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Field capacity from applied water data

Short answer

Effective water stored in soil is 630 m³, equivalent to a depth of 63 mm. The increase in moisture content is 2.625%, so field capacity is 7.62%.

Determine the field capacity of the soil from following data: depth of root zone = 1.5 m; present water content = 5%; dry density of soil = 1600 kg/m³; water applied to the soil = 700 m³; water loss due to evaporation = 10%; area of the plot = 1 hectare.

Given: root zone depth D=1.5D = 1.5 m, present moisture =5%= 5\% (by dry weight), γd=1600\gamma_d = 1600 kg/m3^3 so γd/γw=1.6\gamma_d/\gamma_w=1.6, water applied =700= 700 m3^3, evaporation loss =10%= 10\%, area =1= 1 ha =104= 10^4 m2^2. The applied water brings the soil exactly to field capacity.

Step 1: Water actually stored in the soil

V=700×(1−0.10)=630 m3V = 700\times(1-0.10) = 630\ \text{m}^3

Step 2: Depth of water stored

d=630104=0.063 m=63 mmd = \frac{630}{10^4} = 0.063\ \text{m} = 63\ \text{mm}

Step 3: Increase in moisture content. Using d=Δw⋅γdγw⋅Dd = \Delta w\cdot\dfrac{\gamma_d}{\gamma_w}\cdot D:

Δw=d(γd/γw) D=0.0631.6×1.5=0.02625=2.625%\Delta w = \frac{d}{(\gamma_d/\gamma_w)\,D} = \frac{0.063}{1.6\times1.5} = 0.02625 = 2.625\%

Step 4: Field capacity

FC=5+2.625=7.625%FC = 5 + 2.625 = 7.625\%

Answer: field capacity ≈\approx 7.62%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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