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Suggested solutions to the GCC examination12 June 2009

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Question 1

Question 2

2.3The underground haulages of a mine and some rooms and passages on surface are equipped with 60-W incandescent globes. One of the energy saving drives of the mine is to replace the incandescent globes with compact fluorescent lights (CFL). The cost to change a globe is R15-00 for either and it includes travelling time from globe to globe.

Data:Quantity of incandescent lights: 1 580Power output of CFL’s 11 WPower factor of CFL 0,503Price per energy unit R0,45/kWhPrice per apparent power unit R22,50/kVALamp life of a CFL 6 000 hoursLamp life of an incandescent lamp 1 000 hoursPrice of an incandescent R2,45Price of a CFL R18-00

2.3.1 Calculate the cost in replacing the incandescent globes with the CFL’s.

[7]

Suggested solution:Replacement cost:1 580 (18+15) = R52 140-00

2.3.2 What will the impact be on the total electrical account [1]

Suggested solution:

After 100 hours it starts saving electricity despite the elevated kVA costs

2.3.3 Propose some feasible ways to save energy on a mine[3]

Suggested solution: Lag steam lines to prevent unnecessary heat loss Lag fridge lines to prevent energy loss

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Repair all compressed air leaks Apply power factor correction if p.f is not optimal (± 0,9 or better) Switch off compressor during back shift if compressed air is not required Diversify loads to fill valleys and clip peak consumption periods Negotiate better rate with Eskom. Switch off unnecessary lights and airconditioners when not required. Replace incandescent lamps with CFL’s. Optimise plant equipment – replace oversized motors with correctly sized motors.

[20]

Question 3A mine intends hoisting people up and down a sinking shaft, using a mobile crane. Draw up Code of Practice or procedure, covering the following points: Scope Appointments Responsibilities Design Safety notices Regulatory inspections PPE Maintenance Operational requirements Communication Emergency procedures [20]

Question 66.1 A used 3-phase, 8-pole, 50-Hz motor was salvaged and is to be

considered for a special duty. The following information was obtained after some extensive tests and measurements: Full load slip 3% Rotor resistance 0,0011 Ω/phase Standstill reactance 0,0052 Ω/phase

Find the ratio of maximum to the full load torque and the speed at which maximum torque occurs

[10]

Suggested solution:

p = 4 (8 Poles)f = 50 Hzs = 3%Rrotor = 0,0011 Ω/phase

3

Xo = 0,0052 Ω/phase

When the slip = 3% with R in the circuit, the torque is at maximum.

6.2 A 6 600 V, 3-phase induction motor drawing 200 kW at a 0,8 lagging power factor is coupled in parallel with a 6 600V, 3-phase synchronous motor drawing 250 kVA at 0,9 leading power factor.

Calculate the line current and power factor for the installation.[10]

Suggested solution:

Question 77.1An important induction motor with unique insulation, drives a compressor which is essential to production. The compressor is subjected to varying demands. It is thus important to avoid or minimise damage to the motor which has the following particulars: 4 MW, 6,6 kV, 3-phase, 1 480 r/min with forced ventilation.

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Squirrel cage, star-connected, both ends of each phase winding accessible

Locked rotor stall withstand time = 10 s Starting DOL Normal run-up time = 20 s Outboard white metal pedestal bearings with oil rings.

State all the protective devices you would specify for the motor and its switchgear, briefly describing the purpose and function of the devices.

[14]

Suggested solution: Sensitive earth-fault protection – to detect any earth leakage through the windings

and trip it out before major damage could be caused by a low impedance earth fault. Overcurrent protection – to trip out the motor in the event of unplanned overcurrent

conditions. This facility would have to accommodate the 20 second start-up time. This facility will typically be incorporated in the motor protection relay.

Motor protection relay that would include:o Stall protection in excess of 10 secondso In liaison with the overcurrent protection facility, allow for start-up time

of 20 seconds after which it would trip if the current has not subsided to below 100% of running current

o 3-start per hour lock-out preventing more that 3 starting attempts in a running hour.

o Undervoltage detection – to trip out the machine, should a power dip arise.

Vibration analyser – detecting any vibration on the motor above predetermined vibration levels. This would also have to be able to distinguish between starting-up vibrations and running vibrations.

RTD (heat) sensing on the windings in order to monitor the winding temperatures. When rising above predetermined levels (±95 °C, depending on the type of insulation material used) the motor will trip on over temperature and will have to be interlocked to prevent starting efforts before adequate cooling time of the windings.

Filtering systems to keep the air inside the motor as clean and dry as possible, should the air be blown through the motor.

7.2Describe 3 non-destructive tests on the brake rods and -pins of a man-winder. Briefly describe how these tests are done.

