Belt Conveyor Capacity & Speed Calculation | DGMS Mechanical Notes

🔹 Introduction

One of the most vital design aspects in conveyor engineering is determining the capacity and speed of the belt conveyor. Capacity defines how much material the conveyor can transport per unit time, while belt speed determines the efficiency, stability, and wear rate of the system.
For DGMS exam purposes, engineers must understand how to calculate volumetric capacity (Q) and mass flow rate (M) using key parameters such as belt width, surcharge factor, and material density.
The design formulas —Q=C×b2×vQ = C \times b^2 \times vQ=C×b2×v (for horizontal conveyors)Q=k×b2×vQ = k \times b^2 \times vQ=k×b2×v (for inclined conveyors)
— are essential to determine system throughput.
The blog also explains how material properties, belt inclination, and width affect selection of belt speed. Standard belt widths and surcharge angles used in the industry are summarized for quick reference.
Understanding these parameters helps design conveyors that balance performance with safety and economy — a crucial skill for mining professionals and DGMS aspirants.
The design of a belt conveyor system primarily revolves around determining the capacity and speed required to handle a specific load of material over a fixed distance.
Conveyor capacity defines the quantity of material transported per unit time, while belt speed influences both capacity and mechanical efficiency. DGMS (Directorate General of Mines Safety) requires that every belt conveyor system in mines be designed, operated, and maintained as per safety and mechanical design standards under CMR 2017 – Regulation 88 and 131.

🔹 1. Formula for CapacityFor horizontal conveyors:

Q=Cb2vQ = C b^2 vQ=Cb2v

For inclined conveyors:Q=kb2vQ = k b^2 vQ=kb2vWhere,
C = surcharge factor,
k = flowability factor,
b = effective width,
v = speed.Mass capacity (M):M=ρQM = ρ QM=ρQ

Conveyor Capacity (Q):
It is the volume of material transported by the conveyor per unit time.
It depends on three main factors:

Belt width (B)

Belt speed (v)

Cross-sectional area of material (A)

2.1 Formula for Conveyor Capacity

For a horizontal belt conveyor, $Q = C \times b^2 \times v$ Q=C×b2×v Where,

Q = Capacity (m³/s)

C = Surcharge factor (depends on material type & trough angle)

b = Effective belt width (m) = 0.9B – 0.05

v = Belt speed (m/s)

For mass flow, $M = ρ \times Q$ M=ρ×Q Where ρ = Material bulk density (kg/m³)
→ M = kg/s or Tonnes per Hour (TPH)
2.2 Inclined Conveyor Capacity

When the conveyor is inclined: $Q_i = Q \times \cos(θ)$ Qi=Q×cos(θ) where θ is the angle of inclination.
Recommended Belt Speeds:
Material Belt Speed (m/s)
Fine coal / ore 1.5 – 2.0
Crushed stone 2.0 – 2.5
Sand & Gravel 2.5 – 3.0
Iron ore 3.0 – 4.0
Light materials 1.0 – 1.5
Higher speeds increase capacity but also reduce belt life and safety margin.

🔹 Example

Material	Belt Speed (m/s)
Fine coal / ore	1.5 – 2.0
Crushed stone	2.0 – 2.5
Sand & Gravel	2.5 – 3.0
Iron ore	3.0 – 4.0
Light materials	1.0 – 1.5

Material density = 1200 kg/m³
Belt width = 650 mm,
v = 1.75 m/s,
C = 2.35×10⁻⁴Q=2.35×10−4(0.65)2×1.75=1.6×10−4m3/sQ = 2.35×10^{-4} (0.65)^2 × 1.75 = 1.6×10^{-4} m³/sQ=2.35×10−4(0.65)2×1.75=1.6×10−4m3/s

🔹 3. Factors Affecting Speed
Material type
Inclination angle
Conveyor length
Belt type & power
Maintenance & cost

🔹 5. Factors Affecting Capacity & Speed

Material properties – density, lump size, moisture.

Troughing angle – larger angles increase carrying volume.

Inclination – reduces effective capacity.

Idler spacing – affects belt sag and load support.

Speed – directly affects capacity but influences wear.

🔹 6. DGMS Guidelines for Conveyor Design

All design calculations must be verified by a qualified mechanical engineer.

Overloading beyond rated capacity is prohibited.

Belt sag ≤ 2% of idler spacing.

Fire-resistant belts are compulsory in underground mines.

