Induction motor modelling. 1

[gokulkrish2]

Hi everyone.

I need some help with modelling an induction motor. All i want to do is to get a model of an induction motor which can simulate the starting time(Time it takes to get to the full speed) with no load attached to it. I tried to use matlab simulink but there is a problem. I have the motor parameters like R and X from the manufacturer. They have given me 2 sets of data(Data Attached). I dont know why they have given me 2 sets for full load and Locked rotor. But when i try these values i get two different starting times. I believe its giving the deep bar effect of the motor. But i want to have an idea from you guys. please help.

gokul

 

[gokulkrish2]

Thank you so much for your help. Sorry for the late reply. Somehow the notification went to the spam. I went through your spreadsheet. I have two questions for you on the spreadsheet.

Why do you find the average of the torque from one speed to the next?

Then, you do you multiply torque with voltage square?

Once again thank you so much for the help. It was very helpful.

gokul

 

[electricpete]

Why do you find the average of the torque from one speed to the next?

It is the trapezoidal rule for integration. To integrate we need to find the area under the curve. Column B is snapshots of the curve at specific speeds. The trapezoidal rule is equivalent to drawing a straight line in between each data point and finding the area under that line. Then the average value over the interval is simply the average of the endpoints.

Then, you do you multiply torque with voltage square?

Torque is proportional to voltage squared. We can see this from the equivalent circuit. The power output is given by the power dissipated in the element with resistance
R2*(1-s)/s. Pick a given speed and keep it fixed (s is fixed also). If we multiply the input voltage by 0.9, then every voltage in the system changes by 0.9. The voltage accross the resistance R2*(1-s)/s will change by 0.9. The power consumed by that element (delivered to the load) changes according to V^2/R (where V is voltage accross that element and R = R2*(1-s)/s). So the output power will go to 0.9^2 = 0.81. Since the speed is fixed, the torque (Torque = P/w) will also increase by 0.l9^2 =0.81. So it is seen from the equivalent circuit that torque varies with voltage squared.

I seem to recall for some motors the real exponent is closer to 1.8, but I'm not sure about that last part. Does anyone use 1.8 instead of 2.0?

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[electricpete]

typographical correction:

Since the speed is fixed, the torque (Torque = P/w) will also decrease by 0.9^2 =0.81

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[gokulkrish2]

Thank you so much for the information.

Actually i am trying to theoretically prove a solution to a problem which i found out using trial and error basis.

It is for this project the induction motor time is going to be used.

I would like to discuss the scenario with you.

We have a conveyor which is run by a single drive of 250HP.

This drive is coupled to the load by a Fluid coupling.

Also, This drive is started using a soft starter.

Problem this conveyor was giving to us was that it didnt start when it was started with load on it. But it would start if there is no load on it and it would run with load.

So we figured that it was a starting torque issue.

Now after discussing this issue long back with this forum and having got a lot of inputs from here we figured that we need to have the fluid coupling couple the load to the motor exactly at the maximum torque level.

So we increased the starting voltage applied to the motor and tried to give more time delay to the fluid coupling.

Before this problem the soft starter was set to 4xFLA.

We went to 5xFLA and to the smallest plug possible to give time delay of 5 seconds.

It then started the conveyor with no prob.

Now i want to prove this theoritically.

That is why i am trying to simulate this motor time, to show that at 4xFLA, the time to start the motor is more and the fluid coupling starts coupling the load soon.

So when we increase the voltage the more gets to full speed soon so that it will give the max torque when the load is applied.

i will let you know the outcome after i come with the results.

Let me know if you find any issues with my proof.

thanks


gokul

 

[electricpete]

Beats me. We don't have any fluid couplings or soft-starters at my plant. Maybe someone else can comment on your scenario.

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[gokulkrish2]

Hi electricpete,

I have another question for you. Is there a way i can incorporate the effect of mechanical load added to the motor. I want to increase the Tm(to a value equal to the rated torque of the motor) linearly as the motor start and calculate the time taken by the motor to reach the full speed. Can you help me on this please.

gokul

 

[electricpete]

Attached is an analysis of your motor (based on data provided above) accelerating a load whose torque varies linearly from 0 to 735.6 ft-lbf (rated motor torque) over a speed range from 0 to 1785 (rated motor speed).

The summary is shown on the graphic tab (speed vs time).

The important change made in the calculations tab was simply to use Taccelerating = Tmotor-Tload for computing acceleration, in place of Tmotor that was used in the previous unloaded start case.

There were some housekeeping changes required as well. To develop a good plot I changed the averaging interval from "this-speed-to-the-next" to "last-speed-to-this". That change was a prerequisite to providing a traditional plot since the logical plotting point involves the endpoint of the speed interval rather than the beginning.

As you may notice the speed approaches 1785 rpm asymptotically. We reached 1784 rpm in 0.95 seconds. We cannot continue the numerical solution to 1785 because we always need smaller and smaller intervals since the rate of increase is slowing.

