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AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Basic Mine VentilationBasic Mine Ventilation
Fundamentals of AirflowFundamentals of Airflow
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Agricola (1556) noted “The outer air flows spontaneously into the caverns of the earth, and when it can pass through them it comes out again. This, however comes about in different ways, for in spring and summer it flows into the deeper shafts. . . . and finds its way out of the shallower shafts; but in autumn and winter, it enters the upper tunnel or shafts and comes out the deeper ones. This change of flow in the air currents occurs in the temperate regions at the beginning of spring and the end of autumn.”
Natural Ventilation Pressure (NVP)Natural Ventilation Pressure (NVP)
Mine workings
WINTER
Shaf
t
Shaf
t
Mine workings
SUMMER
Shaf
t
Shaf
t
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AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Air movement caused by conversion of heat to mechanical energy and this was recognised in the 1700’s when fires were used to create air currents
Natural Ventilation Pressure (NVP)Natural Ventilation Pressure (NVP)
Surface
0.5 kg of coal increased the temperature by 30C to 40C per 0.5 m3/s
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Air movement caused by conversion of heat to mechanical energy.
NVP is estimated as the difference of the pressure exerted by the column of air in the shafts
Calculation of NVPCalculation of NVP
Mine workings
Shaf
t Len
gth
= 20
0m
Shaf
t Len
gth
= 10
0m
The mass of a column of air is determined as the density multiplied by the length, and to obtain the pressure this is multiplied by gravitational acceleration (9.81 m/s)
(m) column the of heigh h(m/s) naceleratio nalgravitatio g
)(kg/m air of column the ofdensity mean
(Pa) air of column theby exerted pressure PghP
3C
C
===ρ
=
ρ=
• 3
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Calculation of NVPCalculation of NVP
Mine workings
Shaf
t Len
gth
= 20
0m A
rea
= 10
m2
Shaf
t Len
gth
= 18
0m A
rea
= 10
m2
Density Method
Assume mean density of 1.10 kg/m3 in shaft A and 1.05 kg/m3 in shaft B
NVPA = 1.10 x 9.81 x 200 = 2,158 Pa
NVPB = 1.05 x 9.81 x 180 = 1,854 Pa
The difference between these pressures is the NVP = 2,158 - 1,854 = 304 Pa
Shaft A Shaft B
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
The effect of outside temperature is cancelled if shaft collars are at the same level
Water flowing down the shaft will cause the air to flow in the same direction. In exhausting shaft it increases the resistance and decreasing the flow created by the fan
MINE WORKINGS
Water flowing down an intake shaft decreases resistance
Water flowing down an exhausting shaft
will increase the resistance
NVPNVP
• 4
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
1. For air to flow from one point to another there must be a difference in pressure between the two points.
2. Air will always flow from a high pressure to a low pressure and will continue to flow as long as the pressure is maintained.
3. The greater the difference in pressure the greater the quantity of airflow. (P∝Q)
4. Any resistance to pressure will reduce the quantity of airflow.
5. As the resistance between the two points increases the quantity of airflow decreases
Elementary laws of airflowElementary laws of airflow
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Ventilation Pressure (Pa)
Static pressure (bursting pressure) is the potential energy measured with a side tube normal to the direction of flow
Velocity pressure is the kinetic energy and is calculated v = velocity of the air (m/s)ρ = density of the air (kg/m3)
Total pressure is the sum of the static and velocity pressures. Measured with a facing tube parallel to the flow of air.
2vρ
=P2
V
Atmospheric Pressure • At sea level =10,000 kg/m2
=1 atmosphere =1.013 Bar =101.325 kPa =1013 mb
Barometric Pressure• The pressure measured with a barometer.
PressurePressure
• 5
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Total pressure is the sum of the static and velocity pressures. Measured with a facing tube parallel to the flow of air
Static pressure (bursting) is the potential energy and is measured with a side tube, normal to the direction of airflow.
