PLANT ENGINEERING: MINES AND WORKS - JUNE 2006 MEMORANDUM
MEMORANDUM: June 2006 (Plant Engineering Mines and Works)
QUESTION 1
1.1Distinguish between three classes of fires and the type of fire extinguisher that
should be used in each case. (6)
Solution.
Class A:
Fires that occur in ordinary combustible materials such as paper, wood, etc, √
for which the quenching or cooling effects of water or water solutions are
most effective.√
Soda-acid extinguisher containing a chemical in water solution.
Water tank type, which utilizes a cartridge of pressurized gas to expel the water.
Class B:
√
Fires that involve inflammable liquids, gasses etc., for which a blanketing or
smothering effect is essential.
These extinguishers work on the principle other than cooling.
Foam type spreads a layer of tough foam over the burning material and
extinguishes the fire by excluding oxygen.
The carbon-dioxide extinguisher.
Vapourising-liquid extinguisher.√
Class C:
√
Fires that occur in “live” electrical equipment on which the use of an electrically
non-conductive extinguishing agent is of first importance.
Carbon-dioxide extinguisher.
Dry-chemical extinguisher.
Vapourising-liquid extinguisher.√
1.2Explain the difference between inspections and audits. (6)
Solution:
√
Audit:An audit requires the confirmation of a safety programme by pre inspectionmeetings √ to determine the scope and by field inspections to determine theapplication and the effectiveness of control measures. A report flows on the investigation. √
√
Inspection:An inspection is the confirmation of procedures, technical
standards,instructions etc. √ by inspection and questioning. A report follows on the inspection. √
1.3What are the two main sources of disastrous explosions in fiery coal mines?(2)
Solution.
The two main sources of disastrous explosions in fiery coal mines are coal dust √
and flammable gas such as methane. √ The presence of sufficient concentrations of fine coal dust or methane gas in the air, together with a source of ignition will
inevitably result in an explosion. Very often a coal dust explosion is initiated by
the explosion of methane which raises a cloud of coal dust and ignites it.
1.4What basic precautions are taken to prevent explosions due to QUESTION 1.3
causes?(6)
Solution.
Explosions due to flammable gases may be eliminated by effectively testing the air for dangerous gas concentrations √ and by providing adequate ventilation √ so that by continual dilution and dispersion the accumulation of explosive gas concentrations is prevented. The possible sources of ignition must also be eliminated by using equipment and by preventing the occurrences of naked flames, electric arcs, sparks, etc. √
Effective testing for the presence of flammable gases √ may be through the use of commercially available methanometers, or flame safety lamps, by persons properly trained to conduct such tests. These tests must be carried out in terms of the Regulations as framed under the (Mines and Works Act) Mines Health and Safety Act and Regulations.
Two main lines of defence against coal dust explosions are the prevention of ignitions and the spreading around the mine of incombustible stone dust in sufficient quantity to make sure that any cloud of dust that may be raised in the air will be inexplisive. √
Modern collieries are equipped with sufficient well placed stone dust and water barriers in addition to fully organized systems of cleaning up coal dust and spreading stone dust. √
QUESTION 2
2.1A double shoe of a small hoist is shown in figure 1, diagram sheet 1. The
brake drum rotates clockwise at 100 revolutions per minute. If the coefficient
of friction is 0,3, determine the braking capacity. Determine also the forces
acting at each of the pins A,B,C,D and E. The brake shoe contact angle is 110º. (14)
Solution
2.2Name THREE advantages and THREE disadvantages of using Koepé winders.(6)
Advantages
Can be tower mounted
Inertia of rotating parts less – because of less rotating parts
Excavation cost is less when installed u/g.
No coiling rope – less damage to rope(3)
Any others
Disadvantages
Rope cost mush more – more ropes
Rope examination takes longer
Rope specimen cuts cannot be done as the rope length cannot be reduced
Cannot be used for loading on multiple levels(3)
Any others
QUESTION 3
3.1The maximum efficiency of a 500 kVA transformer occurs at 83% of full load. It has
a full-load efficiency of 97,3% at unity power factor. The 24 hour load cycle is as follows:
No load9 hours
0,6 full load at 0,9 power factor5 hours
0,75 full load at 0,85 power factor3 hours
Full load at 0,8 power factor7 hours
Calculate for the 24 hour period:
3.1.1The total losses in kWh(8)
3.1.2The energy efficiency.(2)
Solution
At Maximum efficiency Pcu = Piwhere Pcu = Copper losses (kW)
Pi = Iron losses (kW)
ηmax occurs at 83% full load
therefore losses = Pi + [ Pi /0,83 ]²Pcu is oroportional to the square of the load
Pi is constant for all loads
η = Pout .
