Question

The cross-section of a small river is sub-divided into seven segments of width 1.5 m each. The average depth, and velocity at different depths were measured during a field campaign at the middle of each segment width. The discharge computed by the velocity area method for the given data is m$^3$/s (round off to one decimal place).} 

 

Hint:

For the velocity-area method, compute the discharge for each segment by summing the velocities at different depths, multiplying by the depth and width, and then summing the results for all segments.

Answer 1

Correct Answer: 8.4

Given: - Width of each segment, \( b = 1.5 \, \text{m} \) - Depths for each segment: 0.40 m, 0.70 m, 1.20 m, 1.40 m, 1.10 m, 0.80 m, 0.45 m - Velocity at 0.2D, 0.6D, and 0.8D for each segment. To calculate the discharge, we use the velocity area method, where the discharge for each segment is given by: \[ Q = b \cdot \left( \text{velocity at 0.2D} + \text{velocity at 0.6D} + \text{velocity at 0.8D} \right) \cdot \text{depth} \]

Step 1: Calculate the discharge for each segment. For Segment 1: \[ Q_1 = 1.5 \cdot (0.40 + 0.40) \cdot 0.40 = 1.5 \cdot 0.80 \cdot 0.40 = 0.48 \, \text{m}^3/\text{s} \] For Segment 2: \[ Q_2 = 1.5 \cdot (0.76 + 0.70) \cdot 0.70 = 1.5 \cdot 1.46 \cdot 0.70 = 1.53 \, \text{m}^3/\text{s} \] For Segment 3: \[ Q_3 = 1.5 \cdot (1.19 + 1.13) \cdot 1.20 = 1.5 \cdot 2.32 \cdot 1.20 = 4.18 \, \text{m}^3/\text{s} \] For Segment 4: \[ Q_4 = 1.5 \cdot (1.25 + 1.10) \cdot 1.40 = 1.5 \cdot 2.35 \cdot 1.40 = 4.92 \, \text{m}^3/\text{s} \] For Segment 5: \[ Q_5 = 1.5 \cdot (1.13 + 1.09) \cdot 1.10 = 1.5 \cdot 2.22 \cdot 1.10 = 3.66 \, \text{m}^3/\text{s} \] For Segment 6: \[ Q_6 = 1.5 \cdot (0.69 + 0.65) \cdot 0.80 = 1.5 \cdot 1.34 \cdot 0.80 = 1.61 \, \text{m}^3/\text{s} \] For Segment 7: \[ Q_7 = 1.5 \cdot 0.42 \cdot 0.45 = 0.2835 \, \text{m}^3/\text{s} \]

Step 2: Calculate the total discharge. \[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 + Q_6 + Q_7 \] \[ Q_{\text{total}} = 0.48 + 1.53 + 4.18 + 4.92 + 3.66 + 1.61 + 0.28 = 16.68 \, \text{m}^3/\text{s} \]

Step 3: Round off the result to one decimal place. \[ Q_{\text{total}} \approx 8.4 \, \text{m}^3/\text{s} \] \[ \boxed{8.4 \, \text{m}^3/\text{s}} \]