Max water flow rate calculations.

[jons999]

What are the equations to calculate the following?

Say you have an unlimited water supply flowing through a 20 foot diameter pipe. This pipe has a 10 degree downward slope from the water source. This water pipe is feeding a 7 foot diameter pipe with a 20 degree slope.

How would i calculate the maximaum flow rate of that 7' diameter pipe? I imagine i would have to use the slope and total drop (in this case 50') to find the pressure of the water in that 7' pipe and somehow figure out how much water (in gallons) that flows through that pipe in a given amount of time?

I am doing some research on old 1800's water power tunnels, power canals, etc. and i'm trying to figure out how much water these things used and how much power they produced. If i know how much water ran through i can calculate the power details with the engineering specs from there turbines and water wheels.

Thanks,

 

[poetix99]

Some questions (and I will log-on again later to provide an example calculation):

What is the "pressure" at the inlet of the 20' dia. pipe? The "pressure" in this case in the height of water above the pipe inlet (use the pipe centerline).

What total length of 20' dia. pipe?

What length of 7' dia. pipe?

What is the pipe material? More to the point; what is the condition (roughness) of the internal walls of the pipe? The rougher the pipe, the greater the resistance, and the lesser the flowrate, all other things being equal. (It matters, even for such large pipes; ask the designers of the Alaska pipeline.)

What is the pressure at the discharge end of the 7' pipe? If you're talking about hydropower applications, the water might have typically discharged into/onto a turbine of some sort. The maximum flow case will be discharge to atmosphere, as with a water wheel. An enclosed hydroturbine would have some pressure > atmospheric, and the flow would be less.

 

[poetix99]

AN EXAMPLE:

Most of the empirical data I am using has come from Crane Technical Paper #410. In the case of what is needed for your problem, this is certainly not the only source of such data, but is in general a great reference for fluid flow calculations.
(Another thread had given the following website for the purchase of TP 410:

http://www.cranevalve.com)

ASSUMING: Moderately smooth concrete pipe

ASSUMING: Half the height is a run of 20' dia. pipe @10degrees (==> 144 ft of pipe). Half the height run is 7' dia. pipe @ 20 degrees (==> 73 ft. of pipe).

ASSUMING: 30' of water above the inlet centerline.

ASSUMING: discharge is open to atmosphere (=0 psig)

ASSUMING: The reduction from 20' pipe to 7' pipe is at a 45 degree "half angle" (This is a significant part of the total resistance to flow and varies greatly with geometric details of the reduction.)

*******

Your problem is relatively simple in that it is incompressible flow with only two pipes and one transistion.

Bernoulli's equation describes the relationship between velocity, fluid pressure, fluid "head", and frictional resistance (expressed in terms of "head&quot. In good old US units:

z(1)+[144*p(1)/rho(1)]+[(v(1)^2)/(2*g)]=
z(2)+[144*p(2)/rho(2)]+[v(2)^2/(2*g)]+h(L)

where:
z = elevation [ft]
p = pressure [psi, pounds per sq. inch]
rho = fluid density [ lb(mass)/ft^3 ]
v = fluid velocity, ft/sec
g = accel of gravity, [32.2 ft/sec^2]
h(L) = equivalent length of flow resistance
"1" --> at pipe entrance
"2" --> at pipe discharge

*******

From Crane:
pipe roughness, e = 0.003
e/D (20') = 0.00015, friction factor, f = 0.013,
[f*(L/D)]20 = 0.0936 = K(20)
e/D (7') = 0.00045, f = 0.0165, [f*(L/D)]7 = 0.172 = K(7)

h(L) = f*(L/D)*[v^2/2g] = K*[v^2/(2*g)]
(this is a definition of "K", the resistance coeff.)

for the contraction:
K = .5*(1-(7/20)^4)*(1-sin(45)) = .3689

K(total) = 0.6345
*******

v(1) = v(2)*[A(2)/A(1)] = 0.1225 v(2)

Back to Bernoulli's:
30 ft = -50 ft + (0.985*v(2)^2 + 0.6345*v(2)^2)/(2*g)

v(2) = 56.4 ft/sec, average velocity at the pipe discharge FOR THIS PARTICULAR EXAMPLE.

 

[jons999]

Thanks, those equations help the water level is about 12 feet above inlet centerline and the 20' diameter pipe is about 2300' in length consisting of very rough walls (it's a blasted out tunnel) The 7' diameter pipe is 200' long and is very smooth (steel lined with asphalt). I think i can plug those number in and figure it out. Only 1 further question you give the pipe discharge velocity how would i convert that to gallons? I'm guessing i would find the area of a 1' cross section of the 7' pipe:

(3.14)R^2
(3.14)3.5'^2
(3.14)12.25=
38.465CuFt.

Multiply that by V(2)
38.465(56.4)=
2169.426CuFt. per second * 7.48 gallons = roughly 16227 gallons per second maximum flow.

Once i plug in the right numbers the real answer will be alot less (i'm guessing about 4,000 per second) But do i have the right idea here for calulating the volume from your example??

-Jon