[6]

5

Suggested solution:Dye-pen testing - X-rayMagnetic field profile

Question 1 (November 1996)

You have been appointed as the engineer and tasked to increase the capacity of a Koepe winder by improving the winding cycle time. The following data is available:

Rope diameter 35 mmNumber of ropes 3Mass of the rope 5,448 kg/mBreaking strength of each rope 918 kNLength of ropes 700 mLength of wind 550 mNominal drum diameter 3 mSheave mass [is this for the tail end? Or tower end?] 6,5 tMass of skips and attachments [is there more than 1 skip?] 12 tMass of load [is this the payload?] 10 tMass of balance weight 17 tLength of balance weight 4 mMass of tail carriage 5 tMoment of inertia of tail carriage sheave wheel 6500 kg.m2 Frictional resistance of skip 3 kNFrictional resistance of balance weight 2 kNCoefficient of friction between ropes and the sheaves 0,18

Calculate:

1.1 The maximum possible acceleration when raising a fully loaded skip from the loading box.

1.2 The factor of safety of the ropes.

Solution

Firstly, mention was made of a ‘drum’, ‘sheave wheel’ and a ‘tail carriage sheave wheel’.

Does this mean that it is a ground mounted Koepe winder????

Secondly, mention is made of ‘skips and attachments’.

Assumptions:- This is a tower mounted Koepe winder The drum mentioned is the Friction Drive

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The sheave mentioned is the Tail Sheave Mass of load is actually the mass of the Payload (not including rope, conveyance,attactments etc.

How many ropes? 2 head ropes and 1 tail rope, or 3head ropes and 3 tail ropes? For this exercise we assume the latter. Now does this mean we have a sheave per rope or is there one sheave with three groves, again we assume the latter for this calculation.

(Please note that this becomes very stressful, as we have made SIX assumptions so far, and this for the already stress candidate to make, is considered unfair. Further a tensioning carriage at the shaft bottom of the shaft is most unusual for South African conditions.)

1.1 The maximum possible acceleration when raising a fully loaded skip from the loading box.

To determine T1 (force in rope on the skip side accelerating upwards)

T1 = Fgravity + Facceleration + Ffriction

Fgravity = Mtotal x g = [Mskip & att + Mrope + Mpayload + ½(Mtail carriage + Mtail sheave)] x g

= [12 000 + (5,448 x 3 x 700) + 10 000 + ½(5 000 + 6 500)] x 9,81

= 39 190,8 x 9,81

= 384 461,748 N = 384,462 kN

Falinear = M x a = [Mskip & att + Mrope + Mpayload + ½Mtail carriage]x a

= [12 000 + (5,448 x 3 x 700) + 10 000 + ½ x 5 000] x a

= 35 940,8a N

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Farotational = ½ Itail sheave x a = 0,5 x 6 500 x a = 1 444,444a N R²tail sheave 1,5²

Fatotal = Falinear + Farotational = 35 940,8a + 1 444,444a = 37 385,244a N

Ffriction = 3 000 N (given)


= 384 461,748 N + 37 385,244a N + 3 000 N = 387 461,78 + 37 385,244a N

To determine T2 (force in rope on the balance weight side accelerating downwards)

T2 = Fgravity - Facceleration - Ffriction

Fgravity = Mtotal x g = [Mbalence weight + Mrope + ½(Mtail carriage + Mtail sheave)] x g

= [17 000 + (5,448 x 3 x 700) + ½(5 000 + 6 500)] x 9,81

= 34 190,8 x 9,81

= 335 411,748 N = 335,412 kN

Falinear = M x a = [Mbalence weight + Mrope + ½Mtail carriage]x a

= [17 000 + (5,448 x 3 x 700) + ½ x 5 000] x a

= 30 940,8a N

Farotational = ½ Itail sheave x a = 0,5 x 6 500 a = 1 444,444a N R²tail sheave 1,5²

Fatotal = Falinear + Farotational = 30 940,8a + 1 444,444a = 32 385,244a N

Ffriction = 2 000 N (given)


= 335 411,748 - 32 385,244a N - 2 000 N = 333411,748 - 32 385,244a N For the maximum acceleration:

T1 = eμθ

T2

387 461,78 + 37 385,244a = e0,18 x π

333411,748 - 32 385,244a

387 461,78 + 37 385,244a = 1,7603[333 411,748 - 32 385,244a]

387 461,78 + 37 385,244a = 586 904,7 - 57 007,7501a

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37 385,244a + 57 007,7501a = 586 904,7 - 387 461,78

a = 2,113 m/s²

This is reasonable, and hence we would restrict the acceleration not to exceed say 1.8 m/sec²to ensure that slip will not occur

1.2 The factor of safety of the ropes.

FOS = Ultimate force in rope = 3 x 918 . T1 [387 461,78 + 37 385,244a] x 10-3

= 3 x 918 . [387 461,78 + 37 385,244 x 2,113] x 10-3

= 5,9This is considered acceptable as it is well over the legal requirements for transporting people. (Again an assumption. Men or mineral?).

