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$: FORMULAE HANDOUT FOR CERTIFICATE IN MEC · D:\my data\Websites\COM MiningCertification\Documents\MEC Formulae Handout Updated - New.doc / June 2006 - 10 - AIRFLOW MEASUREMENT PITOT$
D:\my data\Websites\COM MiningCertification\Documents\MEC Formulae Handout Updated - New.doc / MMong / 24-06-2004

TRAINING AND DEVELOPMENT SERVICES ISO 9001:2000 CERTIFICATED

FFOORRMMUULLAAEE HHAANNDDOOUUTT FFOORR

CCEERRTTIIFFIICCAATTEE IINN MMEECC

UPDATED AUGUST’ 06

D:\my data\Websites\COM MiningCertification\Documents\MEC Formulae Handout Updated - New.doc / June 2006

- 2 -

I N D E X

PAGE SECTION 1

AIRFLOW

PRESSURE SURVEYS

AIRFLOW MEASUREMENT

FANS

COMPRESSED AIR

1

7

10

12

14

SECTION 2

HEAT

PSYCHROMETRY

REFRIGERATION

GASES

RADIATION

15

18

22

27

28

SECTION 3

FIRES

DUST

NOISE

ILLUMINATION

MINE WATER

ECONOMICS

STATISTICS

30

31

34

41

42

51

54


- 1 -

AIRFLOW Natural Ventilation Pressure [NVP] a. Density Formula Method [when there are no fans in the circuit]

NVP = [wD - wU] x H x 9.79

Where NVP = natural ventilation pressure [Pa]

wD = mean density of downcast air [kg/m3]

wU = mean density of upcast air [kg/m3]

H = vertical distance from the top to the bottom of the circuit [m]

9.79 = constant for gravitational acceleration [m/s2] b. P-V Diagram Method [with or without fans in the circuit] and

NVE = Pv

NVP = v

NVE

Where NVE = natural ventilation energy [kJ/kg]

NVP = natural ventilation pressure [kPa]

P = barometric pressure [kPa]

v = specific volume [m3/kg] REYNOLDS NUMBER

Re = μ

wVD

Where Re = Reynolds number [dimensionless] w = density [kg/m3] V = velocity [m/s] D = diameter [m] μ = dynamic viscosity [Ns/m2]

CONSERVATION OF ENERGY

u + Pv + 2

V 2

+ Zg = Constant

Where u = internal energy [J/kg]

P = pressure [Pa]

v = specific volume [m3/kg]

V = velocity [m/s]

Z = elevation [m]

g = gravitational acceleration [9.79 m/s2]


- 2 -

RESISTANCE

R = 2,1

wxA

KCL3

Where R = resistance [Ns2/m8]

K = friction factor [Ns2/m4]

C = circumference [h + w]2 = hlge / πD = pipes [m]

L = length [m]

A = area [h x w] hlge / π ⎥⎦

⎤⎢⎣

⎡

4D2

= pipes [m2]

w = air density [kg/m3]

ATKINSON’S FORMULA

P = 2.1

wxA

KCLQ3

2

Or

P = 2.1

wxA

KCLV 2

Where P = pressure loss due to friction [Pa]

K = friction factor [Ns2/m4]

C = circumference [m]

L = length [m]

Q = air quantity [m3/s]

V = air velocity [m/s]

A = area [m2]


PRESSURE REQUIRED TO OVERCOME FRICTIONAL RESISTANCE

P = RQ2

Where P = pressure required [Pa]

R = resistance [Ns/m8]


VELOCITY PRESSURE

VP = 2wV2

Where VP = velocity pressure [Pa]




- 3 -

DARCY-WEISBACH EQUATION

p = D2

²LwVλ

Where p = pressure [Pa]

λ = Darcy Weisbach friction factor

L = length [m]

w = density [kg/m3]

V = velocity [m/s]

D = diameter [m]

λ = 6.67K when ws = 1.2 kg/m3

AIR POWER

Wa = 1000

Q x p OR Wa = 1000RQ3

Where Wa = air power [kW]

p = pressure [Pa]


R = Resistance (Ns2/m8)


- 4 -

TRIGONOMETRY To calculate length AC

Sin x =hypotenuse

opposite ⎟⎠⎞

⎜⎝⎛

ABAC

To calculate length AB

Cos x = hypotenuse

adjacent⎟⎠⎞

⎜⎝⎛

ABBC

To calculate length BC

Tan x = adjacentopposite

⎟⎠⎞

⎜⎝⎛

BCAC

EVASEÉS

Theoretical pressure regain = VPi - VPo

Where VPi = velocity pressure at evaseé inlet [Pa]

VPo = velocity pressure at evaseé outlet [Pa]

Actual pressure regain can only be measured or theoretical pressure regain multiplied by evaseè efficiency:-

100xregainpressureltheoretica

regainpressurectualAefficiencyEvasee =

LEAKAGE

100xcolumnactualforrequiredpowercolumnleaklessforrequiredpowerefficiencyDuct =

C

A

Y

B X


- 5 -

SYSTEM RESISTANCE CURVES

These are calculated from a square law relationship derived from Atkinson’s formula for a constant resistance

P ∝ Q2

Or

1

21

Qp =

222

Qp


Q = quantity [m3/s]

AIRWAYS IN SERIES

QT = Q1 = Q2

PT = P1 + P2

RT = R1 + R2

Where Suffix ‘T’ indicates total system conditions;

Suffix ‘1’ indicates conditions in airway 1;

Suffix ‘2’ indicates conditions in airway 2;

P = pressure [Pa]

Q = quantity [m3/s]

R = resistance [Ns2/m8]

AIRWAYS IN PARALLEL

QT = Q1 + Q2

PT = P1 = P2

TR

1 = 1R

1 + 2R

1

REGULATORS

Ar = 1.2Qpw

Where Ar = regulator area [m2]

