Universities · vocational schools · apprenticeships
For teaching and education
Piping and plant design to try out: the Plant Engineering Infobase is freely accessible, the calculators show formulas, characteristic values and the governing criterion, and each calculator comes with terms, experiments and an exercise with solution. Everything runs in the browser – no installation and no Plant 3D license needed.
What is free to use
Plant Engineering Infobase
14 chapters, from standards and P&IDs through plant identification, piping specs and drawings to Plant 3D – with sources.
Learning base for the calculators
Terms in everyday language, experiments ("What happens if …?"), typical mistakes and an exercise – at the bottom of each calculator page.
Work aids
Quick references and checklists from practice – project workflow, piping spec definition, scan project.
Calculators
Support spans, pressure drop, weight and thermal expansion free to try; Piping Spec Calculator and PCF import with a customer account. For courses, we set up access on request.
Exercises with solutions
Exercise 1 Nominal size and pressure drop
Water at 20 °C, 36 m³/h, steel pipe DN 100 (114.3 × 3.6 mm), 100 m straight line. What are the velocity, Reynolds number and pressure drop? Is a pump with 2 m head enough to overcome the friction?
Show solution
di = 114.3 − 2 · 3.6 = 107.1 mm; A = π/4 · 0.1071² = 0.00901 m²; Q = 36 m³/h = 0.01 m³/s → v = 0.01 / 0.00901 ≈ 1.11 m/s. Re = 1.11 · 0.1071 / 1.0·10⁻⁶ ≈ 119,000 → turbulent. With k = 0.05 mm, Colebrook-White gives λ ≈ 0.020. Δp = 0.020 · 100 / 0.1071 · 998 · 1.11² / 2 ≈ 11,300 Pa ≈ 0.11 bar ≈ 1.15 m. Yes, 2 m is enough for the friction – but bends, valves and difference in elevation come on top. Reproduce it in the calculator: series "Steel pipe", 36 m³/h, length 100 m, no fittings.
Exercise 2 Thermal expansion
A 40 m steel line is installed at 10 °C and heated to 180 °C in operation. How much longer does it get? And roughly how long would an L-bend leg need to be for a 114.3 mm pipe?
Show solution
α (20 → 180 °C) ≈ 12.5 · 10⁻⁶/K; ΔL = 12.5·10⁻⁶ · 40,000 mm · 170 K ≈ 85 mm – so around 8 to 9 cm. Expansion leg per L = √(3 · E · D · δ / σ) with E ≈ 190,000 N/mm² (hot), D = 114.3 mm, δ = 85 mm, σ = 120 N/mm²: L ≈ √(3 · 190,000 · 114.3 · 85 / 120) ≈ 6.8 m. That is a lot of space – which is why long lines are divided by anchor points or fitted with U-loops or compensators. In the calculator: material steel, 40 m, 10 / −10 / 180 °C, outside diameter 114.3.
Exercise 3 Pipe weight
Steel pipe 114.3 × 3.6 mm filled with water, support span 5 m, no insulation. What load does a support in the middle of a long line carry?
Show solution
Pipe: π/4 · (114.3² − 107.1²) = 1,252 mm² → 1,252 · 10⁻⁶ m² · 7,850 kg/m³ ≈ 9.8 kg/m. Water: π/4 · 107.1² = 9,009 mm² → ≈ 9.0 kg/m. Total ≈ 18.8 kg/m. Per support (tributary length = support span 5 m): 94 kg ≈ 0.92 kN. With an allowance for flanges and fittings, you calculate a bit more.
Exercise 4 Converter DN · Class · material
An American specification says "NPS 4, Sch 40, Class 300, A312 TP316L". What does that mean in German (DIN/EN) terms?
Show solution
NPS 4 ≈ DN 100 with an outside diameter of 114.3 mm; Sch 40 means 6.02 mm wall thickness at this size; Class 300 is the pressure class of the flanges (permissible pressure from the rating table for the material group); A312 TP316L is a seamless or welded stainless steel pipe, comparable to 1.4404. In the converter: open the DN ↔ NPS, Schedule, PN ↔ Class and Materials tabs one after the other.
Exercise 5 Support spans
Why does a DN 200 pipe have a much larger permissible support span than DN 50, even though it is heavier?
Show solution
The second moment of area I (flexural rigidity E · I) grows roughly with D³, the weight per meter only roughly with D to D². With deflection f ~ q · L⁴ / (E · I), stiffness wins – which is why L can grow considerably. Calculate DN 50 and DN 200 one after the other in the Support Span Calculator.
Exercise 6 Wall thickness and piping spec
Steel pipe 114.3 mm, 16 bar, 20 °C, f = 150 N/mm², z = 1. How thick does it need to be by calculation – and why does the catalog pipe still have 3.6 mm?
Show solution
e = 1.6 N/mm² · 114.3 / (2 · 150 · 1 + 1.6) ≈ 0.61 mm. 3.6 mm is ordered because standard series specify minimum wall thicknesses for handling, welding, corrosion and mechanical loads – at low pressure, it is not the internal pressure that determines the wall.
The figures have been double-checked; characteristic values are guide values. For verification, the standards and manufacturer data apply. The calculators provide preliminary sizing – not a piping design calculation and not support design.
For educators
- Tools in class: we set up access to the calculators for your course – just get in touch.
- Guest lecture or workshop: "From P&ID to piping spec" or "Catalogs and piping specs in Plant 3D" – live online or at your institution.
- Theses with a practical focus: topics on piping specs, catalog data and as-built capture on request.
- Show and link pages: the Infobase, learning base and exercises may be shown in class.
Contact
Tell us what you have in mind – we will get back to you within one business day.
Frequently asked questions
Do students need a Plant 3D license?
No. The Infobase, learning base and exercises are web pages; the calculators run in the browser.
Are the calculators free?
Yes. Support spans, pressure drop, weight and thermal expansion without registration; Piping Spec Calculator and PCF import with a customer account. Customer accounts are intended for companies and are activated manually – for courses, we set up access on request.
May we use content in our lecture notes?
Linking and showing in class: yes. For excerpts in your own materials, please check with us briefly.
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