Pout + losses
0,973 = 500 .
500 + losses
√
Losses = 13,8746 kW = Pi + [ Pi/0,83]² and Pi = 5,65943 kW √
And Pcu = 5,65943/0,83² = 8,21517 kW √
Transformer kVA / K1 / PF / Time hrs / kWh500 / 1 / 0,8 / 7 / 500 x 1 x 0,8 x 7 = 2 800
500 / 0.75 / 0,85 / 3 / 500 x 0,75 x 0,85 x 3 = 956,25
500 / 0,6 / 0,9 / 5 / 500 x 0,6 x 0,9 x 5 = 1 350
500 / - / - / 9 / -
Total kWh = 5 106,25 kW
√√
3.1.1The total losses in kWh
% load kW / Pi kW / Pcu kW / Time hrs / Loss kWh100 / 5 65943 / 1² x 8,2157 / 7 / (5,65943 + 8,21517) x 7 = 97,12
75 / 5 65943 / 0,75² x 8,21517 = 4,621 / 3 / (5,65943 + 4,621) x 3 = 30,8413
60 / 5 65943 / 0,62² x 8,21517 = 2,95746 / 5 / (5,65943 + 2,95746) x 5 = 43,08445
No load / 5 65943 / - / 9 / (5,65943 + 0) x 9 = 50,93487
Total power losses = 221,983 kW
√√√
3.1.2The energy efficiency
Energy η = kWh . = 5 106,25 . x 100
kWh + Losses5 v106,25 + 221,98282 1
= 95,834 % √√
3.2Explain why the operation of a plant at a low power factor results in power being
wasted.(5)
Solution
The operation of a plant at low power factor results in power factor being wasted since
the line current is greater than it need be for the transmission of a given amount of √
power. This increase in line current results in an increase in:
(i)transmission losses; √
(ii)Alternator current load and losses; √
(iii)Alternator voltage drop, and √
(iv)Line voltage drop.
The poor voltage regulation resulting from a low power factor will require expensive
appliances to keep up the voltage at the receiving end of the transmission lines since the
effect of a lowered voltage applied to induction motors, for instance, will increase the
load current and decrease the motor torque considerably with the possibility of the
motors falling out of step or stalling. √
3.3What economic criteria apply when considering the extent to which the power
factor of electrical plant may be raised?(5)
Solution
The user of electrical energy will have to take the following criteria into account
when considering the extent to which the power factor may be raised to yield the
greatest economy: √
(i)the savings or decrease in maximum demand charges as levied by the
supply authority: √
(ii)the savings or decrease in energy charges arising out of the corresponding
decrease in system copper losses: √
(iii)the cost of installing static or synchronous power factor correction
apparatus to raise the power factor to the optimum level. √
QUESTION 4
4.1The following information was obtained for an underground mining area that is
to be cooled by chilled service water and air heat exchangers.
Mass of rock mined85 000 t/month
Mass ratio of service water to mined rock2,5:1
Temperature difference of the chilled water across the refrigeration plant19˚ C
Temperature difference of the chilled water across the heat exchangers17˚ C
Quantity of refrigeration to cool the air12 200 kW
Quantity of refrigeration to cool the mined rock 6 280 kW
Determine:
4.1.1The quantity of chilled service water(3)
4.1.2The quantity of chilled water to cool the air(3)
4.1.3The total quantity of chilled water to be provided(3)
4.1.4The capacity of the refrigeration plant(3)
Solution
LetMcsw=Mass of chilled service water
M1=Mass of chilled water supplied to cool air
M2=Mass of chilled water supplied to cool chilled service water
Mtotal=Mass of chilled water supplied to system
24 days production per month
24 hours per day cooling
4.1.1Quantity of chilled service water
Mcsw=Mass ratio of service water/ton of rock mined x mass of rock mined
= 2,5 x 85 000 x 1000 = 102,479 kg/s = 6 148,73 kg/min
24 x 24 x 60 x 60
Or 2,5 x 85 000 = 212 500 m³/month √√ √
4.1.2Quantity of chilled water to cool the air
Energy flow= Mass of chilled water x Specific heat capacity x Temperature change
12 200 = M1x4,187x17
M1=171,4 kg/s √√√
4.1.3The capacity of the refrigeration plant
Mtotal=M1+M2
=171,4+ Energy to cool mine rock/(SHC x temp change)
=171,4+6 280/(4,187 x 19)
=171,4+79,07
=250,47 kg/s √√√
4.1.4The capacity of the refrigeration plant
Capacity of fridge plant=Qair+Qcsw
=12 200 +6 280
=18 480 kW √√√
4.2The data below was taken from a pressure switch mounted on a feeder
breaker in a hazardous location underground in a coal mine.
