Design Proposal for an Ore Rotary Drum Dryer

I. Equipment Design Conditions:

1. Material Conditions:

Material Name: Ore

Initial Moisture Content: 25% (for calculation purposes);

Final Moisture Content: 1% (for calculation purposes);

Throughput: 6,250 kg/h;

Wet Material Temperature: Ambient temperature (θ1 = 20°C);

Hot air temperature: t1 = 450°C;

Outlet air temperature: t2 = 100°C;

Post-drying material temperature: θ2 = 110°C;

Ambient air temperature parameters: Air temperature t0 = 20°C, air humidity d0 = 0.01 k****/kg of dry air;

Specific heat capacity of the material (absolute dry): Cs = 0.4 kcal/kg·°C

Heating method: Direct-fired gas-fired hot air furnace;

Power supply: Voltage 380 V ± 10 V, 50 Hz;

II. Working Principle:

The main body of the rotary drum dryer is a slightly inclined, rotating cylindrical drum. Wet material is continuously and metered into the kiln head via a belt conveyor (provided by the user). As it passes through the interior of the cylinder, it comes into effective contact with the hot air flowing through the cylinder and is dried. The dried product exits from the kiln tail. Since the material particles generate fine dust, a baghouse dust collection system must be installed; the exhaust gas is discharged into the atmosphere by an induced draft fan. During the drying process, the material moves from the higher end to the lower end under the influence of gravity, aided by the slow rotation of the cylinder. The heat carrier used in the drying process is the heat generated by a hot-air furnace. Due to the high quality requirements for the final product, a shut-off feeder is used for continuous feeding. Although the rotary drum dryer is one of the oldest types of drying equipment, it is still widely used today in the metallurgical, building materials, and chemical industries. Since the material requires a final moisture content of nearly zero and has no temperature limitations, a counter-current drying configuration is employed.

III. Features:

1. High production capacity and capable of continuous operation;

2. Simple structure and easy to operate;

3. Few malfunctions and low maintenance costs;

4. Wide range of applications; it can be used to dry granular materials and is particularly effective for highly adhesive materials;

5. High operational flexibility; production allows for significant fluctuations in output without affecting product quality;

6. Easy to clean;

III. Material Balance:

Water removal rate: W_(water) = 2000 kg/h

Raw material supply: W_(raw) = W_(product) + W_(water) = 6250 + 2000 = 8250 kg/h

IV. Heat Balance and Hot Air Mass Flow Rate Calculations:

1. Absolutely dry material flow rate: G = (1 – 0.01) × 6250 = 6288 kg/h

2. Evaporation rate: W = 2000 kg/h H₂O

3. Assuming an ambient temperature of 20°C and a latent heat of evaporation of 595 kcal/kg·°C, the total heat required for the entire system consists of the following four components: Q_(total) = Q₁ + Q₂ + Q₃ + Q₄,

3.1 Qtotal—provided by the drying medium (450°C) air. According to the law of conservation of energy, the heat supplied by the drying medium equals the heat required by the entire system; the required heat is Qtotal (kcal/h);

3.2 Q1—heat required for water evaporation; the mixed temperature is taken as 130°C;

3.3 Q2—heat required for material absorption; the specific heat capacity is taken as 0.40, and the discharge temperature is 110°C;

3.4 Q3—Heat of moisture removal; the moisture removal temperature is taken as 80°C, and the required air flow rate is L (kg/h);

Q1 = W_(water) (595 + 0.45 × t₂ − 1 × t₀) = 2000 (595 + 0.45 × 100 − 1 × 20)

= 1,240,000 kcal/h

Q2 = W_(production) {cs × (1 − ω₂) + 1 × ω₂} × (θ₂ − θ₁)

= 6250 {0.4 × (1 − 1%) + 1 × 1%} × (110 − 20) = 228,375 kcal/h

Q3 = 0.15 × (Q1 + Q2) = 0.15 × (1,240,000 + 228,375) = 220,256 kcal/h

L = (Q1 + Q2 + Q3) ÷ [ (i1 - i2′) - 0.1 × 0.24 × (t2 - t0) ]

i₁ – i₂′ = (0.24 + 0.45 d₁) × (t₁ – t₂) = (0.24 + 0.45 × 0.01) × (450 – 100) = 85.58 kcal/kg

L = 〖(1,240,000 + 228,375 + 220,256)〗 ÷ 〖85.58 - 0.1 × 0.24 × (450 - 20)〗 = 19,804 kg/h

3.5 Heat required for drying, Q_(dry) = G × (i₁ − i₀)

i₁ − i₀ = (0.24 + 0.45 × d₁) × t₁ + 595 × d₁ − (0.24 + 0.45 × d₀) × t₀ − 595 × d₀

= (0.24 + 0.45 × 0.01) × 450 + 595 × 0.01 − (0.24 + 0.45 × 0.01) × 20 − 595 × 0.01 = 105.14 kcal/kg

Q_(dry) = 19,804 × 105.14 = 2,082,094 kcal/h

To ensure production output, a 10% margin is added, so:

