--- title: "Discharge over rectangular and V-notches" subject: "Fluid Mechanics" chapter: 7 key: "semester_4/fluid_mechanics_mechanical:rectangular-weir-and-v-notch-discharge" source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/rectangular-weir-and-v-notch-discharge site: BCT Engineering --- # Discharge over rectangular and V-notches Fluid Mechanics · Chapter 7: Flow Measurement - Asked in: year not given, 6 marks - Video: [Determination of Coefficient of discharge for Rectangular and Triangular Notches | Flow over Notch](https://www.youtube.com/watch?v=8a9hKrAAXUQ) — The Genius Online Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/rectangular-weir-and-v-notch-discharge - Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-7 ## Question Derive the expressions for discharge over a rectangular notch (sharp-crested weir) and a triangular (V) notch. State the advantages of a V-notch for measuring small discharges. Find the discharge over a rectangular weir of crest length 1.5 m under a head of 0.25 m ($C_d = 0.62$), and over a 90° V-notch under a head of 0.15 m ($C_d = 0.6$). ## Short answer Rectangular weir: Q = (2/3) Cd L sqrt(2g) H^(3/2) = 0.343 m³/s. 90° V-notch: Q = (8/15) Cd sqrt(2g) H^(5/2) = 12.35 L/s. The V-notch provides accurate measurement for small discharges. ## Answer #### Rectangular notch Let $L$ be the crest length and $H$ the head over the crest. Consider a horizontal strip of thickness $dh$ at depth $h$ below the free surface. ```text ~~~~~~~~~~~~~~~~~~~~~~ free surface | ^ | | h strip dh | ______|_________ | | L | H |_|_______________|_ crest ``` Velocity through the strip: $\sqrt{2gh}$. Area of strip: $L\,dh$. $$ dQ = C_d\,L\,dh\,\sqrt{2gh} $$ $$ Q = C_d\,L\sqrt{2g}\int_0^H h^{1/2}dh = \frac23\,C_d\,L\sqrt{2g}\;H^{3/2} $$ If the velocity of approach $V_a$ is significant, $H$ is replaced by $(H + h_a)$ with $h_a = V_a^2/2g$ and the equation becomes $Q = \tfrac23 C_d L\sqrt{2g}\,[(H + h_a)^{3/2} - h_a^{3/2}]$. For end contractions, Francis's formula uses an effective length $L - 0.1nH$. #### Triangular (V) notch Let the notch angle be $\theta$. The width of the strip at depth $h$ is $2(H - h)\tan(\theta/2)$. $$ dQ = C_d\,2(H - h)\tan\frac{\theta}{2}\,\sqrt{2gh}\,dh $$ $$ Q = 2C_d\tan\frac{\theta}{2}\sqrt{2g}\int_0^H (H - h)h^{1/2}dh = \frac{8}{15}\,C_d\tan\frac{\theta}{2}\sqrt{2g}\;H^{5/2} $$ #### Advantage of V-notch - For small discharges, the head $H$ is large enough to be measured accurately, since $Q \propto H^{5/2}$. - A single notch covers a wide range of flows. - No end contractions, and no ventilation problem. #### Numerical **Rectangular weir:** $$ Q = \frac23 \times 0.62 \times 4.429 \times 1.5 \times (0.25)^{1.5} = \frac23 \times 0.62 \times 4.429 \times 1.5 \times 0.125 = 0.343\ \text{m}^3/\text{s} $$ **90° V-notch** ($\tan 45^\circ = 1$): $$ Q = \frac{8}{15} \times 0.6 \times 1 \times 4.429 \times (0.15)^{2.5} = 0.01235\ \text{m}^3/\text{s} = 12.35\ \text{L/s} $$ **Answer: rectangular weir $Q = 0.343\ \text{m}^3/\text{s}$; V-notch $Q = 12.35\ \text{L/s}$** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the discharge over a rectangular notch obtained? A. Integrating dQ = C_d L dh sqrt(2gh) from 0 to H, giving (2/3) C_d L sqrt(2g) H^(3/2) B. Using Q = C_d L H sqrt(2gH) with no integration C. Integrating with a width varying linearly with h D. Giving Q proportional to H^(5/2) Answer: A. A strip of area L dh at depth h has velocity sqrt(2gh); integration gives (2/3) C_d L sqrt(2g) H^1.5. (Key point: Rectangular notch derivation) 2. For a V-notch, how does Q depend on head H, and why helps for small flows? A. Q proportional to H^(3/2); head is small B. Q proportional to H^(5/2); head stays large enough to measure accurately at small Q C. Q proportional to H; it is linear D. Q is independent of H Answer: B. Q = (8/15) C_d tan(theta/2) sqrt(2g) H^(5/2); a small Q still needs a measurable head. (Key point: V-notch head dependence) 3. What is the width of the strip at depth h in a V-notch of head H? A. L B. 2(H - h) tan(theta/2) C. 2h tan(theta/2) D. H tan(theta) Answer: B. The width decreases toward the apex: 2(H - h) tan(theta/2). (Key point: Strip width in V-notch) 4. For a 90 degree V-notch, H = 0.15 m, C_d = 0.6, what is the right approach and result? A. tan 45 = 1, Q = (8/15)(0.6)(4.429)(0.15)^2.5 = 12.35 L/s B. tan 90 = infinity, not solvable C. Q = (2/3)(0.6)(4.429)(0.15)^1.5 = 0.15 m^3/s D. tan 45 = 1, Q = 0.343 m^3/s Answer: A. For theta = 90 degrees, tan(theta/2) = 1, giving 0.01235 m^3/s. 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