CERTEX
Ex ib I T3
IEC 60079-11
4.2.1Explain the meaning of the above data.(6)
4.2.2Motivate whether or not the pressure switch is suitable for
use on the feeder breaker.92)
4.2.1Ex ib I T3
Certex: Name of an approved certification body(1)
Ex:Explosion protected apparatus(1)
IbIntrinsically safe approved apparatus type ”b”(1)
(one countable fault) for use in zone 1.
IGas group I
Methane gas ignition temp 537ºC
Methane gas and coal dust ignition temp 150ºC.(2)
T3Maximum operating temperature or temperature class of
apparatus i.e. 200ºC.(2)
4.2 .2 IEC 60079-11
The number of the IEC standard to which the apparatus was tested
to and approved.(1)
The apparatus (pressure switch) may not be used in a location
underground containing methane and coal dust, as it’s operating
temperature exceeds 150ºC.(1)
QUESTION 5
5.1A three-phase load of 3 000 kVA, 0,8 power factor, is supplied at 11 kV from a
step-down transformer having a ratio of 3:1. The primary side of the transformer
is connected to a transmission line, the constants of which are: resistance per
conductor, 2 ohms; reactance per conductor 3 ohms. The resistance and reactance
per phase of the primary windings of the transformer (which are star-connected)
are 5 ohms and 10 ohms respectively, and the corresponding values for the
secondary windings (which are delta-connected) are 1,5 ohms and 3 ohms
respectively. Determine the voltage and power factor at the sending end of the
transmission line.
Solution
The transformer is connected star-delta and has a line voltage ratio of 3:1. Therefore the phase voltage and turns ratios are √3:1.
Let the resistance and reactance of both sides of the transformer be referred to the primary side.
Equivalent resistance per phase = 5 + (√3)² x 1,5 = 9,5 ohms = RT
Equivalent reactance per phase = 10 + (√3)² x 3= 19 ohms = XT
Therefore, including the constants of the transmission line, the total resistance per
Phase= 2 + 9,5 = 11,5 ohms = Ro
Total reactance per phase = 3 + 19 22 ohms Xo
Load phase voltage referred to the primary = 11√3 kV = 19,05 kV = Vph
If the magnetizing current is small enough to be neglected,
Then the primary current I = 300 x 10³ , = 52,49 amperes.
√3 x 33 x 10³
Let the current I be the reference vector I = 52,49 ( 1 + j0) amperes
Now, since the load power factor is 0,8 (assumed lagging), therefore CosØ = 0,8
And sinØ = 0,6, and the load phase voltage Vph = 19,05(0,8 + j0,6) kV
= 15,24 + j11,43) kV
Total impedance drop referred to the primary = IZo
= 52,49 (11,5 + j22) volts
= 603,9 + j1154 volts
= 0,6039 + 1,154 kV
ThenEph= Vph + IZo
= (15,24 + j11,43) + (0,6 + j1,154)
= 15,84 + j12,58 kV
= 20,23 kV
Line voltage at the sending end = Eph x √3 = 20,23 x √3 = 35 kV
Phase angle at sending end= tan-1[12,58/15,84] = 38º 27’ = 0,783 lagging
5.2During the monthly inspection of a 500kVa, 66 000/550V oil cooled transformer
it was discovered that water had entered through the breather pipe into the
transformer and mixed with the oil. Describe the procedure you would follow
to remove the water from the oil. (5)
Solution
5.3A system operating at its maximum current capacity supplies power to four identical machines at a power factor of 0,75. Calculate the power factor improvement required to run a fifth identical machine off the same system. (5)
Solution
Since the supply systems remains unchanged the line voltage VL and current I
remain unchanged.
