2 fundamentals airflow

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AMC - The Business of Mining© 2005 AMC Consultants Pty Ltd

Basic Mine VentilationBasic Mine Ventilation

Fundamentals of AirflowFundamentals of Airflow


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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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


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


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


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

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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


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

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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


Pitot TubePitot Tube

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Static pressure inside the duct holds it open


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

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Pressure loss & flowPressure loss & flow

P∝QLaminar flow

P∝Q2Turbulent Velocity

Pres

sure

loss

Turbulent

Transition

Laminar


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

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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


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

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“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=


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=

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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)


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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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


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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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.


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

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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.


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

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Example Example –– Abrupt ExpansionAbrupt Expansion

Drive A1 - 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) 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


Example Example –– Abrupt ContractionAbrupt Contraction



Airflow 125 m3/s


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


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


Shock losses Shock losses –– other than 90other than 9000 Bends Bends

Mitre Bend other than 900

Xθ = X90 x θ90

θ

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Example Example –– 909000 BendsBends

Mitre bend

5.0m x 5.0m drive

Airflow 125 and 35 m3/s


(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


Velocity pressure

Shock loss pressure


Example Example –– 909000 BendsBends

Mitre bend

5.0m x 5.0m drive

Airflow 125 m3/s


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


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


Velocity pressure

Shock loss pressure(Pa) 15 2

v P2

V

=

ρ=

(Pa) 6 15 x 1.15

P X P V Shock

==

=

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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


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

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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)


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)

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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)


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.)

2 fundamentals airflow

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