Velocity pressure is the kinetic energy can be indirectly measured with apitot tube but is normally calculated from
2vρ
=P2
V
pressure VelocityPressureStaticpressureTotal +=
(m/s) air the of Velocity (kg/m air the of Density
(Pa) pressure Velocity3
==ρ
=
v)
PV
Ventilation PressureVentilation Pressure
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Pitot TubePitot Tube
• 6
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Static pressure inside the duct holds it open
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Mine environmental engineering more frequently use pressure difference than absolute pressure. The differences are indicated on a gauge and measured relative to atmosphere. Often referred to pressure drop
Direction of flow
PS
Static Pressure
Side gauge
PT
Total Pressure
Facing gauge
PV
Velocity Pressure
Combined side & facing gauge
Ventilation PressureVentilation Pressure
• 7
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Pressure loss & flowPressure loss & flow
P∝QLaminar flow
P∝Q2Turbulent Velocity
Pres
sure
loss
Turbulent
Transition
Laminar
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Ventilation EquationVentilation Equation
P = Pressure (Pa)R = Resistance (Ns2/m8)Q = Volume (Quantity) of air (m3/s)
QP
=R
2QP
=R
P∝Q
Holds true for turbulent flowP∝Q2
Holds true for laminar flow
• 8
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Other factorsOther factors
Atkinson postulated that the flow also varied for other reasons such as changes in:
Area
Shape
Length
Roughness of the wall
Density of the gas
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Atkinson's EquationAtkinson's Equation
P = Pressure (Pa)R = Resistance (Ns2/m8)Q = Quantity of air (m3/s)k = Coefficient of friction (Ns2/m4)C = Circumference of Airway (m)L = Length of Airway (m)A = Cross sectional area of airway (m2)ρ = Density of the air (kg/m3)ρStd = Density of “standard” air (1.2 kg/m3)
2
Std3 Q×ρ
ρ×
AkCL
=P
• 9
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
“k” Factor“k” Factor
The calculation ‘k’ is complicated by variables
Velocity
Density
Viscosity
Hydraulic diameter
Reynolds number (itself is complicated by variable velocity, surface roughness, the ration of the roughness to the diameter of the airway).
Estimate using,
but will tend to underestimate for mines
diameter Hydraulic Droughness Relative D2log274.11
H
H
==∈
∈−=
λ0.3 use Blasted
0.01 use Raisebored0.007 lined Concrete
are rfo used be can that values Some ∈
( )
width bheight a
b a2
4ab
diameter Hydraulic DH
==
+=
=
42 m/Ns 6.67
k
=
λ=
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Example ‘k’Example ‘k’
Assume standard density for air (1.2kg/m3)
Blasted rock walled airway 5.5m x 5.0m
Assume relative roughness for rockwalled drive 0.3
Calculate the hydraulic diameter
Calculate Lamda
Calculate k
( )m 5.2
b a2
4ab
diameter Hydraulic DH
=+
=
=
0.076 D2log274.11
H=
∈−=
λ
)m/(Ns 0.011428 6.67
k
42=
λ=
• 10
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Typical “k” factorsTypical “k” factors
0.0030 to 0.0065Flexible ducting0.0030 to 0.0035Normal rigid ducting
0.0035 to 0.0040Concrete surface0.0010 to 0.0200Rock Surfaced0.0035 to 0.0050Raisebored
0.0028Smooth Pipe“k” Factor (Ns2/m4)Airway Type
Typical ranges shown below. (In mines or in doubt err on the high side)
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Example Example -- Pressure differencePressure difference
Calculate the pressure difference (drop) across a concrete lined shaft 3.5 m diameter 350m long. The airflow measure with an anemometer is 175 m3/s. Assume standard density for air (1.2 kg/m3).
What is the pressure drop for the same flow if the shaft was rock walled.