Question 2

2.1 Discuss the advantages and disadvantages of earthing the neutral conductor of a three-phase system. (6)

Suggested solution

Advantages:

a) The voltages created in this system are limited to phase voltages (line-to-ground)

b) The system eliminates arcing ground faults since it is solidly earthed.c) Over-voltages due to lighting are discharged to earth.d) Earthing provides greater safety for persons and equipment since the

earth-fault protection relays operate faster.e) Protective relays in the system may also be used for earth-fault

protection, since any fault-to-earth will cause an unbalance in the system.

Disadvantages:

a) The system experiences higher fault currents than in the insulated neutral system.

b) The earth connection must be made at all points thereby increasing the system capital cost. All points from the substation to the equipment must be earthed.

c) Fast isolation is essential to limit the switching voltages.

2.2 Discuss the influence on the electrical distribution network if the neutral is earthed in more than one point. (3)

Solution

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In a three-phase system earthed at one point, an earth fault in the system causes a large potential to build-up between the earth point and the fault. The resultant fault current is not ‘immediately’ discharged to earth but has to propagate to the earth point before being isolated by the protection system.

So, earthing at multiple points simply limits the earth fault path to very short distances, between the fault and the nearest earth point. The effect is thus quick isolation of the faulty component as soon as possible.

2.3 The underground haulage of a mine and some rooms and passages on surface are equipped with 60 W incandescent globes. One of the energy saving drives of the mine is to replace the incandescent globes with compact fluorescent lights (CFL). The cost to change a globe is R15-00 for either and it includes travelling time from globe to globe.

Data: Quantity of incandescent lights 1 580 Power output of CFL’s 11 W Power factor of CFL 0,503

Price per energy unit R 0-45/kWhPrice per apparent power unit R 22-50/kVALamp life of a CFL 6 000 hoursLamp life incandescent 1 000 hoursPrice of an incandescent R 2-45Price of a CFL R 18-00

2.3.1 Calculate the cost in replacing the incandescent globes with the CFL’s (7)2.3.2 What will the impact be on the total electrical account? (1)2.3.3 Propose some feasible ways to save energy on the mine. (3)

Solution

2.3.1 Calculate the cost in replacing the incandescent globes with the CFL’s

Cost to change globes = R 15-00 x 1 580 = R 23 700-00

Cost to by Fluorescents = R 18-00 x 1 580 = R 28 440-00

Cost in replacing globes = Cost to change globes + Cost to by Fluorescents

= R 23 700-00 + R 28 440-00

= R 52 140-00

2.3.2 What will the impact be on the total electrical account?

Incandescent globe usage = 60 W x 1 000 hours = 60 kWh

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Cost/kWh per bulb = R 0-45 x 60 kWh = R 27-00

Cost/apparent power unit/bulb = 60 VA x R 22-50/kVA = R 1-35

Total cost/bulb = R 27-00 + R 1-35 = R 28-35 over the life of bulb

The cost of the CFL’s = Cost/kWh = 11 W x 6 000 hours = 66 kWh

Cost/kWh = R 0-45 x 66 kWh = R 29-70 (for one)

Cost/apparent power unit = [11 x 10-3/0,503] x R 22-50 = R 0-492/CPL

Total electrical account for incand. bulbs = 1 580 x R28-35 = R 44 793-00 for 1000

hours

Total electrical account for CFL’s = 1 580 x R 30-19 + cost of replacement = R 47 700-20 + R 52 140-00 = R 99 840-20 for 6 000 hours

CFL cost per 1 000 hours = [R 47 700-20/6] + R 52 140-00

= R 60 090-03

This means there will be a slight increase in the cost of electricity (R 60 090-03 – R 44 793-20 = R 15 297-03) the cost includes the initialreplacement cost.

There will be increased savings over the life of the CFL.

2.3.3 Propose some feasible ways to save energy on the mine.

Suggested solution:

a) Proper insulation of the pipes on the refrigeration circuit.b) Detect and repair leaks on compressed air systems.c) Switch off compressor during back shift if compressed air is not requiredd) Introduce solar water heating panels for the hostels and change houses as primary heaters to

geysers.e) Insulate geysers and boilers in change houses and hostels with thermal blankets.f) Control winding times so that no winding should take place during peak times in terms of the

supply authorities.g) Install a wind turbine on the exhaust side of ventilation fans to generate in-house electricity.h) Install motion sensors in offices to ensure that lights and air-conditioners remain off when there

are no persons in offices.i) Install variable speed drives for varying pumping loads.j) Introduce strict water control k) Ensure that most economical use of ventilation air.l) Electric drilling machines in stopes.m) Daylight switching of lighting circuits to switch off unnecessary lights n) Change management in Process Plant wet circuits i.o.w diversify loads to fill valleys and

clip peak consumption periodso) Apply power factor correction if p.f is not optimal (± 0,9 or better)p) Replace incandescent lamps with CFL’s.

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q) Optimise plant equipment – replace oversized motors with correctly sized motors.