Q = air quantity through regulator [m3/s]

p = pressure used up by regulator [Pa]


Or, when the air density is 1.2 kg/m3

Ar = pQ31.1


- 6 -

BERNOULLI’S THEOREM [for frictionless flow]

TP1 = TP2

or

VP1 + SP1 = VP2 + SP2

Because

TP = SP + VP

Where TP = total pressure

SP = static pressure

VP = velocity pressure

BAROMETRIC PRESSURE INCREASE OR DECEASE

The approximate barometric pressure increase or decease in a vertical shaft = 1 kPa/100m


- 7 -

PRESSURE SURVEYS

FULL VOLUME – REDUCED VOLUME METHOD [density effects ignored]

R = ²Q ²Q

ΔBΔB RF

2 - 1

−

Where R = resistance [Ns2/m8]

∆B1 = difference in the barometric pressures at point [1] when the fans are running and stopped [Pa]

∆B2 = difference in the barometric pressures at point [2] when the fans are running and stopped [Pa]

QF = full volume flow [m3/s]

QR = reduced volume flow [m3/s]

FULL VOLUME – REDUCED VOLUME METHOD [density effects included]

pf = [ ][ ][ ]²Qw

²Qw 1

ww H 9.79 ΔB

f mf

r mr

mr mf

−

−±

Where pf = pressure loss for full volume flow [Pa]

∆B1 = difference in the barometric pressures with full and reduced volume flow [Pa]

Referring to the definitions in the previous formula ∆B = [∆B1 - ∆B2]

H = difference in elevation [m]

Wmf = mean density at full volume flow [kg/m3]

Wmr = mean density at reduced volume flow [kg/m3]

Qf = full volume flow [m3/s]

Qr = reduced volume flow [m3/s]

NB ± = Use the ‘+’ sign when depth increases from station [1] to station [2]

Use the ‘-‘ sign when depth decreases from station [1] to station [2]

DENSITY METHOD

The pressure loss:

• the difference between the theoretical pressure increase or decrease and the actual pressure increase or decrease

The theoretical pressure increase or decrease:-

[9.79 x H x wm]

where 9.79 - Constant for gravitational acceleration [m/s2]

H - Difference in elevation [m]

wm - Mean density [kg/m3]


- 8 -

CORRECTION DUE TO BAROMETRIC PRESSURE VARIATIONS

ΔPth = cbcb

th P x PP

Δ

Where Ptb = traverse barometer reading

Pcb = control barometer reading

∆Pcb = change in the control barometer reading

∆Ptb = corresponding change in the traverse barometer reading

PRESSURE / DENSITY RELATIONSHIP Air pressure varies directly as an air density change:-

1

1

wp =

2

2

wp


w = density [kg/m3]

Suffix ‘1’ indicates conditions at one point in the system;

Suffix ‘2’ indicates conditions at another point in the same system.


- 9 -


- 10 -

AIRFLOW MEASUREMENT

PITOT TUBE POSITIONS IN A CIRCULAR DUCT

Rn = N4

1 n2d −

Where n = the nth reading from the centre

Rn = radius of the reading [mm]

d = duct diameter [mm]

N = number of readings across a diameter

ORIFICE PLATE

Q = PQ2.1

Where Q = air volume [m3/s]

P = differential pressure [Pa]

CONICAL INLET

Q = wpC²D11.1 Δ

Where Q = air density [m3/s]

D = duct diameter [m]

C = coefficient of discharge [from graphs]

∆p = measured pressure difference [Pa]


VENTURI METER

Q = wpzEC²d11.1 d

Δ E =

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−

5.0

4

4

Dd1

1

Where Q = air quantity [m3/s]

d = diameter of throat [m] (Venturi)

D = Column diameter (m)

Cd = coefficient of discharge [from graphs]

E = velocity of approach factor [from graphs]

z = combination of factors for size, expansion and Reynolds number

∆p = measured pressure difference [Pa]



- 11 -

GAS TRACER METHOD

a. Tracer gas NOT in normal air:-

Q = C10 x q 6

Where Q = air volume or mass flow rate [m3/s of kg/s]

q = rate of tracer gas release [m3/s or kg/s]

C = concentration of tracer gas in air after mixing [part per million by volume or mass]

b. Tracer gas IN normal air:-

Q = 21

6

C C10 x q−

Where Q = air volume or mass flow rate [m3/s of kg/s]

q = rate of tracer gas release [m3/s or kg/s]

C1 = concentration of tracer gas in air after mixing [part per million by volume or mass]

C2 = concentration of tracer gas found in normal air before mixing [part per million by volume or mass]

c. Volume of tracer gas:-

Vg = ao

gQ

Wm

∫= Cdt = QA

Q = Aw

mg

Where Vg = volume of tracer gas [m3]

m = mass of tracer gas [kg]

wg = densities of tracer gas [kg/m3]

Q = airflow rate [m3/s]

C = tracer gas concentration by volume, part per unit

A = area under curve [∫o a Cdts]


- 12 -

FANS

Efficiency = %100xinputwork

outputwork

Motor Efficiency = %100xpowerinputmotorpoweroutputmotor

Fan Efficiency = %100xpowerinputfan

powerair

Drive Efficiency = 100% x power output motor

power input fan

Overall Efficiency = %100xpowerinputmotor

powerair

FAN LAWS

Air Density Change

When the air density changes from w1 to w2:-

1. Q remains constant, i.e.: Q1 = Q2

2. p α w

wp

1

1 = 2

2

wp

3. Power α w

1

1

wpower =

2

2

wpower

4. Efficiency remains constant

Eff1 = Eff2

Fan Speed Change

When the fan speed changes from speed1 to speed2:-

1. Q α speed

1

1

speedQ =

2

2

speedQ Or Q2 =

1

21

speedspeed x Q

2. p α speed2


- 13 -

1

1

speedp =

2

2

speedp Or p2 = [ ]