350m
3.5 m diameter
175 m3/s
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AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Concrete lined shaft
From tablek = 0.004 (Ns2/m8)and A = 9.62 (m2)C = 11.0 (m)l = 350 (m)Q = 175 (m3/s)
From Atkinson’s Equation
PressureP = 310 (Pa)
Rock wall
From tablek = 0.015 (Ns2/m8)and A = 9.62 (m2)C = 11.0 (m)l = 350 (m)Q = 175 (m3/s)
From Atkinson’s Equation
PressureP = 1,180 (Pa)
Atkinson's EquationP = kCl Q2
A3
350m
3.5 m diameter
175 m3/s
Example Example -- Pressure differencePressure difference
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Shock loss, turbulence, flow separation = Pressure loss due to change in direction
Calculation is complex
Configuration and flow through the element
Angle of the change in direction
Degree of the abruptness of the change
Radius of curvature
Ratio of the radius to width of the airway
Aspect ratio between height and width of the airway
Airway roughness
Shape of the airway immediately before and after the change in direction
Velocity of the air
The number and type of complex elements. (for example any vertical airway - bend, contraction, expansion, bend)
VShock XP=PX = Shock loss factor
PV = Velocity Pressure
The “X” factor i.e. Changing direction of flowThe “X” factor i.e. Changing direction of flow
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AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Shock losses Shock losses -- EntranceEntrance
ConicalX = 0.2
FlangedX = 0.5
PlainX = 0.9
Bell MouthX = 0.05
NotchedX = 0.05
Note that the Notched entry approximates the Bell Mouth and the vortex formed in the notch promotes a smooth flow into the airway.
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Shock losses Shock losses –– Intake RB’sIntake RB’s
X = 0.9
Quantity = 200 m3/s
Diameter = 4.0m
Quantity = 235 m3/s
Velocity Pressure = 210 Pa
X 0.05 = 10 Pa
X 0.9 = 189 Pa
X = 0.05
Quantity = 233 m3/s
• 13
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Shock losses Shock losses –– ExpansionExpansion
Abrupt Expansion
A1
X = ( 1 – [A1 / A2 ])2
A2
Gradual Expansion
A2A1
Note: Expansions are complicated by the regain in static pressure over the length of the change.To keep expansion losses to a minimum the change should take place over the longest available distance and the downstream duct must be at least 4 times the larger diameter to ensure full recovery of the pressure.
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Shock losses Shock losses ––ContractionContraction
Gradual Contraction
A2A1
Note: If the length of contraction is 4 x the smaller diameter then the shock losses are negligible
X = 0.333 [ 1 – (A2 / A1 )]
A2
Abrupt Contraction
X = 0.5 [ 1 – (A2 / A1 )]
A1
• 14
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Example Example –– Abrupt ExpansionAbrupt Expansion
Drive A1 - 4.0m x 4.0m
Drive A2 - 6.0m x 6.0m
Airflow 125 m3/s
Assume standard density for air
Calculate “X”
Calculate the velocity pressure in the airway
Calculate the pressure loss due to shock (Pa) 7.2 2
3.51.2
2v P
2
2 V
=
=
ρ=
0.309 A2A1- 1 X
2
=
=
(Pa) 2 7.2 x 0.3
P X P V Shock
==
=
Abrupt Expansion
A1
X = ( 1 – [A1 / A2 ])2
A2
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Example Example –– Abrupt ContractionAbrupt Contraction
Drive A1 - 6.0m x 6.0m
Drive A2 - 4.0m x 4.0m
Airflow 125 m3/s
Assume standard density for air
Calculate “X”
Calculate the velocity pressure in the airway
Calculate the pressure loss due to shock
(Pa) 36 2
7.81.2
2v P
2
2 V
=
=
ρ=0.278
A1A21 0.5 X
=
−=
(Pa) 10 36 x 0.278
P X P V Shock
==
=
A2
Abrupt Contraction
X = 0.5 [ 1 – (A2 / A1 )]
A1
• 15
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Shock losses Shock losses –– 909000 Bends Bends
X = 1.15
Rectangular Mitre Bend
