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Fluid Mechanics - Split Pipe Flow Question (momentum and Bernoulli eqns)

  1. Sep 28, 2008 #1
    Hi,
    I am stuck on this question as I havent encountered anything similar to it. Normally I have only dealt with pipe with one inlet/outlet.

    1. The problem statement, all variables and given/known data
    I am given a diagram of the pipe shown on the attachment (Sorry for the crude diagram).
    Hosted it while the attachment is "reviewed"
    [​IMG][​IMG]

    I am told that the junction lies in the horizontal plane x-y and the gravity vector is perpendicular to this. Also, the flow rates from each outlet are equal.

    Variables
    Inlet Pressure = 200kPa
    Inlet flowrate = 2m^3/s
    diameters are shown on picture

    Density of water = 1000kg/m^3

    Question,

    (a) I have to select a suitable control volume and apply the momentum equations fo coordinate directions X and Y

    (b) Calculate the outlet pressure assuming no frictional losses

    (c) Calculate the reaction forces in directions X and Y, that must be absorbed by the support system on the junction.

    2. Relevant equations
    mass flow rate = Ro.A.U
    [​IMG]
    Area = Pi.(R^2)


    3. The attempt at a solution

    Lets denote the inlet with suffix "-I", the top outlet "-Ot" the bottom outlet "-Ob"
    (a) I took the control volume at the centre so it includes the inlet and both outlets.

    Before applying the momentum equations i calculated
    Area-inlet = 0.25Pi
    Area-top-outlet = 0.00625Pi
    Area-bottom-outlet = 0.00625Pi

    The mass flowrate is given as 2 so,
    U-inlet = 2/(1000*0.25Pi) = 2.55x10^-3 m/s

    Applying the continuity equation
    U-I*A-I = U-Ot*A-Ot + U-Ob*A-Ob (Density cancels throught)

    We were told that the mass flowrate was the same at each outlet. so Am i right in assuming the speed of water at Ot and Ob are the same? So i can make the RHS 2*U-O*A-O

    Then rearrange and solving for U-O to get the speed of water for both outlets to be 0.0204m/s

    Then using the mass flow equation:
    mass flow rate = Ro.A.U-O
    mass flow rate = 1000*0.0625Pi*0.0204
    mass flow rate = 4 M^3/s

    Then im not sure what im meant to be doing. Do i need to calculate the momentum in the Fx and Fy direction?

    (b)
    To find the pressure im assuming i have to use the bernoulli equations. But im not sure how to apply it to this control volume. I know the value of "z" is zero so they dissapear.

    Can i say:
    P-I/Ro*g + (U-I^2)/2g = [P-Ot/Ro*g + (U-Ot^2)/2g] + [P-Ob/Ro*g + (U-Ob^2)/2g] ???

    then rearrange for on of the pressures on the RHS assuming they will be the same?

    (c)

    I know the method, but havent got this far... need to get the first two bits done. But i would use the:

    F= SQRT[ Fx^2 + Fy^2] equation


    Any help is much appreciated
    Thanks
     

    Attached Files:

    Last edited: Sep 28, 2008
  2. jcsd
  3. Sep 28, 2008 #2
    I should probably add, the flow is going from the LHS pipe (diameter 1m) and splitting between the two pipes (diameter 0.5m)
     
  4. Sep 29, 2008 #3
    Hmm, im starting to think maybe the control volume should be only across the inlet and one outlet pipe. We dont have to take into account friction factor or pipebends (90degree and 45degree etc) because we're not that far in the course. Im unsure whether i can apply the bernoulli and momentum equations between 3 points like i have attempted.

    Like i said, any help would be much appreciated, even a point in the right direction is good enough for me! That way you wont have to type out the long equations :p

    Thanks
     
  5. Sep 29, 2008 #4
    I believe you will need the ANGULAR momentum equation along with the Bernoulli equation. This is a tough problem, does the problem statement mention inviscid flow? You will have a change in momentum from two different affects, the change in direction of the fluid, and its change in speed. Hopefully someone else will have a better idea of what to do for this problem cause I think I'm stuck.
     