Question 3

A mine intends hoisting people up and down a sinking shaft, using a mobile crane. Draw up a Code of Practice or procedure, covering the following points:

Scope, appointments, responsibilities, design, safety notices, regulatory inspections, personal protective equipment, maintenance, operational requirements, communication, emergency procedures.

(20)Suggested solution

3.1 ScopeScope will include the hoisting of people via a mobile crane and pre-sink kibbles in a sinking shaft, as well as the hoisting of material and rock.Permission must be provided by the Engineer to hoist persons with the mobile crane.(Should this not become a licensed winder?)

3.2 AppointmentsThe following appointments will be in place:

3.1 Construction Manager for the overall site

Site Manager – Mine Captain per shaft 2.13.3.1/2.13.3.6 Engineer per shaft Mechanical and Electrical Foreman per shaft Trained and assessed competent Artisans – 4 Fitters, 4 Electricians, 2 Boilermakers per site,

Electricians and Fitters will be on shift. General Foremen per shaft Shift Foremen – 3 per shaft, 1 per shift Appointed Banksman, 3 off, 1 per shift Appointment of trained and assessed competent mobile crane drivers, 4 per shaft, one per shift, 1

standby.

3.3 ResponsibilitiesThe appointed engineer will be responsible for a Hazard Identification and Risk Assessment, and to oversee that the activities associated with the sinking with a mobile crane is conducted in a manner that poses minimum risk to health and safety. The Mine Captain, General Foreman and Shift foremen will be responsible for the activities associated with sinking the shaft, i.e. the mining activities.The appointed artisans will be responsible for inspections (pre-use / operational) and the maintenance of plant and equipment. The mobile crane will be pre-use inspected by the drivers prior to each shift. The drivers will be responsible for the safe operation of the mobile crane.The suppliers of the mobile crane will be on site at least once a week to do a complete inspection and service of the crane. The suppliers will supply a crane which will be safe for the specific application.

3.4 DesignThe Engineer will also ensure that the crane will work within safe load range, given the position of the crane from the centre of the shaft, and also that the crane rope will be able to handle the least favourable load conditions – the falls of rope can be increased to ensure safe working load, given that the rope is long enough to accommodate the depth of the shaft.

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The tripod chain slings of the kibble, the drop nose pins, the kibble palm brackets and the kibbles itself are all to be designed to the requirements of the relevant design standards and law regulations.The two deck stage compartment must be of diameter great enough to allow access through it of the Jumbo Drill Rig, kibbles, 630 Eimco Loaders and all other relevant equipment and materials.The crane must have on it installed and fully functional:

an over wind device (anti two block device)

an under wind device which will indicate to the driver that only 3 turns of rope are left on the drum

a camera on the rope drum which will provide the driver with a visual of the condition of the rope on the drum

overload device which will prevent the driver from overloading the crane

3.5 Safety NoticesThe following notices shall be displayed:

Maximum permitted persons per kibble.

No unauthorised access to the shaft area List of permitted materials Wearing of PPE Signalling Arrangements

The crane is to be barricaded off to prevent other persons from entering the crane area, and vehicles from colliding with the crane.

3.6 Regulatory InspectionsThe Engineer shall inspect at least once a week, not exceeding ten days, the crane and all its safety devices and external parts, and at least once a month, not exceeding 45 days the rope of the crane according to the criteria of the MHSA Regulations Chapter 16.The crane driver shall inspect prior to each shift the entire crane according to a pre-use checklist. The crane driver shall test during each shift the safety devices of the crane when it is safe to do so.The appointed artisans will be responsible for inspections (pre-use / operational) of the plant and equipment, including the Mobile Crane, Jumbo Drill Rig, Eimco 630 Loaders, Bank Area, The Sub-bank Area, Two Deck Pre-Sink Stage, 5 Dymot Winches, Kibbles, Tipping Area, Hauling Trucks, Electrical Reticulation System, the Compressed Air System, Water Supply System.

3.7 Personal Protective EquipmentThe following personal protective equipment shall apply:

Safety Goggles

Gloves Full body overalls Ear plugs Hard hats (with chin strap) Safety Shoes / Gumboots Full Body Harnesses

3.8 MaintenanceThe appointed artisans will be responsible for inspections (pre-use / operational) and the maintenance of plant and equipment, including the Mobile Crane, Jumbo Drill Rig, Eimco 630 Loaders, Bank Area, The Sub-bank Area, Two Deck Pre-Sink Stage, 5 Dymot Winches, Kibbles, Tipping Area, Hauling Trucks, Electrical Reticulation System, the Compressed Air System, Water Supply System. The suppliers of the mobile crane will be on site at least once a week to do a complete service of the crane. 3.9 Operational Requirements

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The kibble height must be low enough to accommodate loading via the 630 Eimco Loader into the kibble but high enough to keep personnel safely in the kibble.5 Dymot winches must be used to suspend the two deck pre-sink stage in the shaft.Operational plant requirements will include:

Mobile Crane

Jumbo Drill Rig

Eimco 630 Loaders Bank Area Sub-bank Area Two Deck Pre-Sink Stage 5 Dymot Winches Kibbles Tipping Area Hauling Trucks Electrical Reticulation System Compressed Air System Water Supply System Diesel Supply for the crane

3.10 CommunicationA compressed air whistle will be situated on the bank which will be operated directly from the bottom and/or two deck stage, the whistle must be clearly observable by the Mobile Crane driver to receive clear distinguishable signals for safe operation.Lights will dip according to the signals given on the two deck stage and to the shaft bottom.Signals from the bank will be given by a appointed banksman.

