H63EMA Wound DC Problems
Exercise 1
A DC shunt generator has armature and field resistances of 0.2and 150,
respectively. The generator supplies 10 kW to a load connected to its terminals
at 230 V. Assuming that the total brush-contact voltage drop is 2V, determine
the induced voltage.
Solution
IL = P/Vt = 10000/230 = 43.48A
If = Vt/Rf = 230/150 = 1.53A
Ia = IL + If = 43.48 + 1.53 = 45A
(summed together as generating power)
Ea = Vt + IaRa + Vbd = 230 + 0.2x45 + 2 = 241V
As we are generating Ea is greater than Vt (as all power including losses is supplied by Ea)
Exercise 2:
Take the same conditions as the machine above, but in a series machine. Why
would this be an issue?
R
fw
IL = If = Ia = P/Vt = 10000/230 = 43.48A, but
Ea = Vt + IaRa + IaRf + Vbd
Ea = 230 + 43.48x0.2 + 43.48x150 + 2 = 6533V!
So the huge field resistance would need an enormous EMF to drive the needed
current
Exercise 3:
A separately excited motor runs at 1045 rpm, with constant field current and an armature
current of 50A at 120V. The armature resistance is 0.1ohm. The motor is further loaded
such that it draws a current of 95A at 120V.
1. What is the motor speed at this load? What is the power and torque delivered?
2. If the armature voltage is now increased to 180V and the load torque remains
the same, what is the new speed of the machine?
3. If the field current is reduced by 25% what will the armature current and speed
of the motor be for the same torque and armature voltage above?
Solution
1. What is the motor speed at this load? What is the power and torque delivered?
V1=V2 = 120 V; Ra =0.1 Ω; Ia1=50 A; Ia2=95 A; N1=1045 rpm.
ω1=N1*2π/60
E1=V1 – Ra∙Ia1 = 120 – 0.1x50 = 115V
E2=V2 – Ra∙Ia2 = 120 – 0.1x95 = 110.5V
Provided that the field current remains constant and that there is no armature reaction effect:
Ea = KaIfm, If and Ka are constant so m must vary in proportion with Ea
ω2/ω1 = E2/E1 → ω2 = ω1*E2/E1 = 105.15 rad/s or N2= ω2*60/2π= 1004 rpm
Pem2 = E2* Ia2 = 10,500 W
P = Tω
T2=P2/ ω2 = 99.84 Nm.
2. If the armature voltage is now increased to 180V and the load torque remains the same,
what is the new speed of the machine?
V3 = 180V; Ia3 = I2 = 95 A
E3 = V3 – Ra∙Ia3 = 170.5 V
ω3 = ω1*E3/E1 = 162.25 rad/s . N3= ω3*60/2π= 1549 rpm
3. If the field current is reduced by 25% what will the armature current and speed of the
motor be for the same torque and armature voltage above?
V4 = V3 = 180V
Torque is the same as in case 2 and 3 (T4 = T2 = 99.84) so a decrease in If has to be
compensated by an increase in Ia to keep the same torque, as T = KaIfIa
Ia4 = 4/3 * Ia3 = 126.67 A
E4=V4 – Ra∙Ia4 = 167.33 V
E4*Ia4 = T4* ω4 → ω4 = E4*Ia4/T4 = 212.29 rad/s. N4= ω4*60/2π = 2027.3 rpm
Note that the EMF constant changes due to the change in the field current ( 𝐸4𝜔4≠𝐸3𝜔3)
Exercise 4:
The armature winding of a 4-pole, DC machine is lap-wound in 46 slots with 18
conductors per slot. The field poles span 70% of the pole pitch and the peak air gap flux
density from them is 0.7T. The machine has an air gap diameter of 15cm and a core
axial length of 25cm.
1. If the armature is designed to carry a continuous line current of 100A, what is
the maximum electromagnetic torque developed by the machine?
2. Calculate the voltage induced in the armature winding when the rotor speed is
1800 rpm.
3. If the machine is supplied from a constant DC voltage source of 380V and the
armature resistance Ra=0.5ohm, what is the rotor speed if a load of 20Nm is
applied? State any assumptions made.
Solution
P = 2 (2 pole pairs/4 poles), N = 46, Z = 18, B = 0.7T, fp = 0.7, D = 0.15m, L = 0.25m, lap
wound
1. If Ia = 100A
As lap wound:
Cond current = Iw = 100/(2P) = 25A
Current loading = A = IWNZ /(pi D) = 25x46x18/(x0.15) = 4.393x104A/m
T=2 fp B A (L ( D2)/4) = 2x0.7x0.7x4.393x104x(0.25(x0.152)/4) = 190.2Nm
2. At 1800 RPM
 = 2xRPM/60 = 188.5rad/s
EaxIa = Tx,
Ea = Tx/Ia = 190.2x188.5/100 = 358.5V
Neglecting brush voltage drop and iron non-linearity, Ra = 0.5, Vt = 380V, load = 20Nm
Ia@20Nm = Ia/T x 20 = 100/190.2x20 = 10.52A
Ea@20Nm = Vt - Ia@20Nm x Ra = 380 – 10.52x0.5 = 374.7V
RPM@20Nm = RPM/Ea x Ea@20Nm = 1800/358.5x374.7 = 1881rpm