Week 7 Flashcards
What is the drainable porosity ~equivalent to in an aquifer?
Specific yield
Moisture content, theta =
vol water in voids/total rock volume
Water saturation, Sw =
vol water in voids/vol voids accepting water
i.e.
moisture content/effective porosity
- how much of the effective porosity you’ve actually used
- better indication of how wet/dry it is
Surface tension, concept
Particle on water/gas interface
Stronger forces from liquid
NET ATTRACTION = SURFACE CONTRACTS
each pore is a water/gas interface
(H) =
contact angle
Smooth surface = low
Rough surface = high
Capillary pressure Pc =
Po-Pw >=Pe
Po = air pressure
Pw = water pressure
Pe = air entry pressure
Pe =
air entry pressure
= pressure needed to increase the air pressure (Po) by to push water back out of the tube i.e. for air to enter
What value must Po have in order for air to enter tube/to empty tube?
Po >=Pw+Pe
Air entry pressure head, w(e) =
Pe/(water density) x g
For a tube of diameter D the air entry pressure, Pe =
(4 x (surface tension) x cos((H)) ) /D
“Cumulative probability distribution of pore sizes within a given pore volume” the plot
y (left) = Sw
y (right) = probability of non-exceedance
x (top) = D (pore size, decreases left to right)
x (bottom) = Pc
“Cumulative probability distribution of pore sizes within a given pore volume” results
Increase Pc = decrease Sw (b/c increasing Pe)
Decrease D = decrease Sw
60% pores <288um
288um Pe = 1000Pa
Pc>=Pe, so 60% of pore volume has Pe >1000Pa
= 60% will remain saturated with water
Even though Po > Pw (indicated by +ve Pc), pores resist air because their Pe’s are so high
Ignoring Po, Pc =
and w =
-Pw
w = -Pc/(water density) x g
Why does w (pressure head) decrease from the saturated zone to the unsaturated zone?
Unsaturated = higher Pc = lower w
Why does z (elevation head) increase from the saturated zone to the unsaturated zone?
h increases with elevation