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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.

I am given a diagram of the pipe shown on the attachment (Sorry for the crude diagram).

Hosted it while the attachment is "reviewed"

http://img443.imageshack.us/img443/2215/fluidmechqn5.th.jpg [Broken]http://g.imageshack.us/thpix.php [Broken]

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.

mass flow rate = Ro.A.U

http://www.it.iitb.ac.in/vweb/engr/civil/fluid_mech/section3/img00094.gif [Broken]

Area = Pi.(R^2)

Lets denote the inlet with suffix "-I", the top outlet "-Ot" the bottom outlet "-Ob"

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

Any help is much appreciated

Thanks

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.

## Homework Statement

I am given a diagram of the pipe shown on the attachment (Sorry for the crude diagram).

Hosted it while the attachment is "reviewed"

http://img443.imageshack.us/img443/2215/fluidmechqn5.th.jpg [Broken]http://g.imageshack.us/thpix.php [Broken]

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.

## Homework Equations

mass flow rate = Ro.A.U

http://www.it.iitb.ac.in/vweb/engr/civil/fluid_mech/section3/img00094.gif [Broken]

Area = Pi.(R^2)

## 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)(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

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