AENG 1 Fundamentls of Agricultural Engineering
Second Lecture Exams
10 September 1997
I. Identification
II. Enumeration (15 points)
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 |