AENG 1 Fundamentls of Agricultural Engineering
Second Lecture Exams
10 September 1997

I. Identification

  1. Ratio of oven dry mass of soil to its volume.
  2. Average length of time in years before a given hydrologic event is equalled or exceeded.
  3. Sum of crop ET, percolation and efficient rainfall.
  4. % MC dry mass basis multiplied by apparent specific gravity.
  5. Areas where static levels are < 6.1 mbgs and well depths are < 20m.
  6. Ratio of the mass of a soil particle to its volume.
  7. Method used to determine the magnitude of the 100-year rainfall or riverflow.
  8. Soil moisture tension (in atm.) at permanent wilting point.
  9. Soil moisture meter that has range from 0 to 0.8 atm moisture tension.
  10. Area drained by a river system.
  11. Percentage ratio of volume of voids in a soil sample to the volume of the sample.
  12. Water tightly absorbed on the surface on individual soil particles.
  13. Streamflow that can be exceeded 80% of the time.
  14. Ratio of irrigation water delivered to the rootzone to that delivered to the field.
  15. Circulation of water from the atmosphere to the earth's surface.
  16. Coefficient which when multiplied by pan evaporation equals potential evapotranspiration.
  17. Potential ET times crop coefficient.
  18. Related to the aggregation of the soil particles.
  19. Groundwater outflow to stream channels.
  20. Most widely used empirical formula for determining PET.
  21. Soil moisture constant when soil moisture tension is zero.
  22. Tanks filled with soil and emplaced in the ground for determining evapotranspiration.
  23. A measure of the suitability of water for irrigation.
  24. Distribution of clay, silt and sand in the soil.
  25. Most accurate and standard method for determining soil moisture content.
  26. Property of water caused by the presence of divalent cations.
  27. Soil moisture content equivalent to 0.8 atm.
  28. Process of water passage through the soil surface.
  29. The constant in Darcy's Law which relates the velocity of soil moisture movement with hydraulic gradient.
  30. Groundwater formation that can be economically developed.

II. Enumeration (15 points)

  1. Two (2) groups of methods for determining crop evapotranspiration.
  2. Two (2) components of ET in Penman equation.
  3. Any three (3) climatic variables in Penman equation.
  4. Two (2) soil infiltration formulas.
  5. Two (2) types of soil moisture.
  6. Three (3) chemical measures of the quality of water.

III. Discussion (15 points)

    Explain the tensiometer method of determining soil moisture content.

IV. Problem Solving (10 points)

    Calculate the equivalent depth of soil moisture in a sample that is 20 cm deep if the apparent specific gravity is 1.5 and the moisture content dry weight basis is 15%. How much irrigation water in cubic meters per hectare will have to be applied to the soil if its moisture content (dwb) is to become 25%.


Answer Key:

 I. Identification

1. apparent density 11. porosity 21. saturation
2. return period 12. hygroscopic water 22. lysimeter
3. crop irrigation requirement 13. dependable streamflow 23. %Na = Na + K
           Ca + Mg + Na + K
4. %MC volume basis 14. application efficiency 24. soil texture
5. shallow well areas 15. precipitation 25. gravimetric method
6. real density 16. pan coefficient 26. hardness
7. hydrologic frequency analysis 17. ETa (actual no. PET) 27. field capacity
8. 15 atmospheres 18. soil structure 28. infiltration
9. tensiometer 19. base flow 29. hydraulic coefficient
10. catchment area 20. Penman equation 30. aquifers

II. Enumeration

1. a) empirical formula
    b) pan evaporation method

2. a) radiation component
    b) aerodynamic component

3. a) solar radiation
    b) temperature
    c) relative humidity

4. a) Morton's Equation     => fp = fc + (fo -fc)e-kt
    b) Philips Equation       =>  fp = bt1/2/2 + a

5. a) hygroscopic soil moisture
    b) capillary soil moisture
    c) gravitational soil moisture

6. a) total dissolved solids
    b) hardness
    c) pH

IV. Problem Solving

D = 20cm
As = 1.5
%MCdb = 15%
d = DPv
PV1 = %Mdwb x As
        = .15 x 1.5
        = 0.225
PV2 = 1.5 x 2.5
       = 0.375
  d1 = 20 (0.225)
     = 4.5 cm
d2 = 20 (0.375)
     = 7.5 cm
Depth of soil at: 15% 25%
Depth of soil moisture d = d2 - d2
   = 7.5 -4.5 cm
   = 3 cm
Vol. of irrigation water   = 3/100 x 10,000m2
  = 300m2