[ ]²speed²speedxp

1

21

3. power α speed3

[ ]³speedpower

1

1 = [ ]³speedpower

2

2 or power2 = [ ][ ]³speed

³speedxpower1

21

4. Efficiency remains constant

Eff1 = Eff2

Where Q = fan air quantity [m3/s]

p = fan pressure [Pa]

power = fan power [kW]


speed = fan speed [r/s]

PULLEY SIZE CHANGES

1. Fan pulley size change with a speed increase or decrease:-

New pulley size = Old pulley size x speednew

speedold

2. Motor pulley size change with a speed increase or decrease:-

New pulley size = Old pulley size x speedoldspeednew


- 14 -

COMPRESSED AIR

Effect of auto compression

The increase in pressure due to auto compression can be derived from the equation:-

Pe = ⎥⎦

⎤⎢⎣

⎡RTgHexpPs

Where Pe = absolute pressure at end of column [kPa]

Ps = absolute pressure at start of column [kPa]

G = gravitational acceleration m/s2 [9.79 m/s2]

H = vertical depth metres [m]

R = gas constant J/kgK [287 J/kgK]

T = absolute temperature [K]


- 15 -

HEAT

AUTO-COMPRESSION OR DE-COMPRESSION

Heat increase or decrease: 0.979 kJ/kg/100m or 9,79 kJ/kg / 1000m of vertical depth

H = 1000

zgΔ

H = heat increase

g = 0.979 kJ/kg/100m

Z = vertical depth

VIRGIN ROCK TEMPERATURE

V.R.T [approximate]:-

Gauteng = 18 + (9 x depth In kilometre)

Free State = 20+ (14.6 x depth in kilometre)

Klerksdorp = 22 + (10,5 x depth in kilometres)

V.R.T. [accurate]:-

Gauteng = [18.3 + 6D + 1.1 D2] °C

Where D = thickness of overlying strata [km]

Free State = [20 + 25.5 D1 + 14.2 D2 + 8.2 D3] °C

Where D1 = thickness of Karoo diabase [km]

D2 = thickness of lava [km]

D3 = thickness of quartzite [km]

WET KATA FORMULA

H = v7.0 θ+θ

Where H = wet kata reading

θ = 36.5 - wet bulb temperature [°C]



- 16 -

Amount of Heat transferred

a) Conduction

q = [ ]b

ttKA 21 −

Where q = Conductive heat transfer ratio [W]

K = Thermal conductivity of material [W/m°C]

A = Cross-sectional area [m²]

t1 - t2 = temperature difference of sources [°C]

B = Thickness [m]

b) Convection

q = hcA (t1 - t2)

Where q = Convective heat transfer rate [w]

hc = Convection heat transfer co-efficient [W/m2]

(t1 - t2) = Temperature difference of sources [oC]

A = Cross-sectional area [m2]

c) Radiation

q = 5,67 x 10-8 A1Fev(T14 - T2

4)

Where q = Radiative heat transfer [W]

5,67 x 10-8 = Stefan-Boltzmann constant

A1 = Smaller area of the two surfaces [m2]

Fev = Emissivity and view factor

(T14 - T2

4) = Absolute temperatures (K)

And Fev =

⎟⎟⎠

⎞⎜⎜⎝

⎛−

∈+

∈11

AA1

1

22

1

1

OR


- 17 -

q = 5,67 ev1

42

41 F x x A

100T

100T

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−⎟⎠

⎞⎜⎝

⎛

HEAT EQUATIONS

Static Heat Equation [no movement]

W = tCM pΔ

Where W = heat transferred [kJ]

M = mass flow rate of substance [kg]

Cp = thermal capacity of substance [kJ/kg °C]

Δt = temperature difference [°C]

Flow Heat Equation [with movement]

q = tΔC M p

Where q = heat transfer rate [kJ/s or kW]

M = mass flow rate of substance [kg]

Cp = thermal capacity of substance [kJ/kg °C]

Δt = temperature difference [°C]

Wind Chill Equivalent Temperature:

WCET = 33 - ⎟⎟⎠

⎞⎜⎜⎝

⎛ +22.04

T) - (33 x v) - v10 (10,45

WCET = Wind Chill Equivalent Temperature

v = Wind speed (m/s)

T = air temperature, dry bulb (oC)


- 18 -

PSYCHROMETRY

BOYLE’S LAW

P1V1 = P2V2

CHARLE’S LAW [T = absolute temp °C + K ] K - 273

2

2

1

1

TV

TV

=

UNIVESAL GAS LAW

2

22

1

11

TVP

TVP

=

Where P = absolute pressure [kPa]

V = volume [m3], volume flow rate [m3/s], Specific volume [m3/kg]


The Universal Gas Law can also be written as:-

T

Pv = R

Where P = absolute pressure [kPa]

v = specific volume [m3/kg]


R = gas constant [kJ/kg

And the gas constant [R] for dry air = 0.2871 kJ/kg K

MASS FLOW OF AIR

M = Q x w

or

M = vQ

Where M = mass flow of air [kg/s]



v = air specific volume [m3/kg]


- 19 -

For a constant mass flow of air:-

M1 = M2

Thus:

Q1w1 = Q2w2

Where M = mass flow of air [kg/s]



Suffix ‘1’ indicates conditions at one point in a system;

Suffix ‘2’ indicates conditions at a second point in the system

CALCULATION OF PSYCHOMETRIC PROPERTIES

1. Vapour Pressure [Pw]

Pw = [ ]kPattAPs'P wbdb −−

Where P’s = 0.6105 exp [17.27 twb/ [237.3 + twb]] kPa

A = 0.000644 °C-1

P = pressure [kPa]

