XX XXX water content on July X is XXX XXXXX XXX us assume was dry that time otherwise question cannot XX solved without the XXXXXX XX saturation
or XXXXX XXXXXXXX or XXXX XXXXX
= 1050kg /m3
the volume of XXXX XXXX the XXXX zone XX equal XX XX x10 x 0.X = 50 mX
mass of XXX XXXX solids =ms
the XXXXXXXXX XXXXX on July 15 can be used XX XXXXX =
XXXXX XXXXXXX July 15 = XX.9% XXXXX XXXXXX XXX XXXXX XXXXXXX below XX %
MW = X.XXX x52500 = 4147.X kg
XXXXXX XX water can XX XXXXXX = = X.15 mX
XXX XX XXXXXX XXX XXXXX content on July X was the XXXXXXX XXXXX, i.e. 22%
bulk XXXXXXX of soil XX XXXXX XX rd= 1050 XX/mX
the volume of the XXXX is XXXXX to 10 x 10x0.X =XXXX
is XXXXXX XX the XXXX ( XXXX solid + water XX + XX% w.c) = XXXXXXX=XXXXX kg
XS + XW = 52500
XXXXXX of XXX XXXX now at 15 July use XXXX water XXXXXXX 29.X -XX is equal to 7.X % MS /MW = 0.XXX MW is XXXXX to X.07 XXXXXX X.79 = 3399.59
value of that XXXXX is
X]
same XXXXXXXX in XXXX case XX XXXX
XX assuming XXX XXXX was dry on 1st July Xs = 50500 kg
weight XX water @ Fc =
on XX July water XXXXXXX is XXXXX to XX.9 %
XXXXXX XX the water XX XXXXX to XX /XX = 0. XXX
XX = X.299 XXXXX = XXXXX.X XX
weight of water XXXXXXXX 24150- S XXXXX.5 = XXXX.5 kg
XXXXXX = 8453.5 / XXXX = 8.45 m3
XX find XXX amount XX rainfall XXXXX, the volume should be divided XX area of the field XX XXXXX XX
XXXXXXXX water XXXXXXX XX XXX XXXX XXXXXXXX point XX XXXX X
MS= XXXXX.79 XX
weight of water @ f c = 4303 X.XXX 0.XX= 1979 5.XX XX
weight of water @ XXXX July = XXX 303 2.X XXX 0.XXX= XXXXX . X kg
amount XX XXXXX XXXXXX= XXXXX.XXXX - 128 66.X= 6928.28 XX
volume XX XXXXX to= XXXX X.28/ 1000= 6.XXXX
in terms of rainfall XXXXX= 6.XX/ XXX= X.XXXXX= X.93 XX XX