sample calculate screw conveyor power
How to Calculate Screw Conveyor Power Requirements
Screw conveyors are widely used in industries for transporting bulk materials horizontally, vertically, or at an incline. Determining the required power ensures efficient operation and prevents motor overload. Below is a step-by-step guide to calculating screw conveyor power.
Key Factors Affecting Power Calculation
Several variables influence the power needed to drive a screw conveyor:
1. Material Characteristics – Bulk density, particle size, and flowability affect resistance.
2. Conveyor Length & Incline – Longer conveyors and steep inclines increase power demand.
3. Screw Diameter & Pitch – Larger screws with higher pitch require more torque.
4. Friction Losses – Internal friction between material and trough impacts efficiency.
Basic Power Calculation Formula
The following empirical formula estimates the required power (*P*) in horsepower (HP):
\[ P = \frac{(C \times L \times W \times F)}{1000} + \frac{(L \times H)}{1000} \]
Where:
- *C* = Capacity (tons/hour)
- *L* = Conveyor length (feet)
- *W* = Material weight per cubic foot (lbs/ft³)
- *F* = Friction factor (typically 1–4 depending on material)
- *H* = Height change (feet for inclined conveyors)
For metric units (kW), adjust constants accordingly or convert HP to kW (*1 HP ≈ 0.746 kW*).
Step-by-Step Calculation Example
Assume a horizontal screw conveyor handling grain (*W = 45 lbs/ft³*) with:
- Capacity (*C*) = 30 tons/hour
- Length (*L*) = 50 feet
- Friction factor (*F*) = 1.5
Using the formula:
\[ P = \frac{(30 \times 50 \times 45 \times 1.5)}{1000} + \frac{(50 \times 0)}{1000} = 101.25 \, HP \, (\text{horizontal}) \]
For a similar inclined conveyor lifting material (*H*) by 15 feet:
\[ P = 101.25 + \frac{(50 \times 15)}{1000} = 101.25 + 0.75 = 102 \, HP \, (\text{inclined}) \]