Calculate speed of water in pipe and drop of pressure 1

[poseilus]

Hello everyone,

I have a problem trying to find the speed of water around 80 centigrade inside a pipe used for central heating.

I know that:

V=(G/(Pi*D^2/4))*(1/3600) in (m/sec)

where G is flow rate of water in m^3/hour
D diameter of pipe
and Pi = 3.14

The results I get are close to the values of tables i have but i cant get the exact values. I tried altering the density of water but thats not the issue. It has to do with the friction inside pipe.

Also after that I want to calculate the friction.
After calculating Reynolds number I have to use Moody diagram in order to find ? which is the friction factor. Moody diagram is almost identical to this equation:

1/(?)^1/2=1,74-2*log(?s/D+18,7/(Re*?^1/2))

which is the colebrook and white equation.

How can i possibly solve this equation?

 

[Zapster]

Take a first guess and iterate to a solution.

Write a program for your calculator to automate the iterative process.

Use a program such as MathCad to solve the equation.

 

[Montemayor]

poseilus:

I'm confused with what you say. You say you "have a problem trying to find the speed of water around 80 centigrade inside a pipe used for central heating." But then you later ask for help on "How can i possibly solve this equation (Colebrook-White)".

I think you basically want to find the friction factor using the Colebrook-White equation and to do that, you must previously have generated the water velocity (not "speed&quot in order to come up with the required Reynolds Number which is required for the Colebrook-White relationship. Am I correct? If so, then consider getting a copy of Dr. Andrew Sleigh's Excel spreadsheet where he resolves the Colebrook-White very easily. Dr. Sleigh is a Civil Engineering Professor at the University of Leeds (

http://www.efm.leeds.ac.uk/~as/).

I downloaded his spreadsheet and added some additional worksheets on the Chen Equation, which is easier and more practical since it does not require reiteration to get the Friction Factor. There are other similar practical equations also. I didn't note Dr. Sleigh's Website address where I downloaded the spreadsheet, but if you are interested give me your email address and I'll send you a copy for your use. I think you'll like the succinct and simple work Sleigh did as well as also using the Chen instead of the Colebrook-White.

Let me know and I'll send it to you.


Art Montemayor
Spring, TX

 

[TBP]

Try Bell & Gossett's website

http://www.bellgossett.com

You can download a program called "System Syzer" that specifically deals with heating loops. It started off life as a circular slide rule in the '50s or '60s.

 

[quark]

This spread sheet does the iteration work for you.

http://me.queensu.ca/courses/mech441/spr/moody.xls

Just go through this link also,

http://me.queensu.ca/courses/mech451/losses.htm

Regards,

 

[BRIS]

Below is an Excel Macro for solution of Colbrook White Darcy formula

Units are SI


ks = roughness mm
L = length m
q = flow m/sec
d = diameter m
vi = kinematic viscosity - for water values are:

Temperature Viscosity
x 10^-6
0 1.795
4 1.568
10 1.31
15 1.131
21 0.984
27 0.864
38 0.687
93 0.317


Function headloss(d, q, l, ks, vi)
ks = ks / 1000
a = (22 / 7) * d * d / 4
q = q
v = q / a
re = v * d / vi
x = ks / (3.7 * d)
y = Application.Log10(x)
root1 = (1 / (-2 * y))
colbrook = root1
z = 1
Do Until z = 5
k = 2.51 / (re * root1)
y = Application.Log10(x + k)
root1 = (1 / (-2 * y))
z = z + 1
Loop
f = root1 * root1
hl = f * l * v * v / (2 * 9.81 * d)
headloss = hl
End Function

 

[25362]

To poseilus. Are you looking for friction losses in water piping ? The Hydraulic Institute Pipe Friction Manual gives, in tables and graphs, the friction drop in ft/100 ft, the velocity, given the flow rate and the pipe dimensions, based on the friction factor as estimated by the Colebrook equation. It even gives information of the losses due to fittings. Would this be sufficient ?

 

[poseilus]

Hello and I really thank you all for your help.

Montemayor, the problem I have is: I have tables with standard types of copper tubes which tables give me velocity of hot water at 80 degrees. They also give me head loss for various diameters of piping.
Because I wanted to do this automatically I search the web and found Colebrook-White and moody diagram. The problem is that I always get values higher than the ones given by the tables I have. eg. for 25mm diameter my tables give 0.5m/s velocity but I get 0.57m/s.
Thats not a big deal ofcourse. I searched inside Dr. Sleigh's website but I did not find that spreadsheet I would really appreciate if you could send me a copy at poseilus@klm.forthnet.gr and thank you again for your help.

TBP i downloaded system sizer and thank you for the info.

quark i downloaded the xls but I could not find how to make it work, but thank you.

BRIS I will try this function right away and let you know how it works. This is actually what I was looking for. But first I have to make it work . Thanks a lot.

25362 I did look inside hydraulic institute I could not find though this pipe friction manual.

 

[poseilus]

BRIS I tried the function you gave me but f is wrong.

Maybe I do something wrong here is the code:
Function HEADLOSS(d, v, l, e, vi)
d = d / 1000
re = (v * d) / vi
x = e / (3.7 * d)
y = Application.Log10(x)
root1 = (1 / (-2 * y))
colbrook = root1
z = 1
Do Until z = 5
k = 2.51 / (re * root1)
y = Application.Log10(x + k)
root1 = (1 / (-2 * y))
z = z + 1
Loop
f = root1 * root1
hl = f * l * v * v / (2 * 9.81 * d)
HEADLOSS = hl
End Function

I would really appreciate to solve this out.

 

[quark]

Poseilus!

You seem to be lazier than me:-(. The implicit functions are generally solved by initially assuming some value. In your case, for example, it is 0.02. Now solve for 'f' (by taking reciprocal and squaring the right side term and substituting 0.02 inplace of 'f' in right hand side of the equation). You will get some value. Now solve the equation again using the new value to get a third value for 'f'. You have to just stop where you are getting (n-1)th and nth values equal. This is what is automatically in the spreadsheet.

You have to input effective roughness in the cell k1 and Reynold's no. in the cell b5 then the cell h5 will give you the frictional coefficient (just see that g5 and h5 values are converging)

PS: Btw, I have a paper (got from TD2K) which gives 7 equivalent expressions of Colebrook equation, but explicit functions of 'f'. If you want to have a look at it, email me at builblock1@yahoo.com

Regards,


Regards,

 

[BRIS]

I have copied your code and it apears to give me the right answers.

Your units are

Diameter in mm
Rouhness e in m
length in m
velocity m/sec

Thus for dia = 25 mm,
e = .00001 m (0.01mm)
velocity = 0.57 m/sec
viscocity = 4 x 10^-7
length = 1.0m
headloss = 0.0105702 m/m

Brian

 

[poseilus]

it was my mistake I was using e in mm instead of m thank you

 

[BRIS]

Good - it is a solution I have used for the past 20 years and it has never failed me:


1/f^0.5 = - 2 log ((ks/(3.7 * D) + 2.51/(Re * f^0.5))

First approximation for f set the viscosity term " 2.51/(re * f^0.5)" to zero and evaluate the equation for f. (f1)

Second approximation recalculate f (f2) using f1 in the viscosity term " 2.51/(re * f1^0.5)"

etc.

f will generally converge within 3 or 4 iterations.

In the code above i have looped 5 times.



Brian