Monday, March 30, 2020

Insulation classification and testing Question & Answer

Insulation classification and testing

1. What is good dielectric break down value for insulating oil?

60 kV.


2. What is the temperature coefficient of insulating materials?

Insulators are negative temperature coefficient materials.

3. What is gap between the electrodes in transformer oil testing kit?

0.1 Inch.

4. What is the life insulation if temperature increased by 10°C?

The life of the machine insulation decreases by half if the temperature of the
insulation increases by 10°C.

5. What is the value of vacuum maintained by vacuum pump in oil filteration machine?
27 Hg.

6. What is the DC HV test voltage range?

1.7* 1.5* rated voltage.

7. What do you mean by term insulating resistance? How it is measured?

Insulating resistance: insulating resistance is the opposition offered by an insulating
material to the flow of current (electrons) through it when an high potential is
applied across it.
Insulating resistance are measured by megger.
First the equipment whose resistance is to be measured is disconnected from supply.
If the machine is a large one, there may be accumulated static charge on the machine.
So we have to discharge it by connecting a wire between the terminals and ground
for 15 minutes. Otherwise megger will give wrong reading.
After this we should remove the wire and we have to connect megger terminals (live
& earth) to the motor terminal and earth. The rating of the megger should be selected
properly. Then rotate the megger at rated speed of 160 rpm and take the readings.

8. What you mean by dielectric absorption test?

Whenever we apply a potential from the megger to test the IR value, initially the
needle of the megger will go to low value of the resistance. This is due the
capacitance effect of the insulation material and after some seconds the needle will
start moving towards the higher value. Because in the insulating material there is
strain on the molecules when the potential is applied. Polarization of the molecules
occurs and they form a Di – pole. The negative charges are attracted to positive
terminal and positive charges are attracted to negative terminal. So there is a strain
on the insulation molecules and they align themselves parallel. This aligning may
take more time. This test is done to know the condition of insulating material.
I
(μ Amps)
(A)
(B)
Time
If the insulation is good the graph is as shown as B and if there is dirt, moisture the
graph will flatten early as shown in A.
After the test terminals to be discharged so that molecules may return to their
unstressed state.

9. Draw the transformer drying out curve and explain each stage.

At this point the heaters are
IR and Switched off.
Temp. Temp.
IR
1st 2nd 3rd
Time in hours
When we start the filtering process initially the temperature will be low, as the
insulation value is high. But on temperature increases the IR value starts to decrease
because the moisture entrapped in the coils are released due to rise in temperature
and this causes the IR value to go down. This is the first stage.
Then comes the point where all the moisture is released and then will be no decrease
in IR value or rise in the temperature. This is the second stage.
At this point the heaters are switched off. Now the moisture is removed by the oil
filters and the IR value goes up and as the heaters are off the temperature decreases.
This is the third stage.

10. The insulation resistance of a DC motor is observed to be 15 MΩ at a temp. of 70°C.
what is its value corrected to 40°C. the correction factor for 70°C is 8.0.
Observed resistance at 70°C – 15 MΩ.
Temperature correction factor – 8.
Rm = kt * Rt kt – correction factor.
Rm = 8 * 15 Rt – resistance measured at +°C.
Rm = 120 MΩ. Rm – corrected value to 40°C
The IR of DC motor corrected to 40°C is 120 MΩ.

11. The armature of a 600 kW, 0.24 k, 1000 rpm DC generator has an indicated IR to
ground of 2 MΩ at a temp. of 30°C. what is the recommended value of insulation? Is
it advisable to put the machine in service? Give reason. Correction factor for 30°C is
0.5.
Data given are
kV – 0.24
Indicated IR – 2MΩ
Temp. - 30°C
Correction factor – 0.5
Recommended value (Rm) = kV + 1 MΩ
= 0.24 + 1
= 1.24 MΩ
Indicated IR at 30°C = 2MΩ
Correction factor – 0.5
So value corrected to 40°C = Rm = kt * Rt
= 0.5 * 2
= 1MΩ
The generator cannot be put in service because the corrected value is lesser than
recommended value. It should be sent for IR re-conditioning .
Q

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