Converting Electrical Power into Mechanical Load/Torque

Electrical Power x motor efficiency = mechanical power

power = torque x Rovolutions (Radians per second.)

Dear friend,

Thanks. But that much I already knew. I need the designer's view.

Regards.

Your question was too abstract. A specific question would have been appreciated and answered better.

Horsepower x 746 = watts

Kilowatts x 1.341 = horsepower

If you are an engineer you will know, as you've already indicated, the basics. i.e. that 746 watts = volts x current x power factor = 1 HP etc.

I can only assume that you need information about the more practical aspects and that therefore you are talking more about easily incorporating torque speed, efficiency and motor types into starting and running real loads.

This is a bit like asking, "how long is a piece of sting", but some general pointers are:

Real power out is equal to electrical power in by efficiency. For optimum systems efficiency will be the maximum for that type of motor and I'd use say 70% for something pretty ordinary to maybe 95% for best possible.

Generalisations include that bigger motors are probably more efficient than smaller ones, but all are subject to build quality. Permanent magnet motors are higher rather than lower efficiency because no field current has to be supplied.

The speed torque characteristics for a motor are absolutely determined by its type. Serial motors for high speed low torque, low speed high torque, induction motors for relatively constant speed, synchronous motors for absolutely constant speed etc.

In real life the load is not always at optimum speed and or maximum HP and while this is probably not critical in most cases as electrical motors are not "peaky" like high performance combustions engines, for example, some allowance should be made and this would normally come within a general safety factor. I am not a practicing electrical engineer so I cannot advise on what is normally used, but I'd be thinking at least 20% if the load was well defined and maybe 100% safety factor if it was not.

Of course care needs to be taken about operating environment temperature and whether the motor capacity specified is for "continuous" or "peak" load.

The other critical area to check is starting load/torque. Many industrial motors are induction motors (squirrel cage). Starting currents for these can be something like seven times running current, and care needs to be taken to ensure that the starting or stall toque is greater than that required. This can be critical when driving something like a reciprocating compressor where the intermittent torque requirement is much higher than the average torque once running.

You get similar voltage drop off and max current matching issues if electrical generators are to supply induction motors.

Of course there is then the three phase power with a discussion on star versus delta wound connections.

There's a lot more could be said, but that's a start.

Dear friends,

I Thank you all for the discussion and the help extended.The facts and factors provided by Mr. Samad and Mr Traver would be enough for my workable rough calculations.

I have problem with 132 kW motor of two stage reciprocating compressor which is drawing 88 Amps current without any load which looks a bit high. The maximum Ampere rating under load is 226 Amps. But when the set pressure crosses 6 bar, the current shoots up and reaches as high as 256 Amps at 7.5 bar which is the maximum delivery pressure of the compressor. Current drawn by 3 phases independently is R- 256 A, Y-254 & B- 256Amps. Connections are permanent Delta.

The compressor is oil lubricated and water cooled with 1200 cu m/hr air delivery rating at 7.5 bar. The inlet strainer has been cleaned and there is sufficient ventilation and open space around compressor. Cooling water temperature remains below 40 deg C and oil level is OK.

Regards

- 88 A no load current for an FLA of 226 looks OK.

- @7.5 bar your current is 256 A that translates to a KVA of 184. Assuming your motor pf at that point is even as low as 0.8, you are consuming almost 150KW , and that is a bit high for the 132KW motor.

Check the power requirements of the compressor. You can improve the current by power factor controllers, but the motor internal current remains same and that will be more than the rated motor FLA (226A) which will destroy the motor quickly.

Without pretending to be the final word on your installation - a quick back of the envelope calculation indicates to me that your motor is too small.

My reasoning is that your air delivered energy (i.e. vol * pressure as specified) is about equal to your electric motor power output if this is equal to the electrical input described by you (I assumed line volts was 410) by 80% efficiency.....a figure that is nearer 250kW than than the 132Kw motor you said you had. So while it might run it will either or both not deliver the specified amount of air or will run hot very quickly.

Then on top of the above there is the matter that the input power to the compressor will also have to offset the friction (and other) losses in the compressor - while 88 amps relating to this does seem a bit high, friction will be quite high and I'd be guessing at least 25% to cover that....so we are getting to over 300kW, if I've not made a mistake, when you have 132 kW available.

I've not double checked my calculations so if I've made a mistake, my apologies in advance.

Houston, I think we have a problem..... !

88A - as per OP is NLA, which is about 40% of FLA and does not look abnormal.

My post above was late at night and I was concerned at the difference between calculated power in and the stated size of the motor. My basic contention that the motor might be too small still stands - but with additional thought I correct as follows:

1. If the stated compressor volume was at 1 atmosphere for delivery at the high pressure (and this I think is the more normal convention) then my calculated power requirement is much less and would be nearer 125kW plus for friction/loses.

2. If the stated current was line current rather than field current, as I now realise was meant/stated then the calculated power out at 75% efficiency is about 130kW (not allowing for any power factor issues) and therefore matches the specified motor capacity.

Putting the two corrections in as above leaves me with the view, still, that the motor is working at or above its capacity when the compressor is at max. If everything else appears normal and its running hot then that would pretty much confirm my view.

Dear all,

Mechanical power

1 N x 1 m/sec = 1 Watt.

or

1 Nm x 1 rad/sec = 1 Watt

Electrical power, we use the same Watt.

With efficiency, which is always less than 1.0, we can convert mechanical power to electrical power or vice verse.

Regards

I am not an engineer. gut I did elecrtical trouble shooting on a lot of motor in 37 years, on thousands of motors.

If your motor is running above FLC and your voltage is within 10 percent of Name plate data and your RPM is at rated speed then your motor is is to small.

And that my friend, if you go back to your basics you will find the answers more quickly.