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[gokulkrish2]

Thank you electricpete. I guess i am closing in to a close good solution. I am talking to the fluid coupling guys to get their input. Possibly will figure out the solution. Will let you know the result once i get all stuffs covered. Thank you once again for the sheet.

gokul

 

[electricpete]

One thing to mention is that I used the motor inertia for the calculation. I don't think you told me any load inertia... I assumed it was small in comparision to the motor inertia (but we all know what they say about assmptions)

If not, the times reported above should be adjusted proportionally to the ratio of total inertia/motor inertia. i.e.
TimeToReach1784rpm=0.95sec*(MotorInertia+LoadInerta)/MotorInertia

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[electricpete]

My integration method should converge to the right solution if step size is small enough. But it is a little clumsy since we are assuming:
Integral (1/Taccel) = 1/Tavg where Tavg is the average value of the torque over the interval. It works well when the change in torque over the interval is a small fraction of the torque. But not so well when the torque is changing.

I think I will try a new approach. On the interval N1 to N2, torque is given by a linear curve:
T = T1 + (N-N1)/(N2-N1) * (T2-T1)

1/T = 1/[T1 + (N-N1)/(N2-N1) * (T2-T1)]

We can integrate the above expression with respect to N over the interval N1 to N2 and should get a result in terms of ln function (since the variable of integration N appears in the denominator). I will do that and post a new spreadsheet shortly.

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[gokulkrish2]

sounds great. Thanks for that.

gokul

 

[electricpete]

Attached is a revised spreadsheet as discussed above. Now the time to get to 1784rpm is 0.89 seconds.

This approach I'm sure is more accurate than my previous approach and also does not require the user to enter as many points in areas where the torque curve changes by a significant fraction (i.e where it approaches 0). This numerical integration is accurate to the extent that the torque changes linearly over the speed range of interest. Thus it does very well in the previously-problematic speed range adjacent to 1785rpm.... since torque changes linearly in this range.

Derivation is shown in the derivation tab. The resulting formula for time dT12 to accelerate thru a given speed range N1..N2 with torque T1 and T2 on the endpoints is:
dT12 = J*delta/(307*(T2-T1)) * [ln(T2/T1)]
where delta = N2-N1

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[electricpete]

One more comment - if your motor is driving a centrifugal pump or fan, the load torque is proportional to speed^2. It would be very easy to modify the spreadsheet to cover that.

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[gokulkrish2]

I have a comment to make. Now when i correct the voltage for reduced voltage ( say .8V) then at some point the load torque becomes more than the motor torque and the resultant goes negative. What should i do then, jus make another column which gives the abs value of the Tacc and proceed. Is that a correct way to do?

gokul

 

[electricpete]

No. The point where the accelerating torque crosses 0 (in direction from positive to negative) is the intersection of the torque speed curve of the motor and load - this is the operating point. Speed will not go higher than this and only the time leading up to this crossover has any physical significance. In the case of parameters you provided where load torque was equal to motor nameplate torque at motor nameplate speed, if we reduce voltage below nameplate then the motor will not make it to nameplate speed, it will be somewhere below nameplate and the load will be somewhere above nameplate (overloaded).

I re-did the simulation using V=0.8 for the data. It is easy to find the point where the accelerating torque turns negative because we get #num (errors) due negative number input to the ln function. I filled in a higher point density just before it turns negative to plot the curve leading up to the operating point. The results are shown in attached. The operating speed is 1777rpm and it takes 3 seconds to get there. The motor is loaded to roughly =(1800-1777)/(1800-1785) = 153% and would probably trip on overload not too long after start.

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[electricpete]

Sorry, forget the part about being overloaded, that was incorrect.. The HP drawn would be less than nameplate if the speed is less than nameplate. And also if we really wanted to determine the operating point we need the operating voltage instead of starting voltage... operating voltage would be higher than starting resulting in higher operating speed. My spreadsheet is not equipped to predict changes in voltage (depends on the system impedance and current drawn by the motor).

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[7anoter4]

I agree with electricpete of course.
The equivalent circuit is the extension of the Steinmetz model for transformer.
Using this model one will get about 10 to 25% errors.
The maximum[break down ,pull out] torque does not depends on rotor resistance but the rotor slip corresponding to this maximum torque does ,then one may increase the rotor resistance in order to increase the starting torque up to maximum.
As in deep slot bar the resistance is elevated at start the deep slot is used -in most of induction motors- in order to improve the starting conditions: limiting the starting [lock out] current and increase the starting torque.
In this case the resistance and the reactance of the rotor are related to the slip in a nonlinear dependence. The conventional model is not applicable entirely.
Some people uses two series of parameters one for start and other for rated speed.
Other problem is the excitation current which depends on supply voltage and emf produced in stator windings and it is strong influenced by core saturation. Emf contains then saturation harmonics but also space harmonics and slot harmonics.

 

[gokulkrish2]

Hi electripete,

I have another question. When i was going through the calculation i found you were using 32.2 constant to arrive at 307. what is this factor 32.2 account for.

gokul

 

[electricpete]

Hi gokulkrish2!

The short answer: 32.2 is a unit conversion associated with British/English/American units. Some people carry around gc but I don't. The way I look at it is simply 1 = [lbf sec^2 / (32.2 lbm ft)] (easy to remember since F=MA => 1=F/MA and since we know that a force on one lbf occurs when a mass of 1 lbm is subjected to acceleration of gravity which is 32.2 ft/sec^2).

