Plant Engineering: Mines and Works - June 2006 Memorandum

PLANT ENGINEERING: MINES AND WORKS - JUNE 2006 MEMORANDUM

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: An inspection is the confirmation of procedures, technical standards, instructions etc. √ by inspection and questioning. A report follows on the inspection. √

1.3 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. (14)

Solution

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 (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

3.1 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 9 hours 0,6 full load at 0,9 power factor 5 hours 0,75 full load at 0,85 power factor 3 hours Full load at 0,8 power factor 7 hours

Calculate for the 24 hour period:

3.1.1 The total losses in kWh (8) 3.1.2 The energy efficiency. (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 Time kWh hrs 500 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 √ √

3.1.1 The total losses in kWh % load Pi kW Pcu kW Time Loss kWh kW hrs 100 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 √ √ √

3.1.2 The energy efficiency

Energy η = kWh . = 5 106,25 . x 100 kWh + Losses 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) 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. √

3.3 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) 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

4.1 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 85 000 t/month Mass ratio of service water to mined rock 2,5:1 Temperature difference of the chilled water across the refrigeration plant 19˚ C Temperature difference of the chilled water across the heat exchangers 17˚ C Quantity of refrigeration to cool the air 12 200 kW Quantity of refrigeration to cool the mined rock 6 280 kW

Determine: 4.1.1 The quantity of chilled service water (3) 4.1.2 The quantity of chilled water to cool the air (3) 4.1.3 The total quantity of chilled water to be provided (3) 4.1.4 The capacity of the refrigeration plant (3)

Solution

Let Mcsw = 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 = M1 x 4,187 x 17

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 √ √ √

4.2 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” (1) (one countable fault) for use in zone 1.

I Gas group I Methane gas ignition temp 537ºC Methane gas and coal dust ignition temp 150 ºC. (2)

T3 Maximum operating temperature or temperature class of apparatus i.e. 200 ºC. (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 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.

R X 0 0 I

R R X X Eph L T L T 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 120 mm Length of stroke 270 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 line 150 mm Depth of shaft 1 000m Volumetric efficiency 95% Pump efficiency 48%

Determine:

6.1.1 The mass of dry solids pumped in kg/s and (8) 6.1.2 The power of the electric driving motor . (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 Assume 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

6.2 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) = 970,63 W = = 3 494,283 kJ/hour Ln (150/100) + . 1 . 0,04 0,15 x 8,3 √ √√

QUESTION 7

7.1 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 The width of the conveyor belt (2) 7.1.2 The power of the driving motor required, and (6) 7.1.3 The mass of the counterweight required. (2)

Solution

V = 1,5 m/s θ= 250º M = 100 000t/month µ = 0,35 ρ = 1,6t/m³ µ = 0,04 m

1 20 m

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 √ √

7.1.2 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.

µθ 0,35 x 250 x π/180 T1/T2 = e = e = 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)

7.1.3 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.

µθ 0,35 x 250 x π/180 T1/T2 = e = e = 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

7.2 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 Inlet Outlet litre/minute Temperature ºC temperature ºC Water jacket of low 350 25 29 pressure cylinder. Water jacket of high 300 26 30 pressure cylinder. Cooling water to inter- 1 200 20 35 cooler

Determine the isothermal efficiency of the compressor (10)

Solution

Q Mw = 350 kg/min Q ∆T = 4ºC Mw = 300 kg/min ∆T = 4ºC Mw = 1 200 kg/s ∆T = 15ºC

Ma = 6 kg/s P = 735 kPa Intercooler 3 P = 85 kPa 1 LP Cylinder T = 403K HP Cylinder 3 T = 300K 1

Work 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 . = 54,134 % √ √ √ Actual WD 2 058, 64

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 40 kPa Condenser pressure 300 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

1 Compressor efficiency = h2 – h1 = 407 – 400 x 100 = 12,28% h2 – h1 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 Normalising (2) 9.2.2 Hardening (2)

9.2.3 Tempering (2)

9.2.4 Nitriding (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- point 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.