3.11 Emergency PreparednessA second mobile crane with the same lifting capacity than that of the service crane must be available on site in case of emergency. This crane must be subjected to the same maintenance and inspection requirements than the service crane.Operations are to be suspended during adverse weather conditions, i.e. high wind and storm conditions.

Question 4.1

On a spring-applied hydraulic-release brake engine of a winding plant, two helical compression springs are nested, one inside the other. Both springs have the same free length G. = 80 x 109 Pa.

Dimensions Spring A Spring BNumber of active coils 12 18Wire diameter 16 mm 12 mmMean coil diameter 100 mm 70 mm

4.1.1 Determine the compression required for a braking force of 15 kN. (4)4.1.2 Calculate the load carried by each spring. (8)

Solution

4.1.2 Calculate the load carried by each spring.

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Since both springs have the same deflection ΔA = ΔB

8FADA³ nA = 8FBDB³ nB Gd4A Gd4B

8 x FA x 0,1³ x 12 = 8 x FB x 0,07³ x 18 G x 0,0164 G x 0,0124

FA = 1,62607 FB But FA + FB = 15 000

1,62607 FB + FB = 15 000

FB = 5 711,949 N and FA = 9 288,05 N

4.1.1 Determine the compression required for a braking force of 15 kN.

deflection Δ = 8FADA³ nA = 8 x 9 288,05 x 0,1³ x 12 = 0,17 m = 170 mm Gd4A 80 x 109 x 0,0164

Question 4.2 (932.6F)

A tripod consists of three steel pipes, is to be used to lift a transformer having a mass of 1,8t. The footings of the tripod are spaced equidistant and are 17 m apart. The pipes, for the purpose of this calculation, may be considered to have pin-jointed struts. Ignore eccentricity and the initial curvature of the pipes.

Factor of safety 5Young’s modulus 200 GPaOutside diameter of the pipes 150 mmWall thickness 6 mmLength of pipes 12 m

Determine the maximum safe load that may be raised without the pipes buckling. (8) Given: Pcr = π²EI (Euler) 4 L²

Solution

Total weight raised = mass x g = 1,8 x 10³ x 9,81 = 17 658 N

Each leg carries = W/3 = 17 658/3 = 5 886 N

Section area of pipe leg = π(D² - d²) = π x (0,15² - 0,138²) = 112,764 x 10-6 m² 4 4

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Second moment of area of section I = π(D 4 – d 4 ) = π(0,15 4 – 0,138 4 ) = 7,047 x 10-6 m4

64 64

Radius of gyration = R = √(Ixx/A) = √(7,047 x 10-6 /112,764 x 10-6) = 0,25 m

(Euler) Pcr = π²EI 4L²

= π ² x 200 x 10 9 x 7,047 x 10 -6 = 24 149,7 N 4 x 12²

FOS = Pcr . = 24 149,7 = 3,5532 Pworking 6 796,6

Question 5 (June 2000)

(a) Explain, with the aid of sketches, where losses occur in centrifugal pumps and describe what can be done to minimize such losses. [5]

SolutionIn a general way we can discern four main possibilities of energy losses. 1. Mechanical friction between fixed and rotating parts in bearings and

shaft gland boxes.2. Hydraulic friction – so called disc friction - between the surfaces,

i.e. the liquid and the external rotating faces of the rotor discs.3. Whenever there is leakage of liquid, energy will be dissipated.4. The main flow of the liquid through the passages of the casing and rotor will be subject to hydraulic loss due to eddying and friction.

To minimize these losses: Bearings – choose sleeve roller or ball Glands – mechanical seal or labyrinth Slip ring – maintain correct clearance Hydraulic – passage design

In the hands of the manufacturer: Disc – surface finishes from very rough to polished Pump rotating element

(b) A 200 mm diameter pipe is fitted with a 30° bend. The pressure at both ends of the bend is similar and equal to 200 kPa. If the bend is horizontal, calculate the resultant force to be exerted by the pipe system for the bend to remain in position if the flow is 0,15 m3/s. [10]

Solution The pressure in the pipe remains the same at 200 kPa

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Volume = Cross-sectional area x Velocity

0,15 = π x 0,2² x velocity 4

0,15 = 0,0314 x velocity Velocity = 4,77465 m/s

F1 = Pressure x Area = 200 x 10³ x 0,0314 = 6,283 kN

Mass/s x velocity/s = kg/s x m/s = kg x m/s²= m x a = Accel. force (Fa)