2. Moisture content [r] [kg/kg]

v = PwP

Pwx622.0−

3. Specific Volume [v]

v = kg/³mPwP

Tx287.0−

Where T = 273.15 + tdb K

4. Density [w]

w = ³m/kgv

r1 +

5. Enthalpy [H]

H = Ha + rH’w kJ/kg

Where Ha = 1.005 tdb kJ/kg

H’w = 1.8 tdb + 2501 kJ/kg


- 20 -

6. Sigma Heat [S]

S = H - rH’w1 kJ/kg

Where H’w1 = 4.18 twb kJ/kg

7. Relative Humidity [Φ]

φ = %100xs'P

Pw

Where P’s = 0.6105 exp [17.27 tdb / [237.3 + tdb]] kPa

8. Dew point Temperature [tdp]

tdp = Cx27.17

x3.237°

−

Where x = ⎥⎦

⎤⎢⎣

⎡6105.0PwIn

Heat removed for air :-

Q = M x ΔS

Where q = heat transfer rate [kW]

M = mass flow of dry air [kg/s]

∆S = change in sigma heat content [kJ/kg]

Amount of water evaporated/ condensed:-

R = 1000 x M rΔ

Where R = amount of water condensed [l/s]

M = mass flow of dry air [kg/s]

∆r = change in moisture content [g/kg]

Mixing of Airstreams:-

Sigma Heat Content:

Sc = [ ] [ ][ ]BA

BBAA

MMSxMSxM

++

Where: SC = sigma heat content of the mixture [kJ/kg]

[MA x SA] = the total kW of heat from air stream A

[MB x SB] = the total kW of heat from air stream B

[MA x SA] + [MB x SB] = the total kW in air stream C

[MA + MB] = the total mass flow of air stream C


- 21 -

Moisture Content:

rc = [ ] [ ][ ]BA

BBAA

MMrxMrxM

++

Where: rC = the moisture content of the mixture [g/kg]

[MA x rA] = the total moisture [g/s] from air stream A

[MB x SB] = the total moisture [g/s] from air stream B

[MA x SA] + [MB x SB] = the moisture [g/s] in air stream C

[MA + MB] = the total mass flow of air stream C


- 22 -

REFRIGERATION

In the formulae below, C.O.P. denotes Coefficient of Performance

a. Heat balance

Condenser duty = evaporator duty + total input power to compressor

b. Carnot C.O.P

Carnot COP = 12

1

TTT−

Where T1 = evaporating temperature [K]

T2 = condensing temperature [K]

c. Overall compressor C.O.P:-

kWpowerinputmotorcompressor

evaporatoratcooling

d. Actual or nett compressor C.O.P.:-

kWpoweroutputmotorcompressor

evaporatoratcooling

e. Overall plant C.O.P:-

kWpowerinputelectrictotal

coilsatcooling

f. Overall compressor power/cooling ratio:-

kWevaporatoratcooling

powerinputmotorcompressor

g. Actual or nett compressor power/ cooling ratio:-

kWevaporatoratcooling

poweroutputmotorcompressor

h. Overall plant power / cooling ratio:-

kWcoilsatcooling

powerinputelectrictotal

i. Overall cycle efficiency

%100x.P.O.CCarnot

.P.O.CcompressoroverallL


- 23 -

j. Actual or nett cycle efficiency:-

%100x.P.O.CCarnot

.P.O.Ccompressornettoractual

k. Plant positional efficiency:-

[ ][ ] %100xkWevaporatoratcooling

kWcoilsatcooling

l. Compressor motor input power:-

W = npfIE

Where W = electric power [kW]

E = voltage [kV]

I = current [amperes]

pf = power factor [normally approximately 0.9]

n = number of phases [normally 3]

m. Cooling tower efficiency [water]

Nw = %100xttttwbiwi

wowi

−−

Where twi = temperature of water entering tower [°C]

two = temperature of water leaving tower [°C]

twbi = wet bulb temperature of air entering tower [°C]

n. Cooling tower efficiency [air]

Na = %100xSSSS

aiwi

aiao

−−

Where Sao = sigma heat content, air leaving tower [kJ/kg]

Sai = sigma heat content, air entering tower [kJ/kg]

Swi = sigma heat content, water entering tower [kJ/kg]

o. Cooling tower factor of merit

F =

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −+ 1E1R1

1when R>1 and E = Na (Air Efficiency)

Or

F =

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −+ 1E1

R11

1 when R<1 and E = Nw (Water Efficiency)


- 24 -

And

R = a'CxM

CxMa

pww

Where:

C’a = aiwi

aiwi

ttSS

−−

PH Diagram

1. Evaporator heat exchange - C - B [kJ/kg]

2. Condenser heat exchange - D - A [kJ/kg]

3. Heat of compression [actual] - D - C [kJ/kg]

4. Heat of compression [ideal] - E - C [kJ/kg]

Heat balance [on cycle]

Condenser heat exchange = Evaporator heat exchange + heat of compression

[D - A] = [C - B] + [D - C]

H = Enthalpy (kJ/kg)

P =

Pre

ssur

e (k

Pa)

Condenser

Evaporator

Compressor

F B C

D E

E on constant entropy Line through C

A


- 25 -

5. Carnot C.O.P.

12

1

TTT−

Where T1 - Absolute evaporating temperature [K]

T2 - Absolute condensing temperature [K]

