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

Evaporation Loss from a Trapezoidal Canal

Step 1: Compute Total Pan Evaporation

Given:

• Number of days in April = 30
• Pan evaporation rate = 0.5 cm/day

Total pan evaporation:

\[ {Pan evaporation} = 0.5 \times 30 = 15 \, \text{cm} = 0.15 \, \text{m}. \]

Step 2: Compute Free Surface Area of the Canal

Given:

• Length of the trapezoidal canal = 10 km
• Average free surface width = 15 m

Surface area:

\[ {Area} = 15 \times 10 \times 1000 = 15 \times 10^5 \, \text{m}^2. \]

Step 3: Compute Effective Evaporation Depth

Evaporation loss considering pan coefficient:

\[ {Evaporation loss} = C_p \times \text{Pan evaporation} \]

Given \( C_p = 0.7 \), we calculate:

\[ {Evaporation loss} = 0.7 \times 0.15 = 0.105 \, \text{m}. \]

Step 4: Compute Volume of Evaporated Water

\[ {Volume} = \text{Evaporation loss} \times \text{Surface area} \]

Substituting values:

\[ {Volume} = 0.105 \times 15 \times 10^5 = 157.5 \times 10^3 \, \text{m}^3. \]

Final Answer:

Correct Answer: \( \mathbf{157.5 \times 10^3} \) m³ (rounded to 1 decimal place).