17 Jan '16 16:342 edits

The 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

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