Question

A symmetrical trapezoidal canal is 100 km long. The bottom width is 10 m and the side slope is 1 Horizontal : 1 Vertical. The average flow depth in the canal is 2.5 m throughout the month of April. The measurement from a Class-A evaporimeter in the vicinity of the canal indicated an average evaporation rate of 0.5 cm/day in April. The volume of water evaporated from the canal (in m³) in the month of April is close to ........ \(\times 10^3\) (rounded off to 1 decimal place).

Hint:

For evaporation loss in canals, multiply the pan evaporation rate by the area of the free surface to determine the volume of water lost.

Answer 1

Correct Answer: 120

In the month of April, the number of days is 30. The pan evaporation (in terms of depth) is: \[ {Pan evaporation} = 0.5 \, {cm/day} \times 30 \, {days} = 15 \, {cm} = 0.15 \, {m}. \] For the trapezoidal canal of length 10 km, the area of the free surface of the canal is: \[ {Area of free surface of canal} = 15 \, {m} \times 100 \, {km} \times 10^3 = 15 \times 10^5 \, {m}^2. \] The evaporation loss in the canal is given by: \[ {Evaporation loss in canal} = C_p \times {Pan evaporation} = 0.7 \times 0.15 \, {m} = 0.105 \, {m}. \] Thus, the evaporation loss from the canal in terms of volume of water is: \[ {Evaporation loss from canal} = 0.105 \, {m} \times 15 \times 10^5 \, {m}^2 = 157.5 \times 10^3 \, {m}^3. \] Thus, the volume of water evaporated from the canal is \( \boxed{157.5 \times 10^3} \) m³ (rounded to 1 decimal place).