---
title: "M1 profile upstream of a free outfall"
subject: "Hydraulics"
chapter: 8
key: "semester_4/hydraulics:m1-profile-free-outfall-bresse"
source: https://www.bctengineer.com/q/semester_4/hydraulics/m1-profile-free-outfall-bresse
site: BCT Engineering
---

# M1 profile upstream of a free outfall

Hydraulics · Chapter 8: Non-uniform gradually varied flow (GVF)

- Asked in: 2077 Chaitra, 2+8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/m1-profile-free-outfall-bresse
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-8

## Question

What is M1 Profile? A rectangular canal is 10 m wide and carries a flow of 50 m³/s. The bottom and sides of the canal are concrete-lined, the longitudinal bottom slope is 0.0006 and the canal ends in a free outfall. If the flow depth is critical at a distance of $4y_c$ upstream of the fall, what is the depth of flow 2 km upstream of the fall? Use either direct integration method (Bresse's method) taking three steps or Standard step method.

## Short answer

M1 is a backwater curve on a mild slope with depth above normal depth, increasing downstream, asymptotic to normal depth upstream and horizontal downstream. For the free outfall the profile is M2; the depth 2 km upstream of the fall is about 2.14 m.

## Answer

#### M1 profile

On a **mild slope** ($y_n > y_c$), the **M1** profile is the backwater curve in zone 1, where $y > y_n > y_c$. Its depth increases downstream. It is asymptotic to the normal depth line upstream and approaches a horizontal water surface downstream. It occurs upstream of dams, weirs and reservoirs. (The profile in the numerical part, upstream of a free outfall, is the drawdown curve **M2**, because the depth falls from $y_n$ to $y_c$.)

#### Numerical part

**Given:** $b = 10\ \text{m}$, $Q = 50\ \text{m}^3/\text{s}$, $S_0 = 0.0006$, free outfall. Concrete lining: $n = 0.014$ assumed.

**Depths** ($q = 5\ \text{m}^2/\text{s}$):

$$
y_c = \left(\frac{5^2}{9.81}\right)^{1/3} = 1.366\ \text{m},\qquad y_n = 2.169\ \text{m}\ (\text{Manning, } R = A/P)
$$

Since $y_n > y_c$ the slope is mild, and the profile is **M2**. The depth is $y_c$ at $4y_c = 5.46$ m upstream of the fall, which is the control section. The point asked about is 2000 m upstream of the fall, i.e. $2000 - 5.46 = 1994.5\ \text{m}$ upstream of the control.

#### Bresse's method (three depth steps)

$$
x = \frac{y_n}{S_0}\left[u - \left(1 - \left(\frac{y_c}{y_n}\right)^3\right)\phi(u)\right],\quad u = \frac{y}{y_n}
$$

$$
\phi(u) = \frac{1}{6}\ln\frac{u^2 + u + 1}{(u-1)^2} + \frac{1}{\sqrt3}\tan^{-1}\frac{2u + 1}{\sqrt3}
$$

$\dfrac{y_n}{S_0} = 3614.5\ \text{m}$ and $1 - (y_c/y_n)^3 = 1 - 0.6298^3 = 0.7502$.

Three equal depth steps from $y_c$ upwards (trial for the end depth):

| $y$ (m) | $u$ | $\phi(u)$ | $x$ from control (m) |
|---|---|---|---|
| 1.366 | 0.6298 | 0.9783 | 0 |
| 1.623 | 0.7483 | 1.1560 | -53.5 |
| 1.880 | 0.8668 | 1.4131 | -322.5 |
| 2.137 | 0.9852 | 2.1877 | -1994.5 |

The last row is the depth at which the distance equals 1994.5 m upstream. A fine numerical integration gives 2.135 m, which confirms it.

**Answer: Depth of flow 2 km upstream of the fall $\approx 2.14\ \text{m}$** (very close to $y_n = 2.17$ m).

## Check yourself

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

1. What defines an M1 profile?
   A. Mild slope with y between yc and yn
   B. Steep slope with y above yc
   C. Mild slope with y > yn > yc, a backwater curve
   D. Mild slope with y < yc
   Answer: C. M1 lies in zone 1 of a mild slope and rises downstream, as upstream of dams. (Key point: M1 definition)

2. What are the asymptotic limits of M1?
   A. Normal depth line upstream and horizontal water surface downstream
   B. Critical depth upstream and normal depth downstream
   C. Bed upstream and yc downstream
   D. Vertical at both ends
   Answer: A. M1 tends to yn far upstream and flattens to a horizontal surface in the pool downstream. (Key point: Asymptotes of M1)

3. Why is the numerical profile upstream of the free outfall M2 rather than M1?
   A. Depth falls from yn to yc, with yn > y > yc on a mild slope
   B. The slope is steep
   C. Depth is above yn
   D. Depth is below yc
   Answer: A. A free outfall on a mild slope draws the water down, giving the drawdown M2 curve. (Key point: Profile at free outfall)

4. Where is the computation started and measured from?
   A. From the normal depth section
   B. From the control section where depth is yc, 4yc upstream of the fall; the 2 km point is 2000 - 4yc from it
   C. From the fall with depth zero
   D. From 2 km upstream
   Answer: B. Depth is yc at 4yc = 5.46 m from the fall, so the target is 1994.5 m upstream of that control. (Key point: Control location)

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