Hi, I just want you to know this video saved my life. I had no clue what was cooking with this hardy cross method, and now I feel like a pro. Please continue with your tutorials!!! :)
Its a great pleasure. I am glad you found the video helpful :) I also struggled with it at the beginning, but once you get the hang of it, it becomes a lot easier.
could you explain an exercise using hardy cross method , but the system has devices as pumps or other things and the head of the device is expressed as 30*Q^2 and 100-80*Q^2 for example , and how is done if you have those devices in series i n one of the branch and other device in other branch. thank you .
+carlos casas I'm actually not entirely sure how one would approach a problem such as that by hand. For such problems normally computer software is used which allows for pumps and reservoirs to be added to the network. My apologies if this was not the answer you were expecting... Have a good day :)
i have a question, can u plus the old q with the q effective just like that? since the q effective is metre cube per second and the old q is litre per second. correct me if i am wrong
Hi, sorry for the late reply. I only worked in litre per second for all q's in the example. Can u please clarify where you spot the metre cube per second?
Hi. Unfortunately I am not, the only free software I know of for networks is EPANET. But when it comes to solvers specifically I don't really know. If you find something please post it here, I would love to take a look :)
nothing yet, but a great paper on the modern methods here: arizona.openrepository.com/arizona/bitstream/10150/192098/1/azu_td_hy_e9791_1997_307_sip1_w.pdf
Hi TM, I've a couple of questions. 1. For the pipe network example you have used in your video, I noticed that the discharges Qa, Qb and QC are all defined. I am wondering if the Hardy Cross method is still applicable if these discharges are unknown, especially in the real world, these values would have been unknown. 2. I presume the example you have shown is a network with the same elevation. if we have branches within the loop with different elevation, do we just simply minus or add a head due to its elevation?
CHIN0094 Thank you for your comment. Q1: I don't think the method would be usable if those discharges were unknown because one would not be able to make a first guess of which pipe carries which discharge. However, in the real world those discharges would not always be unknown. Those discharges will be determined from an analysis of the area each pipe has to serve. Q2: I would suggest the Nodal method for a branched system where different elevations must be taken into consideration because it works directly with the respective heads. I'm unsure if your suggestion of simply adding or subtracting the heads will work, but it might. I hope this reply was helpful, please let me know it it wasn't.
TM'sChannel Hi TM, for Q1 "However, in the real world those discharges would not always be unknown. Those discharges will be determined from an analysis of the area each pipe has to serve", can you show me an example on how to get to a discharge say we have a pump and have discharging pipes of 2 different cross section areas? Q2: I managed to source for a program, so what I am gonna do is to see if the program actually has the same results for flow as compared to a Hardy Cross Excel sheet in just subtracting/adding the heads.
CHIN0094 Good morning. It's a bit difficult to think of an example like that at the moment. I do not fully understand the procedure seeing I only had an introductory class to how those discharges are determined. But as far as I know there are existing models, databases (Historical and stochastic) as well as software that are used to predict a peak discharge that a certain area or land use will require. As a very simple example lets say each household in a certain area requires a peak discharge of 5 m3/s (I made that up but it would be determined with the use of models etc.), then an area which consists of 100 households would need 500 m3/s provided to them during the peak time. This is a very simple scenario, it would be much more complicated when there is an entire network of pipes and more than one pipe is servicing an area. The method to determine such a discharge depends on the method the engineer prefers or is instructed to use, in class we were provided with these values. I hope this answer was helpful, please ask if you are still stuck. Q2: Which program did you find if you don't mind ask me asking. Also please let me know what your findings are, should be interesting.
Hi TM, for Q1 I am not quite sure that is correct. say the household has different levels of elevation, even if the pipes are sized the same, I doubt the discharge flowrate will be the same. that is just my guess. for Q2, I am using flowmaster V7. we can continue our discussion if you would drop me a mail on ray_chin@live.com.sg.
+พิพัฒน์ รักทองจัน Hi, I actually have no idea, but I believe the same principles would apply so maybe it is possible. If you find out please let me know :)
+พิพัฒน์ รักทองจัน Hi, the sign of the flow in each pipe is just an indication of its direction within each respective loop. For Q3 the flow is positive in loop 1, but negative in loop 2 due the the direction of flows chosen for each loop (the orange arrows). Effectively the direction of flow for ex Q3 will be from node E to G in this example.
