Formula for Open Channel Flow in Circular Pipe

[KThomson]

Hello --

I have been trying to figure out a solution for this in Excel and can't figure out how to do it.

I want a formula that can determine velocity in a circular pipe (not flowing full). I have a design flow, capacity of the pipe, and all of that information. I need something that determines the depth of flow (or the angle, as shown in my Lindeburg CERM manual with the equations for area and hydraulic radius) to determine the area of flow, and then use that information to solve for the velocity.

I have charts to find this by hand with the q/Q and v/V ratios, but I want to find or create the formula in excel since I will use it often and save time.

Any ideas??

Thanks in advance -

Kate

 

[MintJulep]

Don't your charts cite the forumla that was used to generate them?

Failing that, by brute force you could attempt to recreate the charts in excel, then use lookup functions on the underlying point pairs.

 

[EGT01]

Kate,

If you are trying to use tabulated flow capacities for a pipe, I think MintJulep makes a good suggestion about understanding how those flow capacities were determined.

In its simplest form
volumetric flow = cross-sectional area * velocity,
Q = A * V.

Knowing any two variables you can determine the other but it sounds like you only know one variable, Q. I don't think you can take a tabulated pipe flow capacity and assume a liquid level in your pipe to determine wetted area then use that to determine the velocity. I believe it is a little more complicated than that. Things like pipe slope, roughness coefficient, etc will also have an impact on pipe flow capacity and therefore on the wetted area and/or velocity.

As far as a relationship between wetted area, pipe diameter and liquid level....

From Perry's Chemical Engineer's Handbook

Pipe diameter = D, ft
Liquid Level = H, ft
Theta = ACOS(1-(2*H/D))
Wetted Area, sqft
Aw = (D^2/4) * (THETA - (SIN(THETA) * COS(THETA)))

Many years before I found this in Perry's, I had derived another form using calculus

Pipe diameter = D, ft
Liquid Level = H, ft
Wetted Area, sqft
Aw = (3.1416*D^2/8) + (D^2/4 * ASIN(2*H/D - 1)) + ((H - D/2)*(D*H - H^2)^0.5)

In both forms, the trig function is in radians. Either should give the same result. You should be able to input the equations as shown directly into Excel (just copy and paste). Just direct the THETA, H and D variables in either equation to the proper cell location.

 

[bltseattle]

Please see my response in the storm/flood forum.

BLTseattle

 

[KThomson]

Thanks -

I have the equations that EGT01 posted from Perry's book - just trying to figure out how to run that in the program to get what I want.

I think BLTSeattle helped (see the other forum for Storm/Flood engineering) - I am not familiar with VBA codes in Excel, but I will try to teach myself tomorrow!

I appreciate everyones help -

Kate

 

[Mala]

I do not think you need to learn VBA to be able to do this.

It is enough to set up the worksheet with an input area comprising the input variables (pipe dia, liquid level etc) and an output area which will contain formulas (for theta, wetted area etc).

The formulas in the output section would refer to the input area cells as required and auto-update whenever one or more of the input cells change(s).


Mala Singh
'Dare to Imagine'

 

[KThomson]

Mala - you may right except I want the spreadsheet to find the depth of water - based on pipe information and flowrate. I don't have that input variable.

I'm sure there is a way to calculate it - I can solve manning's for depth in a trapezoidal channel, but to write it in Excel for a circular channel has me scratching my head....

Thanks for your input -

Kate

 

[blackwed]

Can't do it. You can calculate all sorts of things in the other direction, i.e., if you know the depth and need volume or rate, but you can't calculate depth based on flow. I tried to calculate the time rate of change in the level of fluid in a horizontal cylindrical tank based on the volumetric rate of flow into the tank. The math boiled down to having to solve for alpha in an equation of the form y = K * (alpha - sin (alpha)). It can't be done.

Dan B

 

[tincan]

Solve by trial and error. With a known pipe size, use the depth as a variable, then calculate the remaining factors. By varing the inpur depth, you can determine the flow by trial and error on the input. I use this sheet/method all the time, doesnt take long. If youre frustrated at writing the spread sheet, give me your e-mail and I'll send you a copy.
Best, Tincan.

 

[KThomson]

Thanks Tincan, but I got a spreadsheet that works from someone who posted on the other thread.

I appreciate everyones help - I have what I need. Thanks!

Kate

 

[Gearhead]

If you use a Solver like TK it can be done in seconds

 

[TrimbleUser]

Hi kate, you can use the colebrook-white formula to calculate the velocity in full and part flow. You simply add a shape factor infront of the D part of the formula.
The shape factor formula uses "theta" the same theta angle that is used in working out the partial area. I have this all on a spreadsheet with the different variations of the colebrook-white formula to work out D, v and Sf (hydraulic gradient). The spreadsheet also has worked -out tables for pipes at full and part flow at a huge nr of gradients for pipe design.


(the colebrook-white-formula for velocity, if you dont have it, is widely available on the net and its usually the stuff below that isnt.)

Partial flow depth for d/D = 0.75 say with a 300mm pipe, d = 225mm and D = 300 mm.

theta = 2Cos^-1[1-2d/D]

Shape factor = (theta - sin theta)/theta


For your partial velocity take the colebrook-white formula for full bore flow then add the shape factor infront of the D in each term.

heres the formula for partial area aswell;

Partial area = D^2/8(theta -sin theta)

if you know a formula that things out the opposite way i.e. from a velocity and pipe work out the d/D ratio let me know cos thats the real nightmare to sort out.

remember the basic principles; in pipe design Sf is taken as the physical slope of the pipe, you normally design to a d/D ratio (here in the uk is d/D = 0.75. and you should also design for (vmax > v > vmin) for max and min velocities.

hope this helps

Iain

 

[Denial]

Blackwed and Tincan,

The non-linear implicit equation for Alpha needs to be solved iteratively, as stated. However the iteration process can be automated with Excel's Solver add-in.

Consider the univariate equation f(x) = T where f() is some horribly intractible function. Set up a spreadsheet containing:
(1) An input cell for x
(2) An input cell for T
(3) A calculated cell for f(x)
(4) A calculated cell containing the result [f(x)-T]^2

Enter a reasonably close estimate for x.
Now get the Solver to minimise cell (4) by varying cell (1).
Bingo.

This method is applicable to any non-linear equation. Furthermore it is readily extendable to simultaneous non-linear equations as follows.

Suppose we have to find values for x, y and z such that
f(x,y,z) = A
g(x,y,z) = B
h(x,y,z) = C
Put x, y and z in input cells.
Create a target cell containing
[f()-A]^2 + [g()-B]^2 + [h()-C]^2
Enter a set of reasonably close guesstimates for x, y and z.
Get Solver to minimize the target cell by varying the x,y,z cells.

HTH