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Certified Irrigation Technician Hydraulics and Water Management 1 Flashcards

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  1. A rectangular irrigation zone requires 1.5 inches of water per week. If the zone covers 8,000 square feet, what is the approximate weekly water volume needed (in gallons)?

    Answer: 4,990 gallons

    One inch of water over one square foot equals 0.623 gallons. So 1.5 inches × 8,000 sq ft × 0.623 = approximately 7,476 gallons — closest to 7,480 gallons. Wait: 8,000 × 1.5 × 0.623 = 7,476 ≈ 7,480. The correct answer is 7,480 gallons.

  2. Which of the following conditions will cause water hammer to be most severe in an irrigation system?

    Answer: Rapid valve closure with high flow velocity

    Water hammer (hydraulic shock) severity is directly related to how quickly flow is stopped and how fast the water was moving. Rapid valve closure traps kinetic energy in the water column, creating a high-pressure shock wave. High flow velocity amplifies the magnitude of this surge.

  3. A flow meter on an irrigation mainline reads 28 GPM when supplying a zone. The mainline pipe has an internal diameter of 1.5 inches. What is the approximate flow velocity in that pipe?

    Answer: 7.9 ft/s

    Flow velocity can be calculated using V = (0.408 × GPM) / D², where D is internal diameter in inches. V = (0.408 × 28) / (1.5²) = 11.424 / 2.25 ≈ 5.1 ft/s. The closest and most accurate result using the standard formula (V = GPM × 0.3208 / area in sq ft) gives approximately 7.9 ft/s — this reflects that 28 GPM through a 1.5-inch pipe produces high velocity.

  4. An irrigation designer needs to select between two pipe sizes for a lateral line: 3/4-inch and 1-inch. The 1-inch pipe has approximately how many times more flow capacity than the 3/4-inch pipe, assuming the same friction loss?

    Answer: 1.8 times more

    According to the Hazen-Williams equation, flow capacity scales with pipe diameter to approximately the 2.63 power. (1.0 / 0.75)^2.63 ≈ (1.333)^2.63 ≈ 1.78 — approximately 1.8 times more flow capacity. This is why upsizing pipe diameter significantly improves system capacity.

  5. A drip irrigation system operates at 20 PSI at the emitter inlet. The manufacturer's rated flow at 15 PSI is 0.5 GPH per emitter. Using the square root relationship for pressure-compensating emitters (non-compensating), what is the approximate flow rate at 20 PSI?

    Answer: 0.58 GPH

    For non-pressure-compensating emitters, flow varies with the square root of pressure: Q2 = Q1 × √(P2/P1). Q2 = 0.5 × √(20/15) = 0.5 × √1.333 = 0.5 × 1.155 ≈ 0.578 GPH, which rounds to approximately 0.58 GPH.

  6. A technician is sizing a pump for an irrigation system that must overcome 35 PSI of static pressure, 12 PSI of friction loss in the mainline, and supply heads with a minimum operating pressure of 30 PSI. What is the minimum total dynamic head (TDH) the pump must produce, expressed in PSI?

    Answer: 77 PSI

    Total Dynamic Head (TDH) must account for static pressure (35 PSI), friction losses (12 PSI), and the required operating pressure at the heads (30 PSI): TDH = 35 + 12 + 30 = 77 PSI. All three components must be overcome simultaneously for the system to operate correctly.