Pan evaporation computation with rainfall and withdrawals: A Class-A type pan (diameter 1.5 m) initially had 8.0 cm of water. During the period, rainfall = 5.0 cm, water withdrawn to maintain operation = 3.0 cm, and final water depth = 9.0 cm. If the pan coefficient Kp = 0.6, estimate the reservoir (lake) evaporation loss for the period.

Introduction / Context:
Pan evaporation is routinely converted to reservoir evaporation using a pan coefficient. Correct bookkeeping of pan water balance with rainfall inputs and operational withdrawals is essential for accurate estimates used in reservoir yield studies and irrigation planning.

Given Data / Assumptions:

• Initial pan depth = 8.0 cm.
• Rainfall during the period = 5.0 cm (falls into the pan).
• Water withdrawn (to maintain operation) = 3.0 cm.
• Final pan depth = 9.0 cm.
• Pan coefficient Kp = 0.6 (assume negligible seepage/leakage).

Concept / Approach:
Pan evaporation E_pan is obtained from the pan water balance:
Initial + Rainfall − Withdrawals − Final = Evaporation from panThen convert to reservoir evaporation using E_reservoir = Kp * E_pan.

Step-by-Step Solution:
Compute E_pan = 8.0 + 5.0 − 3.0 − 9.0 = 1.0 cm.Convert to reservoir evaporation: E_res = Kp * E_pan = 0.6 * 1.0 cm = 0.6 cm.Therefore, E_res = 0.6 cm = 6 mm.

Verification / Alternative check:
Consistency check: Despite rainfall, the final depth increased by only 1 cm after withdrawing 3 cm. The remainder must have evaporated: 1 cm pan evaporation. Applying a pan coefficient less than 1 scales to the reservoir surface, yielding 6 mm.

Why Other Options Are Wrong:

• 2 mm, 4 mm, 8 mm, 10 mm: Do not match the calculated 6 mm after applying Kp = 0.6 to the 1.0 cm pan evaporation.

Common Pitfalls:

• Forgetting to subtract final depth and withdrawals correctly in the water balance.
• Applying the inverse of Kp instead of Kp itself.

Final Answer:
6 mm