HELP DC Variable Speed Drive. Experimental Determination of Inertia

[NiCeBoY]

Hi,
I perform an experiment in the lab and i want to calculate the inertia of the drive that i test.

Below is the description of what i did and the results.

Test System Description
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A1 – AC Variable Speed Drive ACS800, 5.5 kW, 415 V, 17.6 A, ‘ABB’
A2 – DC Variable Speed Drive DCS800, 400 V, 25 A, ‘ABB’
M1 – Squirrel–cage induction motor, 3-phase, 415 V, 50 Hz, 1400 rpm, 2.2 kW, 4.7 A
M2 – Separately excited DC motor, 400 V, 1990 rpm, 7.7 A, 2.6 kW.

The shafts of motors M1 and M2 are coupled directly. The AC VSD (A1) is a two-quadrant drive, and hence cannot provide braking. Motor M1, therefore, operates always in the motoring mode. The DC VSD of motor M2 can provide operation in four-quadrants. It is set for torque control (Fig. 2). If the set torque of the DC drive is of the opposite direction to the direction of rotation selected on the AC drive, the DC drive will operate in the braking mode. Consequently, M1 operates as a motor, and M2 – as a
generator.
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Program of Experiment
Experimental Determination of Inertia
1. On the control panel of DC drive, switch to local control (LOC) by depressing the LOC/REM
pushbutton.
2. Start the DC drive by depressing the START p/b.
3. Take the characteristic of the motor no-load torque Tloss vs speed n in the range of 0 -1600 rpm. Vary the reference torque to vary the motor speed. Progress carefully to avoid exceeding the speed limit and tripping the drive. The torque is displayed as a percentage of the motor nominal value stored in the drive memory as parameter (4.23).
4. Plot a graph of Tloss = f(n). If possible, determine a speed interval in which Tloss is approximately constant. (See the graph below.)
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The approximate value of inertia can be calculated from the above expression by substituting the slope of the coasting curve at ω1, with speed expressed in rad/s (see the graph below).
Note: The accuracy of the inertia calculation will be better if ω1 is selected from a speed interval in which Tloss is approximately constant. The coasting curve ω(t) in that interval resembles a straight line, and its slope is approximately constant.
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This is what i get and i plotted on excel below:
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Converter Output Voltage as a Function of Firing Angle
1. Depress the ON pushbutton on the AC drive panel. On the control panel, set the frequency at around 25 Hz. The selected direction of rotation is clockwise (positive).
2. Start the AC motor by depressing the START pushbutton.
3. On the control panel of DC drive, switch to local control (LOC) by depressing the LOC/REM pushbutton.
4. Start the DC drive by depressing the START p/b.
5. Set the reference torque at a level of (-50%) or more (sufficient to make the armature current continuous).
6. Using the oscilloscope, monitor the motor voltage vt(t) and a line supply voltage. On the template graph provided below, record the waveform of vt(t). Record the average armature terminal voltage Vt. Measure the firing angle of the thyristor bridge α using the zero crossing of the line voltage as
a reference point
7. Repeat the above for two other values of Vt. spread over the available range of 0 – 400 V. This requires selecting new values of frequency for the AC drive.

This is what i obtained below:
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So yeah anyone got an idea how i can calculate the inertia of the variable speed drive being tested?

Thanks

 

[MrAl]

Hi,

If you are satisfied with using that equation and you already took all the data, what is the problem...that is, where are you getting stuck?

 

[NiCeBoY]

i mean which value is which and how to use the equation...
thanks

 

[MrAl]

Hi again,

I didnt notice before that it seems that you had not logged the time periods between readings. I assumed you did this already, but since you didnt you'll have to obtain that information. The equation is asking for the time rate of change of w (angular acceleration), and to calculate that we need the time values for each sample.

The equation presented in your first post can be arranged as:

J=-TLP*TNom/(dw(w1)/dt)
where
TLP is the torque loss percentage,
TNom is the nominal torque,
dw(w1)/dt is the time rate of change of angular speed at w1, also called the angular acceleration. The inertia is associated with the acceleration so knowing the acceleration allows calculating the inertia.