X = 1.15
Circular Mitre Bend
0.3151.5
0.3501.0
0.4250.5
0.3800.3250.315X3.02.52.0R/D
R
D
Three Piece Bend 900
Smooth Bend
X = 0.25 x θ2
m2 a0.5 902
m is the radius ratio = Centre line radius / width of the airway
A is the aspect ratio = Height / width
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Shock losses Shock losses –– other than 90other than 9000 Bends Bends
Mitre Bend other than 900
Xθ = X90 x θ90
θ
• 16
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Example Example –– 909000 BendsBends
Mitre bend
5.0m x 5.0m drive
Airflow 125 and 35 m3/s
Assume standard density for air
(Pa) 2
v P2
V
=
ρ=
(Pa) P X P V Shock
=
=
3 piece bend
5.0m x 5.0m drive
Airflow 125 m3/s
Assume standard density for air
Velocity pressure
Shock loss pressure
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Example Example –– 909000 BendsBends
Mitre bend
5.0m x 5.0m drive
Airflow 125 m3/s
Assume standard density for air
Velocity pressure
Shock loss pressure(Pa) 15 2v P
2 V
=
ρ=
(Pa) 17 15 x 1.15
P X P V Shock
==
=
Mitre bend
5.0m x 5.0m drive
Airflow 35 m3/s
Assume standard density for air
Velocity pressure
Shock loss pressure(Pa) 1.2
2v P
2 V
=
ρ=
(Pa) 1.4 15 x 1.15
P X P V Shock
==
=
3 piece bend
5.0m x 5.0m drive
Airflow 125 m3/s
Assume standard density for air
Velocity pressure
Shock loss pressure(Pa) 15 2
v P2
V
=
ρ=
(Pa) 6 15 x 1.15
P X P V Shock
==
=
• 17
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Shock losses Shock losses -- SplitsSplits
Deflected branch
X = (0.5(Q / Qb)2) + ((Q / Qb)-1)2 + Xb
Splitting
θ
Straight Branch
X = 0.5((Q / Qb)-1)2
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Shock losses Shock losses -- JunctionsJunctions
Ideal Junction
θ
θ = 300 & v1 = v 2
v1v2
Q = Total Airflow
Qb =Airflow in the branch being evaluated
Xb= X for the deflected branch
Cc=Coefficient of contraction =Ac / Ao
Ac=Area of the vena Contractor
Ao=Area of the orificeDeflected branch
X = (-0.5 ((Q / Qb)-1)2.5) + Xb
Junction
θ
Straight Branch
X = 3.3((tan θ/2)–0.67).(((Q/Qb).((1/Cc)-1))2
• 18
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Flanged
X=1.0
In all cases X = 1.0 and the velocity pressure is determined by the velocity in the plane of the exit area
Diffused
X=1.0
Plain
X=1.0
Shock losses Shock losses –– Outlet (discharge)Outlet (discharge)
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
System Pressure LossesSystem Pressure Losses
Entry to the system (shock losses)
Frictional losses (roughness, dimensions)
Shock losses (any change in direction of airflow)
Bends
Intersections
Obstructions
Changes in area or shape
Exit from the system(shock losses)
• 19
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
System resistance characteristic (curve)System resistance characteristic (curve)
A system characteristic curve can be generated using the abbreviated ventilation equation
P = RQ2 (Assume R = 0.0746 (Ns2/m8))
1,462
140
1,910
160
2,9842,4171,074746477268120300Pressure (P) (Pa)
200180120100806040200Volume of air (Q) (m3/s)
System resistance characteristic
0
500
1000
1500
2000
2500
3000
3500
0 20 40 60 80 100 120 140 160 180 200
Volume of air (m3/s)
Pres
sure
(Pa)
AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd
Acknowledgements
DALY, B.B., 1978 “Woods Practical Guide to Fan Engineering”(Published by Woods of Colchester 1978)
Le ROUX, W., “Le Roux’s Notes on Mine Environmental Control” Fourth Edition. (The Mine Ventilation Society of South Africa).
JORGENSEN, R. 1983 “Fan Engineering Eighth Edition” (Buffalo Forge Company. Buffalo, New York.)
BURROWS, J., 1989 “Environmental Engineering in South African Mines” (The Mine Ventilation Society of South Africa)
Tien, J.C., 1999 “Practical Mine Ventilation Engineering” (IntertecPublishing Corporation. Chicago, Illinois.)