---
title: "Minimum length of concrete apron of stilling basin"
subject: "Hydropower Engineering"
chapter: 4
key: "semester_7/hydropower_engineering:stilling-basin-apron-length-115cumec"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/stilling-basin-apron-length-115cumec
site: BCT Engineering
---

# Minimum length of concrete apron of stilling basin

Hydropower Engineering · Chapter 4: Headworks of Storage Plants

- Asked in: 2075 Chaitra, 10 marks
- 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-apron-length-115cumec
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4

## Question

Estimate the minimum length of the concrete apron ($S_0 = 0.001$) for a stilling basin downstream from an overflow spillway. The spillway crest is 15 m long and consider a discharge of 115 m³/s. Manning's roughness factor n = 0.025. Assume the stilling basin is the same width as the spillway crest. Assume any other suitable data if necessary. Refer to the figure below.
[Figure: spillway of height 4 m with a stilling basin showing $y_c$, $y_0$, $y_1$, $y_2$, distance $\Delta x$ and jump length $L_j$; below it a curve of $L_j / y_2$ (4 to 6) against Froude number (2 to 5).]

## Short answer

Minimum concrete apron length is 33 m: distance to jump start 14.4 m plus jump length 18.5 m.

## Answer

**Given**: crest length $B=15$ m, $Q=115$ m³/s, height of spillway $P=4$ m, $S_0=0.001$, $n=0.025$, basin width = 15 m. Losses on the spillway face are assumed negligible.

#### Step 1: Unit discharge and critical depth
$$
q=\frac{115}{15}=7.667\ \text{m}^3/\text{s/m},\qquad y_c=\left(\frac{q^2}{g}\right)^{1/3}=1.816\ \text{m}
$$
Specific energy at the crest above the apron: $E_0=P+1.5y_c=4+1.5(1.816)=6.72$ m.

#### Step 2: Supercritical depth $y_1$ at the toe
$$
y_1+\frac{q^2}{2gy_1^2}=6.72\Rightarrow y_1=0.706\ \text{m},\quad V_1=10.87\ \text{m/s},\quad Fr_1=4.13
$$
$$
y_2=\frac{y_1}{2}\left(\sqrt{1+8Fr_1^2}-1\right)=3.78\ \text{m}
$$

#### Step 3: Tail water depth $y_0$ (Manning, rectangular 15 m wide)
$$
Q=\frac1nAR^{2/3}S_0^{1/2}\ \Rightarrow\ y_0=3.43\ \text{m},\quad V=2.24\ \text{m/s},\quad Fr_0=0.386
$$
Since $y_0=3.43\ \text{m}<y_2=3.78\ \text{m}$, the jump does not form at the toe. It moves downstream by a distance $\Delta x$ along the apron, to the section where the supercritical depth $y_1'$ has the conjugate depth $y_0$:
$$
y_1'=\frac{y_0}{2}\left(\sqrt{1+8Fr_0^2}-1\right)=0.823\ \text{m}\quad(Fr_1'=3.28)
$$

#### Step 4: Distance $\Delta x$ (gradually varied flow)
Friction makes the depth rise from $y_1=0.706$ m to $y_1'=0.823$ m. By the direct step method, $\Delta x=\dfrac{\Delta E}{S_0-\overline{S_f}}$, with $S_f=\left(\dfrac{nV}{R^{2/3}}\right)^2$, summed over many small steps:
$$
\Delta x=14.4\ \text{m}
$$

#### Step 5: Length of jump
For $Fr_1'=3.28$, the curve of $L_j/y_2$ against $Fr$ gives $L_j/y_2\approx5.4$ (here $y_2=y_0$ is the tail water depth):
$$
L_j=5.4\times3.43=18.5\ \text{m}
$$

#### Step 6: Minimum length of apron
$$
L=\Delta x+L_j=14.4+18.5=32.9\ \text{m}
$$

**Answer**: the minimum length of the concrete apron is about **33 m** (distance to the start of the jump 14.4 m + jump length 18.5 m).

## Check yourself

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

1. In the apron problem, the tail water depth y0 = 3.43 m is less than the conjugate depth y2 = 3.78 m. What happens to the jump?
   A. It moves downstream along the apron by a distance dx
   B. It forms at the toe as designed
   C. It is drowned and moves upstream onto the spillway face
   D. It forms exactly where the tail water meets the basin wall
   Answer: A. Because y0 < y2 the tail water cannot hold the jump at the toe, so it is swept downstream until the supercritical depth has a conjugate equal to y0. (Key point: Tail water vs conjugate depth)

2. How is the supercritical depth y1 at the toe found?
   A. Set y1 equal to the critical depth yc
   B. Solve y1 + q^2/(2g y1^2) = E0 where E0 = P + 1.5 yc
   C. Use Manning's equation with S0 = 0.001
   D. Take y1 = y2/2 from the tail water
   Answer: B. Energy is conserved down the spillway face (losses neglected), so the toe depth satisfies the specific energy equation with E0 = 4 + 1.5(1.816) = 6.72 m. (Key point: Supercritical depth at toe)

3. How is the distance dx between the toe and the start of the jump obtained?
   A. From the jump length curve L/y2
   B. By dividing the discharge by the crest width
   C. By gradually varied flow (direct step) as depth rises from y1 = 0.706 m to y1' = 0.823 m
   D. By Manning's formula for uniform flow at the tail water
   Answer: C. Friction raises the supercritical depth gradually from y1 to y1' (the depth whose conjugate is y0); the direct step method sums dx = dE/(S0 - Sf). (Key point: Distance dx to the jump)

4. How is the minimum apron length obtained?
   A. L = Lj only = 17.3 m
   B. L = dx only = 14.4 m
   C. L = Lj x Fr = 17.3 x 4.13
   D. L = dx + Lj = 14.4 + 17.3, about 32 m
   Answer: D. The apron must cover both the distance to the start of the jump and the jump length itself. (Key point: Minimum apron length)

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