TLP you have logged, TNom is stored, so you need dw/dt.
To get dw/dt you would select w1 first, say w1=757rpm.
Now select the two samples above and below that sample, w0=577 and w2=878.
Now we have three samples: w0, w1, and w2, and we use the torque loss percentage at w1 which equals -8.6 percent, so TLP=-8.6.
We also need the time values for w0 and w2, calling them t0 and t2 respectively.
Having this information, we can calculate dw(w1)/dt approximately as:
dw(w1)/dt=(w2-w0)/(t2-t0)
Having all the values for the equation now, we can then calculate J.

You'll first have to get the time values for each w value to calculate dw/dt.

 

[NiCeBoY]

Hi again,

I didnt notice before that it seems that you had not logged the time periods between readings. I assumed you did this already, but since you didnt you'll have to obtain that information. The equation is asking for the time rate of change of w (angular acceleration), and to calculate that we need the time values for each sample.

The equation presented in your first post can be arranged as:

J=-TLP*TNom/(dw(w1)/dt)
where
TLP is the torque loss percentage,
TNom is the nominal torque,
dw(w1)/dt is the time rate of change of angular speed at w1, also called the angular acceleration. The inertia is associated with the acceleration so knowing the acceleration allows calculating the inertia.

TLP you have logged, TNom is stored, so you need dw/dt.
To get dw/dt you would select w1 first, say w1=757rpm.
Now select the two samples above and below that sample, w0=577 and w2=878.
Now we have three samples: w0, w1, and w2, and we use the torque loss percentage at w1 which equals -8.6 percent, so TLP=-8.6.
We also need the time values for w0 and w2, calling them t0 and t2 respectively.
Having this information, we can calculate dw(w1)/dt approximately as:
dw(w1)/dt=(w2-w0)/(t2-t0)
Having all the values for the equation now, we can then calculate J.

You'll first have to get the time values for each w value to calculate dw/dt.

The time value is found on the waveforms.. if am not wrong..?

 

[MrAl]

Hi,

Well if the time values were on the waveforms then you can tell me what the time values are for w0 and w2 right?
If you can do that, you can use the formula i gave you previously to calculate dw/dt and that will get you there.
Since those dont look like the right time values, i dont see how you can do this yet although you are more familiar with this project than i am.

If you have or can get the time values for w0 and w2 as i outlined, then you can use the formula i gave you. Does that make sense?
Are those ramping waveforms w vs t or no?

 

[NiCeBoY]

Hi,

Well if the time values were on the waveforms then you can tell me what the time values are for w0 and w2 right?
If you can do that, you can use the formula i gave you previously to calculate dw/dt and that will get you there.
Since those dont look like the right time values, i dont see how you can do this yet although you are more familiar with this project than i am.

If you have or can get the time values for w0 and w2 as i outlined, then you can use the formula i gave you. Does that make sense?
Are those ramping waveforms w vs t or no?

yeah but thats all values i got.. so i am thinking how i can get the rest..

Also do you have an idea what the sign of Vt indicate?
thanks

 

[MrAl]

Hi again,

How did you obtain those ramped curves shown?

Yeah i see that all of the Vt are negative. That sometimes means a phase flip but im not sure if that's what it means there or not.

Part of the way a rotating system like this works is a torque T1 is applied and the motor comes up to speed w1, then a torque T2 is applied at time t=t0 and the motor gradually changes speed and reaches a new speed w2 at time t=t2. It therefore takes the motor a time t2-t1 to get up to the new speed, and it takes this time because of the rotational inertia of the system. In other words, we accelerate the motor and measure the time it takes to reach a new speed which in this case would be somewhere around w1. With larger inertia it will take longer to reach some new speed w2 and with smaller inertia it would take less time to reach the new speed w2, and with zero inertia it would take no time at all to get to the new speed w2 even with a small increase in torque (which of course is not possible in real life).
I almost forgot to mention that the speed has to be in rads/second not rpm.

 

[NiCeBoY]

on the waveforms that i got, anyone know how i can label the time axis?
let say for the 25Hz Frequency waveform,
is the time period of the sine wave 1/25 = 0.04s.
so this is for one cycle of the sinuisoial waveform that i got?

thx

 

[MrAl]

Hi,

Yes that makes sense. If you have a 100Hz wave and it spans 10 divisions, then the period is 1/100=0.01 so each division would be 0.001 second each.