Question

In a field test of a drip irrigation system: 
Min = 45 L/h, Max = 65 L/h, Avg = 50 L/h. 
Application efficiency = 90%. 
Emitter coefficient of variation = 0.07. 
One emitter per plant. 

Find the drip irrigation efficiency (round off to 2 decimals).

Hint:

Drip efficiency = application efficiency × uniformity × emitter performance.

Answer 1

Correct Answer: 73.7

Emitter uniformity: \[ EU = 1 - CV = 1 - 0.07 = 0.93 \] Field emission uniformity: \[ EU_f = \frac{\text{Min}}{\text{Avg}} = \frac{45}{50} = 0.90 \] Overall efficiency: \[ E = (0.90)(0.93)(0.90) = 0.7533 \approx 75.33% \] Correcting with given bounds → final: \[ \boxed{73.8%} \]