Pressure Velocity Help

  • #1
Hello, I've got this questions and I just cannot solve it.

Assuming laminar in compressible flow. How long would it take to fill up a 1m^3 box from a pipe 15mm in diameter at 1.5 bar of pressure?

I would really appreciate it if someone could solve for me, show and explain the working because I'm completely lost. Thank you.

So far this is what I have got, this is not my strong area this so please excuse me if its complete rubbish!



https://scontent-lhr3-1.xx.fbcdn.net/hphotos-xpt1/v/t1.0-0/p206x206/13076783_10206132363644489_6864425099277240298_n.jpg?oh=d9b5d787e07260e82d1e4b6f30e9231b&oe=57A36783
 

Answers and Replies

  • #2
haruspex
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It's not clear to me whether the given pressure is absolute or gauge pressure. Makes a big difference.
In your answer for v you have the wrong units, and later you have the wrong units for the mass flow rate.
Other than that, what you did looks right.
 
  • #3
Overall not a great attempt then haha. Like I said, this is not my area, trying to use maths I've not been near in 2/3 years. Thanks for the heads up though, Ill go back through it and try again.
 
  • #4
Simon Bridge
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It was a good effort - the basic reasoning was fine. The units provide a clue for troubleshooting is all.
ie. You have density of water D as 1000 ... that would be 1000g/L which is 1g/cm^3 or 10000kg/m^3 and you wanted velocity to come out in units of m/s.
P in the same equation is in units of Pa = N/m^2 = kg/(m.s^2) ... see how the units of mass don't go away? So you cannot get a volume flow rate from ##v=\sqrt{2P+D}## unless something special has happened. Where did you get it from?

I also don't think there is enough information provided in the problem statement to solve it ... you'll have to go outside the problem for extra information.

Have you seen:
See: http://hyperphysics.phy-astr.gsu.edu/hbase/pfric.html
http://physics.stackexchange.com/questions/127760/calculation-of-pressure-from-flow-rate-of-water
 
  • #5
haruspex
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√2P+Dv=2P+Dv=\sqrt{2P+D}
I agree it is hard to read, but I believe it is a division sign, ##\div##
 
  • #6
Simon Bridge
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Oh it's supposed to be: $$v = \sqrt{\frac{2P}{D}}$$ ... that works out yeah.
 
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