Design speed must ensure no spillage and smooth loading/unloading.

🔹 25 DGMS-Style MCQs

1️⃣ The capacity of a belt conveyor depends on —
a) Belt width & belt speed
b) Pulley diameter only
c) Motor power
d) Drive type
e) Belt thickness
✅ Answer: a) Belt width & belt speed
Solution: Capacity ∝ B × v × C (cross-section & speed).
2️⃣ The formula for volumetric capacity is —
a) Q = πDL
b) Q = Cb²v
c) Q = T/v
d) Q = M/ρ
e) None
✅ Answer: b) Q = Cb²v
Solution: Standard equation for horizontal conveyors.
3️⃣ For mass capacity, M = —
a) ρ × Q
b) Q/ρ
c) Q × θ
d) Q × 2
e) None
✅ Answer: a) ρ × Q
Solution: Multiply volumetric capacity by bulk density.
4️⃣ Effective belt width (b) = — a) B
b) 0.9B – 0.05
c) B/2
d) 1.2B
e) 0.8B
✅ Answer: b) 0.9B – 0.05
Solution: Corrected for side edge clearances.
5️⃣ For an inclined conveyor, capacity reduces by factor — a) sinθ
b) cosθ
c) tanθ
d) 1/θ
e) 1
✅ Answer: b) cosθ
Solution: Horizontal projection of material flow decreases with inclination.
6️⃣ Unit of volumetric capacity (Q) is — a) kg/s
b) m³/s
c) m/s
d) N·m
e) None
✅ Answer: b) m³/s
Solution: Q represents volume flow rate.
7️⃣ Increasing belt speed by 25% increases capacity by — a) 10%
b) 25%
c) 50%
d) 75%
e) 100%
✅ Answer: b) 25%
Solution: Q ∝ v, hence direct proportionality.
8️⃣ Surcharge factor (C) depends on — a) Belt color
b) Material angle of repose & trough angle
c) Motor power
d) Belt width
e) Pulley size
✅ Answer: b) Material angle of repose & trough angle
Solution: Defines the shape of material cross-section.
9️⃣ Belt speed for coal conveyors is generally — a) 0.5 m/s
b) 1.0–2.5 m/s
c) 4–6 m/s
d) 10 m/s
e) >15 m/s
✅ Answer: b) 1.0–2.5 m/s
Solution: Optimum for coal and medium materials.
🔟 High belt speed results in — a) Lower capacity
b) Higher spillage & wear
c) Reduced power
d) More safety
e) None
✅ Answer: b) Higher spillage & wear
Solution: Excess speed causes dynamic instability.
11️⃣ Capacity of 1 m wide belt at 2 m/s (C=0.075) ≈ a) 0.05 m³/s
b) 0.15 m³/s
c) 0.2 m³/s
d) 0.3 m³/s
e) 1 m³/s
✅ Answer: b) 0.15 m³/s
Solution: Q = Cb²v = 0.075 × (0.85)² × 2 ≈ 0.15.
12️⃣ If density = 1800 kg/m³, mass flow = a) 270 kg/s
b) 324 kg/s
c) 450 kg/s
d) 1800 kg/s
e) 600 kg/s
✅ Answer: b) 324 kg/s
Solution: M = ρ × Q = 1800 × 0.18 = 324.
13️⃣ Belt speed (v) unit is — a) m/s
b) rpm
c) N·m/s
d) kg/h
e) None
✅ Answer: a) m/s
Solution: Linear velocity of belt motion.
14️⃣ Troughing angle for bulk material = a) 10°
b) 20°
c) 35°
d) 60°
e) 80°
✅ Answer: c) 35°
Solution: Standard for 3-roll idler sets.
15️⃣ Belt width for 1000 TPH coal conveyor is — a) 400 mm
b) 650 mm
c) 1000 mm
d) 1600 mm
e) 2400 mm
✅ Answer: c) 1000 mm
Solution: Standard design from DGMS reference chart.
16️⃣ DGMS allows belt sag not exceeding — a) 1%
b) 2%
c) 3%
d) 5%
e) 10%
✅ Answer: b) 2%
Solution: Prevents spillage and misalignment.
17️⃣ Belt load cross-section mainly depends on — a) Belt speed
b) Idler trough angle
c) Belt length
d) Pulley lagging
e) Motor RPM
✅ Answer: b) Idler trough angle
Solution: Larger angle = more carrying volume.
18️⃣ Typical maximum conveyor speed for bulk solids — a) 2 m/s
b) 3.5 m/s
c) 5 m/s
d) 8 m/s
e) 10 m/s
✅ Answer: c) 5 m/s
Solution: Beyond this, spillage and vibrations increase.
19️⃣ Capacity increases linearly with — a) Belt speed
b) Belt weight
c) Belt length
d) Inclination angle
e) Idler spacing
✅ Answer: a) Belt speed
Solution: Direct proportionality in design formula.
20️⃣ Capacity decreases with — a) Higher inclination angle
b) Larger belt width
c) Reduced troughing
d) None
✅ Answer: a) Higher inclination angle
Solution: Due to reduced horizontal projection (cosθ).
21️⃣ For fine material, suitable belt speed — a) 0.5–1.0 m/s
b) 1.5–2.0 m/s
c) 3.0–4.0 m/s
d) 5–6 m/s
e) 7 m/s
✅ Answer: b) 1.5–2.0 m/s
Solution: Prevents dust and spillage.
22️⃣ The capacity formula constant “C” has units of — a) Dimensionless
b) m²
c) kg/m³
d) m³/s
e) N/m
✅ Answer: a) Dimensionless
Solution: Surcharge factor is a non-dimensional coefficient.
23️⃣ Belt conveyor design is based on standard —
a) DGMS & CEMA recommendations
b) BIS 11592 only
c) Manufacturer’s data
d) Operator choice
e) None
✅ Answer: a) DGMS & CEMA recommendations
Solution: These provide design charts and safety limits.
24️⃣ Power required ∝ — a) Belt tension × speed
b) Belt weight × distance
c) Motor current × voltage
d) Torque only
e) None
✅ Answer: a) Belt tension × speed
Solution: P = Te × v.
25️⃣ Over-speed increases —
a) Capacity safely
b) Belt wear, spillage, and power loss
c) Stability
d) Life span
e) None
✅ Answer: b) Belt wear, spillage, and power loss
Solution: Higher dynamic loads cause mechanical strain and safety issues.
🔚 8. Conclusion
Belt conveyor capacity and speed are critical parameters that define the system’s performance and safety.
Understanding design formulas, speed selection, and DGMS recommendations ensures both efficiency and operational safety.
A well-designed system minimizes power loss, spillage, and wear, ensuring long-term reliability in mining environments.
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Belt Conveyor Capacity & Speed Calculation | DGMS Mechanical Notes

