Fluid Mechanics - Split Pipe Flow, Applying momentum equations + bernoulli

[50C]

Hi everyone...

Got this problem to sort out, I havent encountered anything similar to it. Normally I have only dealt with pipe with one inlet/outlet. Wondering if anyone can give me a helping hand

1. The problem statement, all variables and given/known data
I am given a diagram of the pipe shown in the next post (Sorry for the crude diagram).


The flow is going from the 1m diameter pipe and splitting between the two 0.5m pipes.
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

( Calculate the outlet pressure assuming no frictional losses

© 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
Bernoulli Equation
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?

(
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?

©
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

========

After doing all this 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 havent covered that material yet. Im unsure whether i can apply the bernoulli and momentum equations between 3 points like i have attempted.

Any help is much appreciated
Thanks

[50C]

Here is the set of the pipe... couldnt post it in the last post due to board restrictions

img443. imageshack.us/img443/2215/fluidmechqn5.jpg

just add a ht tp : // without spaces to the front!