MEMORANDUM: June 2006 (Plant Engineering Mines and Works)
QUESTION 1
1.1
Distinguish 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.
1.2
√
Explain the difference between inspections and audits.
(6)
Solution:
√
Audit: An audit requires the confirmation of a safety programme by pre
inspection meetings √ to determine the scope and by field inspections to
determine the application and the effectiveness of control measures. A
report flows on the investigation. √
Inspection:
1.3
√
An inspection is the confirmation of procedures, technical
standards, instructions etc. √ by inspection and questioning. A
report follows on the inspection. √
What 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.
1.4
What 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
2.1
A 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.
Solution
(14)
2.2
Name 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
Any others
(3)
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
3.1
The 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 load
0,6 full load at 0,9 power factor
0,75 full load at 0,85 power factor
Full load at 0,8 power factor
9 hours
5 hours
3 hours
7 hours
Calculate for the 24 hour period:
3.1.1
3.1.2
The total losses in kWh
The energy efficiency.
(8)
(2)
Solution
At Maximum efficiency Pcu = Pi
where Pcu = Copper losses (kW)
Pi = Iron losses (kW)
ηmax occurs at 83% full load
therefore losses = Pi + [ Pi /0,83 ]²
Pcu 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
500
500
500
500
1
0.75
0,6
-
0,8
0,85
0,9
-
Time
hrs
7
3
5
9
kWh
500 x 1
x 0,8 x 7 = 2 800
500 x 0,75 x 0,85 x 3 = 956,25
500 x 0,6 x 0,9 x 5 = 1 350
Total kWh = 5 106,25 kW
√√
3.1.1 The total losses in kWh
% load
kW
100
75
60
No load
Pi kW
Pcu kW
5 65943
5 65943
5 65943
5 65943
1² x 8,2157
0,75² x 8,21517 = 4,621
0,62² x 8,21517 = 2,95746
-
Time
hrs
7
3
5
9
Loss kWh
(5,65943 + 8,21517) x 7 = 97,12
(5,65943 + 4,621) x 3 = 30,8413
(5,65943 + 2,95746) x 5 = 43,08445
(5,65943 + 0) x 9
= 50,93487
Total power losses = 221,983 kW
√√√
3.1.2 The energy efficiency
Energy η =
kWh
.
kWh + Losses
=
5 106,25
. x 100
5 v106,25 + 221,98282
1
= 95,834 % √ √
3.2
Explain 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)
(ii)
(iii)
(iv)
transmission losses; √
Alternator current load and losses; √
Alternator voltage drop, and √
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. √
3.3
What 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)
(ii)
(iii)
the savings or decrease in maximum demand charges as levied by the
supply authority: √
the savings or decrease in energy charges arising out of the corresponding
decrease in system copper losses: √
the cost of installing static or synchronous power factor correction
apparatus to raise the power factor to the optimum level. √
QUESTION 4
4.1
The 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 mined
Mass ratio of service water to mined rock
Temperature difference of the chilled water across the refrigeration plant
Temperature difference of the chilled water across the heat exchangers
Quantity of refrigeration to cool the air
Quantity of refrigeration to cool the mined rock
Determine:
4.1.1
The quantity of chilled service water
4.1.2
The quantity of chilled water to cool the air
4.1.3
The total quantity of chilled water to be provided
4.1.4
The capacity of the refrigeration plant
Solution
Let
Mcsw
M1
M2
Mtotal
=
=
=
=
Mass of chilled service water
Mass of chilled water supplied to cool air
Mass of chilled water supplied to cool chilled service water
Mass of chilled water supplied to system
24 days production per month
24 hours per day cooling
85 000 t/month
2,5:1
19˚ C
17˚ C
12 200 kW
6 280 kW
(3)
(3)
(3)
(3)
4.1.1Quantity of chilled service water
Mcsw =
mined
=
Mass ratio of service water/ton of rock mined x mass of rock
2,5 x 85 000 x 1000
24 x 24 x 60 x 60
= 102,479 kg/s = 6 148,73 kg/min
Or 2,5 x 85 000 = 212 500 m³/month √ √ √
4.1.2Quantity of chilled water to cool the air
Energy flow
change
12 200
M1
= Mass of chilled water x Specific heat capacity x Temperature
=
M1
x
4,187 x
=
171,4 kg/s √ √ √
17
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
4.2
=
=
=
Qair
+
Qcsw
12 200 +
6 280
18 480 kW √ √ √
The 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
4.2.1
Explain the meaning of the above data.
(6)
4.2.2
Motivate whether or not the pressure switch is suitable for
use on the feeder breaker.
92)
4.2.1 Ex ib I T3
Certex: Name of an approved certification body
(1)
Ex:
Explosion protected apparatus
(1)
Ib
Intrinsically safe approved apparatus type ”b”
(one countable fault) for use in zone 1.
