View Full Version : Hazen Williams

Cliff Dredge
March 22, 2016, 08:56 PM

What's the Hazen Williams friction factor H2O water uses? I can't read it properly in help file.

Furthermore I'm curious why when flow is increases, the product of the head loss change decreases, example:

0L/s - 0m HL
5L/s - 10m HL
10L/s - 40m HL - product 4
15L/s - 80m HL - product 2
20L/s - 120m HL - product 1.5
and so on...

I thought it would be the other way around, any advice is appreciated.


Cliff Dredge.

Patrick Moore
March 28, 2016, 07:57 AM

Thanks for your question. Could you elaborate just a little on your question, I don't think I fully understand what you are asking and where the results are coming from and want to make sure I fully understand your question before I respond.

Here is an excerpt of the Hazen Williams headloss formula from the User Guide for H2OMap Water

Headloss Formulas: (click for a larger image if necessary)

Headloss for Hazen Williams is as follows:

Hl = 4.2727*C^1.852*d^-4.871*L*Q^1.852

Since the Q (Flowrate) is raised to the power 1.852, headloss has to increase with Flowrate. The resistance coefficient which includes the C factor, diameter, and pipe length are all fixed for a given pipe regardless of the flow which makes the Headloss essentially Hl = a* Q^1.852 where "a" is the resistance coefficient = 4.2727*C^1.852*d^-4.871*L. (NOTE: Q in cfs, L in ft, d is in ft for this formula)

If you can describe more fully what you mean by the "product of the headloss change" it would help us better understand your question. Are you seeing this value somewhere, or is this something you are calculating yourself?

Appreciate the clarification,

Patrick Moore

Cliff Dredge
April 7, 2016, 04:29 PM
Hi Patrick,

Thanks for response and apologies for the delayed reply as I had some time off.

I'm incorrect when I say 'product of the headloss change,' it's really the multiplicand (just looked that one up) or multiplier of the headloss product I'm referring to.

Say we have a 1ft diameter pipe, 1ft long with a C factor of 150.

From HL = aq^b

a = (4.2727)*(150^1.852)*(1^-4.871)*(1) = 45796

So I would like to look at the headloss change as the flow increments by 1.

Flow = 1cf/s. HL = 45796*1^1.852 = 45796ft.

Flow = 2cf/s. HL = 45796*2^1.852 = 165324ft. Increment the flow by 1 and the headloss product is 3.61 times greater.

Flow = 3cf/s. HL = 45796*3^1.852 = 350313ft. Increment the flow by 1 and the headloss product is 2.19 times greater.

Flow = 4cf/s. HL = 45796*4^1.852 = 596819ft. Increment the flow by 1 and the headloss product is 1.7 times greater.

So I was just thinking that multiplier would increase but it's decreasing, and I was curious as to why?

Hope that makes sense.


Cliff Dredge.

Patrick Moore
April 7, 2016, 10:52 PM

A couple of key points I can currently see.

1) The headloss increases as the flow increases. from 45796 ft to 165,324 ft to 350313 ft to 596,819 ft. This is as expected. The headloss should increase as a function of the flow raised to the 1.852 power. So the headloss is increasing.

2) Your "multiplier" is how much greater the headloss is than the previous value. This is not an indicator of the amount of headloss, just the ratio of headloss to the previous point. I'm not sure this is a value that tells you too much. From what you show it appears to decrease with flow, but is that ratio meaningful? Let me know why you are looking at it, I'm just not sure why you would and am a bit confused. I suspect the larger the numbers get the smaller their ratios become and this is why you see this, but I'm not sure how meaningful that ratio is. Let me know if you see something I don't.

3) If you plot the ratio of headloss to the original headloss it goes 1, 3.61, 7.65, and 13.03. This in essence is the q^b factor. These all increase as expected in an exponential manner. Maybe this was the number you were wanting to compare?

Hope this helps.

Patrick Moore

Cliff Dredge
April 7, 2016, 11:29 PM
Hi Patrick,

Yes I'm aware the headloss increases with flow (assuming pipe parameters don't change) I was just curious about that multiplier relationship. But just now I realised the same decreasing relationships occur for all exponential functions so I probably shouldn't be too concerned with it. No worries and thanks again.

Cliff Dredge

Patrick Moore
April 8, 2016, 07:25 AM

That's great. Honestly I had never noticed that personally, but it must just be related to the ratio decreasing because as the numbers get larger and larger the ratio between the numbers on the steep part of the curve will just get smaller and smaller.

Glad you figured out what you were curious about!

Thanks for confirming that on other exponential relationships as well for the benefit of the forum.

Best to you!