Transpiration and Plant Water Uptake
143
demand on hot days, since plants tend to develop resistances and water
potentials which just meet environmental demands.
Example 9.2. Using the same soil properites as in Example 9.1, estimate
the available water in mm in a 1 m deep root zone if the permanent wilting
water potential is - 1500 J kg-'.
Solution.
If we assume that field capacity is at -33 Jlkg, then, from Example 9.1,
Of, = 0.28 m
3 m-3. The available volumetric water content is 0.28 -
0.12 = 0.16 m
3 m-3. With a one meter deep root zone, the available
water in mm is obtained as 0.16(1000 mm)=160 mm of water. If, on the
average, a plant uses 5 &day, then a one meter root zone of this soil
could potentially provide water for about 32 days.
Example 9.3. If a shallow-rooted plant (rooting depth = 200 mm) is
growing in the soil described in Example 9.2 and E, , , ,
= 5 mmlday
every day, how long will it take before E, = 1 &day?
Solution. The total available water in a 200 mm deep profile would be
0.16 x 200 = 32 mm. The available water fraction on the first day is
A , = 1 so E, = 5 mmlday. It can be assumed that E, is appropirate for
an entire day so that on the second day the available water is 32 mm -
5 mm = 27 mm. Thus A, = 27/32 = 0.84 and on the second day the
plant uptake rate can be estimated from Eq. (9.22).
Up = Epmx[l - (1 + 1 . 3 7 ~ , ) - ~
= 5[1 - (1 + 1.37 x 0.84)-~.~] = 4.9mm/day.
On day 3, A, = 22.1132 = 0.69 so that Up = 4.9 &day. This process
can be continued until E, = 1 mdday.
Available
water
mm
mmlday