Fluid flow through branched nozzles

[Marki john]

Homework Statement

I wanted to know if the flow rate and velocity would stay constant or change in nozzles 1,2 and 3 in a incompressible fluid flow through a rectangular duct. (Illustrated image in attachment).

Aim is to have the same amount of fluid flowing out of each of the nozzles.

Homework Equations

Flow Rate (Q) = Velocity (V) x Area (A)

The Attempt at a Solution

From what I know the sum of the flow rates within the nozzles equals the initial flow rate;

Qin = Q1 + Q2 + Q3
I am not fully sure about this.

Flow rate is dependent on velocity of the fluid flowing and the cross-sectional area. Therefore, when flow rate changes velocity would also change and vice versa.

Would the velocity be the same as the flow rate?
Vin = V1 + V2 + V3

Or velocity stays constant throughout?

I just need for this to be clarified or corrected so I am fully sure.

 

[rude man]

Homework Statement

I wanted to know if the flow rate and velocity would stay constant or change in nozzles 1,2 and 3 in a incompressible fluid flow through a rectangular duct. (Illustrated image in attachment).
Aim is to have the same amount of fluid flowing out of each of the nozzles.

Homework Equations

Flow Rate (Q) = Velocity (V) x Area (A)

The Attempt at a Solution

From what I know the sum of the flow rates within the nozzles equals the initial flow rate;
Qin = Q1 + Q2 + Q3
I am not fully sure about this.

Why not? Total in = total out.

Flow rate is dependent on velocity of the fluid flowing and the cross-sectional area. Therefore, when flow rate changes velocity would also change and vice versa.
Would the velocity be the same as the flow rate?

? Velocity and flow rate have different dimensions so this question makes no sense.

Mentally divide the incoming flow into three equal subflows, one going to each nozzle, and apply Bernoulli to each subflow.

 

[Marki john]

Why not? Total in = total out.? Velocity and flow rate have different dimensions so this question makes no sense.

Mentally divide the incoming flow into three equal subflows, one going to each nozzle, and apply Bernoulli to each subflow.

Assume there are no pressure loss in both minor and major losses and perfect ideal world conditions. According to continuity law the total in must equal total out which you are correct but I am not sure about whether the total in stays the same even in the 3 nozzles or the total in splits into 3 flow rates for each nozzles?

so Qin = Q1 = Q2 =Q3 or Qin = Q1 + Q2 + Q3 >> Qin = 3(Qnozzles)

Also what I mean about the velocity is that does it behave similar to flow rate e.g. would the velocity in need to be divided into 3 equal sub velocities or it stays the same?

Hope you could help with this. Thank you

 

[rude man]

Assume there are no pressure loss in both minor and major losses and perfect ideal world conditions. According to continuity law the total in must equal total out which you are correct but I am not sure about whether the total in stays the same even in the 3 nozzles or the total in splits into 3 flow rates for each nozzles?
so Qin = Q1 = Q2 =Q3 or Qin = Q1 + Q2 + Q3 >> Qin = 3(Qnozzles)

If Qin = Q1 = Q2 =Q3 were correct then you would get out 3 times the fluid you put in! A neat trick ...

Also what I mean about the velocity is that does it behave similar to flow rate e.g. would the velocity in need to be divided into 3 equal sub velocities or it stays the same?

To answer this do what i suggested - mentally split the incoming flow Qin into 3 (not necessarily equal) streams, then apply Bernoulli to each stream. You need to do this yourself.

 

[Nidum]

This apparently trivial problem is actually unsolvable using simplistic 1D analysis methods . You can get an answer using Bernoulli but it is unlikely to be the correct answer .

Can you tell us what project you are actually working on ? If we knew the details I'm sure we could help you design a suitable system which not only did what you wanted but was also easier to analyse .

 

[Marki john]

If Qin = Q1 = Q2 =Q3 were correct then you would get out 3 times the fluid you put in! A neat trick ... To answer this do what i suggested - mentally split the incoming flow Qin into 3 (not necessarily equal) streams, then apply Bernoulli to each stream. You need to do this yourself.

Yes you are right it would be 3 times more than the total in and splitting the total in into 3 would make more sense. I am fully sure now thanks to you. Thank you!

 

[Marki john]

View attachment 213982

This apparently trivial problem is actually unsolvable using simplistic 1D analysis methods . You can get an answer using Bernoulli but it is unlikely to be the correct answer .

Can you tell us what project you are actually working on ? If we knew the details I'm sure we could help you design a suitable system which not only did what you wanted but was also easier to analyse .

All I want is for the 3 nozzles to have the same amount of fluid flowing out of it and determine suitable size/area of each nozzles. So now that the total flow rate in can be split into 3 sub flow rates and applying Bernoulli to find out each nozzle velocities which then can be used to determine the suitable nozzle size/area from;
Flow rate = Velocity x Area >> Area = Flow rate / Velocity

Hope this makes sense and it is the correct method. Also why will the answer be unlikely correct when using Bernoulli?

 

[rude man]

All I want is for the 3 nozzles to have the same amount of fluid flowing out of it and determine suitable size/area of each nozzles. So now that the total flow rate in can be split into 3 sub flow rates and applying Bernoulli to find out each nozzle velocities which then can be used to determine the suitable nozzle size/area from;
Flow rate = Velocity x Area >> Area = Flow rate / Velocity

Hope this makes sense and it is the correct method. Also why will the answer be unlikely correct when using Bernoulli?

You have to decide what output velocity you want, then you can compute the nozzle area. Or vice-versa.
All you need is the continuity equation: Q_in = ΣQ_i, i = 1 to 3. You don't need Bernoulli to solve your problem after all (you would if you also wanted to know p₁.)

Your problem states "Assume there are no pressure loss in both minor and major losses and perfect ideal world conditions." So there is no friction anywhere and so the three nozzles are totally equal in Q and V (assuming equal nozzle areas).

 

[Marki john]

You have to decide what output velocity you want, then you can compute the nozzle area. Or vice-versa.
All you need is the continuity equation: Q_in = ΣQ_i, i = 1 to 3. You don't need Bernoulli to solve your problem after all (you would if you also wanted to know p₁.)

Your problem states "Assume there are no pressure loss in both minor and major losses and perfect ideal world conditions." So there is no friction anywhere and so the three nozzles are totally equal in Q and V (assuming equal nozzle areas).

Yes correct, Bernoulli will be used to solve the problem and Q and V and area in all 3 nozzles will be equal. Thank you for your help, much appreciated.

 

[Chestermiller]

Why are you saying that there are no pressure changes? This is a typical flow distribution header, and the fluid velocity down the main channel is decreasing, so the fluid pressure down the main channel must be increasing (toward the dead end). Doesn't Bernoulli tell you this? Bernoulli might give a good approximation to this system if frictional losses are negligible.