If P1 is the total power supplied initially, the connection of a fifth identical machine to the load will increase the total power to P2 such that :
P2 = 5/4 x P1 …………(i)
And for a three phase system: P1 = √3 VICosΦ1 = 0,75 x √3 VI …. (ii) √
Substituting equation (ii) int0 equation (i):
P2 = 5/4 x √3 VI …….(iii) , but √
P2 = √3 VICosΦ2 ….(iv)√
Substituting equation (iv) into equation (iii)
P2 = √3 VICosΦ2 = P2 = 5/4 x √3 VI , whence:
Cos Φ2 = 5/4 x 0,75 = 0,9375 √ √
QUESTION 6
6.1The following are particulars of a three throw single acting belt driven plunger pump used to elevate the settled slime up a vertical shaft
Diameter of plunger120 mm
Length of stroke270 mm
Speed of pump 1,66 rev/second
Relative density of the solids 2,7
Relative density of the slime 1,12
Diameter of pipe line150 mm
Depth of shaft1 000m
Volumetric efficiency 95%
Pump efficiency 48%
Determine:
6.1.1The mass of dry solids pumped in kg/s and(8)
6.1.2The power of the electric driving motor . (2)
Solution
6.1.1The mass of dry solids pumped in kg/s
Flow of pulp Qpulp = Volume moved per second
= X – sectional area x stroke length x number of strokes/sec x ηvolumetric
= 3 x π x 0,12² x 0,27 x 1,66 x 0,95
4
= 0,01445 m³/s √√
Mass of pulp = Mass of water + Mass of solids
ρpulp x Qpulp = ρwater x Qwater + ρsolids x Qsolids
1,12 x 1= 1 x (1 – Y) + 2,7 x Y
Y= 0,0706 m³solids/m³pulp √√
Actual volume of solids = volume solids/ m³pulp x volume of pulp
= 0,0706 x 0,01445
= 0,00102 m³ √√
Mass of solids/s = density of solids x volume flow of solids
= 2,7 x 0,00102
= 0,002754 t/s = 2,754 kg/s √√
6.1.2The power of the electric driving motor .
Htotal= static head + friction head
= 1000 + fLQ²pulpAssume f = 0,006
3,03 d5
= 1 000 + 0,006 x 1 000 x 0,01445²
3,03 x 0,155
= 1 000 +5,445
=1 005,445 m
Power to move pulp = ρ g Htotal x Qpulp = 1,12 x 9,81 x 1 005,445 x 0,01445
= 159,63 kW √
Motor output power = Power to move pulp = 159,63/0,48 = 332,56 kW √
Pump efficiency
6.2A pipe carrying wet steam with a pressure of 1,35 MPa at 193ºC has an external
radius of 100 mm. It is lagged to a radius of 150 mm with insulation with a thermal conductivity of 0,04 W/mºC.
The temperature of the surrounding air is 24ºC and the loss from the surface of the insulation is 8,3W/m² ºC. Neglecting the effect of the conductivity of the pipe itself, calculate the heat loss in kJ/h per 10 m of lagged pipe. (10)
Q through lagging = 2 π L (t1 – t3) .
Ln (r2/r1) + . 1 .√
k r2 α
√ √√√√√
= 2 x π x 10 x (193 – 24) = 970,63 W = = 3 494,283 kJ/hour
Ln (150/100) + . 1 .
0,04 0,15 x 8,3
√√√
QUESTION 7
7.1A conveyor belt runs at 1,5 m/s for 26 days per month, 20 hours per day, to convey
100 000 tons over a horizontal distance of 120 m and a vertical distance of 20 m. The
bulk density of the product conveyed is 1,6 tons/m³. The angle of wrap on the drive
pulley is 250º with a coefficient of friction of 0,35, and a coefficient of friction of the
idlers of 0,04. The drive gearbox has an efficiency of 90%.