🔹 Introduction

🔹 1. Formula for CapacityFor horizontal conveyors:

Q=Cb2vQ = C b^2 vQ=Cb2v

For inclined conveyors:Q=kb2vQ = k b^2 vQ=kb2vWhere,C = surcharge factor,k = flowability factor,b = effective width,v = speed.Mass capacity (M):M=ρQM = ρ QM=ρQ

Conveyor Capacity (Q): It is the volume of material transported by the conveyor per unit time. It depends on three main factors: Belt width (B) Belt speed (v) Cross-sectional area of material (A) 2.1 Formula for Conveyor Capacity

Material density = 1200 kg/m³Belt width = 650 mm,v = 1.75 m/s,C = 2.35×10⁻⁴Q=2.35×10−4(0.65)2×1.75=1.6×10−4m3/sQ = 2.35×10^{-4} (0.65)^2 × 1.75 = 1.6×10^{-4} m³/sQ=2.35×10−4(0.65)2×1.75=1.6×10−4m3/s

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For inclined conveyors:Q=kb2vQ = k b^2 vQ=kb2vWhere,
C = surcharge factor,
k = flowability factor,
b = effective width,
v = speed.Mass capacity (M):M=ρQM = ρ QM=ρQ

Conveyor Capacity (Q):
It is the volume of material transported by the conveyor per unit time.
It depends on three main factors:

Belt width (B)

Belt speed (v)

Cross-sectional area of material (A)

2.1 Formula for Conveyor Capacity

Material density = 1200 kg/m³
Belt width = 650 mm,
v = 1.75 m/s,
C = 2.35×10⁻⁴Q=2.35×10−4(0.65)2×1.75=1.6×10−4m3/sQ = 2.35×10^{-4} (0.65)^2 × 1.75 = 1.6×10^{-4} m³/sQ=2.35×10−4(0.65)2×1.75=1.6×10−4m3/s