Question

The top width and depth of flow in a triangular channel are 3.5 m and 2 m respectively. The velocities at 0.4 m and 1.6 m vertically below the water surface of the channel are 0.8 m/s and 0.4 m/s respectively. Using two point method of velocity measurement, the discharge in the channel is

• A. $1.3m^3/s$
• B. $1.8m^3/s$
• C. $2.1m^3/s$
• D. $2.7m^3/s$

Answer 1

The correct option is C $2.1m^3/s$


Depth of flow in the triangular channel, y = 2m
Area of cross-section of the channel,
$A=\frac{1}{2}\times 3.5\times 2=3.5m^2$

Velocity of flow at 0.4 m depth from the water surface of channel,
$V_{0.2y}=0.8m/s$

Velocity of flow at 1.6 m depth from the water surface of channel,
$V_{0.8y}=0.4m/s$
$\therefore \text{Average velocity}\overline{V}=\frac{V_{0.2y}+V_{0.8}}{2}=\frac{0.8+0.4}{2}=0.6m/s$

Discharge,
$Q=A\overline{V}=3.5\times 0.6=2.1m^3/s$