F2 = F1 + Fa = 6,283 + 0,15 x 1 000 x 4,77465 = 6,999 kN

Fresultant² = 6,283² + 6,999² - 2 x 6,283 x 6,999 x cos 30º

Fresultant = 3,507 kN

(c) The capacity of a centrifugal pump has to be increased from 12 kl/s to 15 kl/s. Calculate the increase in speed required to achieve this and the resultant increase in power consumption. The initial speed of the impeller and the total head is 1 450 rev/min and 300m respectively. [5]

Solution

Q α N then N2 = N1 x Q2/Q1 = 1 450 x 15/12 = 1 812,5 r/m

Increase in speed = 1 812,5 – 1 450 = 362,5 r/m

Power α N³ then P1 = P2 x [N1/N2]³ = [1 450/1 812,5]³x P2 = 0,512P2

P2 – P1 = P2 – 0,512P2 = 0,488P2 = 0,488 x ρgH2Q2

= 0,488 x 1 x 9,81 x 468,75 x 15

= 33,66 MW Question 6: 6.1 A used three phase, 8 poles, 50 Hz motor was salvaged and is to be

considered for a special duty. The following information was obtained after some extensive tests and measurements:

Full-load slip 3%Rotor resistance 0,0011 Ω/phaseStandstill reactance 0,0052 Ω/phase

Find the ratio of the maximum to the full load torque and the speed at which maximum torque occurs. (10

Suggested solution:

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The ratio of the maximum to the full load torque is given by: Tmax = 96,15385 TFL 26,735

= 3,59655

The speed at which maximum torque occurs nmax

Synchronous speed ns = F/P = 50/4 = 12,5 r/s

Then the speed of maximum torque occurs when the slip = 0,21154

i.e. s = ns - nmax ns

so that nmax = ns x [1 – s] = 12,5 x [1 - 0,21154] = 9,85575 r/s = 591,345 r/min

6.2 A 6,6 kV, three phase induction motor drawing 200 kW at 0,8 power factor lagging is coupled in parallel with a 6,6 kV three phase synchronous motor drawing 250 kVA at 0,9 leading power factor.

Calculate the line current and power factor for the installation. (10)

Solution

For the induction motor

The apparent induction power Sind = Induction power = 200 = 250 kVA Cos Φind 0,8

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Reactive power Qind = √[S²ind - P²ind] = √[250² - 200²] = 150 kVAr

Power factor angle Φind = cos-1 0,8 = 36,87º

For the synchronous motor

The active power = ssyn x cos Φsyn = 250 x 0,9 = 225 kW

Reactive power Qsyn = √[S²syn - P²syn] = √[250² - 225²] = 108,97 kVAr

The system phasor diagram becomes:

Pn = Pind + Psyn = 425 kW (line active power)

Qn = Qind – Qsyn = 150 – 108,97 = 41,03 kVAr (new system reactive power)

Tan Φn = Qn/Pn then Φn = tan-1 (41,03/425) 5,514º (new PF angle)

Pn = √3 x VL x IL x cosΦn

425 000 = √3 x 6 600 x IL x cos 5,514º

IL = 37,25 A (line current)

Power factor of the installation = cosΦn = cos 5,514º = 0,995 lagging

Question 7:

7.1 An important induction motor with unique insulation, drives a compressor which is essential to production. The compressor is subjected to varying demands. It

is thus important to avoid or minimise damage to the motor, which has the

followingparticulars:

4 MW, 6,6 kV, 3 phase, 1 480 r/min with force ventilation. Squirrel cage, star connected, both ends of each phase winding accessible. Locked rotor stall withstand time 10 s. Starting direct on line. Normal run-up time 20 s.

Outboard white metal pedestal bearing with oil rings

State ALL the protective devices you would specify for the motor and itsswitchgear, briefly describing the purpose and function of the device.

(14)

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Solution

a) For windings: 1. Fit temperature probes in the windings of each phase, to detect

over-temperature due to overload or single-phase faults.

2. Install an instantaneous earth fault detection relay to detectfaults between the windings and earth. It is normally installed in the residual circuit of the current transformers and set to 20% of full load current. A resistor in series with the relay reduces any currents during start-up.

3. Install a thermal overload device with inverse time characteristics to detect motor thermal withstand capacity which could be exceeded by the positive and negative

sequence currents. These currents are generated by an unbalanced

supply or the failure of one phase.

4. Install a stall relay which operates at 80% of the locked rotorcurrent, but is disconnected during start-up conditions. Thiscould be achieved by installing a timer in the starting circuit which disconnects the relay.

5. Install an instantaneous over current relay which detects anyfaults at the terminals or between phases.

b) For the bearings:

1. Detect bearing temperature with thermometers in each bearing.

Actual temperature, alarm and trip facilities must be available.

