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Aeng1 PS2

This document contains a problem set with multiple parts related to soil-water-plant relationships. It includes calculations to determine various soil properties like bulk density, particle density, and porosity based on measurements of a soil sample. It also calculates moisture content, equivalent depth of water in the soil profile, depth of available water, and maximum volume of water that can be stored in a corn field, based on given measurements and properties of the soil and field.

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0% found this document useful (0 votes)
347 views3 pages

Aeng1 PS2

This document contains a problem set with multiple parts related to soil-water-plant relationships. It includes calculations to determine various soil properties like bulk density, particle density, and porosity based on measurements of a soil sample. It also calculates moisture content, equivalent depth of water in the soil profile, depth of available water, and maximum volume of water that can be stored in a corn field, based on given measurements and properties of the soil and field.

Uploaded by

gigoong
Copyright
© Attribution Non-Commercial (BY-NC)
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX or read online on Scribd
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AENG 1 – Fundamentals of Agricultural Engineering

Problem Set No. 2


Soil-Water-Plant Relationships

1. A cylindrical soil sample with 8 cm diameter and 10 cm height was obtained from the field. It
weighed 875 grams. After ovendrying, the soil sample was reduced to cylindrical column (7.5 cm
diameter and 6.5 cm height) weighing 730 g. density of water is 1 g/cm3. Determine the following
soil properties, and round off your answers to two decimal places.

Given: Dm1 = 8 cm; h1 = 10 cm; FW = 875 g; ODW=730 g =.763kg; Dm2 = 7.5 cm; h2 = 6.2 cm;
ρw=1 g/cm3

a. Bulk density of the soil (kg/m3)

Bulk density, ρb=WSVT ρb=WSVT


Vcyl1=πr12h
Vcyl1= 3.1415926×42cm2×10 ρb=875 g502.654816 cm3
cm ρb=1.740757 g/cm3
Vcyl1= 502.654816 cm3
Vcyl1=VT ρb≈ 1.74 g/cm3

b. Particle density of the soil (kg/m3) ρS=WSVS


Particle density, ρs=WSVS ρb=.73 kg273.9076048125
Vcyl2=πr22h cm3
Vcyl2= ρb=0.0026651323 kg/cm3
3.1415926×3.752cm2×6.2 cm ρb≈ 0.0026 kg/cm3
Vcyl2= 273.9076048125 cm3
Vcyl2=VS ρb≈ 2.6×10-3kg/cm3

c. Apparent specific gravity, AS

AS=WSρwVT
AS=730 g1 AS= 2.665132
gcm3×273.9076048125 cm3 AS≈ 2.67

d. Moisture content dry basis (%), MCdb

MCdb=FW-ODWODW×100
MCdb=875 g-730 g730 g×100 MCdb=23.97%
MCdb=145 g730 g×100

e. Moisture content volume basis (%), MCv

MCv=VwVT×100 or MCv=63.9999
MCv=AS×MCdb MCv=64%
MCv=2.67×23.97

f.
g. Porosity, P
P=VVVT P= 0.455078125
P=VT-VSVT P≈0.46
P=502.654816 cm3-
273.9076048125
cm3502.654816 cm3
2. A field planted with corn was irrigated until saturation. Two (2) days after (when all excess water
has been drained), a soil sample was taken and the following information were obtained:

Fresh weight = 82.1 g Oven dry weight = 66.2 g Total volume = 40


cm3
a. If the effective depth of rootzone is 90 cms, what is the equivalent depth of water (in cm) in
the soil profile during the time of measurement?

MCdb=WwWS×100 AS≈ 1.66

MCdb=FW-ODWFW×100 MCv=As×MCdb
MCv=1.66×19.37
MCdb=82.1-66.282.1×100 MCv=32.1542
MCdb=19.366626 MCv≈32.15
MCdb≈19.37
d=MCv100×drz
AS=WSρwVT d=32.15100×90 cm
AS=66.2 g1 gcm3×40 cm3 d=28.935 cm
d≈28.94 cm
AS= 1.655
b. If the soil moisture content dry basis at permanent wilting point (PWP) is 16.2%, what is
equivalent depth of water (in cm) in the soil profile at PWP?

dPWP=As×MCdb100×drz
dPWP=1.66×16.2100×90 cm
AS=WSρwVT dPWP=24.2028
AS=66.2 g1 gcm3×40 cm3 dPWP≈24.20 cm
c.
AS= 1.655 d. What is the depth of available
AS≈ 1.66 water (in cm) for the corn plants?

dAW=dFC-dPWP dAW=4.74 cm
dAW=28.94 cm-24.20 cm

e. If the rate of water consumption is 15mm per day, how many days will the available water in
the soil last?

DaysAW=dAWRWC DaysAW=4.74 cm1.5 cmday


DaysAW=33.4 cm15 mmday DaysAW=3.16

f. What is the maximum volume of water (in m3) that can be stored in a 2 hectare corn field?

max.dAW=dFC-dPWP
max.dAW=28.94 cm-24.20 cm
max.dAW=4.74 cm

max.VAW=dFC-dPWP×Afield
max.VAW=max.dAW ×Afield
max.VAW=4.74 cm×2 ha
max.VAW=0.0474×20000m2
max.VAW=948m3
Submitted by:
De Leon-de Roxas, Maureen Ceres N.
AENG 1-U2L

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