Water is flowing over a sharp-crested rectangular weir of width 50cm into a tank with cross-sectional area 0.6m2. After a period of 30s the depth of water in the tank is 1.4m. Assuming a discharge coefficient of 0.9, determine the height of the water above the weir. If the rectangular weir is replaced by a 900 notch weir with the same head and a discharge coefficient of 0.8, calculate the depth increase of the water in the tank after 30s.

Sharp-Crested Weir Flow Analysis

Problem Statement

Water is flowing over a sharp-crested rectangular weir of width 50cm into a tank with cross-sectional area 0.6m². After a period of 30s the depth of water in the tank is 1.4m. Assuming a discharge coefficient of 0.9, determine the height of the water above the weir. If the rectangular weir is replaced by a 90° V-notch weir with the same head and a discharge coefficient of 0.8, calculate the depth increase of the water in the tank after 30s.

Given Data

Width of rectangular weir (L)	50 cm = 0.5 m
Tank cross-sectional area (A)	0.6 m²
Time period (t)	30 s
Water depth in tank	1.4 m
Discharge coefficient for rectangular weir (C_d1)	0.9
Discharge coefficient for V-notch weir (C_d2)	0.8
V-notch angle (θ)	90°
Acceleration due to Gravity (g)	9.81 m/s²

1. Part (a): Finding Height of Water Above Rectangular Weir

Volume of water collected in tank:
Volume = A × Depth = 0.6 m² × 1.4 m = 0.84 m³

Discharge calculation:
Q = Volume / Time = 0.84 m³ / 30 s = 0.028 m³/s

For a rectangular weir, the discharge equation is:

Q = (2/3) × C_d × L × H^3/2 × √(2g)

Rearranging to solve for H (head):

H = [(Q) / ((2/3) × C_d × L × √(2g))]^2/3

Substituting values:

H = [0.028 / ((2/3) × 0.9 × 0.5 × √(2 × 9.81))]^2/3

Evaluating step by step:

√(2g) = √(2 × 9.81) = 4.429

(2/3) × C_d × L × √(2g) = (2/3) × 0.9 × 0.5 × 4.429 = 1.329

H = [0.028 / 1.329]^2/3 = [0.021]^2/3 = 0.076 m

The height of water above the rectangular weir is 0.076 m (7.6 cm)

2. Part (b): V-Notch Weir Depth Analysis

For a V-notch weir with angle θ, the discharge equation is:

Q = (8/15) × C_d × tan(θ/2) × H^5/2 × √(2g)

Substituting values with H = 0.076 m (same head as rectangular weir):

Q = (8/15) × 0.8 × tan(45°) × (0.076)^5/2 × √(2 × 9.81)

Evaluating step by step:

tan(45°) = 1

(0.076)^5/2 = (0.076)² × √(0.076) = 0.00578 × 0.2757 = 0.00159

(8/15) × 0.8 × 1 × 0.00159 × 4.429 = 0.003 m³/s

Volume of water flowing in 30 seconds with V-notch weir:
Volume = Q × Time = 0.003 m³/s × 30 s = 0.09 m³

Depth increase in tank:
Depth increase = Volume / Tank area = 0.09 m³ / 0.6 m² = 0.15 m

The depth increase in the tank after 30 seconds with the V-notch weir is 0.15 m (15 cm)

Conclusion

Based on the analysis of the given weir configurations:

(a) The height of water above the rectangular weir required to produce a tank depth of 1.4 m after 30 seconds is 0.076 m (7.6 cm).

(b) When the rectangular weir is replaced with a 90° V-notch weir with the same head (0.076 m) and a discharge coefficient of 0.8, the depth increase in the tank after 30 seconds is reduced to 0.15 m.

This demonstrates how different weir configurations significantly affect flow rates even when operating under the same head conditions.