---
title: "Discharge from hydraulic jump depths 1.5 m to 4 m"
subject: "Hydropower Engineering"
chapter: 4
key: "semester_7/hydropower_engineering:stilling-basin-hydraulic-jump-depth-change-120m"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/stilling-basin-hydraulic-jump-depth-change-120m
site: BCT Engineering
---

# Discharge from hydraulic jump depths 1.5 m to 4 m

Hydropower Engineering · Chapter 4: Headworks of Storage Plants

- Asked in: 2076 Ashwin, 3 marks
- Video: [Part-One: Example on Hydraulic jump stilling basin, Specific energy, length of hydraulic jump](https://www.youtube.com/watch?v=gcPfQwEzdIM) — J-HydroMech Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/stilling-basin-hydraulic-jump-depth-change-120m
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4

## Question

It is proposed to form a hydraulic jump in a stilling basin to dissipate the energy below a spillway. Depth of flow changes from 1.5 m to 4 m. Calculate the discharge over the spillway if the length of the crest is 120 m.

## Short answer

From conjugate depths y1=1.5 m and y2=4 m, Fr1=2.211, unit discharge q=12.72 m³/s/m, total discharge over 120 m crest is 1527 m³/s.

## Answer

**Given**: conjugate depths of the jump $y_1=1.5$ m and $y_2=4$ m, crest length $L=120$ m.

#### Step 1: Froude number before the jump
From the jump equation
$$
\frac{y_2}{y_1}=\frac12\left(\sqrt{1+8Fr_1^2}-1\right)\Rightarrow Fr_1^2=\frac{1}{2}\frac{y_2}{y_1}\left(\frac{y_2}{y_1}+1\right)
$$
$$
Fr_1^2=\tfrac12(2.667)(3.667)=4.889\ \Rightarrow\ Fr_1=2.211
$$

#### Step 2: Velocity and unit discharge
$$
V_1=Fr_1\sqrt{gy_1}=2.211\times\sqrt{9.81\times1.5}=8.48\ \text{m/s}
$$
$$
q=V_1y_1=8.48\times1.5=12.72\ \text{m}^3/\text{s/m}
$$
(Check: $q=\sqrt{\dfrac{g\,y_1y_2(y_1+y_2)}{2}}=12.72$ m³/s/m.)

#### Step 3: Discharge over the spillway
$$
Q=qL=12.72\times120=1526.7\ \text{m}^3/\text{s}
$$

**Answer**: discharge over the spillway $Q\approx1527$ m³/s. The energy lost in the jump is $\dfrac{(y_2-y_1)^3}{4y_1y_2}=0.651$ m.

## Check yourself

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

1. Given y1 = 1.5 m and y2 = 4 m, how is Fr1 found?
   A. Fr1 = y2/y1 = 2.67
   B. Fr1^2 = (1/2)(y2/y1)(y2/y1 + 1) = 4.889, Fr1 = 2.21
   C. Fr1^2 = y2/y1 - 1 = 1.67
   D. Fr1 = sqrt(g y1) = 3.84
   Answer: B. Rearranging the conjugate depth relation y2/y1 = 0.5(sqrt(1+8Fr1^2) - 1) gives the Fr1 expression. (Key point: Froude number from conjugate depths)

2. Once Fr1 is known, how is the unit discharge q obtained?
   A. q = V1 y1 with V1 = Fr1 sqrt(g y1), about 12.72 m^3/s/m
   B. q = V1 y2 using the downstream depth
   C. q = Fr1 x y1
   D. q = Q x L
   Answer: A. Velocity before the jump times the depth before the jump gives discharge per metre width. (Key point: Unit discharge)

3. How is the total discharge over the spillway obtained?
   A. Q = q / L = 0.106 m^3/s
   B. Q = q x y2 = 50.9 m^3/s
   C. Q = q x L = 12.72 x 120, about 1527 m^3/s
   D. Q = V1 x y2 x L
   Answer: C. Total discharge is unit discharge multiplied by crest length. (Key point: Total discharge)

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