The air requirement for the drying system is: L = 19,804 × 1.1 = 21,784 kg/h

Air volume required for drying at 100°C: 19,804 ÷ 0.96 = 20,629 m³/h

Volume of air required for drying at 20°C: 19,804 ÷ 1.2 = 16,503 m³/h

Calculate the enthalpy for the three different states:

Enthalpy of ambient air: i₀ = (0.24 + 0.45 × 0.01) × 20 + 595 × 0.01

= 10.84 kcal/kg of dry air

Enthalpy of the hot air: i1 = (0.24 + 0.45 × 0.01) × 450 + 595 × 0.01

= 115.98 kcal/kg dry air

Enthalpy at exhaust air temperature and supply air humidity:

i2′ = (0.24 + 0.45 × 0.01) × 100 + 595 × 0.01

= 30.4 kcal/kg dry air

V. Calculation of Drum Diameter:

Based on experience, the mass flow rate can be taken as v = 1.5 kg/m²·s. Therefore, the drum diameter is:

D = 2.22 m; rounded to D = 2.2 m

VI. Calculation of Drum Volume:

Determination of drying intensity: Based on the drying intensity parameters of the ore, ω₁ = 25%,

ω₂ = 1%, t₁ = 450 °C, t₂ = 100 °C, and the drying intensity is A = 25–50 kg/m³·h.

Given that the hot air temperature is t₁ = 450 °C and the exhaust air temperature is t₂ = 100 °C, and after comparison, A is taken as 40 kg/m³·°C, the drum volume is:

V = 1.2 × W_(water) / A = 1.2 × 2000 / 25 = 60 m³

VII. Calculation of Drum Length:

Z = V / (0.7854 × D²) = 60 / (0.7854 × 2.22) = 15.8; rounded to Z = 16 m

Therefore, the equipment model is: HZG-2.2×16 (drum diameter is 2.2 m, drying drum length is 16 m)

VIII. Calculation of Drying Time:

For countercurrent conditions, the drying time is calculated using the following formula:

τ = 0.23 × Z / (S × n⁰.⁹ × D) + 8 × Z × v / (G × dp⁰.⁵)

Where: Z — drum length, Z = 16 m;

S — drum inclination, S = tan β, β — angle of inclination, β = 2.5°, S = 0.0437;

D — Drum diameter, D = 2.2 m;

n — Drum rotational speed, n = 4 r/min;

v — Hot air mass flow rate, 20,629 kg of dry air/h = 5.73 kg of dry air/s

G — Throughput, G = 6,250 kg/h = 1.73 kg/s

dp-----Average product particle size, dp = 100 μm

∴ τ = 0.23 × 16 / (0.0437 × 40.9 × 2.2) + 8 × 16 × 5.73 / (1.73 × 1000.5)

τ = 10.99 + 42 = 53 min

IX. Calculation of Cylinder Wall Thickness:

  For unlined cylinders, when the cylinder diameter D is between 2 and 3.6 m, the cylinder wall thickness δ1 = 16 mm, with the material being a carbon steel and SUS316L composite plate (12+4); the wall thickness δ2 below the roller ring is 32 mm.

X. Material-Lifting Plate Design:

1. A 1-meter-long spiral guide plate is welded at the inlet to reduce raw material buildup at the inlet.

2. A 1-meter-long spiral guide plate is also welded at the outlet to reduce product entrainment and accelerate product discharge.

3. Folded-plate-type lifting material-lifting plates are used in other areas.

Number of material-lifting plates: n1 = (6–10) × D; thus, n1 = 6 × 2.2 = 13.2. To facilitate equal circumferential distribution during manufacturing, n1 is set to 12;

Lifting plate height: Based on design data regarding the relationship between the radial height hR of the lifting plates and the drum diameter D, hR/D is set to 0.15, so hR = 0.15 × 2.2 = 0.33 m. Lifting plate thickness: 5 mm; material: SUS316L

XI. Calculation and Selection of Filter Bags:

1. The volumetric flow rate of exhaust air at 80°C is: L1 = 20,629 m³/h. Assuming a filter bag filtration velocity of 1.5 m/min and a filter bag specification of Φ120 × 2000, the number of filter bags required is 259; we will round this up to 260. The filter bags are made of waterproof, oil-resistant, and anti-static membrane-coated material; The filter mesh count is 500 mesh. The housing material in the material-contact areas is SUS316L. Discharge is via a screw conveyor.

Note: Since the equipment configuration includes one rotary kiln for continuous drying, the baghouse dust collector selected is the GMC-260;

XII. Inducer Fan Selection:

Based on the calculated exhaust volume, a 9-26-10D-55Kw inducer fan should be selected.

Airflow: 21,465 m³/h Pressure: 5,920 Pa Power: 55 kW

Speed: 1,450 r/min, equipped with an air control damper and vibration-damping pads;

XIII. Electrical Controls:

① Display of total line voltage and current, as well as fan voltage and current.

② Electrical Start-up:

(1) Induction fan start-up; (2) Drum rotation controlled by variable frequency drive; (3) Feed shut-off fan controlled by variable frequency drive;

③ Temperature and Pressure Measurement:

(1) Hot air outlet temperature measurement; (2) Baghouse inlet temperature measurement.