The long answer is to repeat the full derivation from the beginning, this time accounting explicitly for the units. Here it goes:
Let T(w) = accelerating torque = Telec – Tmech
The rotational equivalent of F = m*a is:
T(w) = J * dw/dt

Rearrange:
dw/dt = T(w)/J
dt = [J/T(w)] dw
t = Int [J/(T(w)] dw [equation 1]

Define the right hand side of equation 1 as "Integral 1"
Integral 1 = Int [J/(T(w)] dw

We know speed N = w/(2*pi)
dN = dw/(2*pi)
Substitute dw = (2*pi)*dN
Integral 1 = Int [J/(T(w)] dw = 2 * Pi * Int [J/(T(w)] dN

Multiply by 1 = [lbf *sec^2 / (32.2 lbm *ft)]
Integral 1 = 2 * Pi * Int [J/(T(w)] dN * lbf *sec^2 / (32.2 lbm *ft)

Multiply by ft/ft
Integral 1 = 2 * Pi * Int [J/(T(w)] dN * ft-lbf *sec^2 / (32.2 lbm *ft^2)

Rearrange:
Integral 1 = 2 * Pi * Int [{/J/<lbm-ft^2>} (T/<ft-lbf>) )] dN * sec^2 / 32.2

multiply by minute/<60*sec>
Integral 1 = 2 * Pi * Int [{/J/lbm-ft^2} (T/ft-lbf) )] dN * minute * sec / (32.2 *60)

Substitute minute = 1/min^-1
Integral 1 = 2 * Pi * Int [{/J/lbm-ft^2} (T/ft-lbf) )] d{N/min^-1} * sec / (32.2 *60)

Put Integral 1 back into equation 1:
t= 2 * Pi * Int [{/J/lbm-ft^2} (T/ft-lbf) )] d{N/min^-1} * sec / (32.2 *60)

Divide both sides by sec
{t/sec}= 2 * Pi * Int [{/J/lbm-ft^2} (T/ft-lbf) )] d{N/min^-1} / (32.2 *60)

Collect Constants at the end
{t/sec}= Int [{/J/lbm-ft^2} (T/ft-lbf) )] d{N/min^-1} *[2*Pi]/ (32.2 * 60)

Now we only need recognize that t/sec is a unitless quantity we would call "time in seconds". J/lbm-ft^2 is a unitless quantity we would call "rotating inertia in lbm-ft^2". T/ft-lbf is a unitless quantity we can call "Torque in ft-lbf". N/min^-1 is a unitless quantity we can call speed in rpm.

Based on the above we can rewrite in a more compact form:
t = Int [J/T)] dN * (2*pi) / (32.2 * 60)
t = Int [J/T)] dN * / 307 [equation 2]
where t in seconds, J in lbm-ft^2, T in lbf-ft, N in rpm.

Now look at some small interval from speed N1 to N2 over which we will perform numerical integration. From here to the end we will ASSUME the torque changes linearly with speed over each interval as given by:
T = T1 + (N-N1)/(N2-N1) * (T2-T1)
Let delta = N2-N1 be the width of the interval
T = T1 + (N-N1)/ delta * (T2-T1)
Call this linear function f(N) = T1 + (N-N1)/ delta * (T2-T1)

Let dt12 be the time to get from N1 to N2
dt12 = J / 307 * Int ( 1/[f(N) ), N=N1..N2 from equation 2.
Evaluate integral by chain rule:
dt12 = J/307 * [ln(f(N) / df/dN] evaluated at N2 minus the same thing evaluated at N1 (provided that df/dN is not itself a function of N)
df/dN = (T2-T1)/delta (we now confimred that df/dN is not a function of N)
dt12 = J*delta/(307*(T2-T1)) * [ln(f(N)] evaluated at N2 minus the same thing evaluated at N1
But we know f(N) evaluated at N2 and N1... it is T2 and T1!
dt12 = J*delta/(307*(T2-T1)) * [ln(T2) -ln(T1)]

Simplify by noting ln(T2)-ln(T1) is ln(T2/T1)
dt12 = J*delta/(307*(T2-T1)) * [ln(T2/T1)] [equation 3]
where
dt12 - time to accelerate through speed range (in seconds)
J - rotating inertia (in lbm-ft^2)
T1, T2 = Torque at beginning and end of speed range (in lbf-ft)
delta = width of the speed range (in rpm)

Should we prefer to avoid British units, instead of having unit factor the factor 307 =(32.2 * 60)/(2*pi) in the denominator, we would not need the 32.2 and instead have a factor (60)/(2*pi) = 9.55 in the denominator.

dt12 = J*delta/(9.55*(T2-T1)) * [ln(T2/T1)]
where
dt12 - time to accelerate through speed range (in seconds)
J - rotating inertia (in kg-m^2)
T1, T2 = Torque at beginning and end of speed range (in N-m)
delta = width of the speed range (in rpm)


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[gokulkrish2]

Thank you so much. I got that.

gokul