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Open channel flow software

Describe the channel. Rischio shapes it, calculates it, and shows its work.

Any cross-section, from a lined trapezoid to a natural channel with floodplains: normal and critical depth, velocity, Froude number and whether the flow fits, with the calculation report ready to hand in. Describe the channel to the agent, or enter it yourself.

Free, no card needed.

Your agent is working on this model
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Or draw it yourself.

The same editor the agent works in. Pick a model, change a number and calculate: the engine answers right here.

What it checks.

Every check with its rule and its limit, Rischio's defaults. In the studio, each result opens into its equation, with your numbers in it.

How it calculates, step by step
CheckLimit

Bank-full capacity

The discharge fits below the lower bank

QQ

≤Qbank\le Q_{bank}

What you can model

  • Any section, vertex by vertex
  • A roughness for every side
  • Compound sections with floodplains
  • Slope and discharge

How it's calculated

  • Uniform flow by Manning
  • The section divided where its roughness changes, each part with its own n
  • Normal depth solved for the discharge; critical depth, Froude number and specific energy
  • The rating curve, from the bed to the bank

The calculation report, ready to sign.

This is the one for the model the editor opens on, as the studio makes it. Read through it.

  • The checks' summary first, each with its margin
  • Every section with its equations and the values they take
  • The assumptions and warnings at the end, no small print

As PDF and Word, with your letterhead, in Professional.

Your engineering firmYour address · your license number

Calculation report

Model
Trapezoidal earth channel
Analysis
Open channel
Date
15 October 2026
Units
SI · kN, m, kPa

1Summary

The model satisfies its check.

CheckValueLimitMarginResult
Bank-full capacity2.00 m³/s≤ 3.41 m³/s+41 %OK

Limits are Rischio's defaults.

2Model

Figure 1. Section of the model.

Section

Trapezoidal earth channelCross-section

Vertexxxmzzmnn
11.001.000.030
22.000.100.030
33.100.100.030
44.000.100.030
55.001.00

xx Horizontal position · zz Elevation · nn Roughness of the segment to the next vertex

Flow

Flow conditionsUniform flow

Longitudinal slope SS0.003 m/m
Discharge QQ2.00 m³/s

3Normal flow

yny_n = 0.670 m
Figure 2. The section with the water surface at the normal depth.

Water surface elevation

Q(yn)=Q(normal depth, Manning)=0.770  m\begin{aligned} Q(y_n) &= Q \quad \text{(normal depth, Manning)} \\ &= 0.770\;\text{m} \end{aligned}

where QQ = 2.00 m³/s

Normal depth

yn=WSE−zbed=0.670  m\begin{aligned} y_n &= WSE - z_{bed} \\ &= 0.670\;\text{m} \end{aligned}

Flow area

A=∫(WSE−z) dx=1.84  m2\begin{aligned} A &= \int (WSE - z) \, dx \\ &= 1.84\;\text{m}^{2}\text{} \end{aligned}

Wetted perimeter

P=ΣPi=4.00  m\begin{aligned} P &= \Sigma P_i \\ &= 4.00\;\text{m} \end{aligned}

Hydraulic radius

R=A/P=0.459  m\begin{aligned} R &= A / P \\ &= 0.459\;\text{m} \end{aligned}

where AA = 1.84 m², PP = 4.00 m

Equivalent roughness

n=(ΣPini1.5P)2/3(Horton–Einstein)=0.030  \begin{aligned} n &= \left( \frac{\Sigma P_i n_i^{1.5}}{P} \right)^{2/3} \quad \text{(Horton–Einstein)} \\ &= 0.030\;\text{} \end{aligned}

where PP = 4.00 m

Mean velocity

V=1nR2/3S1/2=1.09  m/s\begin{aligned} V &= \frac{1}{n} R^{2/3} S^{1/2} \\ &= 1.09\;\text{m/s} \end{aligned}

where nn = 0.030 , RR = 0.459 m, SS = 0.003 m/m

4Flow regime

yny_n = 0.670 mycy_c = 0.430 m
Figure 3. Normal depth (solid) and critical depth (dashed).

Top width

T=3.49  m\begin{aligned} &T \\ &= 3.49\;\text{m} \end{aligned}

Froude number

Fr=VgA/T=0.478  \begin{aligned} Fr &= \frac{V}{\sqrt{g A / T}} \\ &= 0.478\;\text{} \end{aligned}

where VV = 1.09 m/s, AA = 1.84 m², TT = 3.49 m

Critical depth

Q2TcgAc3=1=0.430  m\begin{aligned} \frac{Q^2 T_c}{g A_c^3} &= 1 \\ &= 0.430\;\text{m} \end{aligned}

where QQ = 2.00 m³/s

Specific energy

E=y+V22g=0.731  m\begin{aligned} E &= y + \frac{V^2}{2g} \\ &= 0.731\;\text{m} \end{aligned}

where yy = 0.670 m, VV = 1.09 m/s

5Capacity of the section

00.511.522.530.20.40.60.81Discharge Q (m³/s)Elevation (m)Lower bankQ = 2 m³/s · 0.770 m
Figure 4. Rating curve: water surface elevation against discharge, up to the lower bank.

Bank-full capacity

Qbank=1nAR2/3S1/2(y=ybank)=3.41  m3/s\begin{aligned} Q_{bank} &= \frac{1}{n} A R^{2/3} S^{1/2} \quad (y = y_{bank}) \\ &= 3.41\;\text{m}^{3}\text{/s} \end{aligned}

where ybanky_{bank} = 0.900 m, SS = 0.003 m/m

Check · Bank-full capacity

Q≤QbankQ=2.00 m3/s  ≤3.41 m3/s\begin{aligned} Q &\le Q_{bank} \\ Q &= 2.00\,\text{m}^{3}\text{/s} \; \le 3.41\,\text{m}^{3}\text{/s} \end{aligned}

margin +41 %

where QQ = 2.00 m³/s, QbankQ_{bank} = 3.41 m³/s

OK

Questions, answered.

Which sections can it calculate?

Any cross-section drawn vertex by vertex: trapezoids, rectangles, gutters, compound sections with floodplains and natural channels, each side with its own Manning n.

Which method does it use?

Manning's equation for uniform flow, with the section divided where its roughness changes, as HEC-RAS does: a lined channel and its grassed floodplains each carry their own share. It solves the normal depth for the discharge, and gives the critical depth, the Froude number, the specific energy and the rating curve.

Does it calculate backwater profiles?

Not yet: it calculates uniform flow. Gradually varied flow, culverts and weirs are outside it.

What does the agent do, and what does the engine do?

The agent turns your words into a model and edits it. The engine, the same one behind the editor, does every calculation: no number comes from the agent. You review the model, open every equation and sign the report.

Is it free?

The Free plan has all four tools and up to 3 models. Professional, $49 a month, adds unlimited models, the report as PDF and Word with your letterhead, and your own agent through MCP. Every account starts with 14 days of Professional, no card needed.

What will you design today?

Describe it in a sentence. Rischio builds it, calculates it, and shows you every step.

›Will a 1.5 m concrete channel carry 4 m³/s at 0.2 %?

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