Water is flowing in a rectangular channel of 1m wide and 0.8m deep. Find the discharge over a rectangular weir of crest length 60cm if the head of water over the crest of weir is 30cm and water from channel flows over the weir. Take Cd = 0.62.

Discharge Over a Rectangular Weir

Problem Statement

Water is flowing in a rectangular channel of width 1 m and depth 0.8 m. Find the discharge over a rectangular weir with a crest length of 0.6 m, given that the head of water over the weir is 0.3 m. Consider the velocity of approach.
Take C_d = 0.62.

Given Data

Width of Channel	1 m
Depth of Water	0.8 m
Area of Channel (A)	1 × 0.8 = 0.8 m²
Crest Length of Weir (L)	0.6 m
Head of Water Over Weir (H₁)	0.3 m
Coefficient of Discharge (C_d)	0.62
Acceleration due to Gravity (g)	9.81 m/s²

1. Discharge Without Velocity of Approach

The discharge over a rectangular weir without considering velocity of approach is given by:

Q = (2/3) C_d L √(2g) H₁^3/2

Substituting values:

Q = (2/3) × 0.62 × 0.6 × √(2 × 9.81) × (0.3)^3/2

Calculation gives:

Q ≈ 0.18 m³/s

2. Velocity of Approach

Velocity of approach is given by:

V_a = Q / A = 0.18 / 0.8 = 0.225 m/s

Velocity head is calculated as:

h_a = V_a² / (2g) = (0.225)² / (2 × 9.81) = 0.00258 m

3. Discharge Considering Velocity of Approach

The corrected discharge equation is:

Q = (2/3) C_d L √(2g) [(H₁ + h_a)^3/2 – h_a^3/2]

Substituting values:

Q = (2/3) × 0.62 × 0.6 × √(2 × 9.81) × [(0.3 + 0.00258)^3/2 – (0.00258)^3/2]

Calculation gives:

Q ≈ 0.183 m³/s

Conclusion

The discharge over the rectangular weir without velocity of approach is 0.18 m³/s.
Considering velocity of approach, the corrected discharge is 0.183 m³/s.