Electric Motor Ratings and Transformer Ratings

For your question 1, is the Motor Power

For your question 2, is the input and the output of the transformer.

Is this a homework question or .......

It reminds me my Physics professor-students used to ask him silly question again & again and he kept on replying with smiling face till he is satisfied that the student understood the physical concept in his mind about the answer.This was because professor kenw the answer and he enjoyed putting the correct answer into the minds of even stupit students.

Both indicate the power output of the equipment.

The inefficiency that vary from motor/transformer to the next make,is the extra power to be added to the input power.

In the case of transformer,you can easily check by multiplying the rated current in A (ampere) with rated voltage and 1.732 to get the KVA rating on the name plate.The current and its corresponding voltage should be from the same side.The primary side figure will be slightly higher ( 1 to 2 % higher) compared with the secondary side,if you can get an accurate current rating on the plate or factory test result.

This total loss,usually in KW, of the transformer in the factory test result ( the power that goes into the transformer input is "used up " as losses and is dissipated mainly as heat) plus the rated KVA on the plate is the total input KVA.The output KVA devide by input KVA x 100 give you the efficiency in percentage; in one condition i.e.when power factor,pf= 1.0 at full load. In short,Output in KW/Input in KW x 100 = percent effciency.

Regards,

Khor

Khor,

The definition of the transformer's efficiency is the same regardless of the power factor: Pout/Pin, where Pin is the real power component (kW) of the input power Sin (kVA) and Pout is the output real power component (kW) of the output power Sout (kVA).

The values of the power factors for the 2 parts are different: the input (primary)power factor depends on the current for the transformer magnetization ( nearly constant, regardless of load) while the secondary depends on what character the load connected to it (resistive for pf=1, inductive or capacitive for pf<>1) has.

Input power factor cannot be strictly 1 (because the transformer needs magnetization) but the output power factor can vary, depending on the load connected to it.

tomad,

Thank you.

Best regards,

Khor.

Expanding on Simon's statement-

The HP rating of the motor is the OUTPUT.

The amperage rating is the INPUT. If you divide (the observed amperage multiplied by the observed voltage) by ( the rated amperage times the rated voltage) and them multiply the answer times the rated HP, you will get the operating HP.

NOTE- if the Operating HP is less than 65-70% of the Rated HP, and the current loads are "typical", consider replacing the motor with one better matched to the actual load. Under-utilized motors create significant amounts of power factor problems AND seriously reduce the actual efficiency. A high efficiency (95% or above) motor may be only operating at 65-70% (or even less), causing heat build-up in the motor that reduces its operating live as well as using more power than required. The poor power factor further adds to costs (it can contribute to increased delivery charges) as well as causing the distribution cabling to run hotter than designed- reducing its life and potentially causing a fire hazard (if the power load is high enough).

One more point regarding the transformer- Some times, operating folks will set a transformer output voltage higher than "standard", especially if there are recurring voltage dips, figuring that the higher voltage is better because the amperage will be lower. The higher voltage seriously reduces the capacity of down-stream transformers (due to saturation of the downstream transformer's core), reduces the life of motors due to increased phase-to-phase spike voltages, cuts the life of solid-state ballasts by up to 20% (12-years vs. 15) and reduces the life of other solid-state controls as well as lowering the efficiency of older magnetic ballasts- increasing operating costs. If a power dip is causing specific equipment to have problems, consider installing a voltage-regulating UPS upstream of the critical device.

Whatever usually is writen with "rated"(for instsnce :rated power, rated current) means "actual" ,in the other words, shaft power in an electromotor.Then you should divide output power(power on electeromotor nameplat) by efficiency to get input electrical power.

Question 1: It is the output power

Question 2: It is the Input. However, as a Transformer is a high efficiency machine, it is almost equal to the Input power.

For More info on Transformer Manufacturer, please visit www.transtron.co.in

Thanks a lot for the answers more pointed to the questions as compared to all other answers.However every answer is eduacational.

No one has clearly answered in one line confidently that the Motor & Trafo name plate rating means what----Input or output power ? From the answers( still not so confidently answered ) one can findout why the questions were asked ?

What ever answer that we all had given to you is just a brief general answer and we do believed that, as an engineer or electrician, one should be able to understand it. Could it be possible that you don't understand the answer that we have given to you?.

For your question on the motor KW, the KW is the power deliver at the shaft and also the input power required for the motor to delivery the power to the shaft. That why, these motor are rated in KW and also the KW indicate that the total power required from the main supply point that needed to run this motor to delivery the power to the shaft. Please refer back to the basic induction motor construction.

Transformer rating (KVA) that is printed on the name plate indicated the input power and the output power. There is a simple formula to calculate the input power of the transformer base on the KVA printed on the name plate and the maximum output power of the transformer base on the same KVA printed on the name plate. Please refer to the basic transformer construction and power calculation.

Thanks a lot for the clarification. The question was asked to clarify the basics of name plate rating for calculation of the connected laod in the distribution network to have more correct figures .

Thanks once more to all participants.

Simon,

The rating represents OUTPUT. The same number cannot represent both input and output (measuring units (in this case kVA) can, and have to be the same for a transformer, but not the quantity (always input > output)). When indicated 30 kVA on the nameplate it is 30 kVA output. Definitely the input power can be calculated via efficiency and power factor.

In case of motors, the output power (mechanical at shaft) is rated in kW while the input power (electrical, drawn from the electric grid and that includes magnetization component) has to be rated in kVA.

ok, look , the nameplate rating on the motor in hp is the output the motor can produce with correct volt and frequency . sorry two lines.

the transformer rating in kw is the available ,,or should i say (apparent) power ie: watts ,amps, volts. to do work... so it is capacity or the amount of juice to supply equipment to do work

Motor and transformer ratings on name plates indicates output power .

Regards,

Khor

This is the most correct answer in one shot. This is what is reqiured from the people having clear concept with confidence . Thanks a lot.