Belt Conveyor Design

\[ Q \;(\text{t/h}) = \rho \;(\text{t/m}^3) \times v \;(\text{m/s}) \times A \;(\text{m}^2) \times 3600 \]

Solving for required area:

\[ A = \frac{Q}{\rho \cdot v \cdot 3600} \]

Where \(Q\) is capacity in t/h, \(\rho\) is bulk density in t/m³, \(v\) is belt speed in m/s, and \(A\) is the material cross-sectional area on the belt.

\[ A \approx K(\theta) \cdot B^2 \]

Solving for belt width:

\[ B = \sqrt{\frac{A}{K(\theta)}} \]

Where \(K(\theta)\) is the troughing factor for idler angle \(\theta\), and \(B\) is the belt width in metres. The calculated width is rounded up to the next standard belt size.

Troughing Factors

Troughing Angle (°)	\(K(\theta)\)
20	0.143
30	0.175
35	0.197
45	0.245

Material load at the design capacity (ISO 5048 / DIN 22101):

\[ q_G = \frac{Q}{3.6 \, v} \quad (\text{kg/m}) \]

The belt carries the design tonnage, not a struck-full trough: the troughed area \(K(\theta) B^2\) only sets the belt width and the capacity check.

Belt mass is either user-specified or estimated as:

\[ q_B \approx B \times 15 \quad (\text{kg/m}) \]

Idler rotating mass per metre, carrying and return strands:

\[ q_{RO} = \frac{m_{\text{set,carry}}}{a_0}, \qquad q_{RU} = \frac{m_{\text{set,return}}}{a_u} \]

When not given they are estimated from typical idler sets for the belt width (Rulmeca design guide Tab. 11, lightest roller listed) at \(a_0 = 1.2\) m and \(a_u = 3.0\) m pitch.

\[ F_H = f \cdot L \cdot g \cdot \bigl[q_{RO} + q_{RU} + (2 q_B + q_G) \cos\delta\bigr] \quad (\text{N}) \]

Where \(f\) is the artificial friction factor, \(L\) the conveyor length, \(g = 9.81\) m/s² and \(\delta\) the inclination. The belt term counts both strands; the material only the carrying strand.

\[ F_N = (C - 1) \cdot F_H \]

Feed-zone acceleration and skirt friction, pulley bearings and belt bending around the pulleys. \(C\) from DIN 22101 Table 5 (linear interpolation; constant 1.05 from 2000 m):

L (m)	80	100	150	200	300	400	500	600	700	800	900	1000	1500	≥2000
C	1.92	1.78	1.58	1.45	1.31	1.25	1.20	1.17	1.14	1.12	1.10	1.09	1.06	1.05

DIN 22101 and ISO 5048 permit this method from 80 m with one feed point. Below 80 m the calculator uses the Rulmeca design-guide values (10 m: 4.5, 20 m: 3.2, 30 m: 2.6, 40 m: 2.2, 50 m: 2.1, 60 m: 2.0) and warns that the secondary resistances should be itemised.

\[ F_{St} = q_G \cdot g \cdot H, \qquad H = L \sin\delta \quad (\text{N}) \]

Positive for uphill, negative for downhill (the material helps drive the belt and, on a steep enough decline, overhauls the drive — see Regenerative Declines below).

Note: Only material mass contributes to net lift resistance. Belt mass cancels between the carrying side (going up) and the return side (coming back down) per DIN 22101.

