by Rebecca Shortt, Engineer, Water Quantity, OMAFRA , 30 June 2016
This blog post will show the 2 steps to roughly calculate how long to run a drip system.
First calculate the plant water demand. This was shown in my post from last week.
Plant Water Demand (U.S. gal)
= ET (mm) x Kc x Area occupied by plant (ft2) x conversion factor 0.0245
- ET from vineandtreefruitinnovations.com/ (free but you must register) or www.onpotatoes.com
- Kc (Crop Factors) from the tables below and/or consider extent of canopy
- Area occupied by plant is the row spacing x the plant spacing
- If there is significant space without crop canopy between rows (e.g. apple trees which have been hedged) then this area could be reduced. Keep in mind that the Kc (crop coefficient) may be partially accounting for this already (e.g. strawberries have Kc of 0.75 partly because we assume bare straw between rows). Also keep in mind that living cover crops between rows will be competing with the crop for water.
Second, divide the Gallons/Plant/Day by the number of emitters per plant x emitter flow rate.
Daily Run Time = Daily Plant Water Demand ÷ (# Emitters/plant x Flow Rate)
Strawberry example in Woodstock
Step 1
ET June 23-29 = 30mm/7days = 4.3 mm
Kc = 0.75
Area = 1’ x 4’ = 4 ft2
Plant Water Demand = ET (mm) x Kc x Area (ft2) x conversion factor 0.0245
= 4.3 mm x 0.75 x 4 ft2 x 0.0245
= 0.32 Gallons/Plant/Day
Step 2
# Emitters/plant = plant spacing ÷ emitter spacing = 1 ft ÷ 1ft = 1
Flow rate of Emitter = 0.42 gph (gallons per hour)
Daily Run Time = Daily Plant Water Demand ÷ (# Emitters/plant x Flow Rate)
= 0.32 Gallons/Plant/Day ÷ (1 x 0.42 gph)
= 0.76 hours per Day
Apple 3’ x 12’ (mature with sod) example in Leamington
Step 1
ET June 23-29 = 31mm/7days = 4.4mm
Kc = 1.0
(I’m using a higher Kc because I think we’re closer to full canopy than the chart value would suggest)
Area = 3’ x 12’ = 36 ft2
Plant Water Demand = ET (mm) x Kc x Area (ft2) x conversion factor 0.0245
= 4.4 mm x 1.0 x 36 ft2 x 0.0245
= 3.9 Gallons/Plant/Day
Step 2
# Emitters/plant = plant spacing ÷ emitter spacing = 3 ft ÷ 2 ft = 1.5
Flow rate of Emitter = 0.63 gph (gallons per hour)
Daily Run Time = Daily Plant Water Demand ÷ (# Emitters/plant x Flow Rate)
= 3.9 Gallons/Plant/Day ÷ (1.5 x 0.63 gph)
= 4.1 hours per Day
Crop Factors for Fruit (Kc)
Crop Coefficient |
Approximate Spacing |
|
Bare soil |
0.2 |
N/A |
Early season |
0.4 |
N/A |
Strawberries |
0.75 |
1’ x 4’ |
Raspberries |
0.70 |
2.5’x10’ |
Blueberries |
0.80 |
5’x10’ |
Grapes |
0.75 |
5’x12’ |
Tree Fruit High Density |
1.0 |
3’x10’ to 5’x12’ |
Apples |
0.9 |
7’x12’ to 20’x20’ |
Peaches |
0.8 |
7’x12’ to 20’x20’ |
Pears |
0.8 |
7’x12’ to 20’x20’ |
Cherries |
0.9 |
7’x12’ to 20’x20’ |
Modified from the B.C. Trickle Irrigation Manual by Ted Van der Gulik
Early Spring Crop Factors for Fruit Trees (Kc)
| Month | Permanent sod with herbicide strip | Clean cultivation plus cover crop | ||
| Non-bearing | Mature | Non-bearing | Mature | |
| April | 0.2 | 0.2 | 0.2 | 0.2 |
| May | 0.3 | 0.3 | 0.3 | 0.3 |
| June 1-15 | 0.3 | 0.4 | 0.3 | 0.4 |
| June 16-30 | 0.5 | 0.6 | 0.4 | 0.5 |
| July | 0.6 | 1.0 | 0.5 | 0.65 |
| August | 0.6 | 1.0 | 0.5 | 0.65 |
| September | 0.5 | 0.95 | 0.3 | 0.5 |
Surely the area in a 12x 3 orchard is not the 36sq ft you mention. Three quarters of the area in a high density orchard is roadways for tractors , why are you worried about the grass area, we know very few tree roots will extend into grass sod . A more accurate plant area would be 7 sq ft ( area of 3 ft circle) . Ok let’s say 10sq ft . On your calculations a 12x 3 orchard (1200 trees ) needs 3.9gallons/ plant/ day . Let’s say 4 gallons so 4 x 1200 trees = 4800gollons / acre/ day . On a 75 acre orchard that’s 75x 4800= 360,000 gallons every day! That’s an Olympic swimming pool every day that has to be pumped and filtered or 250gallons/minute every minute of every day. So much for the energy and water conserving of drip.
Hi Paul – thanks for your comment! this reply was prepared by Rebecca Shortt- Yes, with a high density orchard the entire plant spacing area (ex. 12×3) is not covered in canopy so growers can experiment with reducing the value for canopy area in the calculation, which will bring the calculated irrigation demand down. However we have a lot of intercepted solar energy on that “vertical wall” of canopy and that is driving the plant water use. I suggest that growers wanting to reduce the volume of water applied, start with 75% of the area and be extremely cautious about dropping below 30% (the 10sq ft suggested).
Drip irrigation in Ontario usually doesn’t conserve a significant amount of water over overhead. This is primarily because with overhead irrigation fields are generally under-irrigated (not starting early enough and not coming back often enough). This under irrigation leads to crop stress and reduced productivity.
Drip irrigation does make the most effective use of water, often increasing productivity with reduced operating energy costs (lower pressure) amongst other benefits (see video about advantages and disadvantages of drip irrigation http://www.omafra.gov.on.ca/english/engineer/facts/ir-vid.htm)
Irrigation does consume large amounts of water. Robust water supplies are required.
Nicee post thanks for sharing