  6. Sep 29, 2008 #5

    minger

    User Avatar
    Science Advisor

    Just apply conservation of momentum to the entire junction. You have mass flow rates and velocities at each inlet and oulet. Sum momentums in the x and y-directions, the difference of each is the reaction that "holds" the junction in place.

    Let's assume that x is in the horizontal (as its drawn) direction. Summing momentums, you have the contribution from the inlet (the entire flow) and the x-component of the bottom outlet, realizing that there's no component in the top pipe. Sum.

    You can do the same for the y-direction, realizing the only components will be the entire flow through the top outlet and the component of the bottom. Sum.
     
  7. Sep 30, 2008 #6
    Nothing in the problem statement about inviscid flow....

    Ok, thanks for your input... will give it a go, will post up what i get in a little bit :)

    Thanks 2 both of you

    btw, should i be taking density to be 1000 kg/m^3??
     
    Last edited: Sep 30, 2008
  8. Oct 1, 2008 #7
    Right, this is what ive got with what you both said. Is this what you meant i should do?

    [​IMG][​IMG]

    [​IMG][​IMG]

    [​IMG][​IMG]
    The bit cut off on the right is just saying the Area of outlet 2 and 3 = 0.0625*Pi

    [​IMG][​IMG]

    [​IMG][​IMG]

    [​IMG][​IMG]

    i get 199.99Kpa as my answer for the pressure at both outlets... which doesnt really seem correct to me. Surely the pressure should have increased due to the reduced cross sectional area of the pipe? Or is it due to the fact it only has half the mass flow rate passing through it?

    Thanks :)
     
  9. Oct 1, 2008 #8

    minger

    User Avatar
    Science Advisor

    Oh, you haven't learned Bernoulli's well enough yet. More homework problems, haha! The equation states that the total energy across a streamline stays constant. The energy can be transferred to and from pressure, velocity, and elevation. If one goes up or down, the it must be compensated for.

    In your case, the problem is assumed to be planar, so you can cancel the gz (elevation) terms. The question remains: what happens to the flow as the area decreases? Does velocity increase or decrease? Based on this, what would you expect pressure to do?
     
  10. Oct 1, 2008 #9
    Hi,
    lol true, fluid mechanics isnt really my favourite module :p
    In answer to your question:
    As the area of the pipe decreases, for the same mass flow rate... the velocity would increase along with the pressure. right?
    But seeing as the flow is being split in two, the mass flowing through each outlet would be half the total mass entering the control volume so the reduced area is sort or countered? is that right?
     
  11. Oct 1, 2008 #10

    minger

    User Avatar
    Science Advisor

    No no no. OK, ignore the fact that there are two pipes. As you've learned so far (neglecting friction), the geometry of the pipes is of no concern. All we're concerned about is mass flow, pressure, velocity, elevation, density.

    So, what we have is a fluid coming in at 2 m/s at 200 kPa. The AREA of the pipe is 0.25. From this information we can get the mass flow. From continuity and the fact that it's steady-state, the mass flow doesn't change. Mass in = Mass out. Since we have the exit area: 0.0125 (0.00672 x 2), we can get the exit velocity. Again, it doesn't matter what the pipes look like, all we care about is that the flow exits a 0.0125 area tube.

    edit: Areas not right here, didn't see the notation, exit velocity is correct though

    This gives me an exit velocity of 4 m/s. This should make sense, the mass flow stays the same, so velocity must increase as a function of the ratio of areas.

    Now our equation for Bernoulli's looks like this:
    [tex] \frac{P}{\rho g} + \frac{V^2}{2g} = k [/tex]
    Where k is a constant. When looking at a changing flow, we typically say that the initial conditions equal the final conditions, since we both equal the same constant.

