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