2. Motors of this size normally have a closed loop oil flow facility and flow switches in the circuit inhibit the motor from starting an alarm and trip the motor whilst running, if there is low or no oil flow.

c) Ventilation of the motor:

1. Install an air flow switch in the stator to stop the motor from starting when the air flow is too low or sound an alarm when the ventilation stops whilst the motor is running and trips after a predetermined time.

d) Protection devices for the switchgear:

1. A battery condition monitoring system is to be installed if the design employs a separate source for tripping purposes. A

simple system connects a load to the batteries at certain time intervals

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and measures the voltage across the terminals which is recorded

for inspection.

2. Install a switchgear frame leakage system which detects any leakage currents to earth. The relay could sound an alarm and trip at preset values.

3. Install a differential current protection system which operates when unbalanced currents flow in the busbars. These settings are higher than the settings for individual feeders.

7.2 Describe THREE non-destructive tests on the brake rods and pins of a man winder. Briefly describe how these tests are done. (6)

Solution

a) Magnetic particle test

Spray a thin layer of white paint on the brake component. Anelectromagnet is then placed on the component and will create an

electric field around the component. To view the field, an amount of iron filings

is either dusted or sprayed onto the component under test. The field will be uniform for a non-defective brake component. If there are any surface cracks, however there will be an evident distortion of the magnetic field.

c) Die-penetrant test

Also known as the black on white test, it uses two colours on the component under test. An initial layer of the darker colour is applied, then white. It is then viewed under ultraviolet light. If there is any deformative cracks on the surface of the metal, it will show.

c) Ultrasonic test Ultrasonic probes are used together with a thin layer of grease. The

grease is applied between the component under test and the test probe. Calibrate the scope according to the length of the specimen and the frequency rating of the probe. A clean (non-deformed) calibration piece may be used to adjust the waves displayed on the scope. Uniform waves will show on the scope if there is no deformation inside the material. If, however, a defect is in the specimen, distorted waveforms will be “bounced” back to the scope. An estimation of the distance of the defect from the test end can be made, using the location of the wave that is distorted.

Question 8: (972.8)

8.1 A 10 tonne locomotive is used to haul twelve 5 tonne hoppers down an incline

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of 1:40 towards a tip. The coefficient of friction between the locomotive

wheels and the rails is 0,2, the rolling resistance for the train is 200 N/t,

the mass of the locomotive is 10t and only the locomotive has been equipped

with brakes.

8.1.1 Calculate the stopping distance, travelling at 12 km/h towards the tip. (8)

8.1.2 Discuss the feasibility of adding more hoppers to the train without sacrificing the safety standards. (2)

Solution

8.1.1 Calculate the stopping distance, travelling at 12 km/h towards the tip. (8)

Assume that the 5t given is the mass of the hopper plus payload because the train is moving towards a tip.

BE + FRR = Fg + Fd

Braking effort BE = μMloco g = 0,2 x 10 000 x 9,81 = 19 620 N

The frictional resistance is seen differently by various resources

Either Ffriction during deceleration = Mloaded hoppers x d, because the locomotive isstopping the train and the wheels are considered in the rolling resistance.

Then Rolling resistance during deceleration FRR1 = NMhoppers x RR = 12 x 5 x 200 = 12 000 N

Or Ffriction during deceleration = [Mloco + Mloaded hoppers ]x d, because even though

the wheels are braking the train the wheels are still seen as rolling.

Then Rolling resistance during decel FRR2 = [Mloco + NMhoppers ]x RR = [10 + 12 x 5] x 200 = 14 000 N

We will consider both cases in our solution

Gravitational force Fg = [Mloco + NMhoppers] x g x Sinθ = (10 000 + 12 x 5 000) x 9,81 x 1/40

= 17 167,5 N

Deceleration force Fd = [Mloco + Mhoppers] x d = [10 000 + 12 x 5 000] x d = 70 000d N

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BE + FRR1 = Fg + Fd

19 620 + 12 000 = 17 167,5 + 70 000d

d = 0,2065 m/s²

Or BE + FRR2 = Fg + Fd

19 620 + 14 000 = 17 167,5 + 70 000d

d = 0,235 m/s²

Sd1 = 0² – 3,3333² V = 12 km/h = 3,333 m/s 2d

= 0² - 3,333² 2(-0,2065)

Sd1 = 26,9 m

OR Sd2 = 0² – 3,3333² V = 12 km/h = 3,333 m/s 2d

= 0² - 3,333² 2(-0,235)

Sd2 = 23,64 m

8.1.2 Discuss the feasibility of adding more hoppers to the train without sacrificing the safety standards. (2)

Solution

The Mandatory guidelines issued in terms of ‘Railbound Equipment’ states that

the ratio of the braking mass (locomotive) to the unbraked mass(loaded hopper) should not exceed 1:7.

Therefore for a 10t loco the total mass of the loaded hoppers should be equal to or less than 70t.

Thus for this condition, we can add 2 more hoppers.

