Triangular (V-notch) weir – Right-angle notch with Cd = 0.6 For a right-angled (90°) V-notch weir with coefficient of discharge Cd = 0.6, the discharge can be expressed numerically as Q = C * H^(5/2). What is the value of C?

Introduction:
Flow over a triangular V-notch weir is a standard measurement technique. The integrated theoretical discharge for a V-notch is corrected by Cd and depends on the notch angle. For a right-angle notch (θ = 90°), the expression simplifies neatly, and many textbooks tabulate the numerical coefficient for common Cd values.

Given Data / Assumptions:

• V-notch angle θ = 90° so tan(θ/2) = tan(45°) = 1.
• Coefficient of discharge Cd = 0.6 (assumed constant).
• Head H measured above the notch vertex.

Concept / Approach:

The general triangular-notch formula is Q = (8/15) * Cd * sqrt(2 * g) * tan(θ/2) * H^(5/2). For θ = 90°, tan(45°) = 1, giving Q = (8/15) * Cd * sqrt(2 * g) * H^(5/2). Using g ≈ 9.81 m/s^2 yields sqrt(2 * g) ≈ 4.429; with (8/15) ≈ 0.5333 and Cd = 0.6, the numerical coefficient is 0.5333 * 0.6 * 4.429 ≈ 1.417.

Step-by-Step Solution:

Verification / Alternative check:

If Cd were unity, the coefficient would be (8/15) * 4.429 ≈ 2.36. Reducing by Cd = 0.6 gives 1.417, consistent with the calculation.

Why Other Options Are Wrong:

0.417 and 0.1417: Too small; would imply unrealistically low Cd or g.4.171 and 7.141: Too large; would correspond to Cd > 1 or incorrect constants.

Common Pitfalls:

Using degrees vs radians incorrectly or forgetting the tan(θ/2) factor. Also, mixing H^(3/2) (rectangular weirs) with H^(5/2) (triangular weirs) is a frequent mistake.

Final Answer:

1.417 H^(5/2)