Determine:
7.1.1The width of the conveyor belt(2)
7.1.2The power of the driving motor required, and(6)
7.1.3The mass of the counterweight required.(2)
Solution
7.1.1 Width of belt
Tonnage/hour = 300 W² V ρ
W² = 100 000 x 1 .
26 x 20 300 x 1,5 x 1,6
W = 0,5168 m √√
7.1.2Power of the driving motor
Power of the driving motor = Power to elevate load + Power to run empty conveyor
+ Power to overcome loaded belt friction
Power to elevate load = mload g x vertical height = 100 000 . x 9,81 x 20
26 x 20 x 60²
= 10,481 kW √
Assume mass of conveyance = 60 x width of belt = 60 x 0,5168
= 31 kg/m length
Length of conveyor = (120² + 20²)0,5 = 121,65 m
For conveyors longer than 100 m effective length = length of conveyor + 45 m
Power to run empty conveyor = µ mconveyor/m g x effective length x velocity
√
= 0,04 x 31 x 9,81 x (121,65 + 45) x 1,5
= 3,04 kW √
Power to overcome loaded belt friction = µ mload g x L
= 0,04 x 100 000 . x 9,81 x 121,65
26 x 20 x 60²
= 2,549 kW√
Then Power to run conveyor = 10,481 kW + 3,04kW + 2,549 kW
= 16,07 kW √
Power of the driving motor = 16,07/mechanical efficiency
= 15,058/0.9
√
= 17,86 kW select nearest standard size motor greater than this
7.1.3 The mass of the counterweight required.
T1/T2 = eµθ = e0,35 x 250 x π/180 = 4,6051
Then Power to run conveyor = (T1 – T2) x V
16,07 kW = ( 4,6051T2 – T2) x 1,5
T2 = 2,972 kN √
Mass of counter-weight = 2T2 = 2 x 2,972 = 606 kg √
g 9,81
Method 2
7.1.1 Width of belt
Tonnage/hour = 300 W² V ρ
W² = 100 000 x 1 .
26 x 20 300 x 1,5 x 1,6
W = 0,5168 m = 600 mm (nearest standard)
7.1.3Power of the driving motor
Power of the driving motor = Power to elevate load + Power to run empty conveyor
+ Power to overcome loaded belt friction
Power to elevate load = mload g x vertical height = 100 000 . x 9,81 x 20
26 x 20 x 60²
= 10,481 kW
Assume mass of conveyance = 60 x width of belt = 60 x 0,6
= 36 kg/m length
Length of conveyor = (120² + 20²)0,5 = 121,65 m
For conveyors longer than 100 m effective length = length of conveyor + 45 m
Power to run empty conveyor = µ mconveyor/m g x effective length
= 0,04 x 36 x 9,81 x (121,65 + 45) x 1,5
= 3,531 kW
Power to overcome loaded belt friction = µ mload g x L
= 0,04 x 100 000 . x 9,81 x 121,65
26 x 20 x 60²
= 2,549 kW
Then Power to run conveyor = 10,481 kW + 3,531 kW + 2,549 kW
= 16,561 kW
Power of the driving motor = 16,561/mechanical efficiency
= 16,561/0.9
= 18,4 kW select nearest standard size motor greater than this
7.1.3 The mass of the counterweight required.
T1/T2 = eµθ = e0,35 x 250 x π/180 = 4,6051
Then Power to run conveyor = (T1 – T2) x V
16,561 kW = ( 4,6051T2 – T2) x 1,5
T2 = 3,0626kN
Mass of counter-weight = 2T2 = 2 x 3,0626 = 0,6244 t
g 9,81
7.2A two stage reciprocating compressor compresses 6 kg of air per second from
a barometric pressure of 85 kPa abs. to a delivery pressure of 650 kPa gauge.
The temperature of air at inlet is 27º and that of the air leaving the high pressure
cylinder 130ºC.
