---
title: "Hydraulic jump in 90° triangular channel"
subject: "Hydraulics"
chapter: 9
key: "semester_4/hydraulics:jump-90-degree-triangular-channel-flow-rate"
source: https://www.bctengineer.com/q/semester_4/hydraulics/jump-90-degree-triangular-channel-flow-rate
site: BCT Engineering
---

# Hydraulic jump in 90° triangular channel

Hydraulics · Chapter 9: Non-uniform rapidly varied flow (RVF)

- Asked in: 2077 Chaitra, 6+2 marks
- Video: [Hydraulic Jump In Triangular Channel](https://www.youtube.com/watch?v=0ZV-IkOM1cg) — Samrendra Singh
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/jump-90-degree-triangular-channel-flow-rate
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-9

## Question

A hydraulic jump occurs in a 90° triangular channel. Derive an equation relating two depths and the flow rate. If the depths before and after the jump in the above channel are 0.5 m and 1.0 m determine the flow rate and obtain Froude numbers before and after the jump.

## Short answer

Using specific force equality in a 90-degree triangular channel gives Q^2 = (g/3) y1^2 y2^2 (y2^3-y1^3)/(y2^2-y1^2). For y1=0.5 m and y2=1.0 m, Q=0.98 m3/s; Froude numbers are Fr1=2.49 before and Fr2=0.44 after the jump.

## Answer

For a 90° triangular channel the side slopes are 1H : 1V, so $z = 1$.

#### Derivation

Geometry at depth $y$: $A = y^2$, $T = 2y$, and the centroid of the area is at $\bar y = y/3$ below the surface.

Specific force: 

$$
F = \frac{Q^2}{gA} + A\bar y = \frac{Q^2}{gy^2} + \frac{y^3}{3}
$$

For a jump in a horizontal channel, $F_1 = F_2$:

$$
\frac{Q^2}{g}\left(\frac{1}{y_1^2} - \frac{1}{y_2^2}\right) = \frac{y_2^3 - y_1^3}{3}
$$

$$
\frac{Q^2}{g}\cdot\frac{y_2^2 - y_1^2}{y_1^2y_2^2} = \frac{y_2^3 - y_1^3}{3}
$$

$$
\boxed{Q^2 = \frac{g}{3}\cdot\frac{y_1^2y_2^2\,(y_2^3 - y_1^3)}{y_2^2 - y_1^2}}
$$

#### Flow rate

With $y_1 = 0.5$ m and $y_2 = 1.0$ m:

$$
Q^2 = \frac{9.81}{3}\cdot\frac{(0.25)(1)(1 - 0.125)}{1 - 0.25} = 3.27 \times 0.29167 = 0.9538
$$

$$
Q = 0.977\ \text{m}^3/\text{s}
$$

#### Froude numbers

The hydraulic depth is $D = A/T = y/2$.

$$
V_1 = \frac{Q}{y_1^2} = \frac{0.9766}{0.25} = 3.906\ \text{m/s},\qquad Fr_1 = \frac{3.906}{\sqrt{9.81 \times 0.25}} = 2.49
$$

$$
V_2 = \frac{0.9766}{1.0} = 0.977\ \text{m/s},\qquad Fr_2 = \frac{0.977}{\sqrt{9.81 \times 0.5}} = 0.44
$$

**Answer: $Q \approx 0.98\ \text{m}^3/\text{s}$; $Fr_1 \approx 2.49$ (supercritical) and $Fr_2 \approx 0.44$ (subcritical).**

## Check yourself

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

1. What is the specific force for a 90 degree triangular channel (z = 1)?
   A. Q^2/(g y^2) + y^3/3
   B. Q^2/(g y) + y^2/2
   C. Q^2/(g y^2) + y^2/2
   D. Q^2/(2g y^2) + y^3
   Answer: A. A = y^2 and the centroid is y/3 below the surface, so A ybar = y^3/3. (Key point: Specific force triangular)

2. What principle gives the relation between Q and the two depths?
   A. Equal specific energy at the two depths
   B. Equal specific force F1 = F2 (momentum)
   C. Manning's equation
   D. Continuity only
   Answer: B. Momentum is conserved across the jump in a horizontal channel. (Key point: Equation for Q)

3. What hydraulic depth is used for Fr in this channel?
   A. D = y
   B. D = y/2 (A/T)
   C. D = y/3
   D. D = 2y
   Answer: B. A = y^2 and T = 2y, so D = y/2, giving Fr1 = 2.49 and Fr2 = 0.44. (Key point: Froude number)

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