Single pipe equivalent for parallel pipes

[TonyBlair]

Hello there im trying to work out a pipe equivalent for parallel pipes. Once I have worked out a single pipe equilivent I can use the Hazen-Williams formula to calculate the head loss for that single pipe.

I have been told that the formula is:

Dp^2.62 / = D1^2.62 / + D2^2.66 /
Lp^0.54 L1^0.54 L2^0.54

Dp = Diameter of equivalent pipe
D1 = Diameter of pipe 1
D2 = Diameter of pipe 2
Lp = Length of equivalent pipe
L1 = Length of pipe 1
L2 = Length of pipe 2

I understand that you can specify the length of your pipe equivalent, as the diameter should change accordingly, but it doesn’t work.

Lets take two pipes for example:
Pipe 1: Diameter = 0.9m, length = 2km
Pipe 2: Diameter = 1.1m, length = 3km:

Dp^ 2.62 / = d1 ^ 2.62 + d2 ^ 2.62
Lp ^0.54 L1 ^ 0.54 L2 ^ 0.54

Dp^ 2.62 / = 0.9 ^ 2.62 + 1.1 ^ 2.62
3 ^0.54 2 ^ 0.54 3 ^ 0.54


Dp^ 2.62 / 0.758779 + 1.283657
1.809862066 1.453973 1.809862



Dp^ 2.62 / 0.521866 + 0.709257 =1.231123
1.809862066

Dp^ 2.62/ = 1.231123
1.809862066

Dp^ 2.62 = 1.809862066 * 1.231123

2.228^ 2.62 = 1.809862066 * 1.231123


Virtual pipe Dia Cube Root ^2.62 = 1.357701
So a pipe equ of 3 km has a 1.357m dia...
This seems ok… but I know its wrong, as if you set the length of the pipe equivalent to more than that of the longest actual pipe, the equation fails… is this the wrong formula?

Cheers

 

[TonyBlair]

Actually it doesn’t work if you set the equivalent pipe length to less than the biggest pipe. If you set it to 1km it states the length is 1.08m which is obviously wrong.

Any tips chaps?

 

[BigInch]

The problem is not defined. You don't have a flow term, so all its saying is that there is an infinite combination of equivalent Lengths and diameters that will give you the some head loss at the sum of the flowrates of the two parallel pipes. And the formula is only true if the flows in each parallel pipe are equal.

I don't know who told you that formula, so here's another one,

Engineering Toolbox gives, HW formula,

http://www.engineeringtoolbox.com/hazen-williams-water-d_797.html

as something like this. I changed it for 1 foot of pipe.

HL = 0.2083 * (L/c)^1.852 * q^1.852 / d^4.8655

HL = friction head loss in feet of water per 1 feet of pipe (fth20/100 ft pipe)

L = length (feet)
c = Hazen-Williams roughness constant
q = flow (gal/min)
d = diameter (inches)

So, assuming c1=c2 and head loss for all parallel pipes is equal and the sum of the flow in all parallel pipes is Qt, you want to find an equivalent pipe, so its head loss must be equal for the same total flow in the parallel pipes, Qp

Le^1.852 * Qt^1.852 / De^4.8655 = Lp^1.852 * Qp^1.852 / Dp^4.8655

So if flows are equal, you have an infinate combination of possibilities that satisfy,

L1^1.852 / D1^4.8655 = L2^1.852 / D2^4.8655

If lengths are equal, diameters are equal and if diameters are equal, lengths are equal.

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

 

[TonyBlair]

Thankyou biginch, however im getting confused...
I work in metric, my hazen's formula is:

hf=2.777 L / ( C^1.85 * D^4.87) * Q^1.85

hf = head loss in METERS of water
L = length of pipe in KM
C = friction coefficient
Q = Flow (Ml/d)
d = inside diameter of the pipe in Meters

So its basically the same as yours.

But what i was after is basically to use the above formula with the dimensions of the "virtual" pipe equivalent of a series of parallel pipes, to get overall head loss.
For each section of parallel pipe i would greate a pipe equivalent, then i would create a single pipe equivalent for all those series of equivalent pipes, if you see what i mean. But im getting stumped at the first post of "equivalent parallel pipe."

I was hoping flow was only relevant when working out the head loss and not calculating the dimensions for the pipe equ.

You are 100% correct in saying that my original equation (in first post) works for pipes with equal dimensions.... because the pipe equ is correct on pipes with equal dimensions (i have the ans to a couple of pipe examples).

