Engineering guide
The 2448 module provides the dimension series of seamless steel pipes to DIN 2448, a German pipe dimension standard, and from them calculates the inside diameter, the mass per unit length and the total mass. DIN 2448 has been withdrawn; the dimensions and masses per unit length of seamless steel pipes are today standardized in DIN EN 10220, which largely carried over the classic size combinations. For identifying existing pipes, re-assessing older plants and procuring spare parts, the old dimension standard is still the reference document.
In practice the module serves as a quick reference during pipe selection: for a given material and delivery standard, the combination of outside diameter and standard wall thickness is taken from the chosen wall thickness series. From this, the inside diameter for hydraulic design and the weight per metre of the pipe follow directly – the basis for weight estimates of piping systems, support loads and material costing.
The mass per unit length follows from the cross-sectional area of the pipe annulus and the standardized density of steel; together with the pipe length, the module also delivers the total mass of the line or of the pipe lot.
Scope
- Seamless steel pipes
Typical workflow
Select material and delivery standard
The pipe material and the associated delivery standard (technical delivery conditions) are defined – they determine which size combinations are available.
Take the size from the dimension series
From the selected series, the combination of outside diameter and standard wall thickness is taken; alternatively an as-built wall thickness can be specified, for instance for special wall thicknesses.
Determine the inside diameter
The inside diameter results from the outside diameter minus twice the wall thickness and is the governing quantity for flow cross-section and pressure drop calculation.
Calculate the masses
The mass per unit length follows from the annular cross-sectional area π·(D − s)·s and the steel density of 7.85 kg/dm³; multiplied by the pipe length, the total mass is obtained.
Frequently asked questions
What replaces DIN 2448 today?
The dimensions and masses per unit length of seamless steel pipes are governed by DIN EN 10220; the technical delivery conditions are laid down, depending on the application, in EN 10216 (seamless pressure pipes). The common outside diameter/wall thickness combinations of DIN 2448 largely reappear there, so existing pipes can usually be matched unambiguously.
How is the mass per unit length of a steel pipe calculated?
From M = π·(D − s)·s·ρ with the outside diameter D, the wall thickness s and the density ρ = 7.85 kg/dm³. In the numerical form of the standard this reads M = 0.02466·s·(D − s) with D and s in millimetres and M in kg/m. The tabulated values of the standard are rounded on this basis.
What do the wall thickness series of the dimension standard mean?
The series group preferred outside diameter/wall thickness combinations by availability and application: series 1 contains the preferentially available standard combinations, higher series add less common wall thicknesses. For new designs it is advisable to stay within the preferred series to keep delivery times and costs low.
Is the standard wall thickness automatically the wall thickness required for the pressure?
No. The dimension standard only defines the available sizes. The wall thickness required for internal pressure must be calculated separately to the applicable strength code (e.g. DIN EN 13480-3, AD 2000 B 1 or, historically, DIN 2413); the next larger standard wall thickness is then selected, taking the minus tolerance and the corrosion allowance into account.
Worked example
For a seamless steel pipe 88.9 × 3.2 mm, determine the mass per unit length and the total mass of a 6 m pipe length – a worked example of the steel pipe weight calculation.
Given
| Outside diameter D | 88.9 mm |
|---|---|
| Wall thickness s | 3.2 mm |
| Pipe length L | 6 m |
| Density of steel ρ | 7.85 kg/dm³ |
Solution
Inside diameter
di = D − 2·s = 88.9 − 2 · 3.2 = 82.5 mm
Mass per unit length
M = 0.02466 · s · (D − s)
M = 0.02466 · 3.2 · (88.9 − 3.2) = 0.02466 · 3.2 · 85.7 = 6.76 kg/m
Total mass
m = M · L = 6.76 · 6 = 40.6 kg
Result
| Inside diameter | 82.5 mm |
|---|---|
| Mass per unit length | 6.76 kg/m |
| Total mass (6 m) | 40.6 kg |
Technical content source: LV-Solved-Problems-Content 628d313247e7