Subsurface drip irrigation

[Amina]

hello,
I am currently working on the simulation of subsurface drip irrigation. But for the calculation of variable flux for boundary conditions are still not clear. I have studied the PC forum for calculation of flux boundary and it is taken as the volume/the area of the dripper(taken as 60cm/d)but its calculations are still difficult to understand. Please help how to calculate the area of the dripper if we have 18 mm inline drip with emitter spaced at 30cm apart. I have studied few papers and found that it can be considered a sphere as well as a cylinder. What is the difference and which is more appropriate? Also, do we need to double the discharge for axisymmetrical geometry?

Thanks in anticipation
Amina

[Jirka]

It’s simple. First, you need to convert the real 3D case (experiment or reality) into a 2D (plane, or axisymmetrical domain) problem, and then you need to implement that into the model, i.e., to make sure that you get the right amount of water into the simulated transport domain (whether it is half-plane, full-plane, or axisymmetrical domain). See the reference given below, where we tried to explain these issues. J.

Kandelous, M. M., J. Šimůnek, M. Th. van Genuchten, and K. Malek, Soil water content distributions between two emitters of a subsurface drip irrigation system, Soil Science Society of America Journal, 75(2), 488-497, doi: 10.2136/sssaj2010.0181, 2011.

[Amina]

Thanks for the reference given by you. I have studied that paper thoroughly. I was mentioned to take length or surface area of the sphere for flux calculation. Please clarify either we have to take length/cylinder or sphere for 3d consideration of the emitter source. Also, the radius to be used for flux calculation is still unclear. I am having 18mm inline drip tube. then what should be the radius for flux calculations?
Thanks for your help.
Amina

[Jirka]

Amina,

I have tried my best of explaining all this on the HYDRUS website (and in the article that I recommended) and do not know what else I can add to it. Note that we list on the Hydrus website over one hundred papers in which HYDRUS has been successfully used to simulate drip irrigation (and we stopped tracking these applications about two years ago):
https://www.pc-progress.com//Documents/ ... papers.pdf
and thus, the information provided seems to be sufficient for a lot of people. I’m sure that if you think about the problem more (and write it down and fully analyse, which is the only way for you to fully understand it), you will figure it out as well.

J.

[Amina]

Hello,
I am getting the error as Floating Point error-Invalid Number! when I tried to run the software. Can you please help me with the reason for this kind of error.

Amina

[Jirka]

I’m currently on vacation. Just landed in Paris. J.

[hydrus user bgm]

I am working subsurface drip irrigation in Hydrus program. You can see my geometry and boundary conditions in the picture. I run 2 drippers at the same time. There are plants on both drippers. When I do not place the plants, the program calculates but it doesn't calculate when we place the plants. I define the plant from the insert-domain properties- root distribution section. Can you help me ,I can't find where the problem is...Thank you from now...
file:///C:/Users/Beg%C3%BCm%20POLAT/Desktop/hydrus%20picture.PNG
Begüm

[hydrus user bgm]

Also, when I run both drippers, the variable flux 1s look different on the chart. I have also mentioned the picture below. Why does the graphics look like this? is it normal? Thank you again.

VARİABLE FLUX 1..PNG (25.75 KiB) Viewed 70877 times

[Jirka]

I obviously cannot see a picture stored at your computer (file:///C:/Users/Beg%C3%BCm%20POLAT/Desktop/hydrus%20picture.PNG).

I also cannot know if what you do is realistic. I see short irrigation pulses on days 7 and 12, and longer ones on days 20 and 22. However, how should I know if that is what you want to do or not? J.

[hydrus user bgm]

Thank you very much for your reply. Normally my irrigation days are more frequent but in this study I didn't define other days to see if it can calculate. Actually my problem is that before the program calculates, the text in the window, the mesh and the geometric objects say that the root water uptake has different values. I uploaded the image again. I already define the plant as mesh, how do I do it in the geometric objects section? Thank you so much.
Begüm

root water uptake.PNG (170.02 KiB) Viewed 70855 times

[Jirka]

You selected in your project that initial conditions, boundary conditions, and properties are specified on geometrical objects (this selection can be changed on the edit bar to “on FEM”).

The spatial root distribution using the Vrugt function is done on FEM, and thus there is a conflict between those two approaches, and HYDRUS will inform you about it. Hydrus will always give preference to information on FEM, and thus you should be fine.

J.