Electrical Test part 3 - BLOG PELAUT

# Electrical Test part 3

## ELECTRICAL TEST and MEASURING EQUIPMENT 3

(PRACTICAL MARINE ELCTRICAL KNOWLEDGE; Hall, Witherby)

QUESTION 21
What are the likely consequences of attempting to close the incomer's breaker when the generators are not in synchronism?
At the instant of closing the breaker, the voltage phase difference causes a large circulating current between the machines which produces a large magnetic force to 'pull' the generators into synchronism. This means rapi6 acceleration of one rotor and deceleration of the other. The large forces may physically damage the generators and their prime movers and the large circulating current may trip each generator breaker. Result? Blackout, danger and embarrassment!
QUESTION 22
What indication is available to show the optimum synchronised condition?
The incoming generator ammeter will show very little 'kick' when correctly synchronised.

QUESTION 23
How could you monitor the correct instant for synchronising without the aid of a synchroscope or synchronising lamps?
Connect a pair of 500V voltmeter probes across one phase of the incoming machine circuit-breaker. Adjust the generator speed until the voltmeter slowly fluctuates from zero to maximum.

Figure
Close the breaker when the voltmeter passes through zero.
QUESTION 23.1
Gen. 1 delivers 500kW at 0.8 power factdr lag, and Gen.2 delivers 400kW and 35OkVAr lag.
(b) p.f. of Gen.2.
(c) Total bus-bar loading in kW, kVAr and p.f.
Figure
(a) CosΦ  = 0.8, so Φ 1 = 36.9°, and Q1 = P1 . tan Φ 1. 500. tan 36.9° = 375 kVAr
(b) TanΦ 2  = Q2/P2 = 350/400 = 0.875,  so Φ 2 = 41.2°
Then pf2 = cos Φ 2 = cos4l.2° = 0.75 Lag
(c) Total P = 500 + 400 = 900kW, Total Q = 375 + 350 = 725kVAr
TanΦ = Q/P = 725/900 = 0 81
SoΦ = 38.9° and Total p.f. = cos 38.9° = 0.78 Lag.
QUESTION 24
Two generators are load sharing equally in parallel when a total loss of excitation occurs on No.2 machine. What is the likely outcome?
Generator No.2 will run as an induction generator drawing its excitation from No.1. Both generator currents will rise rapidly with No.1 becoming more lagging while No.2 runs with a leading p.f. (indicated on p.f. meter). A 'loss of excitation' trip (if fitted) or an over-current trip should trip No.2 generator possibly causing an overload on No.1. Alternatively, No.1 trips on over-current (current) which deprives No.2 of excitation and its breaker trips out on under-voltage. Result = total power failure!

QUESTION 25
What is the synchronous speed of a 6-pole motor supplied at 60Hz.
20 rev./sec. or 1200 rev./min.
QUESTION 26
How is the rotor direction reversed?
Simply by swopping over any two supply line connections at the stator terminal box. This reverses the direction of the rotating magnetic field.
QUESTION 27
If a 6-pole motor is supplied at 60Hz and runs with a slip of 5%, what is the actual rotor speed?
The synchronous speed is 1200 rpm, and the rotor slips by 5% of 1200, i.e. by 60 rpm so the rotor runs at 1140 rpm.
QUESTION 28
Why is it essential that contactors S and D on page 4/8 are not both closed at the same time?
If you check the starter diagram, you will see that a full short-circuit is applied across the supply in this condition.

QUESTION 29
Estimate and compare the likely starting current surges for a motor that takes 200A on full load when started (a) DOL, (b) Star-Delta, (c) Autotransformer with a 50% tapping.
(a) When starting DOL the initial surge current is about 5 x FL, i.e.1000A.
(b) A star-delta starter reduces the initial starting surge to one-third of the equivalent DOL value, i.e. to about 330A in this case.
(c) The autotransformer method reduces the initial starting surge to (x)2 . IDOL where x =tapping point. In this example x = 0.5. so the surge current level is 0.52 x 1000A = 250A.
The DOL starter is simple and cheap but causes a large starting surge. Star-delta starting reduces the surge but is somewhat more complex, requiring three contactors and a timer. The autotransformer method can be arranged to match the motor surge current and run-up period to meet the supply limitations by a suitable choice of voltage tapping. This starter is considerably more expensive than the other two starter type.

QUESTION 30
What causes the large current surge in open transition starters when going from the start to the run condition?