Fluid Mechanics Velocity at Outlet

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jdawg
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Homework Statement


Water flows through a 1.5 [in] diameter pipe. In a 1.5 [ft] section of the pipe water is injected through the porous walls into the pipe at a velocity of 3 [in/s]. What is the average velocity at the exit if the inlet velocity is 6 [ft/s]?

Homework Equations

The Attempt at a Solution


I used 62.4 lb/ft3 for the density of water.
I calculated the area of the pipe to be A=πR2=0.01227 ft3
m1=ρAv=(62.4)(0.01227)(6)=4.59 lb/s

Aw=DL=(1.5/12)(1.5)=0.1875 ft^2 I'm not super confident that this area is correct...
mw=ρAwvw=(62.4)(0.1875)(3/12)=2.925 lb/s

Then I used the conservation of mass...
m1+mw=m2
(4.59)+(2.925)=m2
m2=7.25 lb/s

Now just plug into formula to get vout
vout=(m2)/(ρA)=(7.52)/(62.4*0.01227)
vout=9.82 ft/s

Which is wrong... My options are 7 ft/s, 18 ft/s, 28 ft/s, and 150 ft/s.
I'm not sure what I did wrong!
 
Attachments
  • Diagram of a horizontal pipe section with labelled inlet velocity Vin entering from the left, outlet velocity Vout leaving at the right, inward arrows Vw through the top wall, diameter D, and length L.
    PorousWallPipe.png
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jdawg said:

Homework Statement


Water flows through a 1.5 [in] diameter pipe. In a 1.5 [ft] section of the pipe water is injected through the porous walls into the pipe at a velocity of 3 [in/s]. What is the average velocity at the exit if the inlet velocity is 6 [ft/s]?

Homework Equations

The Attempt at a Solution


I used 62.4 lb/ft3 for the density of water.
I calculated the area of the pipe to be A=πR2=0.01227 ft3
m1=ρAv=(62.4)(0.01227)(6)=4.59 lb/s
This calculation seems to be OK.
Aw=DL=(1.5/12)(1.5)=0.1875 ft^2 I'm not super confident that this area is correct...
mw=ρAwvw=(62.4)(0.1875)(3/12)=2.925 lb/s
Is the circumference of a circular pipe equal to its diameter? Or is it equal to something else?
 
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Haha I knew that area couldn't be right... I got thrown off because I was given the length of the porous wall. So is that length L just useless information?
 
jdawg said:
Haha I knew that area couldn't be right... I got thrown off because I was given the length of the porous wall. So is that length L just useless information?
No, it's not the length that's the problem with the calculation.

I mentioned specifically the circumference of the pipe and how you calculated it.
 
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Oops sorry. That makes more sense. Circumference=2πr*h=0.0625ft2. Thanks!
 
jdawg said:
Oops sorry. That makes more sense. Circumference=2πr*h=0.0625ft2. Thanks!
What's h supposed to be?