Electrical Machine - Shunt DC Motor Problem

[SK97]

Hi guys, can someone please have a look at the following problem that I am struggling with;

An 8.25 kW shunt DC motor is supplied with a terminal DC voltage of 300 V. The armature resistance of the motor Ra is 0.25 ohms, and the field resistance Rf is 40 ohms. The field winding consists of 1500 turns. An adjustable resistance Radj is connected in series to the field winding. Radj may be varied over the range from 0 to 80 ohms and is currently set at 20 ohms. Armature reaction may be ignored in this machine. The magnetization curve for this motor, taken at a speed of 1200 rpm, is shown in Figure Q1.

(i) What is the speed of the motor when operating at rated load with an armature current of 30 A?

(ii) If the motor is now unloaded with no changes in terminal voltage or Radj, what is the no-load speed of the motor?

(iii) Iron core losses are 250 W. What are the copper losses and mechanical losses in the motor at rated load? (ignore brush and stray losses)

(iv) What is the efficiency of the motor at rated load?

(v) If the terminal voltage is reduced to 260 V and the armature current stays at 30 A, calculate the required Radj in order to restore the speed of the motor to the value in (i).
I have attempted to calculate I(f), which I am getting a value of 5A.
Also the mmf seems to be 7500 A.turns - (1500*5)

But I do not know where to go from here, any help would be appreciated.

 

[cnh1995]

But I do not know where to go from here, any help would be appreciated.

You know the back emf generated with 1500A-T at 1200 rpm from the graph.
Compare it with the actual emf induced in the motor. With constant flux, both the emfs are proportional to the speeds.

 

[SK97]

You know the back emf generated with 1500A-T at 1200 rpm from the graph.
Compare it with the actual emf induced in the motor. With constant flux, both the emfs are proportional to the speeds.

So would it be correct if I used the formula of Ea1/Ea2 = nm1/nm2

And Ea1 in this case would be 75V (reading from the graph at 1500 A-T) and Ea2 being 275V and nm1 is 1200 rpm.
So nm2 would come out to be 4400 rpm?

Thank you for your help so far.

 

[cnh1995]

So would it be correct if I used the formula of Ea1/Ea2 = nm1/nm2

Yes.

And Ea1 in this case would be 75V (reading from the graph at 1500 A-T) and Ea2 being 275V and nm1 is 1200 rpm.
So nm2 would come out to be 4400 rpm?

No. You have E₁=275V (from graph) and N₁=1200 rpm.
You have to calculate E₂ from the data given in the problem and use the above formula to get N₂.

 

[SK97]

Yes.No. You have E₁=275V (from graph) and N₁=1200 rpm.
You have to calculate E₂ from the data given in the problem and use the above formula to get N₂.

okay so using the formula E2 = Vdc - Ia*Ra

I would get a value of 292.5 V for E2.

So therefore my nm2 would be 1276.36 rpm.

Does that look right to you?

 

[cnh1995]

okay so using the formula E2 = Vdc - Ia*Ra

I would get a value of 292.5 V for E2.

So therefore my nm2 would be 1276.36 rpm.

Does that look right to you?

Yes.

 

[SK97]

Yes.

Thank you so much with your help so far!

My next confusion arises with the no load speed of the motor, I'm not sure what no load actually means, and so what assumptions I can make about the system?

Any guidance would be a great help.

 

[cnh1995]

My next confusion arises with the no load speed of the motor, I'm not sure what no load actually means, and so what assumptions I can make about the system?

For a motor, no load means zero load torque. Practically, the motor generates just enough torque so as to overcome friction (and windage.)

To be accurate, you should first determine the frictional torque from the given data and calculate the no-load armature current. But that would come out to be very small. So just take no load armature current=0.

 

[SK97]

For a motor, no load means zero load torque. Practically, the motor generates just enough torque so as to overcome friction (and windage.)

To be accurate, you should first determine the frictional torque from the given data and calculate the no-load armature current. But that would come out to be very small. So just take no load armature current=0.

Okay, so after doing that I am coming to an equation of n2 = (I(f)1*E2/I(f)2*E1) * n1

my problem is how do I get a value of E1 and I(f)1 whilst knowing that n1 is 1200 rpm?

Because I have calculated previosuly that the field current is 5A and I know E2 would be 300 V from the information you have given me.

 

[cnh1995]

Okay, so after doing that I am coming to an equation of n2 = (I(f)1*E2/I(f)2*E1) * n1

my problem is how do I get a value of E1 and I(f)1 whilst knowing that n1 is 1200 rpm?

Because I have calculated previosuly that the field current is 5A and I know E2 would be 300 V from the information you have given me.

