I need to simulate a surface drip irrigation in Hydrus.
I applied 2.4 litres in 40 minutes or 3.6 litres per hour.
My node spacing at the soil surface is 5 cm.
I have an atmospheric boundary.
If I am applying the water through a single node, then what rate of water (cm hr) should I apply - is it 144 cm/hr assuming the surface area of the node is 5x5 cm?
This seems way to high for me. I obviously don't have the conversion from litres to rate right.
Can you help?
Hi Marcus
For HYDRUS simulations of surface drip in 2D, you have to assume that your dripline represents a continuous line source of water. This assumption would allow you to model a 3D problem in 2D.
You need to convert discharge rate (3.6 L/hr)into flux (cm/hr). For that you need to divide the discharge volume by the area. The area is calculated from dripper distance (say 40 cm) multiplied by lateral spread of water in the soil (say 5 cm in your case but it is generally higher in field conditions). Simply divide 3600/40x5 = 18 cm/hr. If you are running the model on daily basis, then multiply it by 24.
I think it would be of some help to you.
Hi Marcus,
This is not an easy question in Hydrus without some initial estimation. The problem is that HYDRUS2D does not have a good solution for surface ponding.
Use the Wooding solution to get some idea of the likely pond size. Then use the radius to define a surface that you can either have as a constant head. This will over estimate the saturated surface at the start but be good at the end. To get better initial flow use a much lower radius and see when this 'blows up'. By playing around with this you will be able to get a reasonable result.
Regards,
Freeman
Marcus,
I understand that you need to convert the emitter discharge to flux in HYDRUS simulation, this depends on two main points:
1- The type of geometry that you selected (Axisymmetrical or vertical plane)
2- The simulated shape of the emitter (a 1cm diameter half-circle is recommended and done in many publications)
In addition you have to bear these points in mind...
1- the out-flux from the emitter should be a negative value, not positive.
2- It is important to refine the FE grid around the emitter in order to avoid calculation problems (set N=20 in the half circle of the emitter)
3- to calculate the flux, assume the water is uniformly flows out of the sphere (in Axisymmetrical geometry) or the cylinder (in vertical plane geometry), then calculate flux as follows:
*For Axisymmetrical geometry:
Flux (LT-1)=Discharge rate (L3T-1)/Surface area of the simulated sphere (L2)
*For vertical plane geometry:
Flux (LT-1)=Discharge rate (L3T-1)/Surface area of the simulated cylinder (L2)
Best regards
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