320-HW8 - BRAE 340 Irrig. Water Management HW 8 -...

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BRAE 340 Irrig. Water Management Name: HW 8 - Irrigation Scheduling (only odd numbered problems will be graded) [A] 1. Soil AWHC = 1.8 in/ft. Crop RZ = 5 ft. MAD = 75 %. ETc = 0.24 in/day. DU = 73 %. PI Losses = 3 %. (ignore salinity, i.e., leaching requirement = 0). What is the maximum allowed SMD? A) 6.75 inches B) 9.26 inches C) 11.57 inches D) 10.67 inches E) 8.04 inches Solution: AWHC = AWHC (in/ft) x RZ = 1.8 in/ft x 5 ft = 9.0 in max allowed SMD = AWHC x MAD/100 = 9.0 in x (75/100) = 6.75 in [C] 2. Soil AWHC = 1.6 in/ft. Crop RZ = 5 ft. MAD = 70 %. ETc = 0.24 in/day. DU = 74 %. PI Losses = 4 %. (ignore salinity, i.e., LR = 0). Irrigation is done with perfect timing. What is Dave for the irrigation? A) 2.72 inches B) 6.23 inches C) 7.57 inches D) 5.14 inches E) 4.09 inches Solution: AWHC = AWHC (in/ft) x RZ = 1.6 in/ft x 5 ft = 8.0 in max allowed SMD = AWHC x MAD/100 = 8.0 in x (70/100) = 5.60 in With perfect timing, Dmin = SMD at the time of irrigation = 5.60 in. Dave = Dmin/(DU/100) = 5.60/(74/100) = 7.57in [B] 3. Soil AWHC = 1.9 in/ft. Crop RZ = 4 ft. MAD = 60 %. ETc = 0.27 in/day. DU = 72 %. PI Losses = 3 %. (ignore salinity, i.e., LR = 0). What is the gross application depth required? Assume perfect timing. A) 8.64 inches B) 6.53 inches C) 4.77 inches D) 7.70 inches E) 5.58 inches Solution: Total available = AWHC (in/ft) x RZ (ft) = 1.9 in/ft x 4 ft = 7.60 inches Max SMD = tot avail (in) x MAD (%) = 7.60 x 60/100 = 4.56 inches Dmin = requirement = 4.56 inches Dave = Dmin/(AE/100) = 4.56/(72/100) = 6.33 inches gross depth = Dave / (1-pre-filtration losses) gross depth = 6.33 / (1-(3/100)) = 6.53 inches [D] 4. Same facts as in #3. What is the AE? A) 62 % B) 94 % C) 51 % D) 70 % E) 83 % Solution: With perfect timing, AE = DU x (1 - PI Loss) AE = 72% x (1 - 3%/100) = 70% [A] 5. Same facts as in #3 and #4. Improving the DU to 76% would improve the AE to what? A) 74 % B) 65 % C) 54 % D) 82 % E) 92 % Solution: With perfect timing, AE = DU x (1 - PI Loss) AE = 76% x (1 - 3%/100) = 74% Solution Key
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[D] 6. This week in the Central Valley, cotton ETc = 0.34 in/day and ETo = 0.25 in/day. This week in the low desert ETo = 0.43 in/day. What is the ETc for cotton in the low desert? A) 0.77 in/day B) 0.67 in/day C) 1.00 in/day D) 0.58 in/day E) 0.91 in/day Solution: Central Valley: Kc = ETc / ETo = 0.34 in/day / 0.25 in/day = 1.36 Low Desert: Use same Kc: ETc = Kc x ETo = 1.36 x 0.43 in/day = 0.5848 in/day [E] 7. What are the four major factors which determine the numerical value of the crop coefficient Kc? A) Crop, stage of growth, weather, and soil texture B) Stage of growth, soil texture, weather and plant moisture stress C) Weather, soil texture, depth of root zone and MAD D) Weather, soil texture, soil surface wetness and plant moisture stress E) Crop, stage of growth, soil surface wetness, and plant moisture stress Explanation Soil texture, depth of root zone do not affect Kc [F] 8. T/F: The USBR operates the CIMIS program to provide ETo and Kc information in Califrornia. Explanation
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320-HW8 - BRAE 340 Irrig. Water Management HW 8 -...

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