Outline
• Announcements
• Where were we?
• Water retention curve
• Hysteresis
Soil Physics 2010
Announcements
• Homework 3 is due now
• Slides or Blackboard?
(blackboard was preferred)
• A brief run through problems 2 & 4
Soil Physics 2010
A
Homework problem 2
5 cm water
B
10 cm soil
C
15 cm soil
D
Pressure
A
0
30
Total
potential
30
B
5
25
30
C
D
Soil Physics 2010
+ Elevation =
15
0
0
0
A
Homework problem 2
5 cm water
B
10 cm soil
C
15 cm soil
D
Pressure
A
0
30
Total
potential
30
B
5
25
30
C
3
15
18
D
0
0
0
Soil Physics 2010
+ Elevation =
Homework problem 4a
Mass in air: 1.1 kg
Density of iron: 5.5 kg / L
Mass
Density 
Volume
Volume of iron = 0.2 L
This is not the
volume of the chunk!
Soil Physics 2010
Homework problem 4
1.1 kg  Voliron  iron  air 
0.8 kg  Volchunk  chunk  water 
? Vol
Vol

iron
Volchunk
chunk  Voliron
Voliron
1.1 kg

 0.2 L
iron   air
0.2 L  iron  water   Mass
Soil Physics 2010
kg
kg 

0.2 L   5.5  1.0   0.9 kg
L
L

The hard way (part 1)
0.8 kg  Volchunk  chunk  water 
Find Volchunk
 1.1 kg 1.0 kg 

0.8 kg  Volchunk  

L 
 Volchunk
1.0 kg 

0.8 kg  1.1 kg   Volchunk 

L 

Vol chunk
Soil Physics 2010
1.0 kg
 1.1 kg  0.8 kg 
 0.3 L
L
The hard way (part 2)
Voliron  0.2 L
Volchunk  0.3 L
b
f  1
p
1.1 kg
2
0
.
3
L
f  1
 1
1.1 kg
3
0.2 L
f = 1/3 or 33.3%
Soil Physics 2010
The easy way
0.8 kg  Volchunk  chunk  water 
kg
kg 

0.9 kg  0.2 L   5.5  1.0 
L
L
Voliron 
Buoyancy for iron only (f = 0):
1.1 kg – 0.9 kg = 0.2 kg
Actual buoyancy:
1.1 kg – 0.8 kg = 0.3 kg
Soil Physics 2010
Volchunk = 1.5 x Voliron
so f = 1/3
Where were we?
Water retention curve
Basic idea:
If the soil were a bunch of capillary tubes, we
could figure out everything about how water
and air move in it…
…if we also knew the size distribution of those
capillary tubes.
The water retention curve is our
best estimate of the soil’s pore
size distribution.
Soil Physics 2010
With that warning, let’s look at water retention
Start with a soil core
that’s saturated:
Atmospheric
pressure
Known
height
Known dry mass
Known porosity
q =f
Soil Physics 2010
So we know the water’s
potential everywhere
So we know the water’s potential everywhere
L
Atmospheric
pressure (0)
5
Known
height L
0
At saturation:
qf
h=0
Soil Physics 2010
If it can drain out
the bottom, then
q0 < f, and
mean h0 = L/2
Then I talked about sponges
Soil Physics 2010
We pull lightly on the water
2 new points:
h1 = Dh1 + L/2
L/2
q1 = f –
(Swater drained/ Vol)
Dh1
Soil Physics 2010
Repeat with a bigger Dh
2 new points:
h2 = Dh2 (+ L/2)
L/2
q 2= f –
(Swater drained/ Vol)
Dh2 > h1
Soil Physics 2010
Potential, h, tension, etc
Suction
Plot the points
Water content
Wetness, q, etc
Soil Physics 2010
Potential, h, tension, etc
Suction
Plot the points
Water content
Wetness, q, etc
Soil Physics 2010
Suction
Height
Potential,
h, tension, etc
Why use this one?
Water content
Soil Physics 2010
Wetness, q, etc
Different regions
Potential, h, tension, etc
Suction
Dry
Middle
Wet
Water content
Wetness, q, etc
Soil Physics 2010
Wet region
Pore only drains if:
Big enough
h
Not isolated
2 cos
r
 w   a  g h
Air can get to it
q
Wet
Air entry
Air access
Structural pores
Soil Physics 2010
A model porous medium being drained
Drainage Pore
allowed: radius:
Big
Small
Soil Physics 2010
A model porous medium being drained
Drainage Pore
allowed: radius:
Big
Small
Soil Physics 2010
A model porous medium being drained
Drainage Pore
allowed: radius:
Big
Small
Soil Physics 2010
A model porous medium being drained
Drainage Pore
allowed: radius:
Big
Small
Soil Physics 2010
A model porous medium being drained
Drainage Pore
allowed: radius:
Big
Small
Soil Physics 2010
Wet region
Pore only drains if:
Big enough
h
Not isolated
2 cos
r
 w   a  g h
Air can get to it
q
Wet
When wetting, air entrapment
limits the final q < f
Soil Physics 2010
Air entry
Air access
Structural pores
Middle region
h
Air and water are both
continuous
Middle
q
Reasonable reflection of
pore size distribution
Mixed textural &
structural pores at wetter
part
Textural pores at drier
part
Hysteresis
Soil Physics 2010
Dry region
Most water is in films
sorbed to solid surface
Water retention mostly
determined by surface area
h
Dry
Little or no hysteresis (if at
equilibrium)
q
Water flow in films is very
slow
q → 0 as h → ∞
(for example, drying at
105° for 24 hrs)
Soil Physics 2010
Hysteresis
• Thermostats
• Speedboats
• “Ink bottle” pores
History-dependent or direction-depedent
Soil Physics 2010