    However, if we look at the equation, we see that the SUM of the two pressure terms (static pressure from the first term, and dynamic pressure [think internal and kinetic energy]) must stay the same.

    We have just determined that velocity increases. That increases the dynamic pressure term. To equal out, that means that the static pressure term MUST decrease. I get a final static pressure of 194kPa.

    p.s. Make sure you use correct units as well, as I messed up the first time. 200kPa needs to be 200,000 Pa in the equations.
     
    Last edited: Oct 1, 2008
  12. Oct 1, 2008 #11
    Ok, thanks for the explanation :)

    About to work through what you said, one thing though...
    The problem statement said the mass flow rate was 2 m^3/s, so is it correct to take the velocity as 2m/s?

    I tried to calculate it as ashown here:
    [​IMG]

    But i get a very small speed!
     
  13. Oct 1, 2008 #12
    I have got different velocity and as a result different pressure. Area is decreased which in turn results in increase in velocity and therefore higher pressure. This makes sense!

    I am not too bothered about getting the right answer! I think i have got correct method so hopefully I will get at least 15+.

    PS: you are getting small speed because you have made one mistake! everything else is correct! look back at continuity.
     
  14. Oct 1, 2008 #13
    Thanks for the reply.
    hmm.. i can't spot my mistake..

    For the velocity at the inlet (U1) I used the equation
    mass flowrate = Velocity x Density x Area

    We're given the mass flowrate 2 m^3/s. Density of water is 1000 kg/m^3 and the area of the inlet pipe is 0.25*Pi

    so,
    Velocity = (m-dot)/ (Density x Area)
    = (2)/(1000)*(0.25*Pi)

    Which gives us U1= 2.55 x10^-3

    The only variable that i might be using wrong is the density! Because m-dot, and area are pretty much given.

    The putting that into my continuity equation (After canceling density on both sides). I said that U2 = U3 as it says in the question there is an equal flow rate out of each outlet. Then rearranging i consequently get a small value for U2 and U3. Where did i go wrong?
     
    Last edited: Oct 1, 2008
  15. Oct 1, 2008 #14

    stewartcs

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    No it doesn't. An increase in velocity will result in a decrease in pressure, not increase.

    u2 > u1 ==> P2 < P1

    CS
     
  16. Oct 1, 2008 #15
    mass flow rate has units of kg/sec!!! (i think)

    Look back at your first year notes and/or work out the units for --> dm/dt = density x area x velocity

    flow rate is 2m^3/sec

    i am sure you can do it now!

    taking Pv = RT into consideration we can say that pressure decreases as velocity increase if T is constant... but....ARGHHHHHHHHHHHHHHHHHHHHH

    SOMEONE KILL ME!
     
  17. Oct 1, 2008 #16
    Ahhh right! Yeah got it... they gave us the volume flow rate in the question...so

    [​IMG]

    So i get U1 = 2.55 m/s

    Is that what you got?

    As i remember, the "v" in that equation stands for volume and not velocity :)
     
  18. Oct 1, 2008 #17
    OK here are my final answers... i get:
    U1 = 2.55 m/s
    U2 = 5.1 m/s
    U3 = 5.1 m/s
    P2 = P3 = 190.2 kPa
    Fx = -1495N
    Fy = 1495N
     
  19. Oct 1, 2008 #18
    yes! I know quite a few people who got the same.


    and yes of course it stands for volume! -silly me! too much work, too much drinking, too much smoking! Its all having a big effect on my brain!
     
  20. Oct 1, 2008 #19
    Cool, thanks for you help... appreciate it.

    One last thing, the Fx and Fy we worked out is for the force on the control volume right?
    So is the force that must be absorbed by the support system the opposite?

    i.e
    Fx = 1495
    Fy = -1495

    Might be worth a mark... not sure...
     
  21. Oct 1, 2008 #20
    I get one of them to be negative aswell :)
     
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