# Electrical Test part 3

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**ELECTRICAL TEST and MEASURING EQUIPMENT**** 3**

**ELECTRICAL TEST and MEASURING EQUIPMENT**

(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?

**ANSWER 21**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?

**ANSWER 22**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?

**ANSWER 23**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.

Gen. 1 delivers 500kW at 0.8 power factdr lag, and Gen.2 delivers 400kW and 35OkVAr lag.

Estimate (a) kVAr loading of Gen.1.

(b) p.f. of Gen.2.

(c) Total bus-bar loading in kW, kVAr and p.f.

**QUESTION 23.1**Gen. 1 delivers 500kW at 0.8 power factdr lag, and Gen.2 delivers 400kW and 35OkVAr lag.

Estimate (a) kVAr loading of Gen.1.

(b) p.f. of Gen.2.

(c) Total bus-bar loading in kW, kVAr and p.f.

(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.

(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?

**ANSWER 24**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.

**ANSWER 25**20 rev./sec. or 1200 rev./min.

**QUESTION 26**How is the rotor direction reversed?

**ANSWER 26**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?

**ANSWER 27**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?

**ANSWER 28**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.

**ANSWER 29**(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}. I

_{DOL}where x =tapping point. In this example x = 0.5. so the surge current level is 0.5

^{2}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?

**ANSWER 30**
All motors generate a back emf against the supply voltage when they are running. When the supply is removed from a running induction motor the magnetic field does not immediately collapse. The motor begins to slow down but still generates an emf. When the supply is reconnected, the supply voltage and motor emf are not necessarily in phase (the condition is similar to synchronising a generator onto the bus-bars). So each time the starter is operated a different surge current will be produced - sometimes very large, sometimes quite small. Large surges could cause appreciable voltage dip on the supply and so affect other equipment. Closed transition starters overcome this because the motor is never disconnected from the supply during the starting cycle.

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