(1)
I
Gas group I
Methane gas ignition temp 537ºC
Methane gas and coal dust ignition temp 150 ºC.
(2)
Maximum operating temperature or temperature class of
apparatus i.e. 200 ºC.
(2)
T3
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 150 ºC.
(1)
QUESTION 5
5.1
A 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.
R0
X0
I
Eph
RL
RT
XL
XT
Vph
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
Then
Eph
= 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
5.2
During 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
5.3
A 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
6.1
The 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 plunger
Length of stroke
Speed of pump
Relative density of the solids
Relative density of the slime
Diameter of pipe line
Depth of shaft
Volumetric efficiency
Pump efficiency
120 mm
270 mm
1,66 rev/second
2,7
1,12
150 mm
1 000m
95%
48%
Determine:
6.1.1
6.1.2
The mass of dry solids pumped in kg/s and
The power of the electric driving motor .
(8)
(2)
Solution
6.1.1 The 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 √ √
6.1.2 The power of the electric driving motor .
Htotal
= static head
+ friction head
= 1 000
+ fLQ²pulp
3,03 d5
= 1 000
+ 0,006 x 1 000 x 0,01445²
3,03 x 0,155
= 1 000
+
=
Assume f = 0,006
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
6.2
= Power to move pulp = 159,63/0,48 = 332,56 kW √
Pump efficiency
A 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)
Ln (150/100) + . 1 .
0,04
0,15 x 8,3
√
√√
=
√√√
970,63 W = = 3 494,283 kJ/hour
QUESTION 7
A 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:
7.1.1
7.1.2
7.1.3
The width of the conveyor belt
The power of the driving motor required, and
The mass of the counterweight required.
(2)
(6)
(2)
Solution
V = 1,5 m/s
M = 100 000t/month
ρ = 1,6t/m³
µ = 0,04
θ= 250º
µ = 0,35
20 m
7.1
1 20 m
7.1.1 Width of belt
Tonnage/hour = 300 W² V ρ
7.1.2
W²
= 100 000 x
1
.
26 x 20
300 x 1,5 x 1,6
W
= 0,5168 m √ √
Power 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
16,07 kW
T2
Mass of counter-weight
= (T1 – T2) x V
= ( 4,6051T2 – T2) x 1,5
= 2,972 kN √
= 2T2 = 2 x 2,972 = 606 kg √
g
9,81
Method 2
7.1.1 Width of belt
Tonnage/hour = 300 W² V ρ
7.1.3
W²
= 100 000 x
1
.
26 x 20
300 x 1,5 x 1,6
W
= 0,5168 m = 600 mm (nearest standard)
Power 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
16,561 kW
T2
Mass of counter-weight
= (T1 – T2) x V
= ( 4,6051T2 – T2) x 1,5
= 3,0626kN
= 2T2 = 2 x 3,0626 = 0,6244 t
g
9,81
7.2
A 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:Quantity
litre/minute
low
350
Water jacket of
pressure cylinder.
Water jacket of high
pressure cylinder.
Cooling water to intercooler
Inlet
Temperature ºC
25
Outlet
temperature ºC
29
300
26
30
1 200
20
35
Determine the isothermal efficiency of the compressor
(10)
Solution
Q
Mw = 350 kg/min
∆T = 4ºC
Q
Mw = 300 kg/min
∆T = 4ºC
Mw = 1 200 kg/s
∆T = 15ºC
Ma = 6 kg/s
P1 = 85 kPa
T1 = 300K
Intercooler
LP Cylinder
Work
HP Cylinder
P3 = 735 kPa
T3 = 403K
Work
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 .
Actual WD
2 058, 64
= 54,134 % √ √ √
QUESTION 8
8.1 The 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
Condenser pressure
Temperature after compression
40 kPa
300 kPa
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
Work on refrigerant by compressor
= 400 – 193
457 – 400
=
h1 – h2
h2 – h1
= 3,63
Compressor efficiency = h21 – h1
h2 – h1
= 407 – 400 x 100 = 12,28%
452 – 400
1
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
9.1
A production shaft with 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 .
9.2
Describe how the following heat treatments of steel are carried out, and why
they are used:
9.2.1
9.2.2
9.2.3
9.2.4
Normalising
Hardening
Tempering
Nitriding
(2)
(2)
(2)
(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 changepoint with the objective of reducing the hardness or increasing the toughness to a
grater or lesser degree to prevent the steel from cracking or chipping .
Nitriding or Nitrogen case-hardening:
The process is based on the fact that certain steels will absorb nitrogen, enabling
an extreamely hard surface to be produced. It consists of maintaining the steel at
a carefully controlled temperature, normally in the range of 490ºC to 530 ºC and
subject it to the action of nascent or active nitrogen produced a surface of the
work by the decomposition of gaseous ammonia. The nitrided surface is
exceptionally hard but correspondingly brittle, and usually extremely thin.