- 17 Jan '16 16:34 / 2 editsThe problem is an application of Uniform Flow in Open Channels. Specifically, I have a 10" diameter drain used in a cooling process. The depth of flow in the drain is apprx. 4". Virtuallly, the same process is to be installed with another machine, with the exception that the drain is to be downsized to 6" diameter.

The Question: What will the depth of flow be in the 6" drain?

The traditional method to determine this in engineering ( for normal sized channels carrying water) is by use of the Manning Equation (for more details see: https://en.wikipedia.org/wiki/Manning_formula)

Q = (k/n) * A * R_h^(2/3) * S^(1/2) The Manning Equation ( Note: The equation is not dimensionally homogeneous)

"Q" is the flow rate

"k/n" is a coefficient for roughness and system of units

"A" is the cross-sectional area of the flow

"R_h" is the Hydraulic Radius and is equal to (cross-sectional area/wetted perimeter)

"S" is the slope of the channel.

Assumptions: Q, k/n, and S are constant between the 10" pipe and 6" pipe.

Solutions Governing Equation:

(A_10)*(R_10)^(2/3) = (A_6)*(R_6)^(2/3)

or

(A_10)*(R_10)^(2/3) - (A_6)*(R_6)^(2/3) = 0 Equation(1)

So in General:

A ( secant area) = D²/4*( ß - sinß*cosß ) Equation(2) where ß is the half angle subtended by the secant arc in radians.

P ( wetted perimeter ) = D*ß Equation(3)

R ( hydraulic radius) = A/P = D/4*( 1 - (sinß*cosß )/ ß ) Equation(4)

y ( depth of flow ) = D/2* (1-cosß ) Equation(5)

Next get A(D,y) and R(D,y) and solve Equation(5) as:

A(D_1, y_1)* R(D_1, y_1)^(2/3) - A(D_2, y_2)* R(D_2, y_2)^(2/3) = 0

where the only unknown parameter is either y_1, or y_2

From Equation(5)

cosß = 1- 2*y/D

sinß = √( 1 - cos²ß )

ß = arccos( cosß )

The funny this is when I do this I can get different results depending on which "y" I vary.

For Instance: What depth in the 10" pipe will completely fill the 6" pipe? The appx. solution I get when I vary the flow depth in the 10" pipe to satisfy the equation comes to 3.4512 inches, but when I set the parameters to (10, 3.4512) on the left and vary the 6" flow depth I expect the end result to be y = 6" ( full pipe), but I solves to 4.9164. An obvious contradiction...if anyone has any input on this I'd greatly appriciate it?

Thanks - 17 Jan '16 16:41 / 2 editsWhat does that mean, the depth of flow is 4 inches? Is that how far the gravity drop is, like 4 inches in ten feet or something like that?

Is that cooling water going back to be recooled after it does its cooling job? I think you said the cooling medium is water so that limits how many BTU you can drag off the unit being cooled, having to stay under 100 C and all that.

Do you know the approximate BTU cooling requirement? - 17 Jan '16 16:44 / 2 edits

It means how much of the pipe is full for a given flow rate. Take a cylinder, turn it on its side, fill it up partially. Draw a line perpendicular to the liquids free surface ( through the liquid) to the intersection of the pipe. The depth of flow is the "longest line" you can draw under these conditions.*Originally posted by sonhouse***What does that mean, the depth of flow is 4 inches? Is that how far the gravity drop is, like 4 inches in ten feet or something like that?**

Is that cooling water going back to be recooled after it does its cooling job? I think you said the cooling medium is water so that limits how many BTU you can drag off the unit being cooled, having to stay under 100 C and all that.

Do you know the approximate BTU cooling requirement?

Yes, I could find the BTU/hr requirement but its not part of my problem so I don't want to muddy it up the agenda with tangent topics at this time. - 18 Jan '16 19:06

Our resident air conditioning/electronics engineer here gave me this link, may be of use?*Originally posted by joe shmo***It means how much of the pipe is full for a given flow rate. Take a cylinder, turn it on its side, fill it up partially. Draw a line perpendicular to the liquids free surface ( through the liquid) to the intersection of the pipe. The depth of flow is the "longest line" you can draw under these conditions.**

Yes, I could find the BTU/hr requirement but it ...[text shortened]... t part of my problem so I don't want to muddy it up the agenda with tangent topics at this time.

http://www.tlv.com/global/TI/calculator/water-pipe-sizing-pressure-loss.html - 19 Jan '16 12:26

Thanks Sonhouse, on the right track, but it doesn't help with the specific problem. I'm really just concerned with why the mathematics model I described above behaves strangely. Unfortunately, someone has to try to solve the equation for themselves to really see what I'm saying. I may try and describe the behavior of the model in more detail later.*Originally posted by sonhouse***Our resident air conditioning/electronics engineer here gave me this link, may be of use?**

http://www.tlv.com/global/TI/calculator/water-pipe-sizing-pressure-loss.html - 19 Jan '16 23:31

Ah, I thought you were after a solution to a problem but really you are after why the model doesn't seem to work right. Is there a way you can diagram what you are talking about with what I think is a pipe within a pipe thing? Or am I totally wrong about that?*Originally posted by joe shmo***Thanks Sonhouse, on the right track, but it doesn't help with the specific problem. I'm really just concerned with why the mathematics model I described above behaves strangely. Unfortunately, someone has to try to solve the equation for themselves to really see what I'm saying. I may try and describe the behavior of the model in more detail later.**