Hardy Cross Part 1

[@hectorvalencia123]

Thank you for sharing your knowledge that is helping me a lot. An example with a grid sprayer system?

[@kennethwlamb]

A grid sprayer system, or any other system for that matter, would use the same approach. You would just have smaller diameter pipes and lower flowrates.

[@hectorvalencia123]

Kenneth Lamb I'm doing an exercise comparing the result with hass software. can i send it to you for you revision?

[@kennethwlamb]

I've never used Hass software so I doubt I would be much help.

[@hectorvalencia123]

Kenneth Lamb Ok I'll continue to exercise, I'll surely i has a catch error. Best regards from Mexico City

[@mitchvermeire4391]

Do you have to have an outflow from every junction to use this process? For instance in your example you have a demand of 52 gpm leaving all your junctions. I have a scenario where not every junction has a flow leaving the system.

[@kennethwlamb]

You do not have to have an outflow from each junction. Some junctions may be present only to join two pipelines together.

[@yusufrashid248]

Thank you for sharing this. Super helpful. You estimate the demand at the beginning. Can you do this same type of analysis with an existing system where we know the flow leaving the pumping system but we're unsure of the discharges at the various junctions? Do we do the same guesstimation and then just continue going forward or do you recommend something else? I have an existing piping system and I'm a little unsure of the discharges at the various locations and that's what I'm hoping to calculate

[@kennethwlamb]

COrrect, if you know the amount of flow leaving the pump station then you essentially know the total demand of your system. Then you'd have to guess the flowrates going in each pipeline and use this approach to refine those initial guesses.

[@nourelislamlounis3328]

u're not using wood charles mothod ? some call it linear method u'll work on matrices and ofc with iterations i use excel on it actually it converge faster than hardy cross method ( usually )

[@kennethwlamb]

I'm not using the Wood Charles method because Hardy Cross is what we teach. A stupid reason right? but Hardy Cross appears more when talking about solving for flowrate and residual pressure in water networks. To be honest, I'd never do a problem like this by hand. It's far better to just build the model in EPANET or some other software and hit the "run" button.

[@arielnitro84]

Thank you, sir, for sharing this very helpful presentation of hardy cross method for water system. I would just like to ask about some approach on ways to increase pipe velocity that falls below 1ft per second. Altering the size of the affected pipe sometimes improve the flow and velocity but is not that effective. Is there another way to solve this problem? Thanks again, and more power to you.

[@kennethwlamb]

To increase pipe velocity, your only tool is decreasing the diameter of the pipeline. However, you can not decrease it too much or it will restrict the flow too much and cause more flow to be directed into other pipes in the network.

[@arielnitro84]

@@kennethwlamb thank you. maybe i can use a cistern + pump sized accordingly to improve the flow on the low pressured service area.

[@4rgetit]

Hi! I'm just wondering, if ever i use Darcy-Weisbach's Formula, how can i determine the value of r?

[@kennethwlamb]

For Darcy-Weisbach (DW), r = (f L) / (2 D g A^2). In order to use DW in Hardy Cross you need to add one more iteration step. You must first make a guess for the friction factor and compute your 'r' value. Next, run the iterations shown on my videos. Once the delta-Q is '0' and the headloss in each loop is '0' then you use the flow rates in each pipe to compute the Reynolds number. From that result, you compute the new friction factor. After that you start the whole process over again with the revised 'r' value. In the old days, before software and this was all done by hand, it was standard practice to assume that the flow in a small diameter closed conduit pipe was fully turbulent. This means that the friction factor was only a function of the relative roughness of the pipe (i.e. f = epsilon / D). This assumption helped simplify the Hardy Cross solution process to something similar to the Hazen-Williams process.

[@4rgetit]

Thank you :)

[@hadjerjojo6460]

Thank you so much

[@javiernarciso1189]

cachame