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6kV vs 6.6kV
2

6kV vs 6.6kV

6kV vs 6.6kV

(OP)
Recently a contractor installed three conveyor motors for my  company. The motors are 230kW, 160kW & 150kW all 4 pole,three phase, 50Hz but the motor voltage is 6.6kV whilst our generated supply is 6kV with a power factor of 0.8 lagging. What will be the effect on the motors? Extra current draw, more heat generated in the windings, any effect on the motor output?
Remedies anyone? We cannot replace the motors easily as the lead time from Toshiba is more thsn 6 months.

RE: 6kV vs 6.6kV

There are many existing threads in this forum discussing the effects of low voltage on the operation of AC induction motors. I would recommend a search of this forum or a check of the FAQ's for an immediate answer to questions of a general nature.

More important though, you are looking at a new installation that is not correct. I would look to the organization responsible for the discrepancy, whether it be motor manufacturer, contractor, or both to explain the reason for the discrepancy and the remedy that they propose.  

Responsibility aside, the motor manufacturer should be able to provide detailed design data for motor performance at reduced voltages that would help you assess the effects on the motor.  
 

RE: 6kV vs 6.6kV

If your motors are not fully loaded and starting torque requirement is not severe, then I do not see any problems running them at 6 KV. As a matter of fact, you may have better efficiency and pf with reduced voltage with reduced loads.

RE: 6kV vs 6.6kV

This is a common occurrence with IEC voltages, with system and transformer nominal voltages existing at 6, 6.3 and 6.6 KV in different parts of the world. Especially when persons work in a different geographic location from what they are used to.

In addition to the 6 KV generated, there is also a system voltage drop to the motor, making a bad situation worse.

I assume you mean 0.8 PF as a rating of the generator?

I guess you should derate the motor's output (shaft) capability by more than 10%. Hopefully your load (process) does not need all the power?

RE: 6kV vs 6.6kV

Your motors will have 90.9% of nameplate voltage. Most of the standards for Motor Manufacture request satisfactory operation of electric motors under +/_ 10% voltage variation.

Motor torque will drop with the squared ratio of voltages.

The full load current will increase at least with the inverse proportion of voltage drop. The power factor could have a slight increase but the efficiency will be lower.

IFL = HPout*0.754/( 1.73 * VLL * Eff * PF)

The core losses will decrease due to lower flux densities in the magnetic circuit.

The winding losses will increase due to the current increase  CUloss= R* I^2

Overall efficiency will be probably lower with the same load as compared to that operating at 6.6 kV.

Your motors will be in constant challenge, since you have no allowance for regular voltage drops or slight phase voltage unbalance.

A Olalde

RE: 6kV vs 6.6kV

(OP)
Thanks everyone for your replies. We have decided ( a forced decision) to carry on with the installation as is but will change out the 160kW motor as it is way too small for the applied load. The contractor ( MSc (Mech Eng)) has given up guessing and has been dismissed. Indonesia is a land of challenges.

RE: 6kV vs 6.6kV

Suggestions:
1. The remedies may include an autotransformer increasing the voltage from the generated 6.0kV to the motor terminal voltage appropriate voltage such that the motor will receive 6.6kV across its terminals.
2. If possible, the motor with 6.0kV or lower terminal voltage should have had decreased its shaft load.

RE: 6kV vs 6.6kV

2
A while back someone recommended the Brook Crompton "little red book" on electric motors, which I have since obtained.  Therein is a table on the effects of voltage variation on DoL induction motors, based on +6% to -6% (the permissible supply variation in the UK at the time), I hope eng-tippers will find it of interest:

                       VOLTAGE 6% UP   VOLTAGE 6% DOWN
Full load speed         up to 0.5%      down 0.75%
Starting torque         up 12%           down 11%
Starting current        up to 6%         down 5%
Full load current       down 4%         up 5%
Efficiency 1/2 FL       down 1.5%      up 2%
Efficiency 3/4 FL       down 1.0%      no change
Efficiency 1/1 FL       no change        down 1%
Power factor 1/2 FL  down 4%         up to 4%
Power factor 3/4 FL  down 3%         up to 2%
Power factor 1/1 FL  down 2%         up 1%
Temperature rise     down 4%         up 6%

In the context of Jeffer's problem (voltage down 10%), the danger is overheating of the windings - initially due to a slow start, and steady state due to increased current.  If the steady state current doesn't exceed the FL nameplate value, and the starting isn't onerous, then no problem.

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