Yes, so you know E1=275V, N1=1200 rpm.
Now you have E2=300V, what is N2?
The field current (and mmf) is unchanged.

 

[SK97]

Yes, so you know E1=275V, N1=1200 rpm.
Now you have E2=300V, what is N2?
The field current (and mmf) is unchanged.

Therefore n2 = (300/275) * 1200
Which comes out to be 1309.09 rpm
Correct?

 

[cnh1995]

Therefore n2 = (300/275) * 1200
Which comes out to be 1309.09 rpm
Correct?

Yes.

 

[SK97]

Yes.

Thank you so much for your help, I feel indebted to you now!

I will attempt (iii) now, will you be able to verify if my answer is correct please?

 

[SK97]

Here is my attempt at the next question:

The rated load armature copper loss is given as: Pacl = Ia^2 * Ra

Which comes out to a value of 225 W

The field copper loss is: Pfcl = I(f)^2 * (Rf + Radj)

Which comes out to be 1500 W

Now I'm not sure what to do with the information of Iron Core losses are 250 W?

My attempt to get the mechanical loss is as follows:

Power absorbed by the machine is: Pind = Ea * Ia ----> (267.5) * (30)

And that comes out to be 8025 W

Now I would subtract that by 8.25 kW but it would get me a negative answer, so I'm not sure what's going on.

 

[cnh1995]

Here is my attempt at the next question:

The rated load armature copper loss is given as: Pacl = Ia^2 * Ra

Which comes out to a value of 225 W

The field copper loss is: Pfcl = I(f)^2 * (Rf + Radj)

Which comes out to be 1500 W

Now I'm not sure what to do with the information of Iron Core losses are 250 W?

My attempt to get the mechanical loss is as follows:

Power absorbed by the machine is: Pind = Ea * Ia ----> (267.5) * (30)

And that comes out to be 8025 W

Now I would subtract that by 8.25 kW but it would get me a negative answer, so I'm not sure what's going on.

Leave the field loss out of the calculations, for it matters only when calculating the efficiency.

Armature voltage is 300V, armature current is 30A. What is the back emf?(you've already done that part). What is the gross mechanical power? How much is the copper loss?

 

[SK97]

Leave the field loss out of the calculations, for it matters only when calculating the efficiency.

Armature voltage is 300V, armature current is 30A. What is the back emf?(you've already done that part). What is the gross mechanical power? How much is the copper loss?

The back emf value is 292.5 V?

If so the power absorbed by the machine would be Ea * Ia = (292.5)(30) = 8775 W

Copper loss would just be as above, no? Ia^2 * Ra = (30)^2 * (0.25) = 225 W

Mechanical loss would be 8775 - 8250 = 525 W

How do I calculate gross mechanical power?

 

[cnh1995]

If so the power absorbed by the machine would be Ea * Ia = (292.5)(30) = 8775 W

This is the gross mechical power.

Copper loss would just be as above, no? Ia^2 * Ra = (30)^2 * (0.25) = 225 W

Yes.

Mechanical loss would be 8775 - 8250 = 525 W

No, 525W is the fixed (or constant) loss. It is the sum of iron loss and friction loss.
You can subtract the iron loss from the fixed loss to get the mechanical loss.

 

[SK97]

This is the gross mechical power.

Yes.

No, 525W is the fixed (or constant) loss. It is the sum of iron loss and friction loss.
You can subtract the iron loss from the fixed loss to get the mechanical loss.

Thank you!

So for efficiency, as you stated we need to use field copper loss, which would be what I calculated above as 1500 W.

But what I'm confused about is that I've seen an equation saying Power in can be calculated as follows: Vdc * (Ia + If)

With this equation it seems like i don't need the other values?

Can you please clarify this.

Thanks again for helping me so much!

 

[cnh1995]

But what I'm confused about is that I've seen an equation saying Power in can be calculated as follows: Vdc * (Ia + If)

That is the total input power to the motor, and 8.25kW is the output power. You can calculate efficiency from this.

 

[SK97]

That is the total input power to the motor, and 8.25kW is the output power. You can calculate efficiency from this.

Thanks again!

For the last part of the question i have to find a suitable Radj value so as the speed of the motor would be 1276.36 rpm when terminal voltage is 260 V instead of 300V.

Would you point me in the right direction for the final time please.

 

[cnh1995]

Thanks again!

For the last part of the question i have to find a suitable Radj value so as the speed of the motor would be 1276.36 rpm when terminal voltage is 260 V instead of 300V.

Would you point me in the right direction for the final time please.

