---
title: "Percentage error in notch discharge"
subject: "Fluid Mechanics"
chapter: 6
key: "semester_3/fluid_mechanics:notch-head-error-percentage"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/notch-head-error-percentage
site: BCT Engineering
---

# Percentage error in notch discharge

Fluid Mechanics · Chapter 6: Flow measurement

- Asked in: 2072 Chaitra, 5 marks
- Video: [Effect on discharge over a notch or weir due to error in measurement of head](https://www.youtube.com/watch?v=i-A0mfXgT5E) — Pavika fluid says Tawi
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/notch-head-error-percentage
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-6

## Question

A sharp edged rectangular notch 30 cm long and a right-angled triangular notch are to be used alternatively for gauging a discharge estimated to be about 20 lit/s. Find in each cases the percentage error in computing the discharge that would be introduced by an error of 1 mm in observing the head over the notch.

## Short answer

For a 1 mm head error, percentage discharge error is n dH/H. Rectangular notch: H=0.110 m, n=1.5, error 1.37%. Triangular notch: H=0.182 m, n=2.5, error 1.37%. Both are about 1.37%.

## Answer

**Given:** $Q = 20$ L/s $= 0.02$ m³/s; error in head $dH = 1$ mm $= 0.001$ m. The coefficients are not given, so take $C_d = 0.62$ for the rectangular notch and $C_d = 0.60$ for the right-angled triangular notch.

**Percentage error formula:** since $Q \propto H^n$,

$$
\frac{dQ}{Q} = n\frac{dH}{H}
$$

#### Rectangular notch ($L = 0.30$ m, $n = 3/2$)

$$
H = \left[\frac{Q}{\frac{2}{3}C_dL\sqrt{2g}}\right]^{2/3} = \left[\frac{0.02}{\frac{2}{3}(0.62)(0.3)(4.4294)}\right]^{2/3} = 0.1099\ \text{m}
$$

$$
\frac{dQ}{Q} = 1.5\times\frac{0.001}{0.1099} = 0.01365 = 1.37\ \%
$$

#### Right-angled triangular notch ($\theta = 90^\circ$, $\tan 45^\circ = 1$, $n = 5/2$)

$$
H = \left[\frac{Q}{\frac{8}{15}C_d\sqrt{2g}}\right]^{2/5} = \left[\frac{0.02}{\frac{8}{15}(0.6)(4.4294)}\right]^{0.4} = 0.1819\ \text{m}
$$

$$
\frac{dQ}{Q} = 2.5\times\frac{0.001}{0.1819} = 0.01374 = 1.37\ \%
$$

**Answer: the error in discharge is about 1.37 % for the rectangular notch and about 1.37 % for the triangular notch.** Both give nearly the same error for 1 mm error in head; the result depends only slightly on the assumed $C_d$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which relation links head error to discharge error?
   A. dQ/Q = dH/(nH)
   B. dQ/Q = dH/H
   C. dQ/Q = n dH
   D. dQ/Q = n dH/H
   Answer: D. Since Q is proportional to H^n, differentiating gives dQ/Q = n dH/H. (Key point: Error relation)

2. What are the exponents n for rectangular and right-angled triangular notches?
   A. 3/2 and 5/2
   B. 1/2 and 5/2
   C. 5/2 and 3/2
   D. 3/2 and 3/2
   Answer: A. Rectangular Q is proportional to H^(3/2) and triangular to H^(5/2). (Key point: Exponents)

3. How is the head found for a given Q?
   A. Assume H = 0.1 m
   B. Use the 1 mm error
   C. Rearrange the notch formula for H
   D. Use Q = a sqrt(2gH)
   Answer: C. H is solved from Q = formula, giving 0.1099 m (rectangular) and 0.1819 m (triangular). (Key point: Head from discharge)

4. What are the percentage errors?
   A. 1.5% and 2.5%
   B. About 1.37% for both
   C. About 2.5% for both
   D. 0.5% and 1%
   Answer: B. 1.5 x 0.001/0.1099 = 1.37% and 2.5 x 0.001/0.1819 = 1.37%. (Key point: Result)

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- [Notch discharge coefficient error](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Anotch-coefficient-discharge-error) — Fluid Mechanics, chapter 6

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