Fans and Auxiliary Ventilation

Classification of Fans in Mining Operations

Fans are categorized based on two primary criteria: their location and role within the ventilation circuit, and their mechanical design.

Categorization by Location and Role

  • Main Fans: These fans are responsible for handling the entirety of the air passing through the underground mine. They are the primary movers of the ventilation system.

  • Booster Fans: These are installed within the mine to assist the through-flow of air in specific, discrete areas where the main fan pressure is insufficient to overcome local resistance.

  • Auxiliary Fans: These fans provide ventilation to blind headings or dead-end workings. They typically operate in conjunction with ducting to ensure fresh air reaches the working face.

Categorization by Mechanical Design

  • Centrifugal Fans: These fans move air radially. Key components include the collection scroll or volute, diffuser, suction eye (air inlet), impeller, and blades. They can utilize various blade types such as aerofoil or plate blades and may include variable inlet vanes for control. Discharge is available at various angles.

  • Axial Fans: These fans move air parallel to the axis of rotation. They are characterized by a more compact design compared to centrifugal fans.

  • Mixed Flow Fans: These represent a hybrid design featuring both impeller blades and guide vanes to manage airflow.

Comparative Analysis of Fan Types

Feature

Axial Fans

Centrifugal Fans

Size and Mass

Smaller and more compact

Larger impeller and profile

Pressure

Single-stage limit of approximately 5kPa5\,kPa

Capable of very high pressures

Stall Point

Prone to severe stalling

Features a more robust impeller

Efficiency

High efficiency maintained over a wide range

Smaller range for peak efficiency

Control

Flexible through blade pitch adjustment

Inlet guide vanes used, but they reduce efficiency

Reversibility

Can achieve 40%60%40\%-60\% flow; low efficiency

Requires complex duct and door arrangements

Reliability

Generally less reliable

Generally more reliable

Maintenance

Motor is located in the air stream

Easier access to the motor

Wear

Susceptible to wear

More robust and reliable in high-wear conditions

Noise

Higher noise levels

Quieter performance at the same duty

Note: If speed control is available, there is no significant difference in control efficiency between axial and centrifugal fans.

Fan Pressures and Measurement

Pressure Definitions

  • Fan Total Pressure (TPfanTP_{fan}): Calculated as the difference between the total pressure at the outlet and the inlet.     TPfan=TPoutTPinTP_{fan} = TP_{out} - TP_{in}

  • Fan Velocity Pressure (VPfanVP_{fan}): Defined by the velocity pressure at the fan outlet.     VPfan=VPoutVP_{fan} = VP_{out}

  • Fan Static Pressure (SPfanSP_{fan}): The difference between fan total pressure and fan velocity pressure.     SPfan=TPfanVPfanSP_{fan} = TP_{fan} - VP_{fan}     SPfan=(TPoutTPin)VPoutSP_{fan} = (TP_{out} - TP_{in}) - VP_{out}

Measurement Scenarios for Fan Static Pressure

Fan at the Start of a Duct
  1. Definition: SPfan=TPoutTPinVPoutSP_{fan} = TP_{out} - TP_{in} - VP_{out}

  2. In this scenario, TPin=0TP_{in} = 0 as the inlet is open to the atmosphere.

  3. Expansion: SPfan=(SPout+VPout)VPoutSP_{fan} = (SP_{out} + VP_{out}) - VP_{out}

  4. Result: SPfan=SPoutSP_{fan} = SP_{out}. Measurement is taken at the fan outlet.

Fan at the End of a Duct
  1. Definition: SPfan=TPoutTPinVPoutSP_{fan} = TP_{out} - TP_{in} - VP_{out}

  2. In this scenario, TPout=VPoutTP_{out} = VP_{out}, assuming the air exits to the atmosphere (outlets static pressure is zero).

  3. Result: SPfan=TPinSP_{fan} = -TP_{in}.

  4. Measurement: This is recorded as the total pressure at the fan inlet (which will be a negative value in an exhaust system).

