Engineering Finding Torque and Half Rated Torque in DC Shunt Motors

[Joe85]

Homework Statement

Hi All, I'm struggling a little bit with the last part of the question and would really appreciate a bit of help on how best to approach and solve. I've written below what i think is the answer but it just doesnt feel right to me so would be grateful for some guidnance.


i) A 20 kW, 500 V d.c. shunt wound motor draws a current of 45A when running at full load with a speed of 600 rev min–1. On no load, the current drawn from the supply is 5 A. If the armature resistanceis 0.3Ω and the shunt field resistance is 220Ω, calculate the armature current, speed and efficiency of the motor when the torque falls to half of its full load value, assuming that the flux per pole doesnot change.(Hint: don’t forget that all armature current is supplying torque. Onemust decide at some points what is the armature current producingtorque and at others, to determine generator emfs, the total armaturecurrent).

(ii) State what you would expect to happen to the speed if a resistance isinserted into the field circuit.

(iii) If a resistance of 50 Ωis inserted in series with the shunt field,calculate the armature current required to maintain half full-loadtorque. Calculate also the speed of the motor when supplying this torque.

Relevant Equations

Below:

Homework Equations:

E=V-I_aR_a
n₂ = (n₁×E₂)/E₁
I_f = V/R_f
V×I_a
T∝ΦI_a
E∝ΦN

i)
V = 500v, N₁ = 600RPM, I_t = 45A, R_f = 220Ω, R_a = 0.3Ω

I_t = I_a + I_f
I_f = V/R_f = 500/220 = 2.273A
I_a = I_t - I_f = 45 - 2.273 = 42.727A

Now: T∝ΦI_a and if the flux per pole is constant then T∝I_a

So I_a when the Torque falls to half it's full load value is: I_a/2 = 42.727/2 = 21.3635A.

Since n₂ = (n₁×E₂)/E₁, we will need to find the back EMF at full load and half load.

E₁=V-I_aFLR_a
∴ E₁ = 500 - (42.727×0.3) = 487.182v

&

E₂=V-I_aHLR_a
∴ E₂ = 500 - (21.3635×0.3) = 493.591v
n₂ = (600x493.591)/ 487.182 = 608RPM.

Total Power at Half Torque :
P_HL = VI_aHL = 500×21.3635 = 10681.75 Watts

Total Power at No Load:
P_NL = VI_aNL = 500×5 = 2500 WattsField Loss:
P_f = I_f²R_f = 2.273 × 220 = 1136.64 Watts

Fixed Loss:
2500 - 1136.64 = 1363.36

Armature Loss:
P_a = I_a²R_a = 21.3635 × 0.3 = 136.92 Watts

10684.75/(10681.75+2500+136.92 = 0.80 = 80%
Speed = 608RPM, I_a = 21.3635, Effeciency = 80%

ii)

An increase in field resistance would result in the in the reduction of current flowing in the field circuit and thus a reduction in flux. Since speed is inversely proportional to flux, if the flux is reduced the speed will increase. And since T∝ΦI_a, the armature current will need to increase to maintain the same Torque output to compensate for the reduction in flux.
iii)

This is where i have hit a brick wall. I'm not sure how to find the torque or the half rated torque to begin solving this problem. Would appreciate nudge in the right direction.Thanks,

Joe.

 

[cnh1995]

For solving iii), use your reasoning for part ii).
What is the half load armature current before adding extra field resistance?

 

[Joe85]

Thanks.
I_aHL=21.3635A?

If we add the resistance to the the field circuit then:

R_f=270Ω and I_f=500/270 = 1.852A.

If I_t=45A and full load then I_aFL= 45-1.852

& and the required current at half load will be I_aHL = (45-1.852)/2?

Seems too simple!

 

[cnh1995]

IaHL=21.3635A

Right.

and the required current at half load will be IaHL = (45-1.852)/2?

No.
Hint: To maintain half load torque,
the product of field current and armature current after adding field resistance should be equal to the product of field current and armature current before adding field resistance.

 

[Joe85]

Ok so:

21.365 x 2.273 = 48.558A
500/270 = 1.852A.

48.558/1.852A= 26.22A?

 

[cnh1995]

48.558/1.852A= 26.22A

Right.

 

[Joe85]

Thank you cnh.

And the speed will just be the same as above: n2 = (n1xe2)/e1 = 606RPM?

 

[cnh1995]

n2 = (n1xe2)/e1

Only if the flux were constant in both cases..

 

[Joe85]

Thanks again, So i should begin by trying to work out the flux before the resistance inserted and then after it's inserted to be able to know what the change in speed is?

 

[cnh1995]

Thanks again, So i should begin by trying to work out the flux before the resistance inserted and then after it's inserted to be able to know what the change in speed is?

Yes.
Speed is proportional to back emf/flux.

 

[Joe85]

Right and the only equation that seems of any use to be able to resolve for flux is: E₁/E₂=Φn₁/Φn₂. I feel like a complete idiot but I cannot see how to isolate Φ.

Edit: Ok so i believe by using the following equation i can isolate for n₂:

n∝V/Φ
and Φ∝I_f

So n∝E/I_f and n=(E/I_f)*K

Which gives:

n₂=( n₁E₂I_f1) )/( E₁I_f2)

 

[cnh1995]

I cannot see how to isolate Φ.

Φ is proportional to field current.

 

[Joe85]

Just before you replied i edited this in above, honest :D

Edit: Ok so i believe by using the following equation i can isolate for n2:

n∝V/Φ
and Φ∝If

So n∝E/If and n=(E/If)*K

Which gives:

n2=( n1E2If1) )/( E1If2)

 

[cnh1995]

n2=( n1E2If1) )/( E1If2)

Right.

 

[Joe85]

Thank you. Your assistance and patience is very much appreciated.