---
title: "Contraction with step down, 4.25 m³/s"
subject: "Hydraulics"
chapter: 7
key: "semester_4/hydraulics:contraction-gradual-and-step-down-4-25-m3s"
source: https://www.bctengineer.com/q/semester_4/hydraulics/contraction-gradual-and-step-down-4-25-m3s
site: BCT Engineering
---

# Contraction with step down, 4.25 m³/s

Hydraulics · Chapter 7: Energy and Momentum Principles in Open channel flow

- Asked in: 2072 Magh, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/contraction-gradual-and-step-down-4-25-m3s
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7

## Question

An open rectangular channel carrying a discharge of 4.25 m³/s is flowing at a depth of 1.15 m with energy of 1.2 m and a width of 3 m. The flow encounters a simultaneous gradual contraction to a width of 1.5 m and a smooth downwards step of 0.6 m. With these flow conditions, determine the depth of the downstream flow.

## Short answer

The 0.6 m step down adds energy, so E2 = 1.8 m. Solving y2 + 0.4092/y2² = 1.8 for the subcritical root yields downstream depth y2 = 1.65 m. Using exact E1 = 1.227 m gives 1.68 m.

## Answer

**Given:** $Q = 4.25\ \text{m}^3/\text{s}$, $b_1 = 3\ \text{m}$, $y_1 = 1.15\ \text{m}$, $E_1 = 1.2\ \text{m}$, $b_2 = 1.5\ \text{m}$, step down $\Delta z = 0.6\ \text{m}$. Losses are neglected.

**Step 1: Upstream flow type**

$$
\begin{aligned} V_1 &= \frac{4.25}{3 \times 1.15} = 1.232\ \text{m/s} \\ Fr_1 &= \frac{1.232}{\sqrt{9.81 \times 1.15}} = 0.367 < 1 \end{aligned}
$$

The approach flow is subcritical, so the downstream flow is also taken on the subcritical branch.

**Step 2: Critical conditions in the contracted section**

$$
\begin{aligned} q_2 &= \frac{4.25}{1.5} = 2.833\ \text{m}^2/\text{s} \\ y_c &= \left(\frac{q_2^2}{g}\right)^{1/3} = \left(\frac{2.833^2}{9.81}\right)^{1/3} = 0.935\ \text{m} \\ E_c &= 1.5\,y_c = 1.403\ \text{m} \end{aligned}
$$

**Step 3: Energy at the downstream section**

The bed falls by 0.6 m, so the specific energy increases by 0.6 m:

$$
E_2 = E_1 + \Delta z = 1.2 + 0.6 = 1.8\ \text{m} > E_c = 1.403\ \text{m}
$$

So no choking occurs and the upstream depth is unchanged.

**Step 4: Solve for $y_2$**

$$
y_2 + \frac{q_2^2}{2 g y_2^2} = 1.8 \quad\Rightarrow\quad y_2 + \frac{0.4092}{y_2^2} = 1.8
$$

Trial: $y_2 = 1.65$ gives $1.65 + 0.150 = 1.800$. Checked by bisection: $y_2 = 1.650\ \text{m}$ (subcritical; $V_2 = 1.72\ \text{m/s}$).

**Answer: Downstream depth $y_2 \approx 1.65\ \text{m}$.** (Using the computed $E_1 = 1.227\ \text{m}$ instead of the rounded 1.2 m gives 1.68 m.)

## Check yourself

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

1. Why is the downstream depth taken on the subcritical branch?
   A. Because the contraction always forces subcritical flow
   B. Because the step down makes flow subcritical
   C. Because Fr1 = 0.367 < 1, so the approach flow is subcritical
   D. Because the channel width is reduced
   Answer: C. Fr1 = V1/sqrt(g y1) = 0.367, so approach flow is subcritical and stays on that branch if not choked. (Key point: Upstream flow regime)

2. How is it decided that the contraction does not choke?
   A. E2 = 1.8 m exceeds the minimum energy Ec = 1.403 m at width 1.5 m
   B. The depth y1 is larger than yc
   C. The velocity is below 2 m/s
   D. The width ratio is exactly 1/2
   Answer: A. Available energy after the step down is larger than the minimum needed for q2, so upstream depth is unaffected. (Key point: Choking check)

3. How does a smooth downward step of 0.6 m affect the specific energy at the downstream section (no losses)?
   A. It decreases by 0.6 m
   B. It stays the same
   C. It doubles
   D. It increases by 0.6 m, giving E2 = 1.8 m
   Answer: D. Total head is constant, so a lower bed leaves 0.6 m more specific energy. (Key point: Effect of downward step)

4. Which equation is solved by trial to get y2?
   A. y2 + q2^2/(2 g y2^2) = 1.8 with q2 = 4.25/1.5
   B. y2 + q1^2/(2 g y2^2) = 1.2
   C. y2 = (q2^2/g)^(1/3)
   D. y2 = 2/3 x 1.8
   Answer: A. Specific energy at section 2 with the contracted discharge per unit width equals E2 = 1.8 m, giving y2 near 1.65 m. (Key point: Solving for y2)

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