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Velocity of the water flowing through the pipe

  • Thread starter Struggling
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  • #1
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Hi,

i have a rotameter reading, flow rate (lit/min), Head Loss(mm) and diameter of a pipe.

from this i need to find out the Velocity of the water flowing through the pipe.

iam confused as to what formula to use to find this out, i have searched through 2 textbooks and cannot find anything.

can anyone help me in the right direction as to which formula i should be using.

thanks
 
Last edited:

Answers and Replies

  • #2
Doc Al
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Read this: http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Pressure/VolumeFlowRate.html [Broken]
 
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  • #3
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i know those formulas but i struggle to see where my data fits into the equation.
i think i must be lost with the meaning of the rotameter and flow rates given.

*** edit thought about it longer

sorry hang on is it saying

Flow rate = cross sectional area x velocity??????
 
  • #4
Doc Al
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Struggling said:
Flow rate = cross sectional area x velocity??????
That's the one.
 
  • #5
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do you mind if i ask, if my flow rate is 0.3 lit/min, and my cross sectional area is A = pi/4(19.23) = 15.103 mm^2

V = av
v = a/V
v = 15.103/0.3 = 50.34?

or was it ment to be v = V/a?
my basic math is shocking :blushing:

*** never mind im 90% sure its v = V/a so the answer would be 0.02m/s or 2 cm/s which makes more sense than 50.34 m
 
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  • #6
Doc Al
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I'm suspicious of your area formula: what data are you actually given?

Before using: Flow rate = (Area)*(speed), or speed = (Flow rate)/(area), be sure to convert everything into standard units:

Area in m^2; flow rate in m^3/sec; speed in m/sec.

To convert from liters/minute, realize that:

1 (liter)/(minute) = (10^-3 m^3)/(60 seconds)

(or you can look up a unit conversion chart)
 
  • #7
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we did some tests using rotameters. The rotameters would return results such as 50mm on rotameter 10 (i forget what the sizes mean) we then had a graph for the size rotameter we used and we would look at the graph and find 50mm to have a flow rate of 0.3 (lit/min).

so eg for a size 18 rotameter with a reading of 151mm the flow rate was found to be on the graph 5.75 (lit/min) giving me the velocity of 0.381 m/s

i have the diameter of the tube inside which is 19.23mm. by memory i thought that the cross sectional area of a tube to be A = pi/4 x (Diameter)
 
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  • #8
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or is the area supposed to be A = pi x diameter x diameter. ?
 
  • #9
Doc Al
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Struggling said:
i have the diameter of the tube inside which is 19.23mm. by memory i thought that the cross sectional area of a tube to be A = pi/4 x (Diameter)
The area of a circle is:
Area = pi*radius^2 = (pi/4)*(Diameter)^2

where (Diameter)^2 = (Diameter)x(Diameter)
 
  • #10
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ah ok so its...

A = pi/4*19.23^2 = 290.435 ???

so for a rate of flow of 0.3 lit/min

the answer would be

0.3 = 290.435*v
0.3/290.435 = v
v = 0.00103m/s
 
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  • #11
Doc Al
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Put you data into standard units: length in meters, not mm; area in m^2, not mm^2. (Once you've used the formula to find the answer, you can convert the answer to any units you like.)
 
  • #12
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the only data i have is 19.23mm(0.1923 m) diameter and the flow rate which is lit/min.

Area in m^2; flow rate in m^3/sec; speed in m/sec.

To convert from liters/minute, realize that:

1 (liter)/(minute) = (10^-3 m^3)/(60 seconds)
to convert this into m^3/sec is totally baffaling me.
We got a rotameter reading in mm, looked at a graph which gave us the flow rate in lit/min.
why does this have to be changed?
unless it doesnt i had got

v = 0.3/0.02904
v = 10.39 m/s

thanks
 
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  • #13
Doc Al
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Struggling said:
We got a rotameter reading in mm, looked at a graph which gave us the flow rate in lit/min.
why does this have to be changed?
If you don't change to standard units, your expression for speed will have units of:

(lit/min)/(m^2) :yuck: , which is certainly not equivalent to m/s.

But if you expressed the flow rate in m^3/s, your speed will have units of:

(m^3/s)/(m^2) = m/s o:)
 
  • #14
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Argghhhhh i dont know how to do it. my heads aching lol.
its very late maybe i should sleep on it.
any other info or help on how to convert it would be much appreciated.
thanks!

thanks Doc Al, youve been alot of help :smile:
 
  • #15
Doc Al
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Struggling said:
any other info or help on how to convert it would be much appreciated.
I thought I showed how to do the conversion in post #6? Do the indicated arithmetic! (Or you can just Google it. Try it.)
 

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