354

You will need some other piece of information to calculate mass flow rate. Think of this practical example  how much water flows though the nozzle attached to my garden hose? It depends on whether the spigot is open or closed, right? Could be anywhere from 0 gallons/minute to perhaps 5 gallons/minute, depending on how wide open the spigot is. So unless I tell you how fast the water flows in the hose, or how fast it comes out of the nozzle, or the pressure in the hose you can't figure out the mass flow rate.
Last edited by ChipB; Dec 10th 2012 at 12:58 PM.

 
Dec 10th 2012, 02:05 PM

#5  Junior Member
Join Date: Dec 2012
Posts: 6

Thanks, ill see if I can get any more figures as there may be an error with the question.

 
Dec 15th 2012, 07:04 AM

#6  Junior Member
Join Date: Dec 2012
Posts: 6

I have now got the flow rate through the conical tube. So I can now work out the actual answer. Thanks for pointing that out! The flow rate is 10 m/s flowing from left to right
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. But that doesn't seem right. If you knew the flow rate was constant through the pipe what is the point of calculation?
Or have a missed a step?

 
Dec 15th 2012, 07:45 AM

#7  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,354

The flow rate is measured in either mass per second, or for an incompressible fluid like water it may be expressed in volume per second. So I would assume that it's not 10 m/s, but rather 10m^3/s (meters cubed per second). This is constant throughout the length of the nozzle, but the veloity of flow (in meters per second) is not constant, byt rather varies inversely to the crosssectional area. At the middle of the nozzle the velocity of flow in m/s is what you calculated: (10m^3/s)/(1.227m^2) = 8.149 m/s. Similarly for the end where the nozzle is 0.5m in diameter, and hence A = 0.1963m^2, the velocity is v = 10/0.1963 = 50.9m/s. At the other end where the diameter is 2m and the area is 3.1416m^2 the velocity is 3.18 m/s. This effect is quite common  if you run water out of a garden hose and put your thumb over the end of the hose to restrict the area of flow you can get the water to flow at a much higher rate.

 
Dec 15th 2012, 08:09 AM

#8  Junior Member
Join Date: Dec 2012
Posts: 6

Originally Posted by ChipB The flow rate is measured in either mass per second, or for an incompressible fluid like water it may be expressed in volume per second. So I would assume that it's not 10 m/s, but rather 10m^3/s (meters cubed per second). This is constant throughout the length of the nozzle, but the veloity of flow (in meters per second) is not constant, byt rather varies inversely to the crosssectional area. At the middle of the nozzle the velocity of flow in m/s is what you calculated: (10m^3/s)/(1.227m^2) = 8.149 m/s. Similarly for the end where the nozzle is 0.5m in diameter, and hence A = 0.1963m^2, the velocity is v = 10/0.1963 = 50.9m/s. At the other end where the diameter is 2m and the area is 3.1416m^2 the velocity is 3.18 m/s. This effect is quite common  if you run water out of a garden hose and put your thumb over the end of the hose to restrict the area of flow you can get the water to flow at a much higher rate. 
So what would the answer to question A be then? The 9.99 or the 8.149? I'm having trouble determining which is which!

 
Dec 17th 2012, 05:48 AM

#9  Junior Member
Join Date: Dec 2012
Posts: 6

I've found mother method of working out the flow rate in my notes. It gives me a totally different answer to the 8.149.
The question is wrong. I think it is asking for velocity of flow through the midpoint and not the flow rate.
so considering the fact that the flow rate is 10m/s through the 0.5m diameter.
I would work out the surface area of the 0.5 diameter
∏X0.5^2/4= 0.1963m^2
And then divide the surface area of the midpoint. 1.227m^2 into the surface area of the 0.5 diameter 0.1963m^2
1.227/0.1963=6.250
And then multiply this answer by the original flow rate though 0.5
6.250x10=62.5
So the flow rate through the midpoint is 62.5m/s?
I'm confused how this answer is different? Which one is right. Bear in mind I took this from a different example so this might have been applied in the wrong way

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