CONTROL SYSTEMS ENGINEERING D227
S.A.E. SOLUTIONS
TUTORIAL 2 – ELECTRIC ACTUATORS
SELF ASSESSMENT EXERCISE No.1
1. A series motor is connected to a 500 V supply. The motor power output is 10 kW at 1200 rev/min. The efficiency of the motor is 92 %.
Determine the current and torque.
Electric Power = V I = 10/92% = 10.87 kW
I = 10870/500 = 21.74 A
T = 60P/2
π
N = 60 x 10000/(2 x
π
x 1200) = 79.58 Nm
2. A series motor has an armature resistance of 0.5 ohms and a field resistance of 1 ohm. It is connected to a 300 V supply and draws 50 Amperes from it. The motor runs at 1000 rev/min. The flux per pole is 0.03 Wb. The current is dropped to 25 Amperes and the flux per pole drops to 0.02
Wb. Determine the speed of the motor under these conditions.
Initial conditions.
Electric Power = Volts x Amps = 300 x 50 = 15 000 W
Resistance loss = I
2
(R a
R f
) = 50
2
(0.5 + 1) = 3750 W
Mechanical power = 15000 – 3750 = 11250 W
T = (11250 x 60)/(2
π
x 1000) = 107.4 N m.
T = K1
φ
Ia
K
1
= T/
φ
Ia = 107.4/(0.03 x 50) = 71.6
Final conditions. Use the same value for K
1
.
T = K1
φ
Ia = 71.6 x 0.02 x 25 = 35.8 N m
Electric Power = 300 x 25 = 7500 W
Resistance loss = I
2
(R a
R f
) = 25
2
(0.5 + 1) = 937.5 W
Mechanical power = 7500 – 937.5 = 6562.5 W
Mechanical Power = 2
π
NT/60
N = 60 x 6562.5/(2 x π x 35.8) = 1750 rev/min

SELF ASSESSMENT EXERCISE No.2
1. A 250 V, 30 kW series motor runs at 800 rev/min on full load. A resistance of 0.5
Ω
is connected in series with the motor and an output torque of 200 Nm is produced. Calculate the speed at this condition. It may be assumed that there are no energy losses and that the field flux is directly proportional to current.
Initial conditions.
Motor current I = Power/Volts = 30 000/250 = 120 A
Mechanical power = 2
π
NT/60 = 30 000 W
T = (30 000 x 60)/(2
π
x 800) = 358 N m.
T = K1
φ
Ia Since flux is proportional to current and the field current is the armature current then
T = K Ia2
358 = K x 1202 K = 0.02486
Final conditions. Use the same value for K. T = 200 = 0.02486 Ia2
Ia = 89.7 Amps Input Power = V Ia = 250 x 89.7 = 22 423 Watts
Loss in 0.5
Ω
resistance is Ia2 R = 89.72 x 0.5 = 4 023 W
Useful Power = 22 423 - 4 023 = 18 400 Watts = 2
π
NT/60
N = (18 400 x 60)/(2
π
x 200) = 878.5 rev/min
2. The armature of a 200 V d.c. motor has a resistance of 0.4
Ω
. The no load armature current is
2.0 A. When a torque is applied, the armature current increases to 50 A, and the speed is 1200 rev/min. Find the no load speed stating any assumptions made.
ON LOAD
E a
= 200 – 50 x 0.4 = 180 V
Power = 180 x 50 = 9000 W
T = 60P/(2
π
N) = 60 x 9000/(2
π
x 1200) = 71.62 Nm
T = K I a
2
K = 71.62/50 =0.02865
OFF LOAD
E a
= 200 – 2 x 0.4 = 199.2 V
P = 199.2 x 2 = 398.4 W
T = K I a
2
= 0.02865 x 2
2
= 0.1146 Nm
N = 60P/(2
π
T) = 60 x 398.4 /(2
π
x 0.1146) = 33200 rev/min

