Chain Conveyor Design (Trough/Scraper Chain Conveyor)
Design a trough or scraper chain conveyor for bulk material using a simplified resistance method: from mass flow, conveying speed, conveying length and height, and the friction coefficients of the material and the chain, the horizontal resistances, gradient resistance, driving force at the drive shaft, drive power and chain tension follow, together with a check against the allowable chain tension.
Calculation
Design rating is an approximation: simplified resistance method without start-up/wear and sprocket polygon effect - manufacturer catalogue is authoritative.
Conveyed material and incline
- Conveyed material mass m_L'
- 92.59 kg/m
- Material/trough friction coefficient µ_G
- 0.6
- Chain/rail friction coefficient µ_K
- 0.25
- Incline angle δ
- 0 °
Resistances and driving force
- Horizontal resistance material F_material
- 16,350 N
- Horizontal resistance chain F_chain
- 3,678.8 N
- Main resistance F_H
- 20,028.8 N
- Gradient resistance F_grade
- 0 N
- Driving force F_U
- 23,033.1 N
Drive power and chain tension
- Drive power P
- 8.129 kW
- Chain tension F_chain,max
- 23,033.1 N
Sketch: trough chain conveyor with drive/idler sprocket, chain with flights and conveyed material (not to scale)
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Formulas and fundamentals
Mass per Metre of Conveyed Material
The mass flow ṁ is spread over the conveying length at conveying speed v; the conveyed material mass per metre is:
With ṁ in t/h and v in m/s, m_L' follows directly in kg/m. Trough and scraper chain conveyors run considerably slower than belt conveyors (guideline 0.05 … 0.6 m/s) so the bulk material is not thrown over the scraper flights.
Horizontal Resistance of the Material
The conveyed material is dragged through the trough; the friction coefficient µ_G between material and trough (guideline 0.4 for light, 0.6 for medium, 0.8 for heavy/abrasive bulk material) gives the horizontal resistance:
Horizontal Resistance of the Chain
The chain itself rubs its own weight against the sliding rail or roller track - on both the carrying and the return strand, hence the factor 2 in front of the chain mass per metre and strand m_K':
The friction coefficient µ_K depends on the guide: a sliding rail (guideline 0.25) or a roller chain running on rollers (guideline 0.15, considerably lower friction).
Main Resistance and Gradient Resistance
The main resistance and gradient resistance follow from the horizontal resistances and lifting the conveyed material over the conveying height H (the chain itself rises and falls oppositely on the carrying and return strand, so its contribution is conservatively neglected):
Driving Force, Drive Power and Chain Tension
Via the secondary resistance factor C (deflection, drive; guideline 1.15), the driving force at the drive shaft follows, and with the drive efficiency η (guideline 0.85) the drive power:
The maximum chain tension in the pulling strand is approximated (without pretension) by the driving force, F_chain,max ≈ F_U. If an allowable chain tension F_allow is known from the chain catalogue, the calculator checks the utilisation F_chain,max/F_allow.
Maximum Incline
Scraper chain conveyors can convey at considerably steeper angles than belt conveyors, since the flights prevent the material from sliding back; the guideline maximum incline is around 30°, beyond which the risk of material sliding back increases and a bucket elevator or steep conveyor is usually the better choice.
Worked example
Reference example: a horizontal trough chain conveyor transports ṁ = 100 t/h at a conveying speed v = 0.3 m/s over a conveying length L = 30 m (H = 0). The chain mass per metre and strand is m_K' = 25 kg/m, the material/trough friction coefficient µ_G = 0.6 (medium bulk material), the chain/rail friction coefficient µ_K = 0.25 (sliding rail). Secondary resistance factor C = 1.15, drive efficiency η = 0.85.
The conveyed material mass per metre is m_L' = 100/(3.6·0.3) = 92.59 kg/m. The horizontal resistance of the material is F_material = 9.81·30·0.6·92.59 = 16350 N, the horizontal resistance of the chain F_chain = 9.81·30·0.25·2·25 = 3678.75 N. The main resistance is F_H = 16350 + 3678.75 = 20028.75 N; since conveying is horizontal, the gradient resistance is zero (F_grade = 0).
The driving force at the drive shaft is F_U = 1.15·20028.75 + 0 = 23033.1 N. The drive power follows as P = 23033.1·0.3/(0.85·1000) = 8.13 kW. The chain tension is approximated as F_chain,max ≈ F_U = 23033.1 N - with an allowable chain tension F_allow = 30 kN this would give a utilisation of about 77% (pass). Conveying speed and incline are both within the usual guideline range.
Frequently asked questions
How do I calculate the drive power of a chain conveyor?
From mass flow, conveying speed, conveying length, and the friction coefficients of the material (µ_G) and the chain (µ_K), the horizontal resistances of material and chain follow, giving the main resistance F_H. Together with the gradient resistance F_grade (from the conveying height) and the secondary resistance factor C, the driving force F_U = C·F_H + F_grade at the drive shaft follows; the drive power is P = F_U·v/(η·1000) with the drive efficiency η.
What is the difference between a scraper chain conveyor and a belt conveyor?
In a scraper chain conveyor, the bulk material is dragged through a fixed trough by flights on a circulating chain, instead of resting on a moving belt. This allows considerably steeper inclines (guideline up to about 30° instead of 18 … 20° for a belt conveyor) and the conveyor is less sensitive to coarse or hot bulk material, but it runs slower (0.05 … 0.6 m/s) and with more wear on the trough and chain.
How is the chain tension verified?
The maximum chain tension in the pulling strand is approximated (without pretension, start-up or shock allowances) as equal to the driving force at the drive shaft, F_chain,max ≈ F_U. If the allowable chain tension F_allow is known from the manufacturer's chain catalogue, the calculator computes the utilisation F_chain,max/F_allow: up to 80% pass, up to 100% a note, above that the selected chain is too weak and a stronger size or a multi-strand drive is required.
What is the maximum incline for a chain conveyor?
Trough and scraper chain conveyors, thanks to their positive-drive flights, can convey at considerably steeper angles than belt conveyors; a guideline maximum incline of around 30° applies. Above this value, the risk increases that bulk material slides back over the flights or jams - for steeper or vertical conveying, bucket elevators or steep conveyors with special flight profiles are the more suitable choice.
How do I choose the friction coefficients µ_G and µ_K?
µ_G (material/trough) depends on the bulk material: free-sliding, non-abrasive materials around 0.4, medium materials around 0.6, heavy or abrasive materials around 0.8. µ_K (chain/rail) depends on the guide: a simple sliding rail is around 0.25, a roller chain running on rollers considerably lower at around 0.15. For the final design, manufacturer data or test values should be consulted.
Why are chain conveyors typically so slow?
The conveying speed of trough and scraper chain conveyors is usually between 0.05 and 0.6 m/s - considerably below that of a belt conveyor. Higher speeds would throw the bulk material over the flights or fling it backwards and greatly increase wear on the chain and trough; for larger conveying rates the trough cross-section is therefore increased rather than the speed.
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