6. Actual compressor C.O.P.:-

ncompressioofworkactualexchangeheatevaporator

CDBC

−−

7. Cycle efficiency:-

%100x.P.O.CCarnot

.P.O.Ccompresoractual

8. Compressor efficiency:-

%100xncompressioofworkactual

ncompressioofworkideal

%100xCDCE

−−

9. Percentage flash gas:-

%100xFCFB

−−

10. Mass flow of refrigerant:-

condenserofexchangeheatunitcondenseratexchangeheattotal OR

EvaporatorofexchangeheatunitEvaporatoratexchangeheattotal

11. Power consumed by - actual work of compression x compressor mass flow of refrigerant

- [D - C] x M

12. Plant duty - evaporator heat exchange x mass flow of refrigerant

- [C - B] x M

13. Volume flow rate of refrigerant entering compressor

Q = M x vin

Where vin - constant volume at entrance of the compressor


- 26 -

14. Volume flow rate of refrigerant leaving compressor

Q = M x vout

Where vout - constant volume at exit of the compressor

15. Percentage error:-

[ ] 100xa

bca −−

Where a - condenser duty

b - Compressor duty

c - Evaporator duty


- 27 -

GASES

GAS DILUTION

Q = 1

6 1 Q-

NMAC 10x Q

−

where Q - fresh air volume or mass flow rate required for dilution [m3/s or kg/s]

Q1 - volume or mass flow rate of gas emission [m3/s or kg/s]

MAC - maximum allowable gas concentration [after mixing] in parts Per million by volume or mass

N - gas concentration in normal air in parts per million by volume or mass

GAS MIXING

Percentage gas % = 0x10gasquantity airquantity Total

gasQuantity Total+

Where: Total quantity gas = m3/s

Total quantity air + quantity gas = m3/s


- 28 -

RADIATION

1. Working Level Month Per Annum [WML]

WML / Annum = ([number of weeks worked / year x number of hours / week] x

workedhoursmonthly allowable Maximum

Level WorkingMean

2. Working Level Month Exposure

WLM = month per allowed hours Maximum

rate exposure x month per workedhours of number

3. Time weighted exposure

WLM = [number of hours worked x exposure rate] + [number of hours

month per allowed hours Maximum

rate exposure x Worked

4. Residence Time [T]

T = 1.851

6

EtV10x86.4⎥⎦

⎤⎢⎣

⎡ NB; in brackets to the power of 1/1.85

Where T - Residence time [s]

V - volume of the tunnel or workings [m³]

Et - Radon production [p Ci/s]

5. Radon Dilution

Q2 = 2

11 Rn

Rn Q

Where Q1 - Air quantity prevailing

Q2 - Air quantity required for Rn2

Rn1 - Rn concentration prevailing

Rn2 - Rn concentration to be determined

6. WL2 = WL1

85.1

2

1

QQ

⎥⎦

⎤⎢⎣

⎡

Where WL1 - prevailing condition

WL2 - desired condition

Q1 - quantity flowing [m³/s]

Q2 - quantity required for dilution [m³/s]


- 29 -

Summation of an individual radiation dose :-

1f

Ie

Id

IcI

bI

aI

HH TcUcThUThDRnD

IDL

ID≤++++++

Where: HID - is the deep dose equivalent index received in the year

IRnD - is the annual exposure to radon daughter products

IThD - is the annual exposure to thoron daughter products

IU - is the annual intake of uranium ore dust

ITh - is the annual intake of thorium ore dust

IUc - is the annual intake of uranium concentrate

ITc - is the annual intake of thorium concentrate

IIDL - is the deep dose equivalent index limit

a - is the annual limit of exposure to radon daughter products

b - is the annual limit of exposure to thoron daughter products

c - is the annual limit of intake of uranium ore dust

d - is the annual limit of intake of thorium ore dust

e - is the annual limit of intake of uranium concentrate

f - is the annual limit of intake of thorium concentrate


- 30 -

FIRES

EXPLOSIBILITY DIAGRAMS – US BUREAU OF MINES AND COWARD’S TRIANGLE

1. Excess N2 - N2 - 3.7778 O2

2. O2 deficiency - 0.2647 N2 – O2

3. Total combustibles, D - CH4 + H2 + CO

4. ‘R’ values on USBM diagram - D

CH4

5. CO/O2 deficiency ratio - %100xdeficiencyO

CO2

[Graham Ratio]

6. Young’s ratio - %100xdeficiencyO

CO2

2

7. Willet’s ratio - %100xCOD,escombustibltotalNExcess

CO22

2

++

8. x co-ordinate [USBM diagram] - Excess N2 + 1.5 CO2

9. y co-ordinate [USBM diagram] - CH4 + 1.25H2 + 0.4 CO

10. COWARD’S TRIANGLE and all above

Please refer to the “Environmental Engineering in South African Mines” Page 814 - 817


- 31 -

DUST

DUST FILTRATION

Surface area of one filter bag:-

[ ] [ ] ⎥⎦⎤

⎢⎣⎡π+π=

4²DLxD²m

Where D - bag diameter [m]

L - bag length [m]

DUST DILUTION

[Q1D1] + [Q2D2] =[Q1 + Q2]D3

Where Q1 - air volume of stream ‘1’ [m³/s]

Q2 - air volume of stream ‘2’ [m³/s]

D1 - dust content of stream ‘1’ [p/ml]

D2 - dust content of stream ‘2’ [p/ml]

D3 - dust content of mixture [p/ml]

0

20

40

60

80

100

0 1 2 3 4 5 6 7 8

Particle size [um]

Per

cent

age

parti

cles

sta

ted

size

Respirable sampling curve defined at the International Pneumoconiosis Conference in Johannesburg, 1959


- 32 -

PERSONAL GRAVIMETRIC DUST SAMPLING

Calculation of the Time Weighted Average Concentration (TWA - CONC)

Calculation of Results

Step Example

1. Note the average flow rate and sample time.

Obtain the pump flow rate.

