Mid-July, corn at full canopy. The weather station's reference evapotranspiration is running 6.2 mm/day and the crop coefficient for this stage is 1.15. The block is 32 ha under a system that applies water at 85 % efficiency and can realistically run 90 % of the hours in a day. Sets are watered 4 ha at a time, and the schedule calls for a 30 mm application per set. Find the crop's peak water use, the capacity the system has to sustain, and the run time each 4 ha set needs.
Every number in this problem is editable — change any value below and the whole chain recalculates.
Given
ET₀ = 6.2 mm/day — Reference evapotranspiration
K_c = 1.15 — — Crop coefficient, mid-season corn
A = 32 ha — Irrigated block
E_a = 85 % — Application efficiency
f = 90 % — Operating fraction of the day
d = 30 mm — Application depth per set
A_set = 4 ha — Area per set
Determine
(a)the crop's peak daily water use
(b)the system capacity the block demands
(c)the run time per 4 ha set
Step 1 of 3(a) · solve for Crop evapotranspiration
Reference ET is what a well-watered grass sward would use; the crop coefficient converts it to this crop at this stage. Corn at full canopy outdrinks the reference — 1.15 × 6.2 is 7.13 mm/day, a 71,300 L drink per hectare per day.
Rearranged for ET_c
ETc=Kc×ET0
Your values, in your units
ETc=(1.15)×(6.2mm/day)
Answer
ETc=7.13mm/day
Carried onward at full precision, not this rounded figure.
Capacity is peak use grossed up twice — once because 15 % of pumped water never reaches the root zone, once because no system runs 24 hours forever. 7.13 mm/day over 32 ha at 85 % efficiency and a 90 % duty needs 34.5 L/s sustained. A system sized to average-week ET instead of peak-week is the one that falls behind in exactly the week that costs yield.
Rearranged for Q
Q=EafAETp
7.13 mm/daycarried from step 1
Your values, in your units
Q=(85%)⋅(90%)(32ha)⋅(8.25231e−08m/s)
Converted to base units
Q=(85%)⋅(90%)(32ha)⋅(7.13mm/day)
Answer
Q=2.0712m3/min
Carried onward at full precision, not this rounded figure.
The set time is volume over flow: 30 mm across 4 ha is 1,200 m³, and at 34.5 L/s that takes 9.66 hours — call it a ten-hour set. Eight sets of ten hours is most of four days, which is the real reason peak-week capacity has so little slack in it.
Rearranged for t
t=QdA
2.0712 m³/mincarried from step 2
Your values, in your units
t=(0.0345195m3/s)(30mm)⋅(4ha)
Converted to base units
t=(34.5195L/s)(30mm)⋅(4ha)
Answer
t=9.6564h
Carried onward at full precision, not this rounded figure.
Therefore the corn is using 7.13 mm/day at peak, the block needs 34.5 L/s of sustained capacity to keep up, and each 4 ha set runs just under ten hours — the whole rotation takes most of four days, arriving back at the first set almost exactly as its 30 mm is used up.
Why this order
Irrigation scheduling is a supply chain with the atmosphere as the customer. The order here is the load calculation any utility would run: demand per unit (ET_c), total demand grossed up for losses and uptime (capacity), then the dispatch schedule (run time). The two gross-ups in step (b) deserve their own respect — application efficiency is physics and hardware, while the operating fraction is honesty about breakdowns, moves and wind days. Multiplying them under the demand is what separates a system that keeps up in the hot week from one that only keeps up on average.
The closing arithmetic is the discipline: depth per set divided by daily use is the return interval the schedule can actually deliver — here 30 mm at 7.13 mm/day is a hair over four days, and the rotation takes four days to complete. That is a system running flat out with nothing in reserve; the trap is reading it as comfortable because every individual number looked reasonable.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.
