Measuring DC motors...misterious values

[spacebeer13]

Hi,

I've performed some measurements in motors as i always do with a energy analyzer.
My intention was to evaluate the efficiency of several DC motors in a factory. Some of them are of 78, 160 and 400 kW, but here i'm going to use the example of a 160 kW DC motor, with the following characteristics:
- P = 160 kW
- 2200rpm
- 50 Hz
- 695 N.m (constant torque until 2200 rpm red from the torque vs rpm chart)

I measured the energy consumption at the inlet of the circuit-breaker (so in a 3 ph system without N) and at the same time de rotational speed of the motor to use in the following equations:

a) Mechanical power = Torque (Nm) x rpm / 9550
b) Efficiency = Mec. power / Elect. power

The results were:
a) rpm = 1550
b) Active power = 39 kW
c) Power factor = 0,41

So, applying the formulas:
a) Mechanical power = 695 x 1550 / 9550 = 112,8 kW (it seems OK assuming the speed)
b) Efficiency = 112,8 / 39 = 2,9 !!!!!! WRONG!!!

Can anyone help me??

NOTE: i checked all the data from the analyzer and it seems that the measurement was done well)

 

[Skogsgurra]

One obvious problem is that you state 50 Hz for a DC motor. Also, how do you know what the torque is? The "chart" is probaly the characteristics of the motor when run in a dyno. The chart does not give you the true torque of the load. I think that you need to concentrate more on what the DC motor actually does than on the measurements. That will get you on track.

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[spacebeer13]

One obvious problem is that you state 50 Hz for a DC motor. Also, how do you know what the torque is? The "chart" is probaly the characteristics of the motor when run in a dyno. The chart does not give you the true torque of the load. I think that you need to concentrate more on what the DC motor actually does than on the measurements. That will get you on track.

Thanks very much for the reply!

Regarding the frequency that is the frequency of the electrig grid.
Please see the attachment with the motor curve provided by the manufacturer.

 

[Skogsgurra]

That curve shows the load vs speed characteristics. That is how much you can thermally load the motor at different speeds.

It does not show the actual torque at the speed you are running. To know that torque, you either has to measure the torque or trust the manufacturer and measure speed, DC current and DC voltage and get your actual torque and Power consumed indirectly. If you really want to get it right, you should measure the torque with a good torque transducer. There are several available.



Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[mikekilroy]

You should be able to get correct answer to within motor constant tolerance, typically +/-15% by using T=Kt*I where Torque for your mech power equation (not 695) equals motor torque constant Kt * amps at the moment of measurement.

Don't know Kt value? You re running 1550rpm, below rated 2200rpm, so Kt is constant = 695nm/nameplate_amp_rating.

Now just measure amps into armature and you can figure torque to use for your mechanical calc.

http://www.KilroyWasHere<dot>com

 

[spacebeer13]

Dear Skogsgurra,

I think you ar wright.
It's not possible...

Dear mikekilroy,

The current measured was 140A vs 372A of the nameplate. So, 140/372 x 695 = 261,6 Nm
therefore the Mechanical power should be = 261,6 x 1550/9550 = 42,5 kW > 39 kW (measured). So still we get a value for efficiency greater than 1....
Thanks anyway

 

[waross]

There is not enough information given to even calculate the efficiency at full load.
The efficiency is the ratio of work done over energy used, or kW out over kW in.
OR
kW out over kW in + (armature losses + field losses + mechanical losses)
Field losses: up to base speed, the field losses are constant. However, as part of the ratio to total power, the percentage will increase and the efficiency will drop as the kW out drops. (Neglect field weakening.)
Armature losses: This depends on I²R. As the kW out drops, the armature losses drop as the root of the current. Due to the square factor, many motors, both AC and DC, have their greatest efficiency at less than full load.
Some of the information that is missing;
Voltage? Without the rated voltage we can't calculate the efficiency at full load.
Field current; Without the field current we can't determine the field losses.
Torque; Without a measurement of the actual output torque of the motor we really can't power out or any efficiency.
Mechanical losses; Usually small. Not linear with speed.
You need to find some way to measure the actual torque the motor is developing.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[iop95]

In this equation "Mechanical power = 695 x 1550 / 9550 = 112,8 kW (it seems OK assuming the speed)", torque value is not rated torque, is developed torque (= load torque), so result is wrong...
Motor develop a torque if there are a load attached; rated torque is a value that may be developed by motor, it's not an imposed value.
Need to know developed torque.

 

[mikekilroy]

Obviously as said above, knowing lots more stuff would help you understand WHERE the losses come from.

Since you added the nameplate current of 372amp, and nameplate torque rating 695nm at this amperage, then your motor Kt should be 1.87nm/amp

Since you measured 140amps, and you are below rated speed, your motor output should be 261.6nm, +/- the tolerance on Kt. Again, tolerance on Kt is not untypically +/15%. Kt is obviously very dependent on the field amps being right on. Not knowing the field amps (or field voltage and resistance at the temperature you made your measurement), I cannot state 100% that your motor Kt is accurate per the math.

Assuming your Kt is 15% low and your field is dead nuts on, then your ACTUAL output torque is not 261.6 but rather 222.4nm, for mech output of rather 36kw; less than your 39kw input value....

I have seen old DC motors of this size with very widely varying Kt values that no longer meet factory specs. Even after being rebuilt. So it appears to rely on this scheme is probably not worth much. Seems only a precision torque transducer in line with the 1550rpm load will give you an accurate mech power output number. Or a torque reading device on the locked motor shaft with 140amps into it maybe.... Neither of which probably makes any sense for your project.

http://www.KilroyWasHere<dot>com

 

[lukin1977]

Efficiency = Mec. power / Elect. power

Mec. Power = Torque x speed = kt x Iarm x speed = 1,86 x 140 x (1550rpm x 0,1047 rad/s/rpm) = 42260 W

Elect. Power = Varm x Iarm = Varm x 140

Eff = 301,85 / Varm

Do you have the Varm at Iarm = 140A?

 

[lukin1977]

Motor power = 160.000 W

FLA=327A ..... 160000/372 = 430,1 V so I estimate that nominal Varm=440V

at 440 V speed is 2200rpm so at 1550 rpm, Varm is 310V

eff = 301,85/310 = 0,97

 

[mikekilroy]

He measured power in via a power meter (how accurate? what type? how measured?) at 39kw....

This type motor is typically about 75-80% efficient at this kind of speed below base....

Therefore I assume these measurements are based on too many variables that are not correct to be of value?

http://www.KilroyWasHere<dot>com