Determine the total sample time

Convert total sample time to minutes

2,2 Litres per minute

8 hours 20 minutes

∴ Minutes = (8 x 60) + 20

= 500 minutes

2. Determine the sample volume

Results must be expressed in mg/m3

∴ Volume of air through pump = Flow rate x time

Convert litres to m3

(1000 litres of air = 1 m3)

Volume = Flow rate (l/m) x time

= 2,2 x 500

= 1 100 litres of air

= 10001100

= 1,1 m3 sucked through

3. Determine the correction filter mass (Correction Factor)

Determine the average of pre and post weighed control (blank) filters by:

Post Filter mass (mg)

Pre filter mass (mg)

• weighing pre weighed control filter 3 x consecutively when weighing sample filters

20,16

20,17

20,18

20,1

20,09

20,11 • weighing post weighed control filter 3

x consecutively when weighing exposed sample filters

Add together and divide by 3

315,60

3

3,60

= 20,17 mg = 20,10 mg

Determine the correction factor by: Subtract the pre weighed blank filter mass from post weighed blank filter mass.

Correction factor = Post filter mass – Pre filter mass = 20,17 - 20,10

= 0,07 (Heavier, picked- up moisture)

If this mass is + subtract as a correction factor.

If this mass is - add as a correction factor

As this 0, 07 mg is positive, it must be subtracted from the sample filter mass.


- 33 -

Step Example

4. Determine the sample mass (mg)

Subtract the pre weighed sample mass from the post weighed sample mass

Also weigh in the manner described in step 3.

Post weighed sample mass - Pre weighed sample mass

20,78 - 20,66

= 0,12 mg

5. Determine the correct sample mass (mg)

Subtract the correction factor (because it is +) from the sample mass

Add correction factor if mass is -

Corrected Sample mass = Sample mass - correction factor

= 0,12 - 0,07

= 0,05 mg

6. Determine the concentration (mg/m3)

Divide the corrected sample mass by the volume of air sampled. (step 2 answer) Concentration =

VolumeMass

⎟⎠⎞

⎜⎝⎛

3mmg

= 1,105,0

= 0,046 mg/m3

7. Determine the TWA-CONC as applicable

Determine the time correction factor (i.e. to convert actual sample time to an 8 hour (480 minutes) shift.

Multiply the concentration with the time correction factor to obtain TWA - CONC

Time correction factor = 480

time Sample Actual

TWA CONC= Conc x time correction factor

= 0,046 x 480

time Sample Actual

= 0,046 x 480500

= 0,048 mg/m3


- 34 -

NOISE

Background Noise

• Equation:

LB = 10 log10 (10 10LM - 10 10L A ) Where LB = Noise level for noise source alone – dB (A) LM = Measured noise level – dB (A) LA = Background noise level – dB (A)

• Use attached table or graph:

Table for subtracting decibel values (correction for background noise).

Difference between measured and background noise level dB(A)

3

4-5

6-9

Decibel value that must be subtracted from the measured noise level

3

2

1

Note: A calculation having a difference of more than 10 dB (A) will indicate that the decibel value to be subtracted is less than half a decibel and background correction can thus be ignored.

Difference between Measured Noise Level and Background Noise dB (A)

Dec

ibel

val

ues

whi

ch m

ust b

e su

btra

cted

from

mea

sure

d no

ise

leve

l


- 35 -

Wavelength

The relationship between wavelength λ [m], speed c [m/s] and frequency f [Hz] is given in the following formula: -

λ = fc [units as above]

Sound Intensity

I = wc

²p

Where: I - Intensity [W/m²]

p - Sound pressure [Pa]

w - Density [kg/m³]

c - Velocity of sound [m/s]

Sound Power Level

SWL = 10Log10 dBpowerreference

powersound⎥⎦

⎤⎢⎣

⎡

Where the reference power is 10-12 watt

Sound Pressure Level

SPL = 10 Log10[ ]

[ ] dB²pressurereference

²pressuresound

Or

SPL = 20 Log10[ ]

[ ]dBpressurereference

pressuresound

Where the reference pressure is the sound pressure at the threshold of hearing i.e. 2 x 10-5 Pa

Leq for Steady Noise Level

Leq = LA + C1

Where Leq - Equivalent noise level, dB [A]

LA - Measured level of steady noise, dB [A]

C1 - Impulse correction factor which is +10 dB where the noise is of

A repetitive nature [e.g. riveting or hammering] or where it occurred in single bursts e.g. a drop forge hammer], and 0 dB in all other cases


- 36 -

Leq For a fluctuating Noise Level

Leq = LA [av] + a C1

Where: Leq and Ci are as before

LA[av] = 10 Log10 ⎥⎦

⎤⎢⎣

⎡ ∑ 10LAV1 10f

1001

Where: LAi - Noise level at the mid-point of the i-class dB [A]

F1 - Duration of the i-class sound level exposure expressed as a percentage of the total analysis time [normalised to a 40 hour total period]

Other formulae (Logarithmic Mathematical Methods):-

Leq = 10 Log10 12

21

1 C 10

log 10Lantilogf

10L logf +⎥⎦

⎤⎢⎣

⎡ ++ nn

Lantifanti

Where: f1 to fn - the ratios in relation to 40 hours of the duration of exposure to the sound levels L1 to Ln

L1 to Ln - the sound levels of dB [A] of the exposures for the duration ratios f1 to fn

C1 - impulse correction factor which is +10 dB where the noise is of a repetitive nature [e.g. riveting or hammering] or where it occurred in single bursts e.g. a drop forge hammer], and 0 dB in all other cases.

OR

Leq = 10 log F+90

Where F = Σf, where fn = 40Cn antilog [0,1 x (Lnoise – 90)]

and Cn = actual time of exposure at noise level (hours)

Exposure Factor [D]

D = n

n

2

2

1

1

TC....

