PLSOIL 105 & 106
Water problems: Problem set
The method of solution suggested for this problem set appears on the back of this page. Please
do not turn in this page! Keep it, as a study guide for the first exam. Put your solutions on a
separate sheet of paper with your name written clearly at the top. Note: It is much safer to
mimic the examples (on back) and show your full solution with the units and all, so you can
get at least partial credit if something goes wrong, rather than just one number as the answer.
Data. A 180-cm3 soil sample was collected three days after a drenching rain when the soil was
assumed to be at field capacity (Not Saturated!). The wet weight of the sample was 274.1 g
and after drying over night in an oven the sample weighed 214.7 g.
1. What was the gravimetric moisture content of this soil at field capacity? (I’m cocking my
gun and aiming it at my head, and I swear I’ll fire if you put the wet soil weight in the
denominator!)
2. If this soil retains 11.67% water by weight at the permanent wilting point what is the
gravimetric plant available water that this soil can store?
3. What is the dry bulk density of this sample?
4. What is the porosity of this sample as a percent of total volume
5. What is the volumetric water content of the soil at field capacity?
6. What fraction of the soil’s pore space is filled with water at field capacity?
PLSOIL 105 & 106
Water problems: Worked example
Data. A 205-cm3 soil sample was collected three days after a drenching rain when the soil was
assumed to be at field capacity (Not Saturated!). The wet weight of the sample was 361.6 g
and after drying overnight in an oven the sample weighed 298.8 g.
1. What was the gravimetric moisture content of this soil at field capacity?
361.6 gA(water+drysoil) - 298.8 gAdrysoil = 62.8 gAwater
(62.8 g ⋅ water )
× 100 = 21.0% (Make sure dry soil is on bottom)
(298.8 g ⋅ drysoil )
2. If this soil retains 11% water by weight at the permanent wilting point what is the
gravimetric plant available water that this soil can store?
FC - PWP = PAW = 21% - 11% = 10% (Not too good!)
3. What is the dry bulk density of this sample?
(dry soil weight)/(soil volume) = (298.8 g)/(205 cm3) = 1.458 gAcm-3
4. What is the porosity of this sample as a percent of total volume
g
1.458 cm3
Bulk density
(1 −
) × 100 = (1 −
g ) × 100 = 45%
Particle density
2.65 cm3
5. What is the volumetric water content of the soil at field capacity?
(0.21
gw
⋅ ater
gdrysoil
⋅
)(1.458
gd
⋅ rysoil
)(1
cm ⋅soil
3
cm3 ⋅water
gw
⋅ ater
) × 100 = 30.65%
6. What fraction of the soil’s pore space is filled with water at field capacity? (Let’s see, now.
If we know that 45% of the soil’s volume is pore space and that 30.65% of the soil’s
volume is water when it’s at field capacity, then surely...)
30.65%/45% = 0.681 = fraction of pore space filled by water