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Estimating the Water Requirements of Crops

The water requirement of crops is the amount of water that is required to meet the evapotranspiration rate so that crops may thrive. The evapotranspiration rate is the amount of water that is lost to the atmosphere through the leaves of the plant, as well as the soil surface.

Therefore, in order to estimate the water requirement of a crop we first need to measure the evapotranspiration rate. The reference rate, ET0, is the estimate of the amount of water that is used by a well-watered grass surface that is roughly 8 to 15 centimeters in height. Once ET0 is known, the water requirement of the crop can be calculated.

Many meteorological stations publish ET0  values on daily basis.


There are at least two methods that can be used to measure or predict the evapotranspiration rate of crops:

  •   Evaporation pan
  •   Using equations that predict the evapotranspiration rate based on climatic parameters.


In this method, pan is filled with water and the loss of water from the pan is measured. Provided that there is no rainfall, the evaporation rate, which is recorded as millimeters per day, is quite easy to measure.

This method of measurement takes into account wind, temperature, radiation and humidity, which are the same factors that affect crop transpiration rate.

However, there are a few factors that prevent this recording from being entirely accurate. For one, the solar radiation results in heat storage in the pan. This can lead to increased reading of the evaporation rates at night, when transpiration usually does not occur. In addition, temperature and humidity levels above the pan surface will vary from what would naturally occur.


Evaporation Pan
In this method, different kinds of pans are used to measure the water requirements of crops, with Kp representing the pan coefficient, according to the kind of pan, solar radiation, wind, humidity and the surroundings.

ETo = K pan × E pan


The reference rate, ET0, is calculated using the Penman Equation, which takes into account the climatic parameters of temperature, solar radiation, wind speed and humidity.

A variation of this equation, published by the FAO is:



ETo reference evapotranspiration [mm day-1],
Rn net radiation at the crop surface [MJ m-2 day-1],
G soil heat flux density [MJ m-2 day-1],
T air temperature at 2 m height [°C],
u2 wind speed at 2 m height [m s-1],
es saturation vapour pressure [kPa],
ea actual vapour pressure [kPa],
es – ea saturation vapour pressure deficit [kPa],
D slope vapour pressure curve [kPa °C-1],
g psychrometric constant [kPa °C-1].

Another method of calculating the reference crop evapotranspiration, ET0.

ETo = p (0.46 T mean +8)

ETo = Reference crop evapotranspiration (mm/day)

Tmean = mean daily temperature (°C)

p = mean daily percentage of annual daytime hours.

As we can notice from the equation above, this method takes into account only the measurement of temperature. Therefore, it is not a very accurate estimation of the reference evapotranspiration rate.


ET0 represents the maximum, or potential, evapotranspiration rate that can occur. However, the water requirement of the crop is usually less than ET0, as there are factors of the crop itself that have to be taken into account.

These include the growth stage of the plant, the leaf coverage that provides shade to the ground, and other particulars of the crops that make them vary from each other. With these factors taken into account, ET0 is converted into ETc, through the crop-specific coefficient, Kc.

ETc represents the evapotranspiration rate of the crop under standard conditions (no stress conditions).

When calculating ETc, one must identify the growth stages of the crop, their duration and select the proper Kc coefficient that need to be used.

ETc = Kc*ET0


Climatic effects are incorporated into ET0, while the effect of the crop characteristics are incorporated into Kc.


Crop: potato

Growth stage: Initial growth

Kc for initial stage: 0.45

ET0 (measured by a local meteorological station): 9 mm/day

ETc = Kc*ET0  = 0.45 X 9 = 4.05 mm/day 

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