Suggested Solutions to the GCC Examination

Suggested solutions to the GCC examination

12 June 2009

Question 1

Question 2

2.3

The 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:1 580

Power output of CFL’s11 W

Power factor of CFL0,503

Price per energy unitR0,45/kWh

Price per apparent power unitR22,50/kVA

Lamp life of a CFL6 000 hours

Lamp life of an incandescent lamp1 000 hours

Price of an incandescentR2,45

Price of a CFLR18-00

2.3.1Calculate the cost in replacing the incandescent globes with the CFL’s. [7]

Suggested solution:

Replacement cost:

1 580 (18+15) = R52 140-00

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

Suggested solution:

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

2.3.3Propose 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
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 3

A 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 6

6.1A 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 slip3%
Rotor resistance0,0011 Ω/phase
Standstill reactance0,0052 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 Hz

s = 3%

Rrotor = 0,0011 Ω/phase

Xo = 0,0052 Ω/phase

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

6.2A 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 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 following particulars:

4 MW, 6,6 kV, 3-phase, 1 480 r/min with forced ventilation.
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:
Stall protection in excess of 10 seconds
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
3-start per hour lock-out preventing more that 3 starting attempts in a running hour.
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.

7.2

Describe 3 non-destructive tests on the brake rods and -pins of a man-winder. Briefly describe how these tests are done. [6]

Suggested solution:

Dye-pen testing -

X-ray

Magnetic 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 mm
Number of ropes / 3
Mass of the rope / 5,448 kg/m
Breaking strength of each rope / 918 kN
Length of ropes / 700 m
Length of wind / 550 m
Nominal drum diameter / 3 m
Sheave mass [is this for the tail end? Or tower end?] / 6,5 t
Mass of skips and attachments [is there more than 1 skip?] / 12 t
Mass of load[is this the payload?] / 10 t
Mass of balance weight / 17 t
Length of balance weight / 4 m
Mass of tail carriage / 5 t
Moment of inertia of tail carriage sheave wheel / 6500 kg.m2
Frictional resistance of skip / 3 kN
Frictional resistance of balance weight / 2 kN
Coefficient of friction between ropes and the sheaves / 0,18

Calculate:

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

1.2The 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

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.)

1.1The 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

Farotational =½ Itail sheavex a =0,5 x6 500 x a = 1 444,444a N

R²tail sheave1,5²

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

Ffriction = 3 000 N (given)

T1 = Fgravity + Facceleration + Ffriction

= 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 sheavex a =0,5 x6 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)

T2 = Fgravity + Facceleration + Ffriction

= 335 411,748 - 32 385,244a N - 2 000 N = 333411,748 - 32 385,244a N

For the maximum acceleration:

T1= eμθ

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

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

1.2The 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

= 5,9

This is considered acceptable as it is well over the legal requirements for transporting

people. (Again an assumption. Men or mineral?).

Question 2

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