Help with flow rates / mass flow rates.

[Ross891]

1. The problem statement, all variables and given/known


A conical Nozzle has a diameter on one end of 0.5m and 2m on the other. The nozzle is 6m long. The flow rate is 10 m/s. or 10m^3/s

We are assuming that the liquid is in compressible.

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

Homework Equations

Flow rate = velocity x surface area

The Attempt at a Solution

All I have done so Far on part a 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


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?

Have I missed a step? I'm only doing part a at the moment. Once I solve this I will move onto B

 

[mfb]

The flow rate is 10 m/s. or 10m^3/s

10m/s or 10m^3/s? That is not the same.

Then the surface area of the midpoint

PI x 1.25^2/4 = 1.227

You should work with units here. I agree with the numerical value.

The flow rate is 10 m/s flowing from left to right

It cannot be 10m/s everywhere, so you have to specify where you have 10m/s flow.
I think that you have 10m^3/s everywhere, that is a completely different statement.

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

10m/s is a velocity. Again, I think you mean 10m^3/s as flow, in that case 8.149m/s (note the units!) is the velocity there.

Can you give the full problem statement?

 

[Ross891]

In my notes. I have flow rate written in m/s this is how we have been taught not m^3/s.

The question

A conical nozzle expands from 0.5m to 2m in a distance of 6m, the flow rate is 10m/s

A) calculate the flow rate at the midpoint along the nozzle
B) if the fluid has a relative density of 0.866 calculate the mass flow rate.


So surely the question is worded wrong? We want the velocity of flow otherwise it would be pointless?

 

[mfb]

You need the flow rate as volume per time or the flow velocity at some specific point to solve it. Maybe the 10m/s refer to the input with 0.5m as diameter. This would give something to calculate.

 

[Ross891]

You need the flow rate as volume per time or the flow velocity at some specific point to solve it. Maybe the 10m/s refer to the input with 0.5m as diameter. This would give something to calculate.

I think you are correct there. Am I right in saying it would be flowing at 10m/s through the 0.5m diameter.

If that's correct there is no labeled diagram but that's what we'll assume.

 

[mfb]

I would write that assumption down (as part of the solution), and work with it afterwards.

 

[Ross891]

Ok 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?

Is that any better?

 

[haruspex]

As far as I am aware, you can have volumetric flow rate, mass flow rate, and velocity of flow. "Flow rate" should not be used for velocity. But to make a reasonable question out of the wording given, I have to assume it contains two errors: calling velocity of flow flow rate; failing to specify where in the nozzle the 10m/s applies.

Ok 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?

Yes, but by leaping into numerics you've made it a bit harder than it needed to be.
Input area = πr_in²; velocity in = v_in; vol flow rate = πr_in²v_in = πr_mid²v_mid
v_mid = v_in (r_in/r_mid)²
= 10(1.25/.5)² m/s

 

[Ross891]

As far as I am aware, you can have volumetric flow rate, mass flow rate, and velocity of flow. "Flow rate" should not be used for velocity. But to make a reasonable question out of the wording given, I have to assume it contains two errors: calling velocity of flow flow rate; failing to specify where in the nozzle the 10m/s applies.

Thanks for clarifying that. If I had the right information to start I night not be struggling as much! So instead of flow rate, it's velocity of flow. In this case. Thanks!

Yes, but by leaping into numerics you've made it a bit harder than it needed to be.
Input area = πr_in²; velocity in = v_in; vol flow rate = πr_in²v_in = πr_mid²v_mid
v_mid = v_in (r_in/r_mid)²
= 10(1.25/.5)² m/s

Could you just explain this par not quite sure what's going on!

Thanks a lot

 

[mfb]

He did basically the same as you, just with parameters instead of numeric values.
If you are confused by the last line only: haruspex swapped numerator and denominator there, it should be
= 10(.5/1.25)² m/s

A larger diameter has to correspond to a lower velocity.

 

[Ross891]

He did basically the same as you, just with parameters instead of numeric values.
If you are confused by the last line only: haruspex swapped numerator and denominator there, it should be
= 10(.5/1.25)² m/s

A larger diameter has to correspond to a lower velocity.

Ok That make sense. I think I'll stick with numerical values for now. It might be longer but my head may explode If I do anything else!

 

[Ross891]

Now onto part B.

This should be simple. I think I'm ok with mass flow rates... Famous last words!


So to get the mass flow rate I take the relative density given as 0.866 and multiply it by the velocity I worked out. Presumably at the midpoint. And then multiply it by the surface area of the midpoint.

So


0.866x62.5x1.227=66.411 (I'm not sure on units since none were given on the relative density.)


Does that seem ok guys?

 

[haruspex]

If you are confused by the last line only: haruspex swapped numerator and denominator there, it should be
= 10(.5/1.25)² m/s

Whoops. Thanks for picking that up.

 

[mfb]

0.866x62.5x1.227=66.411 (I'm not sure on units since none were given on the relative density.)

You used a wrong velocity value here. Apart from that, it looks fine. "Relative density"... well, ok. I would give m^3/s as unit in that case. This can be multiplied with a density (kg/m^3) to get mass flow (kg/s).

 

[Ross891]

You used a wrong velocity value here. Apart from that, it looks fine. "Relative density"... well, ok. I would give m^3/s as unit in that case. This can be multiplied with a density (kg/m^3) to get mass flow (kg/s).

Ah yes let me correct that

0.866x8.1499x1.227=8.65 Kg/s


The 62.5 I used first by mistake, i know it sounds silly but what is that? I worked it out above, I referred to some previous notes I had. it was from an example wit a tube that had multiple diameters and we had to work out the flow rates through these, this is what is probably confusing me the most!

Thanks

 

[Ross891]

Just a bump,

If somone can clarify the reason whyI have two answers for the velocity of flow 8.1499 and 62.5? And which method of getting there is correct!

Im nearly there!

 

[mfb]

You got two answers because you have one correct and one wrong calculation.

62.5m/s would be the speed at the point of 0.5m diameter if the fluid would flow with 10m/s where it has a diameter of 1.25m.

 

[Ross891]

You got two answers because you have one correct and one wrong calculation.

62.5m/s would be the speed at the point of 0.5m diameter if the fluid would flow with 10m/s where it has a diameter of 1.25m.

That makes sense. I used the wrong method there. Thanks for your help in solving this anyhow. I have anther question that ill post up that I think I may have messed up on we shall see.