as the water content XX July 1 XX XXX XXXXX XXX XX XXXXXX XXX XXX XXXX XXXX otherwise XXXXXXXX XXXXXX be XXXXXX XXXXXXX the XXXXXX of saturation
or water contains or XXXX XXXXX
= 1050kg /m3
the XXXXXX XX XXXX upto XXX root XXXX is equal XX 10 XXX x 0.5 = 50 mX
mass XX XXX XXXX XXXXXX =ms =1050x XX = 52500kg
the XXXXXXXXX water on XXXX XX can be used by XXXXX =
XXXXX content July 15 = 29.X% plant cannot XXX XXXXX XXXXXXX XXXXX 22 %
XX = 0.079 XXXXXX = XXXX.5 XX
volume of water XXX be uptake = = X.XX m3
XXX XX XXXXXX XXX water content XX XXXX X was XXX wilting point, i.e. XX%
bulk density of soil is equal XX rd= 1050 kg/m3
the XXXXXX XX the XXXX is XXXXX XX XX x 10x0.X =50mX
XX weight XX XXX XXXX ( XXXX solid + water wt + XX% w.c) = 1050x50=XXXXX kg
XS + XX = 52500
XXXXXX of the soil now at XX July XXX open water content XX.X -22 is XXXXX XX X.X % XS /XX = 0.079 XX XX XXXXX to X.07 XXXXXX X.XX = 3399.59
XXXXX XX XXXX XXXXX is
C]
same XXXXXXXX in XXXX XXXX as well
XX XXXXXXXX XXX soil was XXX XX Xst XXXX Ms = XXXXX XX
weight of XXXXX @ XX =
XX 15 XXXX XXXXX XXXXXXX is XXXXX to 29.9 %
XXXXXX XX the XXXXX XX equal XX MW /MS = 0. XXX
XX = X.XXX 52500 = XXXXX.X XX
XXXXXX of water XXXXXXXX 24150- S XXXXX.5 = 8452.5 XX
volume = 8453.X / XXXX = X.45 m3
XX find the XXXXXX of rainfall depth, the XXXXXX should be XXXXXXX XX XXXX of the field XX XXXXX XX
assuming XXXXX XXXXXXX at XXX XXXX building point on July 1
MS= XXXXX.XX kg
XXXXXX of XXXXX @ f c = XXXX X.XXX X.46= XXXX 5.XX XX
XXXXXX XX water @ 15th XXXX = XXX XXX 2.X XXX 0.XXX= XXXXX . 8 kg
XXXXXX XX XXXXX XXXXXX= 19795.XXXX - 128 XX.X= 6928.XX kg
volume is equal XX= XXXX 8.XX/ XXXX= X.93m3
in XXXXX of rainfall XXXXX= 6.93/ 100= X.XXXXX= 6.93 XX CM