Fan in the Middle of a Duct
  1. At the middle position: SPfan=SPoutTPinSP_{fan} = SP_{out} - TP_{in}.

  2. This is calculated as the difference between the duct outlet static pressure and the inlet fan total pressure.

Manufacturers often use static pressure curves because they have no control over the specific inlet/outlet duct fittings or environmental conditions at the entry/exit points of the installation.

Fan Characteristic Curves and Performance

Pressure-Quantity (P-Q) Relationship

A fan's performance is mapped on a P-Q curve. The theoretical curve is adjusted by accounting for frictional losses and shock losses to produce the actual curve.

  • Stall Zone: A region on the curve where airflow becomes unstable.

  • Open Circuit Capacity: The maximum quantity of air the fan can move when there is zero resistance.

  • Checking for Stall:

    • For a forcing fan: A glove should stick to the guard.

    • For an exhaust fan: A glove should flap away from the guard.

Power and Efficiency Calculations

  • Air Power (WW): Calculated in kilowatts (kWkW).     W=P×Q1000W = \frac{P \times Q}{1000}     where PP is pressure in Pascals (PaPa) and QQ is air quantity in m3/sm^3/s.

  • Fan Efficiency (η\eta):     η=Air Power (for specific pressure P)Consumed Power×100%\eta = \frac{\text{Air Power (for specific pressure } P)}{\text{Consumed Power}} \times 100\%

Operating Point

The operating point of a ventilation system is determined by the intersection of the Fan Characteristic Curve and the Mine Resistance (Circuit) Curve.

Fan Laws: Speed and Density Changes

Impact of Speed Changes (nn)

  • Quantity (QQ): Directly proportional to speed.     Q1Q2=n1n2\frac{Q_1}{Q_2} = \frac{n_1}{n_2}

  • Pressure (PP): Proportional to the square of the speed.     P1P2=(n1n2)2\frac{P_1}{P_2} = (\frac{n_1}{n_2})^2

  • Power (EE): Proportional to the cube of the speed.     E1E2=(n1n2)3\frac{E_1}{E_2} = (\frac{n_1}{n_2})^3

Impact of Density Changes (ρ\rho)

  • Quantity (QQ): Remains unchanged (Q1=Q2Q_1 = Q_2).

  • Pressure (PP): Directly proportional to density.     P1P2=ρ1ρ2\frac{P_1}{P_2} = \frac{\rho_1}{\rho_2}

  • Power (EE): Directly proportional to density.     E1E2=ρ1ρ2\frac{E_1}{E_2} = \frac{\rho_1}{\rho_2}

Fans in Combination

Series Combination

  • Quantity: The same quantity of air passes through both fans.

  • Pressure: The total pressure produced is the sum of the pressures of each fan (Ptotal=PA+PBP_{total} = P_A + P_B).

  • Application: Usually used with axial fans in long duct runs. Fans should have a similar duty curve.

Parallel Combination

  • Pressure: The pressure drop across both fans is identical.

  • Quantity: The total quantity is the sum of the quantities of each fan (Qtotal=QA+QBQ_{total} = Q_A + Q_B).

  • Application: Used with axial or centrifugal fans for surface or underground booster applications. Curves should be of similar duty.

Natural Ventilation Pressure (NVP) Correction

Natural ventilation can assist or hinder the mechanical fan. The system resistance curve is adjusted to account for NVP:

  • Standard Circuit: P=RQ2P = R \cdot Q^2

  • With Natural Ventilation: P=RQ2NVPP = R \cdot Q^2 - NVP

If NVP is positive, it reduces the pressure the fan must provide to achieve a certain quantity; if negative, it increases it.