You need to find the ratio of torques in the two cases.
How can you express gross mechanical power in terms of torque and speed? Take the ratio of the mechanical powers in the two cases to find the torque ratio.

 

[SK97]

You need to find the ratio of torques in the two cases.
How can you express gross mechanical power in terms of torque and speed? Take the ratio of the mechanical powers in the two cases to find the torque ratio.

So I found the relating equation to be: P = Ea * Ia = Tind * Wm

Therefore torque would be Tind = Ea * Ia / Wm

but I am still unsure about what to do with this equation?

 

[cnh1995]

So I found the relating equation to be: P = Ea * Ia = Tind * Wm

Therefore torque would be Tind = Ea * Ia / Wm

but I am still unsure about what to do with this equation?

What is the ratio of torques in the two cases then?
Torque is proportional to the product of armature current and field current. You can find the new field current when the voltage is 260V, and find R_adj accordingly.

 

[SK97]

What is the ratio of torques in the two cases then?
Torque is proportional to the product of armature current and field current. You can find the new field current when the voltage is 260V, and find R_adj accordingly.

So working through this I obtained a value of 58.36Nm for the torque when the voltage is 260V by applying the equation.

I(f) would be Vdc/R(f) but wouldn't I need Radj to get the value of I(f).

I'm still quite confused on how this is working, but thank you for all the help so far!

 

[cnh1995]

I(f) would be Vdc/R(f) but wouldn't I need Radj to get the value of I(f).

You know both the torques (at 300V, 30A and at 260V, 30A).
Hint: As I said earlier, torque is proportional to the product of field current and armature current.

 

[SK97]

You know both the torques (at 300V, 30A and at 260V, 30A).
Hint: As I said earlier, torque is proportional to the product of field current and armature current.

so i found the torque for 300v and did the following:

T(300) / T(260) = I(f)1 / I(f)2 (as Ia cancel out because they are the same)

and then I(f) = Vdc / Rf + Radj

but i end up getting 20 ohms for my answer which was the same as the original?

What am i doing wrong here?

 

[cnh1995]

If you show your calculations, I could see what is (or maybe isn't) wrong. I haven't calculated any of it.

 

[SK97]

If you show your calculations, I could see what is (or maybe isn't) wrong. I haven't calculated any of it.

okay so to get T (260) is did: (260) (30) / (1276.36) (2pi/60)
the 1276.36 was the original speed of of the motor that i found, the speed the motor was when operating at 30 A.

and to get T (300): (300) (30) / (1276.36) ( 2pi/60)

then since you said T is proportional to I(a) * I(f)

I use the ratio of the two torques as : T (300) / T (260) = 5A / I (f)2

then i get I(f) 2 to equal = 5 / 1.15 which would give me I(f) 2 to equal 4.33 A

then I(f)2 = Vdc / Rf + Radj

so 4.33 A = 260 / 40 + Radj

So Radj = 19.99 which is 20 ohms

the same as the original Radj in the question.

 

[cnh1995]

okay so to get T (260) is did: (260) (30) /

You have equated the 'total' input power to the gross mechanical power. You need to use the value of the back emf generated at 260V, 30A.

 

[SK97]

You have equated the 'total' input power to the gross mechanical power. You need to use the value of the back emf generated at 260V, 30A.

would i do that for both voltages? use the back emf generated to get the torques?

 

[cnh1995]

would i do that for both voltages? use the back emf generated to get the torques?

Yes. Isn't mechanical power E_b*I_a?

 

[SK97]

Yes. Isn't mechanical power E_b*I_a?

I think my confusion is lying in what i exactly equate at which step.

For the first step of the solution do i use the equation T = Ea * Ia / Wm

For both 300V and 260V to get the torques for both?

Then i would need to relate that to field current I(f).

So where does mechanical power come into play?

Sorry for taking this long to understand.

 

[cnh1995]

So where does mechanical power come into play?

To get the magnitudes of torques and the torque ratio, mechanical power is needed.

For the first step of the solution do i use the equation T = Ea * Ia / Wm

Yes, where E_a*I_a is the mechanical power.

 

[SK97]

To get the magnitudes of torques and the torque ratio, mechanical power is needed.

Yes, where E_a*I_a is the mechanical power.

Okay so applying that method;

After i get the torques in the two cases is it correct for me to do this:

T(300V) / T (260V) = I (f) 1 / I (f) 2

where I (f) 1 is 5A and I (f) 2 is unknown.

Also is the Wm (mechanical speed in radians) the same for both volatges?

 

[cnh1995]

Also is the Wm (mechanical speed in radians) the same for both volatges?

Yes, isn't that mentioned in the problem?