Is there any other way i can calculate the pipe equ without having to deal with flow?
The reason for this is because i want to calculate the head loss for a range of flows through for a simplified system (one virtual pipe equ) from a load of parallel pipes that all start and join together at the end.

Is this impossible?

 

[BigInch]

Don't think so, because without flow there is no head loss. You're just stringing together an infinite combination of possibilities that all equal HL1 + HL2+... +Hn, any of which could equal any other, depending on its flow, hoping eventually to be able to solve for one to determine the others.

You are not using the electrical equivalent of "Kirchoff" Law, summing voltage drops and then describing voltage as a function of current and wire resistance with IR = E. Then, with resistance known, you could find current, or with current and voltage known, you could determine the value of resistance. If you had the system of equations describing Q for each pipe as a function of HL and could solve them similarly to the electrical method, simultaneously (but hydraulics requires an iteration since the equations are nonlinear). But HL depends on D and L and viscosity and pipe roughness... all things you are not specifying. You are trying to solve for 3 primary unknowns, Q, L & D, using 1 equation, by writing it in terms of 1 variable containing L and D, eliminating it by calling it Q=f(U f(D,L)) and you still don't know the value of D or L. All you get that way is 1=1.

Its like trying to solve for current using only voltage drop, with some variable (K) containing both I & R, without specifying resistance or current. Or some combination like that. All you get is V = unKnown. I think you might be able to get to the point where you could find the ratio of flow in parallel pipe 1 vs flow in parallel pipe 2, given head loss, but never determine the actual value of either flow.... if you're lucky, but I don't think that's the idea. Or at least I don't see the benefit of knowing that information, which would be the equivalent of, hey we have this voltage drop, lets see how many pipe diameter and length combinations we can use to make all possible flowrates, right?

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

 

[vzeos]

TonyBlair,
The addition of pipes in parallel is accomplished by adding their flow capacities, Q_i. The easiest and probably the best way to do this is by using a flow calculator. You can also do this algebraically. Assume you need to add a 4” pipe and an 8” pipe using a flow calculator. The first thing you do is to assume an arbitrary length, say L = 1000 feet, and an arbitrary pressure drop, say dP = 200 psig. Now calculate the flow rate in the 4” and 8” pipes with L=1000 feet and dP=200 psig. I get for 4”, Q_4” = 966.64 gpm, and for 8”, Q_8” = 5888.92 gpm. Now, Q_Total = Q_4”+ Q_8”= 6855.56 gpm. Use Q_Total along with with L=1000 feet and dP=200 psig and solve for diameter. This is the equivalent diameter. I get 8.4811”. Note that the arbitrarily selected L and dP are the same in all of the previous calculations thus they cancel out.

If you want to do this algebraically, pick a flow equation. I like the Darcy equation:

Q= C x [√]((dP / (Ro x L)) x (D ^2.5 ) x 1/[√]f

C = 68.04138 for Darcy
f = friction factor, Darcy
Q = gpm
L = feet
D = inches
dP = psid
Ro = lb/cu ft,1


The objective is to get Q_Total = Q₁+ Q₂ then solve for D_Total in terms of D₁ and D₂

Q ₁= C x [√]((dP / (Ro x L)) x (D ₁ ^2.5 ) x 1/[√]f
Q ₂= C x [√]((dP / (Ro x L)) x (D ₂ ^2.5 ) x 1/[√]f

C x [√]((dP / (Ro x L)) x (D_Total ^2.5 ) x 1/[√]f = C x [√]((dP / (Ro x L)) x (D ₁ ^2.5 ) x 1/[√]f + C x [√]((dP / (Ro x L)) x (D ₂ ^2.5 ) x 1/[√]f

Cancel out like terms and solve for D_Total.

Note that the quantity, 1/[√]f, is a function of diameter but is customarily assumed to be constant for this analysis.

Now, If you want to add two pipes that have different lengths, you need to adjust the length and diameter of one of the pipes. You assume the new length is equal to the length of the other pipe then you calculate the new diameter. Add the new diameter to the diameter of the other pipe as was done above. The criteria for establishing the new length and new diameter are: 1)The flow rate must be the same as the old pipe and 2)the pressure drop must be the same as the old pipe.

Algebraically,
C x [√]((dP / (Ro x L_New)) x (D_New ^2.5 ) x 1/[√]f = C x [√]((dP / (Ro x L_Old)) x (D_Old ^2.5 ) x 1/[√]f

Assume L_New has the same value as the other pipe and solve for D_New.