8.2 (052.8)

A monorope installation is used to convey ore at the rate of 80 t/h over a distance of 2 km The load is elevated through a vertical height of 200 m from the loading point to the tip.

Information:

Rope mass 3,5 kg/mSpeed of rope 4 m/s. Friction factor 0,02 kg/N

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Mass of empty containers 300 kgLoading interval between containers 25 seconds.Efficiency of drive train 70%

Determine the following:

8.2.1 Spacing distance between containers (1) 8.2.2 Number of containers/minute moving on the ropeway past the

loading point (1) 8.2.3 The minimum number of containers on the ropeway (1) 8.2.4 The load carried by each container (1)

8.2.5 The power of the electric motor to drive the monorope. (6) SolutionAssumptions:- 1) Does ‘Mass of empty containers’ mean total mass of empty

containers or mass of one empty container. 2) Should the friction factor be = 0,02 unitless or is the unit

kg/N correct

8.2.1 Spacing = loading interval x velocity = 25 x 4 = 100m

8.2.2 Number of containers = rope speed = 4 x 60 = 2,4 containers per minute Spacing 100

8.2.3 Minimum number of containers on ropeway = rope length = 2 x 2 000 Spacing 100

= 40

8.2.4 Load conveyed/container = load conveyed/min = 80 x 1000

Number of containers/min 60 x 2,4

= 555,555 kg 8.2.5 Power of the electric motor required to drive the ropeway if an

efficiency of 70% is assumed for the drive.

Work done to overcome friction = friction coeff x masstotal x g x distance moved/s = µ x [ mrope + mcontainer + mload ] x g x velocity

= 0,02 x [ 3,5 x 2 x 2000 + 40 x 300 + 20 x 555,55] x 9,81 x 4

= 29,1248 kJ/s

Work done to elevate the load = mload x g x vertical height = 80 x 1000 x 9,81 x 200 60² = 43,6 kJ/s

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Motor poweroutput = Word done per sec = 29,1248 +43,6 = 103,893 kW drive efficiency 0,7 0,7

If the unit for friction must be considered as kg/N then:

Work done to overcome friction = friction coeff x masstotal x g x distance moved/s

≠ kg/N x kg x g x m/s

≠ kg/N x N x m/s

≠ kg x m/s

= kg x g x m/s

Then W to overcome friction = µ x g x [ mrope + mcontainer + mload ] x g x velocity

= 0,02 x 9,81 x [ 3,5 x 2 x 2000 + 40 x 300 + 20 x 555,55] x 9,81 x 4

= 285,7143 kJ/s

Work done to elevate the load = mload x g x vertical height = 80 x 1000 x 9,81 x 200 60² = 43,6 kJ/s

Motor poweroutput = Word done per sec = 285,7143 + 43,6 = 470,45 kW drive efficiency 0,7 0,7

A considerable difference if the units given are incorrect !!!!Question 9 (981.8)(091.9)

(a) Describe the potential danger of an explosion occurring in a rotary type compressor.

[5](b) Two tenders for a rotary type compressor were received to

provide 40 m3/s free air at a pressure of 450 kPa. The first tender is for a compressor which is uncooled, that is the compression process is irreversible adiabatic (pV1,49 = constant). The second tender is considerably more expensive and is for a water cooled compressor, that is the compression process is reversible polytropic (pV1,3 = constant). The atmosphere pressure is 103,5 kPa, the ambient temperature is 32°C and Cp = 1,004 kJ/kg.K.

9.2.1 Compare the two tenders on the basis of power consumption. (10)

9.2.2 Briefly discuss other elements which may influence your recommendations. (5)

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= . -

T2 = T1[P2/P1](n-1)/2

=

a) Describe the potential danger of an explosion occurring in a rotary type compressor.

Solution

b) Consider the first tender for a compressor which is uncooled, that is the compression process is irreversible adiabatic (pV1,49 = constant). A rotary compressor.

P1V1 = mRT1 R = Cp – Cv = Cp – Cp/ γ

103,5 x 40 = m x (1,004 - 1,004/1,4) x 305

m = 47,319 kg/s

W = m Cp [T2 – T1]

= 47,319 x 1,004 x [494,54 - 305]

= 9 004,72 kW

Consider the first tender for a reversible polytropic (PV1,3 = constant).

It should be noted that this compressor is a reciprocating compressor becausen < 1,4.

Then = . - = .P1V1 -

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W = 1,3 x 103,5 x 40 x [(450/103,5)0,3/1,3 – 1] 1,3 - 1

= 7 243,5 kW

It would be obvious to purchase the reciprocating compressor when considering the electrical consumption (second tender).

The second tender will need auxiliary equipment such as a cooling tower, cooling water pump and valves, intercoolers and possibly after-coolers. More instrumentation than a centrifugal compressor will be required. More maintenance will be required making the overall cost a lot greater than that of a centrifugal compressor.

The choice would be to determine the capital and running cost over an equal life span for both tenders if the air delivery is sufficient in both cases.

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