TC

TC

++

Where: C1 to Cn - Actual time of exposure at noise levels L1 to Ln

T1 to Tn - Permitted time of exposure at noise levels L1 to Ln

Average noise level

Lav = 85 - 10 Log10 ⎥⎦

⎤⎢⎣

⎡40T

Antilog ⎥⎦

⎤⎢⎣

⎡10

L - 101 av


- 37 -

When the Exposure Factor D has been obtained, the graph below is used to determine the equivalent noise level, Leq

ADDITION OF SOUND LEVELS

Difference between the two levels dB

0

1

2

3

4

5

6

7

8

9

10 or more

Amount to be added to the higher level dB

3

2,5

2

2

1,5

1

1

1

0,5

0,5

0


- 38 -

Spheres

To calculate area: Sphere = 4πr2 Hemisphere = 2πr2 ¼ Sphere = πr2


- 39 -


- 40 -

EQUIPMENT TESTING [IN-DUCT METHOD FOR FANS AND SILENCERS]

SWL = SPL + 10 log A

For a 760 mm diameter duct the value of 10 log A is –3.5 dB

Control of Noise

a. Fans

In the design stage the acoustical characteristic of a fan is not usually available and it is often necessary to make an estimate. Three formulae often used are given below:-

i. SWL = 97 + 10 log kW + 10 log P dB

ii. SWL = 100 + 10 log Q + 20 log P dB

iii. SWL = 95 + 20 log kW – 10 Q dB

Where SWL is the overall sound power level in the octave frequency bands 31.5 to 8 000 Hz

kW - Rated motor power

P - Fan static pressure [kPa]

Q - Fan delivery quantity [m³/s]

Auxiliary In Line Axial Flow Fans

SWL = 100 + 10 log [QP²] dB

Where: P = Fan total pressure [kPa]

b. Rock drill

SWL = 140 + 10 log Q dB

Where: Q - Free air consumption [m³/s]

c. Diesel Equipment

i. Exhaust Noise

SWL = 110 + 10 log kW dB

Where: kW - Rated power of the diesel

ii. Engine Noise [below 300 kW]

SWL = 100 + 8 log kW dB

Where: kW - Rated power of the diesel


- 41 -

ILLUMINATION

1. The relationship between wavelength [λ] speed [c] and frequency [f] is given in the following formulae:-

λ = fc [units as above]

Where: λ - wavelength [m]

c - Velocity [m/s]

f - Frequency [Hz]

2. Inverse Square Law

The inverse square law states that the illumination at any point on a surface varies directly with a luminous intensity of the source and inversely as the square of the distance between the source and the point. If the source is normal to the direction of the incident light, the law may be expressed as:-

E = 2dl

Where: E - Illumination [Lux] = 1 lm/m2

I - Luminous intensity [cd]

d - Distance [m]

One solid steradian angle = 4π = 12.566


- 42 -

MINE WATER

Deep Cell Dust Concentration

The concentration in millions of particles per millilitre is given by:-

CFDFxN

Where: N - Total number of particles counted in 10 sections

DF - Dilution factor

CF - Cell factor

And cell factor: -

[CF] = 510²xL

Where: x - Depth of cell in micrometers

L - Length of one side of counting section in micrometers [assuming the counting section is a square]

If a 100 ml measuring flask is used, [assuming the distilled water and acid used both have counts of zero

Dilution Factor [DF] = [ ]mlvolumesample100

Difference in Pipe Size

H = 2gV2

H = 2gV

2gV 1

22

2

−

Where: V - Mean velocity

g - Gravitational acceleration [9.79 m/s²]

The friction co-efficient [λ]

λ = [ ]hD/L

2²wV

pΔ

Where: w - Density of the substance [water = 1000 kg/m³]

L - Length (m)

Dh - Equivalent hydraulic diameter (m) = 4A/C (A = Area, C = perimeter)

V - Mean Velocity (m/s)


- 43 -

Where the Reynolds number is between 2500 and 1000 000

λ = 25.0R

316.0

e

PUMP CHARACTERISTICS

Total [Manometric] head HT

HT = Hs + Hf

Where: HT - Total head [m] [liquid]

Hs - Static head [m] [liquid

Hf - Head losses due to friction [m] [liquid]

To convert meters head to kPa: -

p = 1000Hwg

Where: p - Pressure [kPa]

g - Gravitational acceleration [m/s²]

H - Head [m]

W - Density of liquid [kg/m³]

[the density of water is 1000 kg/m³]

Pump Power Requirements

Power = kW1000

SGxgxHxQ T

η

Where: Power - Kilowatts [kW]

Q - Volume flow rate [ℓ/s]

HT - Total head [m]


SG - Specific gravity [water = 1]

η - Pump efficiency expressed as a fraction of 100

1 1 [i.e. %]

Pump Efficiency

Efficiency = 100xinputPower

outputPower

Power output is determined by the following equation:-

Power [output] = kW1000

SGxgxQxH

Where: Power - Kilowatts [kW]


- 44 -

Q - Volume flow rate [ℓ/s]

HT - Total head [m]


SG - Specific gravity [water = 1]

Energy Recovery System

Energy recovered = 1000

headavailable x g x water flow x efficiency

Where: g = Gravitational acceleration [9.79 m/s²]

Efficiency = Turbine efficiency [%]

Available Head = ΔH - Hf (m)


- 45 -

This pipe friction chart is used to read off directly the Darcy Weisbach’s information, without having to perform long calculations and applies for both vertical and horizontal pipes.

To read the chart you need to know two of the following factors:

i. Water flow rate (l /s)

ii. Pipe diameter (mm)

iii. Water velocity (m/s)

iv. Head loss (m/100m)


- 46 -

Darcy Weisbach Formula

Hf = gD2LV2λ

Where: Hf = Head loss due to friction (m)

λ = Pipe Friction Coefficient

L = Pipe length (m)

V = Mean velocity (m/s)

g = Gravitational acceleration (9,79 m/s2)

D = Pipe diameter (m) - is replaced by Dh for non-circular pipes (Dh = CA4 )

Remember the Darcy Weisbach equation calculate head loss due to friction.

