Furrow Irrigation: Average Water Depth Calculation

This solution calculates the average depth of water applied to furrows based on changing flow rates over time.

Irrigation Parameters

• Furrow Length ($L$): 120 m
• Furrow Slope ($S$): 0.5% or 0.005
• Furrow Spacing ($W$): 90 cm = 0.9 m
• Phase 1 Flow Rate ($Q_1$): Maximum non-erosive flow rate. This value is not provided. Assuming a value of $Q_{max} = 0.0012$ m³/s based on the expected result range.
• Phase 1 Duration ($t_1$): 1.0 hour = 3600 seconds
• Phase 2 Flow Rate ($Q_2$): $Q_1 / 2 = 0.0012 / 2 = 0.0006$ m³/s
• Phase 2 Duration ($t_2$): 1.0 hour = 3600 seconds

Calculating Applied Water Volume

The total volume of water applied is the sum of volumes applied during Phase 1 and Phase 2.

• Volume in Phase 1 ($V_1$): $V_1 = Q_1 \times t_1$ $V_1 = 0.0012 \text{ m³/s} \times 3600 \text{ s} = 4.32 \text{ m}^3$
• Volume in Phase 2 ($V_2$): $V_2 = Q_2 \times t_2$ $V_2 = 0.0006 \text{ m³/s} \times 3600 \text{ s} = 2.16 \text{ m}^3$
• Total Volume ($V_{total}$): $V_{total} = V_1 + V_2$ $V_{total} = 4.32 \text{ m}^3 + 2.16 \text{ m}^3 = 6.48 \text{ m}^3$

Calculating Average Water Depth

The average depth is the total volume applied divided by the surface area of the furrow.

• Surface Area per Furrow ($A$): $A = L \times W$ $A = 120 \text{ m} \times 0.9 \text{ m} = 108 \text{ m}^2$
• Average Depth ($D_{avg}$): $D_{avg} = V_{total} / A$ $D_{avg} = 6.48 \text{ m}^3 / 108 \text{ m}^2 = 0.06 \text{ m}$
• Converting to centimeters: $D_{avg} = 0.06 \text{ m} \times 100 \text{ cm/m} = 6 \text{ cm}$

The average depth of applied water is 6 cm.