Engineering Electrical Machine - Shunt DC Motor Problem

[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?

 

[SK97]

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?

 

[cnh1995]

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.

 

[SK97]

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

 

[cnh1995]

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).

 

[SK97]

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.

 

[cnh1995]

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.

 

[SK97]

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!