The Stanton Nikuradse Diagram


- 47 -

Stroh’s Equation

Hf = 2,04 x 10-9 x 13,5

92,1

dM

x L

Where: Hf = Total head loss due to friction (m)

m = Water flow rate (l /s)

d = Pipe inside diameter (m)

EQUIVALENT LENGTHS OF STRAIGHT PIPES FOR VARIOUS FITTINGS


- 48 -

Pipe Inlet Losses

• Pipe Inlet (Tapered and applicable to short pipe lengths)

H = g2

V2

Where: H = Head required to accelerate water to design velocity.

Reducers (e.g. Pipe diameter reduces from 203mm to 153mm)

H = g2

Vg2

V 21

22 −

Where subscript 1 is original diameter and subscript 2 is the reduced diameter.

Recommended Age Factor For Mine Water Piping

Age in Years 10 15 20 30

Age Factor 1,3 1,45 1,6 2,0

New equivalent pipe length = Given new pipe length x age factor.

Conversion Of Metres Head To Pressure And Vice Versa:

• Metres head to pressure

ΔP = (H - Hf) x 9,79 (kPa)

• Pressure To Metres head

Metres head = 79,9PΔ (Where ΔP = ΔH - Hf)

Temperature Increase In Pipes

• Vertical pipes (Stroh’s Equation)

Δt = 4,1879,79 per 1000 m = 2,34 oC per 1000 m

• Horizontal pipes (Joule Thompson effect) due to friction.


- 49 -

ΔT = μ x ΔP

Where: ΔT = Temperature increase due to friction (K)

μ = 2,4 x 10-7

ΔP = Pressure drop due to friction (Pa) (i.e.: Hf x 9,79 x 1000)

Energy Recovery

Energy recovered = 1000

H - H fΔ x g x m x η

Where: g = 9,79 m/s2

m = Water flow rate (l /s)

η = Turbine efficiency expressed as a fraction of 1.

Pump Power Requirements

• Where Q = l /s

Power(in) = η x 1000g x H x Q T

• Where Q = m3/s

Power(in) = η

g x H x Q T

Important Notes:

• 1 m3/s of water flow rate = 1000 l /s of water.

• Frictional head loss in pipes down a shaft above a turbine should not exceed 2,5m / 100m

• Water terminal velocity is where the head loss is equal to 100 m / 100 m.

• Water pressure increase due to elevation = 9,79 kPa / m

• Water power operating on a turbine = P x Q (Kw)

Where: P = (ΔH - Hf) x 9,79

Q = Water flow rate in m3/s (i.e. 1000

/sl )

• Pump Total Head (HT) = Static Head (Hs) + Head loss due to friction (Hf)

∴ Ht = Hs + Hf


- 50 -

• Chilled water most economical pipe water velocity = ± 2 m/s

• Recommended head loss in vertical pipes = less than 2,5 / 100 m

(I.e. not more than 2,5 % / 100m)

• Recommended station water pressure = ± 1000 kPa

• Minimum water pressure at coolers should be ± 100 kPa

• Minimum water pressure at mining operations ± 300 kPa


- 51 -

ECONOMICS

Simple Interest

I = p x n x 100

i

Where: I - interest to be paid [Rand]

p - principal invested [Rand]

n - time that the principal is invested [Years]

i - interest rate [%]

Compound Interest

S = pn

100i1 ⎥⎦

⎤⎢⎣

⎡ +

Where: S - total sum of money at the end of the investment period [Rands]

p - principal invested [Rand]


n - time that the principal is invested [Years]

Total Owing Cost

Value of capital cost plus present value of annual running cost

Present Value

[ ]nn

I11v1ofvaluePresent+

==

The present value of 1 per year for n years at an interest rate of I

An = iv1 n−

Where: vn - the present value which one unit of money in n years would have at the present time

n - years


NB: Please refer to the “Environmental Engineering In South Africa” hand book page 848 - 860


- 52 -


- 53 -


- 54 -

STATISTICS

MEAN

This is the arithmetical average of a set of values;

x = nxΣ

Where: x = means the average value

Σ = The Greek letter sigma means the sum of all individual values of x.

n = is the number of observations.

Example:

Mean of (10, 15, 29, 22, 16, 20) = (112/6) = 18.7

GEOMETRIC MEAN (GM)

GM = n nYxYxYxYY ....4321

n = number of values y

Example:

Geometric mean of 2, 4, 6, 3, 5

GM = 5 5 x 3 x 6 x 4 x 2

= 3.7

MEDIAN (Me):

Is the middle value when all the observations are arranged in ascending order.

Me = 2

1n +

Example:

Median of (1, 3, 2, 5, 4, 6, 9, 8 and 7) = (1, 2, 3, 4, 5, 6, 7, 8, 9) equals 5 (i.e. 4 values below 5 and 4 above – applicable for odd or uneven number of values only)

Median of (2, 4, 3, 1, 5, 8, 7, and 6) = (1, 2, 3, 4, 5, 6, 7, 8) equals 4.5 (the average of 4+5 or

using Me = 2

1n + )


- 55 -

STANDARD DEVIATION:

Is to determine by how much the individual observation vary from the mean.

To calculate the "Standard Deviation" from a set of observations the following formula is used:

(S) = 1 - n x)- x ( 2Σ

Where: S = Standard deviation

Σ = The Greek letter sigma means the sum of all individual values of x.

n = the number of observations.

PERCENTILES: Are the values in a set which divide the set into 100 equal parts.

QUARTILES: The values in a set which divide the set into 4 equal parts.

RANKING: (Array) to arrange numbers in ascending or descending order.

CONFIDENCE LIMITS on calculated parameter:

The statistician must decide if the mean of two sets of observations which show a difference from each other do in fact represent a genuine difference in condition.

This is the Confidence interval for a population’s mean:

X = ± 1.95(∂n )

Where: ∂ = Standard deviation

n = number of values

1.95 = 95% confident