- #1
Joe85
- 29
- 3
- 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-IaRa
n2 = (n1×E2)/E1
If = V/Rf
V×Ia
T∝ΦIa
E∝ΦN
i)
V = 500v, N1 = 600RPM, It = 45A, Rf = 220Ω, Ra = 0.3Ω
It = Ia + If
If = V/Rf = 500/220 = 2.273A
Ia = It - If = 45 - 2.273 = 42.727A
Now: T∝ΦIa and if the flux per pole is constant then T∝Ia
So Ia when the Torque falls to half it's full load value is: Ia/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-IaFLRa
∴ E1 = 500 - (42.727×0.3) = 487.182v
&
E2=V-IaHLRa
∴ E2 = 500 - (21.3635×0.3) = 493.591v
n2 = (600x493.591)/ 487.182 = 608RPM.
Total Power at Half Torque :
PHL = VIaHL = 500×21.3635 = 10681.75 Watts
Total Power at No Load:
PNL = VIaNL = 500×5 = 2500 WattsField Loss:
Pf = If2Rf = 2.273 × 220 = 1136.64 Watts
Fixed Loss:
2500 - 1136.64 = 1363.36
Armature Loss:
Pa = Ia2Ra = 21.3635 × 0.3 = 136.92 Watts
10684.75/(10681.75+2500+136.92 = 0.80 = 80%
Speed = 608RPM, Ia = 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∝ΦIa, 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.
E=V-IaRa
n2 = (n1×E2)/E1
If = V/Rf
V×Ia
T∝ΦIa
E∝ΦN
i)
V = 500v, N1 = 600RPM, It = 45A, Rf = 220Ω, Ra = 0.3Ω
It = Ia + If
If = V/Rf = 500/220 = 2.273A
Ia = It - If = 45 - 2.273 = 42.727A
Now: T∝ΦIa and if the flux per pole is constant then T∝Ia
So Ia when the Torque falls to half it's full load value is: Ia/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-IaFLRa
∴ E1 = 500 - (42.727×0.3) = 487.182v
&
E2=V-IaHLRa
∴ E2 = 500 - (21.3635×0.3) = 493.591v
n2 = (600x493.591)/ 487.182 = 608RPM.
Total Power at Half Torque :
PHL = VIaHL = 500×21.3635 = 10681.75 Watts
Total Power at No Load:
PNL = VIaNL = 500×5 = 2500 WattsField Loss:
Pf = If2Rf = 2.273 × 220 = 1136.64 Watts
Fixed Loss:
2500 - 1136.64 = 1363.36
Armature Loss:
Pa = Ia2Ra = 21.3635 × 0.3 = 136.92 Watts
10684.75/(10681.75+2500+136.92 = 0.80 = 80%
Speed = 608RPM, Ia = 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∝ΦIa, 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.
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