---
title: "Jump relation in horizontal triangular channel"
subject: "Hydraulics"
chapter: 9
key: "semester_4/hydraulics:jump-triangular-channel-froude-relation"
source: https://www.bctengineer.com/q/semester_4/hydraulics/jump-triangular-channel-froude-relation
site: BCT Engineering
---

# Jump relation in horizontal triangular channel

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

- Asked in: 2070 Bhadra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/jump-triangular-channel-froude-relation
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-9

## Question

For a hydraulic jump in a horizontal triangular channel show that $3Fr_1^2 = \frac{r^2(r^3-1)}{r^2-1}$, where $Fr_1^2 = \frac{v_1^2}{gy_1}$ and $r = \frac{y_2}{y_1}$.

## Short answer

From the momentum/specific force equation for a triangular channel, using A=zy^2 and centroid y/3, the Froude number relation becomes 3Fr1^2 = r^2(r^3-1)/(r^2-1). This holds for horizontal channels with negligible bed friction.

## Answer

**Geometry (triangular, side slope $z$):** $A = zy^2$, $T = 2zy$, centroid depth $\bar y = y/3$.

**Specific force** (momentum function):

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

**Momentum equation** for a horizontal channel, $F_1 = F_2$:

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

$$
\frac{Q^2}{gz}\cdot\frac{y_2^2 - y_1^2}{y_1^2y_2^2} = \frac{z}{3}\left(y_2^3 - y_1^3\right)
$$

Put $y_2 = ry_1$:

$$
\frac{Q^2}{gz}\cdot\frac{y_1^2(r^2 - 1)}{y_1^4r^2} = \frac{z\,y_1^3}{3}(r^3 - 1)
$$

$$
\frac{Q^2}{gz^2y_1^5} = \frac{1}{3}\cdot\frac{r^2(r^3 - 1)}{r^2 - 1}
$$

**Froude number at section 1.** For a triangular channel, $V_1 = Q/(zy_1^2)$ and the hydraulic depth is $D_1 = A/T = y_1/2$. The problem defines $Fr_1^2 = V_1^2/(gy_1)$, so

$$
Fr_1^2 = \frac{V_1^2}{gy_1} = \frac{Q^2}{gz^2y_1^5}
$$

which is exactly the left side of the equation above. Hence

$$
\boxed{3Fr_1^2 = \frac{r^2(r^3 - 1)}{r^2 - 1}}
$$

**Check:** for $y_1 = 0.5$, $y_2 = 1.0$ ($r = 2$), the right side is $\dfrac{4 \times 7}{3} = 9.33$, so $Fr_1^2 = 3.11$ and $Fr_1 = 1.76$ (with $Fr_1 = V_1/\sqrt{gy_1}$, which is $\sqrt2$ times smaller than the usual $V_1/\sqrt{gD_1} = 2.49$).

## Check yourself

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

1. What is the starting equation for the proof?
   A. Equal specific energy E1 = E2
   B. Equal specific force F1 = F2 with F = Q^2/(g z y^2) + z y^3/3
   C. Continuity Q = A V only
   D. Bernoulli between the two sections
   Answer: B. Momentum is conserved in a horizontal jump, so the specific forces are equal. (Key point: Starting equation)

2. What is the centroid depth of a triangular section of depth y?
   A. y/2
   B. y
   C. 2y/3
   D. y/3
   Answer: D. For a triangle with apex at the bottom the centroid is y/3 below the surface, so A ybar = z y^3/3. (Key point: Centroid)

3. Using the problem's definition Fr1^2 = V1^2/(g y1), what does Q^2/(g z^2 y1^5) equal?
   A. Fr1^2, since V1 = Q/(z y1^2)
   B. 2 Fr1^2
   C. Fr1^2/3
   D. 1/Fr1^2
   Answer: A. Substituting V1 = Q/(z y1^2) gives exactly Fr1^2, so 3Fr1^2 = r^2(r^3-1)/(r^2-1). (Key point: Link to Fr1)

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