How Does Field Current Affect DC Motor Speed?

[Michael Neo]

Homework Statement

A 250 V shunt motor has an armature resistance of 0.4 Ω and runs at a speed of 750 rev min –1 when taking a full load current of 25 A.

Estimate the speed of the motor at no load when the armature current is 3 A, assuming that the flux per pole remains constant.

Relevant Equations

Back EMF:
E=V-Ia*Ra

Field Current:
If=Il-Ia

Field Resistance:
Rf=V/If

Change in Speed:
Old speed = (New speed)*(V-If*Ia)/E

Field Current

If=25-3
If=22 A

Field Resistance

Rf=250/22
Rf=11.36363636 Ohms

Back EMF

E=250-3x0.4
E=248.8 V

New Speed

New Speed = (Old speed)*(V-If*Ra))/E
New Speed = (750)*(250-22*0.4)]/248.8
New Speed = 727.0900322 rpm

Where can I find the appropriate formula for the change in speed in this scenario?
I have searched several books.

 

[NascentOxygen]

Field is shunt connected, so I_f is determined by supply voltage, V.

On no load, we usually consider I_a to be zero.

The shunt motor under load will be slowed down from its unloaded speed.

 

[Michael Neo]

So, at zero load the shunt motor will speed up.

The numbers substituted in are correct; therefore, the formula is incorrect.

A different approach is required - but which one. I cannot find this formula anywhere.

...

After a sleepless night:

Motor speed N = K*(V - IR)/φ

In this case, where flux is constant,

N2/N1 = (V - Ia*Ra)/(V - Il*Ra)

So,

N2 = N1 * (V - Ia*Ra)/(V - Il*Ra)
N2 = 750 * (250 - 3*0.4) / (250 - 25*0.4) = 777.5 rev/min

 

[cnh1995]

N2 = 750 * (250 - 3*0.4) / (250 - 25*0.4) = 777.5 rev/min

That is correct.

 

[Michael Neo]

Thank you.