Hardy Cross Method

[@rasoulkhoshravan5912]

This video explained neatly in short time. specially the shared pipe between two loops was explained completely which I haven't seen in previous videos

[@ucheanyanwu5972]

Thank God for software 🤲

[@marcopadilla1884]

Thanks a lot, you clarified to me the common pipe analysis

[@Neuf436]

Thanks a lot sir, I’ve been struggle in step 8 for a whole day!

[@mattcamota]

This helped me a lot, Thanks!

[@paintwithja4107]

This video is so much helpful thankyou so much....

[@melodiewilliams]

This is great! Thank you. It's really clear.

[@techhunt4922]

What if i put random directions and get negative flow rates and then consider that oh , i assumed the direction wrong which should be otherwise?

[@kennethwlamb]

Correct. If you assume one direction, and the flowrate turns out to be negative, then the result is just telling you the flow rate is opposite to the direction that you chose.

[@tadnashgunz2355]

Is there an excel file example? my iteration is not converging

[@kennethwlamb]

I have another series of videos where I work an example in Excel: kzbin.info/www/bejne/ppu4qYaFeLeJpdE

[@tadnashgunz2355]

@@kennethwlamb I found it thank you so much these videos are really helpful

[@ahsan.t1897]

Thanks a lot, sir ... I want to know how do we analyze the dead end/tree distribution system? are they the same as grid-iron system?

[@kennethwlamb]

When analyzing a dead end/tree system (system with no looping), then you can just analyze headloss, directly, because you should know the flow in each pipeline. Let's say there are 4 homes on a dead-end pipeline. The main pipe is sized/analyzed to convey the flow for all homes (Q-total), then the laterals from the main are sized/analyzed using the demand for each house (Q1, Q2, Q3, Q4). If you know the lengths of each pipeline then you can determine the headloss (and therefore the residual pressure) at each house. No need for Hardy cross in this situation. Let me know if you need a worked example.

[@ahsan.t1897]

@@kennethwlamb sir it would be really helpful if u upload an example video... Thank you

[@vangelissilver4]

What happens if in the mid-section of a sector we have a pump that gives variable flow. Like for example a pump in the 2-3 section with equation H(pump)=40-0.3*Q23-0.9Q23^(2).

[@kennethwlamb]

If you add a source of energy to the system, such as a pump or tank, then you account for this with a "virtual loop". If, as you suggested, the pump is added between J-2 and J-3, then the virtual loop will be loop around the outside of the whole system (J-3, -4, -5, -1, -2). I'd probably need to make another video to illustrate this. It's a bit unrealistic to do that by hand, so you might as well download EPANET and use that to analyze your network.

[@rasoulkhoshravan5912]

@@kennethwlamb Looking forward that video

[@carlosnoriega7296]

can we use those same formulas with SI units?

[@kennethwlamb]

Yes. The difference is that r(SI) = 10.67 L / (C^1.85 D^4.87); L ~ m, D ~ m, and Q ~ m^3/s. You can find the equation on Wikipedia: en.wikipedia.org/wiki/Hazen%E2%80%93Williams_equation

[@kirinrobinson5006]

what does q stand for

[@kennethwlamb]

There is Q, which is the flowrate (gal/min, or similar unit) flowing through, or out of, the system. Is that what you are referring to?

[@kirinrobinson5006]

Ya actually exactly thank you

[@ipohlucy884]

good

[@ABHISHEKJHA_832]

Can u tell me the software name for it

[@kennethwlamb]

I am assuming you are referring to the software that can do the network analysis. The best free version is EPANET, developed by the US Environemental Protection Agency. www.epa.gov/water-research/epanet. There are other tools out there that are more sophisticated (WaterCAD, H2ONET) but those are expensive.

[@shazibkhan3221]

Asansol engineer college

[@samuelduchesne]

Can the Hardy-Cross method be adapted to solve for the diameters instead of the flow rates? eg.: change the diameters in each loops so that the sum of head losses is zero...

[@kennethwlamb]

I do not think so because there are too many unknowns. You still would need to 'guess' one of the unknowns. In the example here, we guess the pipe diameter and run the analysis. If that is unknown then you would have to guess the headloss or another one of the inputs.

[@rasoulkhoshravan5912]

@@kennethwlamb Very good question and good answer.

[@peterdegloma8661]

Can you please explain the 10.48 HW head loss coefficient? I am used to seeing 4.52 in this formula. I could use more information. Is head loss in psi? Feet of head? Using a psi to feet conversion factor, I am arriving at 10.42-10.44 depending on rounding.

[@kennethwlamb]

I computed 10.48 by starting with the unit headloss Hazen-Williams equation that I've found in Water Resource Engineering, by Larry Mays & Pumping Station Design 3rd Edition. hf = [ (149 Q) / (C D^2.63) ]^1.85 In this equation, Q is in gal/min and D is in inches. To account for the length of the pipe I multiply this term by (L/1000) to get the following: hL = (L/1000) * [ (149 Q) / (C D^2.63) ]^1.85 If you compute 149^1.85 then divide by 1000 you get 10.48. Perhaps the difference is because we are starting from different places ...?

[@hlongwakhulekani405]

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