Engineering · Diploma thesis
Coal Conveyor Belt System
Introduction
Design of a complete coal transport system: the conveying section — belt, drums, idlers and frame — and the drive, including the choice of electric motor and gearbox. Calculated and modelled with computer methods.
Details
- Type
- Engineering · Diploma thesis (2019)
- Capacity
- 276.48 t/h
- Drive power
- 24 kW (30 kW motor selected)
- Software
- SolidWorks · KissSoft · KeyShot
Project & visuals: MayCo.
Mechanical engineering
A conveyor for a 600 MW power plant.
01 / The brief
Move 276 tonnes of coal an hour.
A power plant of 600 MW, split into three 200 MW blocks, has to be supplied with coal. The first step is to turn the plant output into a coal flow, and then to design the belt conveyor and its drive around that number.
- Plant
- 600 MW · 3 blocks × 200 MW
- Plant efficiency
- 39% (coal 20 MJ/kg → 7.8 MJ/kg usable)
- Coal per MW
- 460.8 kg/h
- Per 200 MW block
- 92.16 t/h
- Whole plant
- 276.48 t/h
- Route
- 10° incline · 20 m lift · 115.21 m + 25 m
- Material
- Coal · 0.8 t/m³ · lumps 70–150 mm
02 / Belt and idlers
Sizing the belt.
From the load tables the permitted belt speed for this material is 2.5 m/s. With three-roll troughed idlers at 30° and a 20° surcharge angle, a 650 mm belt carries the required flow. The belt was first assumed at 315 N/mm; the final check shows a maximum tension of only about 18 N/mm, so a 200 N/mm belt is enough.
- Belt speed
- 2.5 m/s
- Belt width
- 650 mm
- Idler set
- 3 rolls · 30° trough
- Idler spacing
- 1.65 m carrying · 3 m return
- Idlers
- Ø108 mm · bearings 6204 · 250 mm carrying / 750 mm return
- Carrying idler load
- 65.74 daN static · 81.25 daN dynamic (centre roller 52.81 daN)
- Return idler load
- 29.2 daN static · 33.98 daN dynamic
03 / Forces and drive power
From resistance to kilowatts.
The tangential force at the drive drum is the resistance of the carrying side (970 daN) plus the return side (−141 daN, helped by gravity). Together with the 200° wrap around a rubber-lagged drum, this sets the belt tensions and the counterweight.
Fu = Fa + Fr = 970 − 141 = 829 daN P = Fu · v / (100 · η) = 829 · 2.5 / (100 · 0.86) ≈ 24 kW T2 = Fu · Cw = 829 · 0.42 = 348 daN T1 = Fu + T2 = 1177 daN T3 = T2 + Fr = 207 daN Counterweight force Tg = 2 · T0 = 814.6 daN
04 / Drums and shafts
Drive and return drums.
Both drums are Ø315 mm. The drive drum turns at 151 rpm and carries a torque of 1512 Nm. Its shaft is sized on the combined bending and torsion moment; the return drum is only bent. The drive shaft needs at least 61 mm, the return shaft 35 mm, so the 60 mm catalogue bore is sufficient.
Drive drum: Cp = √((T1+T2)² + qt²) = 1525 daN
Mf = Cp/2 · ag = 114 daNm Mt = P/n · 954.9 = 151.8 daNm
Mif = √(Mf² + 0.75·Mt²) = 173 daNm → W = 22 122 mm³ → d ≈ 61 mm
Return drum: Cp = 416 daN · Mf = 31.2 daNm → W = 3 989 mm³ → d ≈ 35 mm05 / Motor and gearbox
Selecting the drive.
A 30 kW (40 HP) four-pole induction motor covers the 24 kW requirement, and a helical gearbox with a ratio of 8.88 brings the speed down to about 158 rpm at the drum shaft.
- Motor
- 30 kW (40 HP) · 4 poles · 1770 rpm at 60 Hz · 159 Nm · efficiency 94.1% · IP55
- Gearbox
- Helical · ratio 8.88 · output 157.66 rpm · nominal torque 2000 Nm
- Gearbox details
- Service factor 1.3 · efficiency 0.94 · Ø60 output shaft · mounting M1
06 / Result
A complete, calculated system.
The supporting frame is left open on purpose: it is designed per installation, once the terrain and the structural loads are known. The 3D model shows the section that matters most — idlers, drums, belt and the drive arrangement.
- 2.5 m/sBelt speed
- 650 mmBelt width, 315 N/mm assumed
- Ø108 mmIdlers at 1.65 m / 3 m
- Ø315 mmDrive and return drums
- ≥ 24 kWRequired drive power