 

[rconner]

[See also e.g. sometimes convenient tables on pages 17-10&11 @

http://www.acipco.com/adip/products/Sect17.pdf

]

 

[katmar]

TonyBlair, it seems to me that your formula does give the correct answers. I do not understand the line in your first post where you said
"Virtual pipe Dia Cube Root ^2.62 = 1.357701"

I come to the same answer of 1.357 m diameter but I got it by this route:-
Dp^2.62 = 1.8099 * 1.2311 = 2.228
Dp = 2.228^(1/2.62) = 1.357

Similarly, for your second example where you were looking for the equivalent diameter for a 1 km line I would calculate it as
(Dp^2.62)/(1.0^0.54) = 1.2311
Dp = 1.2311^(1/2.62) = 1.083, which is what you got as well.

I tested these results using software which is based on the Darcy-Weisbach formula, and using a pipe roughness of 0.1 mm. As vzeos has noted, for your formula to be valid there has to be the same pressure drop across each of the parallel legs, and the same pressure drop would be across the equivalent single pipe.

Using a pressure drop of 200 kPa, and ignoring all fittings and end effects (i.e. taking only friction into account) I get the following flowrates
Pipe ID = 0.9m, length = 2 km, calculated flow = 2395 litre/sec
Pipe ID = 1.1m, length = 3 km, calculated flow = 3287 litre/sec
Total flow for two parallel legs = 2395 + 3287 = 5682 litre/sec

For your first equivalent pipe
Pipe ID = 1.357m, length = 3 km, calculated flow = 5678 litre/sec
These two flowrates (i.e. 5682 and 5678) are better than identical in my book.

For your second equivalent pipe
Pipe ID = 1.083m, length = 1 km, calculated flow = 5512 litre/sec
This is still within 3% of the total flow calculated before, and is probably a smaller error than is introduced by other assumptions and measurements.

Overall, I cannot understand why you have concluded that your calculated equivalent diameters are "obviously wrong".

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[BigInch]

Sorry Tony. I think I was blinded by the light at the destination. I see where Katmar is correct as usual, so I guess I just didn't see the reason for running through that process. You can solve these systems for all pipes simultaneously, quite easily with some comuter power, and without any substitution of equivalent pipes. Sorry that I might have discouraged you from your actual intent.

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

 

[vzeos]

Katmar,
Congratulations on deciphering the computations in the original post. But it seems that the computed value was not his problem. I think he thinks that the equivalent diameter is supposed to be constant because he says “…if you set the length of the pipe equivalent to more than that of the longest actual pipe, the equation fails... is this the wrong formula? ” You showed that his formula is correct but it may be worthwhile to point out that you can vary the length and diameter of any pipe so long as the pipe resistance remains constant. Such that

R_p = R₁ + R₂

Where
R_p = D_p^2.62/L_p ^0.54

R₁ = D₁^2.62/L₁ ^0.54

R₂ = D₂^2.62/L₂ ^0.54

To solve for D_p you can write:

D_p = (L_p ^0.54 * (D₁^2.62/L₁ ^0.54 + D₂^2.62/L₂ ^0.54)) ^1/2.62

Using this formula you can calculate a number of valid solutions, e.g.:
Lp = 3
Dp = 1.357700997
or
Lp = 5
Dp = 1.508442251
or
Lp = 7
Dp = 1.616764187

 

[BigInch]

That's what I got caught up in. Don't see why you would want to define something in terms of infinite possibilities.

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

 

[vzeos]

Yes, I know. That’s why I went through the customary two step approach. I guess this might be an educational exercise; just to flex the equations.

 

[katmar]

BigInch and Vzeos,

I have not seen it approached this way before but I suppose somebody somewhere must have seen a need for it to have gone to the trouble of recasting the Hazen-Williams formula into the form used by TonyBlair. It probably had more applicability in the days when pipe network software was not available and this formula is a hang-over from those days.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[TonyBlair]

Guys thanks so much for your help, and yes I can use my formula’s, basically you are right. I was getting confused about the lengths of parallel pipes, where they join i.e you have to split the pipes up into segments if they join or split, then you can use the above formula, combined with:
Ls = L1 + L2 …etc
Ds4.87 D14.87 D24.87

You may query why I wanted to do this, well I wanted a way for simplifying a complex of water pipes in a supply system.
I know there are calculators to do the job for me, but my theory is if you don’t know why something works, then don’t use it.
Hence women cant drive (joke).

Cheers chaps for all you hard work

 

[BigInch]

Now go download a copy of EPANET.

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain