CAIS_SampleQuestions-1

# CAIS_SampleQuestions-1 - CERTIFIED AGRICULTURAL IRRIGATION...

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2003 CAIS Example Questions Page 1 CERTIFIED AGRICULTURAL IRRIGATION SPECIALIST SAMPLE OF CAIS MATH QUESTIONS THESE ARE IRRIGATION MANAGEMENT ORIENTED QUESTIONS THAT REQUIRE A GOOD UNDERSTANDING OF HOW AGRICULTURAL IRRIGATION IS DONE IN THE FIELD. D 1. Given: Soil moisture depletion = 4 inches, Average infiltration = 5.2 inches, DUmin = 0.60. How would you characterize the adequacy of irrigation on this field? A) entire field is over irrigated B) entire field is under irrigated C) perfect irrigation timing D) field has both under and over irrigation SOLUTION Note: the following example demonstrates how to construct and use Water Destination Diagrams. However, they simplify the procedure because they assume that the DU is not the DUlq, but rather a DU that uses the absolute minimum. Dave = 5.2 Dmin = Dave x DU = 5.2 in x 0.60 = 3.12 in Dave - Dmin = 5.2 in - 3.12 in = 2.08 in Dmax = Dave + 2.08 in =5.2 in + 2.08 in = 7.28 in Since SMD = 4 in, SMD is between Dmin and Dave => some over irrigation, some under irrigation Dmin = 3.12 in Dave = 5.2 in Dmax = 7.28 in SMD = 4 in C 2. Given: Soil moisture depletion = 4 inches, Average infiltration = 5.2 inches, DU = 0.77. How would you characterize the adequacy of irrigation on this field? A) field is over irrigated B) field is under irrigated C) perfect irrigation timing D) field has both under and over irrigation SOLUTION Note: the following example demonstrates how to construct and use Water Destination Diagrams. However, they simplify the procedure because they assume that the DU is not the DUlq, but rather a DU that uses the absolute minimum. Dave = 5.2 Dmin = Dave x DU = 5.2 in x 0.77 = 4.0 in Dave - Dmin = 5.2 in - 4.0 in = 1.2 in Dmax = Dave + 1.2 in =5.2 in + 1.2 in = 6.4 in Since SMD = 4 in, SMD = Dmin => perfect irrigation timing Dmin = 4.0 in Dave = 5.2 in Dmax = 6.4 in SMD = 4 in

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2003 CAIS Example Questions Page 2 B 3. FC is 38%. What is FC in inches per foot? A) 5.24 in/ft B) 4.56 in/ft C) 3.84 in/ft D) 3.16 in/ft 38% x 12 in/ft = 4.56 in/ft D 4. PWP is 1.85 inches per foot. What if PWP in %? A) 28.3 % B) 25.8 % C) 20.1 % D) 15.4 % 1.85 in/ft ÷ 12 in/ft = 15.4% D 5. Clay soil, FC = 48%, PWP = 28% moisture by volume. What is AW (in/ft)? A) 5.24 in/ft B) 4.56 in/ft C) 3.84 in/ft D) 2.40 in/ft AW = FC - PWP = 48% - 28% = 20% 20% x 12 in/ft = 2.40 in/ft A 6. Clay loam soil, PWP = 1.8 in/ft, AW =1.6 in/ft. What is FC (%) A) 28.3 % B) 25.8 % C) 20.1 % D) 15.4 % AW = FC - PWP => FC = PWP + AW FC = 1.8 in/ft + 1.6 in/ft = 3.4 in/ft 3.4 in/ft ÷ 12 in/ft = 28.3% A 7. Clay loam soil, saturation = 4.8 in/ft, AW = 1.6 in/ft, PWP = 2.0 in/ft, air dry = 1.5 in/ft. What is FC (in/ft)?
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CAIS_SampleQuestions-1 - CERTIFIED AGRICULTURAL IRRIGATION...

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