Numerical Examples

Determining Operating Point

Given:

  • Fan P-Q Data: (100m3/s,2200Pa)(100\,m^3/s, 2200\,Pa), (145m3/s,2000Pa)(145\,m^3/s, 2000\,Pa), (190m3/s,1500Pa)(190\,m^3/s, 1500\,Pa), (215m3/s,1000Pa)(215\,m^3/s, 1000\,Pa), (235m3/s,500Pa)(235\,m^3/s, 500\,Pa), (250m3/s,0Pa)(250\,m^3/s, 0\,Pa).

  • Scenario A: Mine Resistance R=0.034Ns2/m8R = 0.034\,Ns^2/m^8.

  • Scenario B: Mine Resistance R=0.020Ns2/m8R = 0.020\,Ns^2/m^8.

Steps:

  1. Calculate system pressures using P=RQ2P = R \cdot Q^2.

    • For R=0.034R = 0.034: At 200m3/s200\,m^3/s, P=0.034×(200)2=1360PaP = 0.034 \times (200)^2 = 1360\,Pa.

  2. Plot these against the fan curve.

  3. The intersection for R=0.034R = 0.034 is approximately 200m3/s200\,m^3/s at 1360Pa1360\,Pa.

Speed and Power Adjustment

Given:

  • Original Duty: 120m3/s120\,m^3/s at 1400Pa1400\,Pa.

  • Original Speed: 360rpm360\,rpm.

  • Efficiency (η\eta): 81%81\%.

  • New Speed: 515rpm515\,rpm.

Calculation:

  1. New Quantity: Q2=120×515360=171.7m3/sQ_2 = 120 \times \frac{515}{360} = 171.7\,m^3/s.\n2. New Pressure: P2=1400×(515360)2=2865PaP_2 = 1400 \times (\frac{515}{360})^2 = 2865\,Pa.

  2. Original Power: W1=120×14001000×0.81=207.4kWW_1 = \frac{120 \times 1400}{1000 \times 0.81} = 207.4\,kW.

  3. New Power: E2=207.4×(515360)3=607kWE_2 = 207.4 \times (\frac{515}{360})^3 = 607\,kW.

Multi-Factor Correction and Parallel Operation

Scenario: A fan operates at 750rpm750\,rpm with density ρ1=1.2kg/m3\rho_1 = 1.2\,kg/m^3. Change to 850rpm850\,rpm and ρ2=1.16kg/m3\rho_2 = 1.16\,kg/m^3, with two fans in parallel. Mine resistance R=0.015Ns2/m8R = 0.015\,Ns^2/m^8, efficiency 76%76\%, and NVP=+250PaNVP = +250\,Pa.

Correcting Single Fan (Point: 100m3/s,2200Pa100\,m^3/s, 2200\,Pa):

  1. Quantity for Speed: Q2=100×850750=113.3m3/sQ_2 = 100 \times \frac{850}{750} = 113.3\,m^3/s.

  2. Pressure for Speed: Ptemp=2200×(850750)2=2825.78PaP_{temp} = 2200 \times (\frac{850}{750})^2 = 2825.78\,Pa.

  3. Pressure for Density: Pfinal=2825.78×1.161.2=2732.59PaP_{final} = 2825.78 \times \frac{1.16}{1.2} = 2732.59\,Pa.

Parallel Calculation:

  • At the corrected pressure of 2732Pa2732\,Pa, two fans provide 113×2=226m3/s113 \times 2 = 226\,m^3/s.

Operating Point with NVP:

  1. Resistance Data (P=0.015Q2250P = 0.015 \cdot Q^2 - 250):

    • At 400m3/s400\,m^3/s: P=(0.015×4002)250=2400250=2150PaP = (0.015 \times 400^2) - 250 = 2400 - 250 = 2150\,Pa.

  2. The resulting operating point for two parallel corrected fans and the NVP-adjusted resistance curve is calculated by plotting or solving for the intersection.

  3. Power Consumed (at Q=385m3/sQ = 385\,m^3/s, P=2250PaP = 2250\,Pa, η=76%\eta = 76\%):     W=2250×3851000×0.76=1139.8kWW = \frac{2250 \times 385}{1000 \times 0.76} = 1139.8\,kW