Conveyor Belt Calculator
Calculate conveyor belt speed, capacity, and motor power from load and dimensions. Enter values for instant results with step-by-step formulas.
Formula
Capacity = A x V x 3600 x density | Power = (Friction + Lift) x V / 1000
Where A = material cross-section area (m2), V = belt speed (m/s), density = bulk density (kg/m3). Power calculation sums friction resistance and gravitational lift force, then multiplies by belt speed.
Worked Examples
Example 1: Mining Conveyor Belt Sizing
Problem: Design a conveyor for 800 t/h of crushed stone (density 1600 kg/m3) over 30m length with 5m lift. Belt width 600mm, speed 1.5 m/s, friction factor 0.03, belt mass 15 kg/m.
Solution: Cross-section area needed = 800 / (3600 x 1.5 x 1.6) = 0.0926 m2\nMaterial mass/meter = 0.0926 x 1600 = 148.1 kg/m\nFriction force = 0.03 x (148.1 + 30) x 9.81 x 30 = 1,568 N\nLift force = 148.1 x 9.81 x 5 = 7,264 N\nTotal force = 1,568 + 7,264 = 8,832 N\nPower = 8,832 x 1.5 / 1000 = 13.25 kW\nMotor (90% eff) = 14.7 kW, select 15 kW motor
Result: Capacity: 800 t/h | Power: 13.25 kW | Motor: 15 kW | Belt tension: 8,832 N
Example 2: Horizontal Package Conveyor
Problem: A 400mm wide flat belt conveyor moves packages at 0.5 m/s over 15m with no lift. Load cross-section 0.01 m2, package density 400 kg/m3, belt mass 8 kg/m.
Solution: Capacity = 0.01 x 0.5 x 3600 = 18 m3/h = 7.2 t/h\nMaterial mass/m = 0.01 x 400 = 4 kg/m\nTotal mass/m = 4 + 16 = 20 kg/m\nFriction = 0.03 x 20 x 9.81 x 15 = 88.3 N\nLift = 0 N (horizontal)\nPower = 88.3 x 0.5 / 1000 = 0.044 kW\nSelect 0.12 kW (1/6 HP) motor minimum
Result: Capacity: 7.2 t/h | Power: 0.044 kW | Motor: 0.12 kW minimum
Frequently Asked Questions
How do you calculate conveyor belt capacity in tons per hour?
Conveyor belt capacity in tons per hour is calculated by multiplying the cross-sectional area of the material on the belt (in square meters) by the belt speed (in meters per second) by 3600 (seconds per hour) by the material bulk density (in kg per cubic meter) and dividing by 1000 to convert to metric tons. The cross-sectional area depends on the belt width, the trough angle of the idlers, and the surcharge angle of the material. For a standard three-roll troughing idler at 20 degrees, the cross-sectional area is approximately 10 percent of the belt width squared. The CEMA standard provides detailed lookup tables for cross-sectional areas based on belt width, idler configuration, and material properties.
How is conveyor belt motor power calculated?
Conveyor belt motor power is determined by calculating the total resistive force that the belt must overcome and multiplying by the belt speed. The total force includes friction resistance from the idlers and belt flexing (proportional to the conveyor length and total weight), the force needed to lift material vertically (material weight times lift height), and any additional resistances from belt cleaners, skirtboards, and loading impact. The formula is Power (kW) = Total Force (N) x Belt Speed (m/s) / 1000. A motor efficiency factor (typically 85-92 percent for AC motors with gearbox) is applied to determine the required motor nameplate rating. The motor is then selected from the next larger standard frame size.
What factors determine the optimal belt speed for a conveyor?
Optimal belt speed depends on the material being conveyed, the belt width, and the application requirements. Typical belt speeds range from 0.5 to 6 meters per second. Abrasive or heavy materials like crushed rock require lower speeds (1.0-2.5 m/s) to reduce belt and idler wear. Fine materials like flour or cement use lower speeds (0.5-1.5 m/s) to prevent dusting. Lighter materials like grain can use higher speeds (2.5-4.0 m/s). Wide belts can run faster than narrow belts because the material load is better distributed. Transfer points and loading zones limit speed because high-speed material impact damages the belt and creates spillage. CEMA guidelines provide recommended maximum speeds for various material types.
What is the CEMA standard for conveyor belt design?
CEMA (Conveyor Equipment Manufacturers Association) publishes the definitive engineering standard for belt conveyor design, currently in its seventh edition. The CEMA standard provides comprehensive methods for calculating belt tensions, power requirements, idler selection, pulley sizing, and belt specification. It defines standard belt widths (18 to 96 inches or 450 to 2400mm), idler spacing guidelines, and material handling factors. The standard includes correction factors for surcharge angle, conveyor inclination, belt sag between idlers, acceleration and deceleration forces, and environmental conditions. Most conveyor manufacturers and engineering firms worldwide use the CEMA methodology as the basis for belt conveyor design, and many building codes reference it for safety requirements.
How does conveyor inclination affect capacity and design?
Conveyor inclination reduces the effective capacity because material tends to slide back on steep inclines, and the cross-sectional area of material that can be carried decreases. Most bulk materials can be conveyed on inclines up to 15-18 degrees on smooth belts. Beyond that angle, special measures are needed such as cleated or corrugated sidewall belts (up to 90 degrees), high-friction belt covers, or reduced loading. Inclination also significantly increases the power requirement because of the gravitational component: lifting force equals material mass per meter times gravitational acceleration times the vertical lift height. A horizontal conveyor carrying 500 tons per hour might need 15 kW, but the same conveyor inclined to lift material 20 meters vertically would need an additional 27 kW just for the lifting component.
What belt width should be selected for a given capacity?
Belt width selection depends on the required capacity, material lump size, and belt speed. As a general guideline, the maximum lump size should not exceed one-third of the belt width for unsorted material or one-fifth for sized material. Standard belt widths are 400, 500, 600, 650, 800, 1000, 1200, 1400, 1600, 1800, and 2000mm. For a given capacity in tons per hour, a wider belt running slower is generally preferable to a narrow belt running fast because it reduces wear, spillage, and dust generation. The CEMA capacity tables relate belt width, speed, troughing angle, and material density to tons per hour capacity. Always select the next larger standard belt width when the calculated requirement falls between standard sizes.