3 A 230 V d.c. motor has a separate field coil with a constant current of 2.2 A. The armature resistance is 0.15
Ω
. The motor was tested with no load applied to the shaft. The armature current was found to vary with armature voltage as follows.
Current 4.9 A 4.5 A 4.0 A 3.6 A
230 180 130 80 Volts
The speed at 230 V was 1150 rev/min. Determine the speeds at the other voltages.
Since the field current is constant the torque is T = K I a
For the given condition :
Ea = 230 – 4.9 x 0.15 = 229.265 V Power Output = 229.265 x 4.9 = 1123.4 W
T = 60P/2
π
N = 60 x 1123.4/(2
π
x 1150) = 9.328 Nm hence K = 9.328/4.9 = 1.904
At 180 V
Ea = 180 – 4.5 x 0.15 = 179.325 V Power Output = 179.325 x 4.5 = 807 W
T = K I a
= 1.904 x 4.5 = 8.568 Nm
N = 60P/2
π
T = 60 x 807/(2
π
x 8.568) = 899 rev/min
At 130 V
Ea = 130 – 4 x 0.15 = 129.4 V Power Output = 129.4 x 4 = 517.6 W
T = K I a
= 1.904 x 4 = 7.616 Nm
N = 60P/2
π
T = 60 x 517.6/(2
π
x 7.616) = 649 rev/min
At 80 V
Ea = 80 – 3.6 x 0.15 = 79.46 V Power Output = 79.46 x 3.6 = 286 W
T = K I a
= 1.904 x 3.6 = 6.854Nm
N = 60P/2
π
T = 60 x 286/(2
π
x 6.854) = 398 rev/min
4. The armature resistance of a 250 V d.c. shunt wound motor is 0.7
Ω
and the armature current is
2.0 A when operating under no load conditions. A torque is then applied to the shaft and the armature current rises to 60 A and the speed falls to 1000 rev/min. Determine the no load speed.
ON LOAD
Ea = V – IaRa = 250 – 60 x 0.7 = 208 V
Power = 208 x 60 = 12480 W
T = 60P/(2
π
N) = 60 x 12480/(2
π
x 1000) = 119.17 Nm
T = k1
φ
Ia but since the flux is constant T = K I a
K = 119.17/60 = 1.986
OFF LOAD
Ea = V – IaRa = 250 – 2 x 0.7 = 248.6 V
Power = 248.6 x 2 = 497.2 W
T = K I a
= 1.986 x 2 = 3.972 Nm
N = 60P/(2
π
T) = 60 x 497.2/(2
π
x 3.972) = 1195 rev/min

5. Sketch the Speed - Torque characteristic for a series motor and shunt motor. Explain why the series motor is more suited to traction and winching operations.
Series is shown on left. At low torque it is likely to over speed but it has high torque at low speed so it is suitable for high torque low speed applications such as winching. The shunt is shown on the right.
SELF ASSESSMENT EXERCISE No.3
Using the manufacturers data sheet determine the voltage and current needed to run the motor
GR16C at 3000 rev/min with a load torque of 40 N cm.
From the table we find k t
= 23.7 N cm per amp k
e
= 24.8 V per 1000 rev/min k
d
= 3.57 N cm per 1000 rev/min
T f
= 7.7 N cm R m
= 0.95 Ω
Td = (N/1000) x kd = (3000/1000) x 3.57 = 10.71 N cm
I = (TL + Tf + Td)/kt = (40 + 7.7 + 10.71)/23.7 = 2.465A
V = (N ke/1000) + (IaRm)
V = (3000/1000) x 24.8 + (2.465 x 0.95) = 76.7 V
SELF ASSESSMENT EXERCISE No.4
1. Identify which of the diagrams shows a series, shunt and compound wound motor.
2. Identify which torque - speed graph represents series and shunt wound motors.
The left is for shunt and the right is for series.

3. A DC servo motor has the following constants. ke = 7.1 V per 1000 rev/min kt = 6.7 Ncm /A kd = 0.78 Ncm per 1000 rev/min
Tf = 3.2 Ncm
Ra = 0.85 Ohm
Calculate the current, terminal voltage and electric power when the motor speed is 3000 rev/min and the load torque is 45 N cm.
Td = (N/1000) x kd = (3000/1000) x 0.78 = 2.34 N cm
I = (TL + Tf + Td)/kt = (45 + 3.2 + 2.34)/6.7 = 7.543A
V = (N ke/1000) + (IaRm)
V = (3000/1000) x 7.1 + (7.543 x 0.85) = 27.71 V
Power = 27.71 x 7.543 = 209 W
4. A 500V shunt wound d.c. motor has an armature resistance of 0.5
Ω
and a field coil resistance of 250
Ω
. Calculate the total current taken from the source at start up. What value resistance must be placed in series with the armature to reduce the starting current to 50A?
Starting current = 500/0.5 + 500/250 = 1002 A
If the current is limited to 50 A then
50 = 500/(R + 0.5) + 500/250
48 = 500/(R + 0.5)
(R + 0.5) = 500/48 = 10.417
Ω
R = 9.917 Ω
5. A d.c. servo motor must accelerate from rest to 15 rad/s in 2 seconds. The rotating mass has a moment of inertia of 2 kg m2. The load torque is 12 Nm and the friction torque is 1.5 Nm. The damping coefficient is 0.08 Nm s/radian. The torque constant is 90 Nm per ampere.
Determine the required current just before acceleration ends. (0.33 A)
A d.c. servo motor must accelerate from rest to 15 rad/s in 2 seconds. The rotating mass has a moment of inertia of 2 kg m2. The load torque is 12 Nm and the friction torque is 1.5 Nm. The damping coefficient is 0.08 Nm s/radian. The torque constant is 90 Nm per ampere. Determine the required current just before acceleration ends.
ω
= 15 rad/s
T
L
= 12 Nm
T f
= 1.5 Nm
Td = Kd
ω
= 0.08 x 15 = 1.2 Nm
α
= (15 – 0)/2 = 7.5 rad/s
2
Ti = I
α
= 2 x 7.5 = 15 Nm
T = TL + Tf + Kd d
θ
/dt + I d2
θ
/dt2 = 12 + 1.5 + 1.2 + 15 = 29.7 Nm
I = 29.7/90 = 0.33 Amperes