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Screw conveyor design

Design a trough screw conveyor for bulk material: screw diameter, pitch, speed and fill level give the volume and mass flow rate, while conveying length, conveying height and the resistance factor yield drive power and torque plus a guideline rating for critical speed and incline angle.

Calculation

Fill level phi0.3 · OK
n·√(D/1000)30 · OK
Incline angle0 ° · OK

The rating is a guideline: critical speed and incline limit depend on the specific bulk material and are purely informational, not a strength or load-capacity verification.

Throughput

Fill level phi
0.3
Volume flow rate Q
13.25 m³/h
Mass flow rate ṁ
10.6 t/h
Axial conveying speed v_ax
0.25 m/s

Drive

Resistance factor lambda
2.5
Drive power P
0.722 kW
Drive torque M
115 Nm
Incline angle
0 °

Sketch: trough screw conveyor with conveying direction

L = 10 m, H = 0 m
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Formulas and fundamentals

Volume Flow Rate

The volume flow rate follows from the helix cross-section (annular area from the screw diameter D and shaft diameter d), the pitch S per revolution, the speed n and the fill level phi (the fraction of the trough cross-section actually filled with bulk material):

Q = (pi/4)·(D² − d²)·S·n·phi·60 / 1e9

D, d and S are entered in mm, n in rpm; the factor 60 converts minutes to hours, and dividing by 1e9 converts mm³ to m³ (giving Q in m³/h). Without a shaft (d = 0), the annular area simplifies to the full circular area.

Mass Flow Rate

With the bulk density rho of the conveyed material, the mass flow rate follows directly from the volume flow rate:

ṁ = Q·rho

Drive Power and Torque

The drive power is made up of a friction component (conveying length L times the resistance factor lambda) and a lift component (conveying height H for inclined or vertical conveying):

P = ṁ·(L·lambda + H) / 367

ṁ is entered in t/h, L and H in m; the factor 367 = 3600/9.81 converts the units directly into kW. The resistance factor lambda combines friction between the bulk material, the helix and the trough with the stirring work in the material, and is an empirical value per bulk material class. The drive torque follows from the power:

M = P·9550 / n

Conveying Speed and Critical Speed

The axial conveying speed of the bulk material equals the helix advance per unit time:

v_ax = S·n / 60000

Above a critical speed, centrifugal force at the material exceeds gravity, the material rotates with the helix instead of being pushed forward ("tumbling"), and throughput collapses. A rough guideline is:

n·√(D/1000) <= limit (guideline ~45, material-dependent)

D is entered in mm (D/1000 converts it to metres). The calculator flags an exceedance as a note - the exact limit depends on the bulk material (flow behaviour, particle shape) and should be checked against manufacturer data if in doubt.

Inclined Conveying

With an inclined installation (conveying height H > 0), efficiency drops noticeably because the bulk material slides back along the helix - markedly so above about 20° of incline, and above roughly 45° a trough screw conveyor is practically no longer used (the domain of dedicated vertical screw conveyors with their own design logic). For steeply inclined or vertical conveying, comparison with a belt conveyor or bucket elevator calculator is recommended.

Worked example

Reference example: A horizontal trough screw conveyor with D = 250 mm screw diameter (full cross-section, d = 0), pitch S = 250 mm (pitch ratio 1) and speed n = 60 rpm conveys a medium-weight bulk material with fill level phi = 0.3 and bulk density rho = 0.8 t/m³ over a conveying length L = 10 m horizontally (H = 0).

The volume flow rate is Q = (pi/4)·250²·250·60·0.3·60/1e9 = 13.25 m³/h, giving a mass flow rate of ṁ = 13.25·0.8 = 10.60 t/h. With a resistance factor of lambda = 2.5 (medium-weight material), the drive power is P = 10.60·(10·2.5+0)/367 = 0.72 kW and the drive torque M = 0.72·9550/60 = 115.0 Nm.

The axial conveying speed is v_ax = 250·60/60000 = 0.25 m/s. The critical speed check gives n·√(D/1000) = 60·√0.25 = 30 - well below the guideline of 45 (green); since conveying is horizontal, the incline rating is skipped (0°, also green).

Frequently asked questions

How much material does a screw conveyor move?

The volume flow rate follows from Q = (pi/4)·(D² − d²)·S·n·phi·60/1e9 [m³/h]: it grows with the square of the screw diameter D, and linearly with pitch S, speed n and fill level phi. For mass flow rate in t/h, Q is multiplied by the bulk density of the conveyed material. In practice, critical speed and the maximum sensible helix size limit the achievable throughput.

How is the fill level chosen?

The fill level phi indicates what fraction of the trough cross-section is filled with bulk material. Guidelines: light, non-abrasive material (e.g. grain) phi ≈ 0.45, medium-weight material (e.g. sand, gravel) phi ≈ 0.30, heavy or abrasive material (e.g. sharp-edged bulk solids) phi ≈ 0.15. Too high a fill level increases material sliding back and trough wear - the calculator warns above 0.45.

How is the drive power of a screw conveyor calculated?

The drive power P = ṁ·(L·lambda + H)/367 [kW] is made up of a friction component (conveying length L times resistance factor lambda for friction and stirring work) and a lift component (conveying height H for inclined conveying). The resistance factor lambda is an empirical value: light material lambda ≈ 1.2, medium-weight material lambda ≈ 2.5, heavy or abrasive material lambda ≈ 4.0. The drive torque follows from M = P·9550/n.

Can a screw conveyor convey on an incline?

Yes, up to a certain angle - the calculator accounts for the conveying height H in the lift component of the power. Efficiency, however, drops noticeably as the incline increases because the bulk material slides back along the helix; the calculator warns above about 20°. For steeply inclined or vertical conveying, dedicated vertical screw conveyors with different design logic are the usual choice.

What is the critical speed of a screw conveyor?

Above a critical speed, centrifugal force at the bulk material exceeds gravity, the material rotates with the helix instead of being pushed forward, and throughput collapses. A rough guideline is n·√(D/1000) <= 45 (D in mm); the calculator flags an exceedance as a note. The exact limit is material-dependent (flow behaviour, particle shape, cohesion) and should be checked against a manufacturer's catalogue if in doubt.

What does the resistance factor lambda mean?

The resistance factor lambda combines friction between the bulk material, the helix and the trough, plus the stirring work in the material, into a single empirical value that feeds directly into the power formula. It is not a physically well-defined friction coefficient, but a design factor derived from practical experience per bulk material class - for final sizing, checking against manufacturer data or test values is recommended.

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