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Hi,
I am stuck on this question as I haven't 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 http://g.imageshack.us/thpix.php
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
Area = Pi.(R^2)
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 I am not sure what I am meant to be doing. Do i need to calculate the momentum in the Fx and Fy direction?
(b)
To find the pressure I am assuming i have to use the bernoulli equations. But I am 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 haven't got this far... need to get the first two bits done. But i would use the:
F= SQRT[ Fx^2 + Fy^2] equationAny help is much appreciated
Thanks
I am stuck on this question as I haven't 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 http://g.imageshack.us/thpix.php
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
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 I am not sure what I am meant to be doing. Do i need to calculate the momentum in the Fx and Fy direction?
(b)
To find the pressure I am assuming i have to use the bernoulli equations. But I am 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 haven't got this far... need to get the first two bits done. But i would use the:
F= SQRT[ Fx^2 + Fy^2] equationAny help is much appreciated
Thanks
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