You are required to know how to calculate the amount of water in a soil given various water potentials. This will help you understand that it is not the total amount of water in a soil that determines if water is available to plants, but is the plant available water. Seek help if you do not understand how to do these problems.
DATA:
Soil Core Volume = 250 cc (for each soil core below) Weight of soil core at -1/3 bar (field capacity) = 420 g (July 4, 19??) Weight of soil core at -15 bar (wilt point) = 350 g Weight of soil core at present field condition = 395 g (on July 10) Weight of Oven dry soil core = 300 g
Questions:
1) What is the Bulk Density?
Answer: B.D. = 300 g/250 cc = 1.2g/cc (remember, always use oven dry weight)
2)What is the % water by weight at field capacity?
FC - Oven dry
Answer: 420 g - 300 g = 120 g of water
120 g water/300* g soil = .4 (*use oven dry weight)
.4 x 100= 40% water by weight at field capacity
3) What is the % water by volume at field capacity?
FC - Soil Volume =
Answer: 420 g - 300 g = 120 g of water
120 g water/250 cc soil =.48
.48 x 100= 48% water by volume at field capacity
(Another way to calculate this is: BD X % water wt.= % water by volume):
1.2 X 40% water by weight = 48% water by volume
4) What is the total possible % Available Water-holding Capacity (AWC) by volume?
(AWC = FC - WP)
FC - WP / Soil Volume =
Answer: (420-350) / 250 =.28 x 100 = 28%
available water [(FC-WP) / vol.]
5) How many inches of AWC are in the upper 5 ft. of soil?
inches of soil x % AWC =
Answer: 5ft. X (12"/ft.) X .28 = 16.8" of AWC in upper 5 ft. of soil.
6) How many inches of available water are left in the soil at
present field condition?
Field Cond. = 395 and Wilt Pt. = 350; therefore:
(395-350)/250 = 45/250 = .18 (%AWC by Vol.)
and
0.18 X 60" (of soil) = 10.8" of water available in upper 5 feet.
In other words, the soil has lost 6" of water (16.8-10.8) since it was at field capacity.
7) What is the depth of wetting for a 1.5 inch rainfall event for the soil at Field Condition?
Answer : Field Capacity - Field Cond./ vol x 100 = % water by vol between field
condition and Field Capacity. or 420 - 395/250 = 25/250 = .1 or 10%
thus 1.5 inches of rain will infiltrate (1.5 / .1) = 15 inches. of soil because
Inches of water = % Water vol x Soil Depth. or 1.5 = .1 x ?inches of soil
Soil Water Chapter 6