Loaded belt (signed; negative means the belt overhauls the drive):

\[ F_{\text{total}} = \bigl(F_H + F_N + F_{St} + F_S\bigr) \times k_{\text{impact}} \]

Empty belt (no material term, no lift, no loading impact):

\[ F_{\text{empty}} = C \cdot f \cdot L \cdot g \cdot \bigl[q_{RO} + q_{RU} + 2 q_B \cos\delta\bigr] + F_S \]

Design effective tension for the belt tensions, drum torque and take-up:

\[ F_{\text{design}} = \max\bigl(|F_{\text{total}}|, |F_{\text{empty}}|\bigr) \]

Where \(F_S\) (additional_resistance_n) accounts for scrapers, ploughs and skirtboards beyond the feed zone, and \(k_{\text{impact}}\) is the impact factor. CEMA (7th ed.) and DIN 22101 require the empty, loaded and part-loaded operating cases to be checked. On a single slope the force is linear in the load fraction, so part-loaded belts lie between the empty and full-load values and these two cases govern. On any decline with \(C f \cos\delta + \sin\delta < 0\) the empty belt needs more motoring force than the loaded one.

Power at drum periphery:

\[ P_{\text{drum}} = F_{\text{total}} \times v \quad (\text{W}) \]

Design power with service factor:

\[ P_{\text{design}} = P_{\text{drum}} \times \text{SF} \]

Required motor power for each load case, \(\eta = \eta_{\text{gearbox}} \times \eta_{\text{coupling}} \times \eta_{\text{motor}}\):

\[ P_{\text{motor}} = \frac{|P_{\text{drum}}| \times \text{SF}}{\eta} \quad \text{(motoring, } F \ge 0\text{)} \] \[ P_{\text{motor}} = |P_{\text{drum}}| \times \text{SF} \times \eta \quad \text{(regenerative, } F < 0\text{)} \]

The larger of the loaded and empty requirements sets the motor. In regenerative operation the drive-train losses absorb part of the braking power before it reaches the motor, which runs as a generator (CEMA 7th ed., declined conveyors; DIN 22101).

When a load case is regenerative the calculation flags it and reports the braking power fed back through the motor:

\[ P_{\text{regen}} = |P_{\text{drum}}| \times \eta \]

The drive must absorb this power continuously: use a regenerative (active front end) VFD or a braking resistor, plus a fail-safe mechanical brake sized for the loaded decline, because the belt runs away on loss of power. A decline that is regenerative loaded and motoring empty needs a drive that controls torque in both directions. Regenerative power is largest with low running resistance, so re-check with a lower-bound friction factor (DIN 22101 suggests \(f \approx 0.012\)–0.016 for braked downhill conveyors). In regenerative operation the tight and slack strands swap at the drive pulley: \(T_1\) is on the strand leaving the drive, so locate the take-up at the low-tension point for that case.

The ratio of tight-side to slack-side tension around the drive drum:

\[ \frac{T_1}{T_2} = e^{\mu \cdot \alpha} \]

Combined with the traction force relationship \(T_1 - T_2 = F_{\text{design}}\):

\[ T_2 = \frac{F_{\text{design}}}{e^{\mu \cdot \alpha} - 1} \] \[ T_1 = T_2 + F_{\text{design}} \]

Where \(\mu\) is belt–pulley friction coefficient and \(\alpha\) is wrap angle in radians.

\[ T_{\text{drum}} = F_{\text{design}} \times \frac{D_{\text{drum}}}{2} \quad (\text{N·m}) \]

The drum and gearbox torque reverse in regenerative operation; the rating uses the magnitude.

\[ n_{\text{drum}} = \frac{v}{\pi \cdot D_{\text{drum}}} \times 60 \quad (\text{rpm}) \] \[ \text{GR} = \frac{n_{\text{motor}}}{n_{\text{drum}}} \]

Take-up travel for thermal expansion and elastic elongation:

\[ \Delta L = 0.02 \cdot L + 0.005 \cdot L = 0.025 \cdot L \quad (\text{m, min 0.15 m}) \]

Take-up counterweight force:

\[ F_{\text{take\text{-}up}} = 1.5 \times T_2 \]

\(T_2\) is from the design (largest \(|F|\)) load case.

pytest calcs/components/c0026_conveyor_design/ -v
pytest calcs/test_c0026* -v
pytest -m validation mcp_public/tests/test_validation_corpus.py -k c0026 -q