The following readings have been obtained:-
Quantitylitre/minute / Inlet
Temperature ºC / Outlet temperature ºC
Water jacket of low pressure cylinder. / 350 / 25 / 29
Water jacket of high pressure cylinder. / 300 / 26 / 30
Cooling water to inter-cooler / 1 200 / 20 / 35
Determine the isothermal efficiency of the compressor (10)
Solution
P1V1 = mRT1 then V1 = 6 x 0,287 x 300 = 6,078 m³/s√
85
Isothermal Work = P1V1Ln(P2/P1) = 85 x 6,078 x Ln [(650 + 85)/85]
= 1 114,42 kW√
Energy supplied to air during compression = m x cp x ∆t
= 6 x 1,005 x (130 -27)
= 621,1 kW√
Energy supplied to the water in the low pressure water jacket/sec = m cp ∆t
= 350/60 x 4,19 x (29 – 25) = 97,7 kW
Energy supplied to the water in the high pressure wate√r jacket/sec = m cp ∆t
= 300/60 x 4,19 x (30 – 26) = 83,74 kW√
Energy supplied to cooling water in the inter-cooler/sec = m cp ∆
= 1 200/60 x 4,19 x (35 – 20) = 1 256,1 kW √
Neglecting radiation
Actual WD/s = Qwater from LP cylinder + Qwater from intercooler + Qwater from HP cylinder + Qexcess in air
= 97,7 kW + 1 256,1 kW + 83,74 kW + 621,1 kW
= 2 058,64 kW √
Isothermal Efficiency = Iso WD . = 1 114,42 . = 54,134 % √ √ √
Actual WD 2 058, 64
QUESTION 8
8.1The manufacturer claims that a refrigeration plant using Freon 22 has a coefficient of 3:1. The user monitors the plant and records the following:
Suction pressure 40 kPa
Condenser pressure300 kPa
Temperature after compression 70ºC
The refrigerant vapour is dry at the compressor inlet with no under-cooling of
condensate.
Determine from the attached Freon 22 chart the actual coefficient of performance and
the compressor efficiency.(12)
Solution
Due to a typing error on the examination paper, the following assumptions were
made:
Condenser pressure = 400 kPa
Evaporator pressure = 300 kPa
Then:COP= Refrigeration effect =h1 – h2
Work on refrigerant by compressor h2 – h1
= 400 – 193 = 3,63
457 – 400
Compressor efficiency = h21– h1 = 407 – 400 x 100 = 12,28%
h2 – h1452 – 4001
8.2 A solenoid-operated hydraulic valve fails to operate. List FOUR possible causes
with appropriate corrective actions.(8)
Possible cause of failure:
- Power loss to solenoid -check electrical supply
- Coil burned – replace coil
- Blockage in solenoid – strip and clean
- Blockage to solenoid – strip and clean
- Dirt in vlve blocking operation – clean mesh
- Incorrect fluid viscosity – check with supplier
- Corrosion on valve as a result of moisture – remove valve and clean with fine sandpaper
- Any other logical description
QUESTION 9
9.1A production shaftwith a total workforce of 200 people is to be established
remote from any services or townships. List and describe briefly the services,
plant and amenities that must be provided on surface to bring the shaft into
production.(12)
Solution
Services: Fire brigade, water, electricity, communication, roads, lamp room,
stores, waste dump(operational), refuse dump (domestic), timber yard, explosive
yard, medical station, training centre, schools, shopping centre, petrol (filling)
station, securities, recreational facilities, canteen, dams fresh and mine water,
control centre, offices, general mine support i.e. time office, HR etc, crèche,
information centre, etc
Plant:Fans, sewage, fans(ventilation), winder, headgear, compressor, workshops, refrigeration plant(if required), batch plant, process plant, rescue points, general equipment i.e. conveyors, stand-by generators etc.
Amenities: Ablution block, housing, hostel, change house, laundry, etc .
9.2Describe how the following heat treatments of steel are carried out, and why
they are used:
9.2.1Normalising (2)
9.2.2Hardening (2)
9.2.3Tempering (2)
9.2.4Nitriding (2)
Solution:
Normalising:This means heating a steel (however previously treated) to a
temperature exceeding its upper critical range, and allowing it to cool freely in the
air. It is desirable that the temperature of the steel shall be maintained for not
less than two minutes per mm thickness, and shall not exceed the upper limit of
the critical range by more than 50ºC. This will ensure that the steel will return to
its natural state free of stresses.
Hardening: This means heating a steel to its normalizing or required
temperature, and cooling rapidly in a suitable medium, eg water, oil or air, so that
it becomes hard and can resist scratches.
Tempering: This means heating a steel to a temperature below its lower change-