Electrical Machine - Shunt DC Motor Problem

  • #26
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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?
 
  • #27
cnh1995
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If you show your calculations, I could see what is (or maybe isn't) wrong. I haven't calculated any of it.
 
  • #28
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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.
 
  • #29
cnh1995
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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.
 
  • #30
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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?
 
  • #31
cnh1995
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would i do that for both voltages? use the back emf generated to get the torques?
Yes. Isn't mechanical power Eb*Ia?
 
  • #32
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Yes. Isn't mechanical power Eb*Ia?

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.
 
  • #33
cnh1995
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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 Ea*Ia is the mechanical power.
 
  • #34
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To get the magnitudes of torques and the torque ratio, mechanical power is needed.

Yes, where Ea*Ia 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?
 
  • #35
cnh1995
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Also is the Wm (mechanical speed in radians) the same for both volatges?
Yes, isn't that mentioned in the problem?
 
  • #36
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Yes, isn't that mentioned in the problem?
Yes so after doing the method i posted earlier, I am still obtaining an answer of 20 ohms.

So something I'm doing must not be right, but i can't understand what?
 
  • #37
cnh1995
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Yes so after doing the method i posted earlier, I am still obtaining an answer of 20 ohms.

So something I'm doing must not be right, but i can't understand what?
Post your working. Maybe there is some arithmatic mistake.
 
  • #38
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Post your working. Maybe there is some arithmatic mistake.
ok here it is:

Ea for 260V: Vdc - IaRa

260 - 30(0.25) = 252.5V

Ea for 300V as calculated before is 292.5V

T (260V) = 252.5 (30) / (1276.36) (2pi/60) = 65.65 Nm
T (300V) = 292.5 (30) / (1276.36) (2pi/ 60) = 56.67 Nm

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

65.65 / 56.67 = 5/ I(f)2
1.158 = 5 / I (f)2

I (f) 2 = 5 / 1.158
I ( f) 2 = 4.32 A

then I (f) = Vdc / Rf + Radj

so: 4.32 = 260 / 40 + Radj

Radj = 20.24 ohms
 
  • #39
cnh1995
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ok here it is:

Ea for 260V: Vdc - IaRa

260 - 30(0.25) = 252.5V

Ea for 300V as calculated before is 292.5V

T (260V) = 252.5 (30) / (1276.36) (2pi/60) = 65.65 Nm
T (300V) = 292.5 (30) / (1276.36) (2pi/ 60) = 56.67 Nm

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

65.65 / 56.67 = 5/ I(f)2
1.158 = 5 / I (f)2

I (f) 2 = 5 / 1.158
I ( f) 2 = 4.32 A

then I (f) = Vdc / Rf + Radj

so: 4.32 = 260 / 40 + Radj

Radj = 20.24 ohms
Your calculations look correct to me.
When you used terminal voltages (300V and 260V) instead of back emfs, you got Radj=20 ohms and the new value for Radj=20.24 ohms, which is very close to the previous value. This is because the ratio of terminal voltages (260/300) is almost equal to the ratio of back emfs (252.5/292.5).
 
  • #40
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Your calculations look correct to me.
When you used terminal voltages (300V and 260V) instead of back emfs, you got Radj=20 ohms and the new value for Radj=20.24 ohms, which is very close to the previous value. This is because the ratio of terminal voltages (260/300) is almost equal to the ratio of back emfs (252.5/292.5).
So it's just the nature of the question that made both values close to each other?

All in all thank you so much for your thorough help in this problem, I can't tell you how much I appreciate you and am thankful for helping me understand this topic clearer.
 
  • #41
cnh1995
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So it's just the nature of the question that made both values close to each other?
I think so.

To summarize,
the motor develops 65 Nm torque with 300V supply voltage and 30A armature current, and the speed is 1276 rpm. If you reduce the terminal voltage to 260V with the load unchanged, the field current will reduce and the speed will also decrease in order to allow for more current to flow in the armature, so as to maintain the torque at 65Nm. If you want to bring the speed back to 1276 rpm and armature current to 30A, you'll have to reduce the load torque from 65 Nm to 56 Nm and increase Radj from 20 ohm to 20.24 ohm.
 
  • #42
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I think so.

To summarize,
the motor develops 65 Nm torque with 300V supply voltage and 30A armature current, and the speed is 1276 rpm. If you reduce the terminal voltage to 260V with the load unchanged, the field current will reduce and the speed will also decrease in order to allow for more current to flow in the armature, so as to maintain the torque at 65Nm. If you want to bring the speed back to 1276 rpm and armature current to 30A, you'll have to reduce the load torque from 65 Nm to 56 Nm and increase Radj from 20 ohm to 20.24 ohm.
Yeah that seems to make sense.

Thank you again for all your hard work!
 

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