Testing of Machines Questions and Answers

Answer: C

Coupling two identical machines and letting one drive the other as a generator returns the bulk of the power electrically, so the supply provides only the losses. Two identical machines have to be available, which is its limitation.

Answer: D

Correct answer: produce rated temperature rise without rated load

The mixed frequency method superimposes two supplies of different frequency so that the machine draws currents producing the rated losses while delivering no useful output. The heating is then correct without a load to absorb full power, which is what makes the temperature rise of a very large machine testable at all.

Answer: A

A megohmmeter applies perhaps five hundred or a thousand volts and the reading is taken after a stated interval, since the current falls as the dielectric polarises. Temperature has a large effect and the reading is corrected for it.

Answer: B

Correct answer: ten minutes to that at one minute

\[ PI = \frac{R_{10\ \mathrm{min}}}{R_{1\ \mathrm{min}}}. \] Clean dry insulation polarises slowly and its measured resistance rises steadily, giving an index of two or more. Wet or contaminated insulation carries a steady conduction current that does not decay, so the resistance barely rises and the ratio approaches one.

Answer: C

Correct answer: moisture or contamination in the insulation

A low index means the current is not decaying with time, which happens when a conductive path - moisture, dirt, carbon tracking - carries a steady leakage that swamps the polarisation current. The remedy is drying and cleaning, and a machine failing the index should be dried before any high-voltage test is applied.

Answer: D

The value is set by the standard as a multiple of rated voltage plus a margin, and passing it proves the insulation was built correctly. Repeating it on aged insulation risks damaging it, which is why service testing uses lower voltages.

Answer: A

Correct answer: the dielectric losses and their growth with voltage

Tan delta measures how much of the current through the insulation is in phase with the voltage, which is the loss component. More revealing than the absolute value is the tip-up: a sharp rise as the voltage is increased indicates voids ionising internally, and therefore deterioration in progress rather than merely a lossy material.

Answer: B

Correct answer: small discharges in voids within the insulation

A void within the insulation sees a higher stress than the surrounding material and breaks down repeatedly at a fraction of the applied voltage, each event a nanosecond pulse of a few picocoulombs. The discharges progressively erode the surrounding insulation, so measuring them detects deterioration long before any test of resistance or withstand would.

Answer: C

Correct answer: a short-circuited coil

A growler applies an alternating field to the armature, and a shorted coil carries a large circulating current in response. The classic detection is a steel strip held over the slots, which vibrates and buzzes - the growl the instrument is named for - above the faulty coil.

Answer: D

Measuring the resistance between successive segments round the commutator reveals the pair whose coil is faulty as a departure from the otherwise uniform readings. Blackening of the same segments in service is the corroborating evidence.