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ECE 320
DC Machines Problems
Lesson 26
1, Questions 8.2, 8.3, 8.4, 8.5, and 8.6 on page 552.
8.2 How can the speed of a shunt dc motor be controlled? Explain in detail.
A shunt DC motor speed can be controlled by one of three methods:
1. Adjusting the field resistance: Increasing the field resistance decreases the field current. Field
weakening increases the steady state speed.
2. Adjusting the terminal voltage adjusts the field current and the armature current. This adjusts torque
which leads to a corresponding adjustment in steady state speed.
3. Inserting a resistance in series with the armature makes the machine more sensitive to load,
decreasing its steady state speed if it is under load. This method is not popular because a resistance in
series with the main energy-carrying current wastes a lot of energy.
Details on these methods are explained on pages 479-491 of the textbook.
8.3 What is the practical difference between a separately excited and a shunt dc motor?
The main difference is the way that the field is connnected to its source of energy. In the shunt
machine, the same source serves both armature and field; in the separately excited machine, a
separate source serves each machine winding. The practical effect is less complicated control of
speed with greater flexibility in the separately excited machine.
8.4 What effect does armature reaction have on the torque-speed characteristic of a
shunt dc motor? Can the effects of armature reaction be serious? What can be done to
remedy this problem?
Armature reaction shifts the neutral plane of the machine and reduces the magnetic flux within
the field; in other words, it weakens the field. Shifting of the neutral plane causes commutation
under load to occur when voltage remains on the commutator. This leads to sparking and arcing
at the brushes; in severe cases, it can lead to flashover between commutator segments.
The field weakening aspect of armature reaction gives greater than expected speed for a given
terminal condition of voltage and current. In severe cases, this may cause runaway in a manner
similar to what occurs with poorly designed differential compounding.
See pages 433-444 of the textbook for details.
8.5 What are the desirable characteristics of the permanent magnets in a PMDC machine?
Permanent magnet dc machines have no field circuit. There is no requirement for a source of energy for
the field, saving the expense of a separate excitation or the slower control of shunt or series machines.
Consequently, PMDC machines tend to be smaller than comparably rated wound field motors There is
little energy loss in the field circuit of a PMDC machine. These advantages tend to be more significant in
smaller machines.
8.6 What are the principal characteristics of a series dc motor? What are its uses?
The series connection of the field winding leads to a torque-speed characteristic that has the speed
proportional to the reciprocal of the square of the torque, shifted a little lower by the effect of the
winding resistances. This leads to an enormous starting torque and a need to always have some
load applied to prevent overspeed. Winding resistances are quite small.
A series dc machine has an advantage where high starting torque is needed. Applications include
starter motors, elevator motors, and traction motors in locomotives.
2. Problems 8.1, 8.2 and 8.3 on page 553 of the textbook.
ILrated  110  A
VT  240  V
NF  2700
NSE  14
RA  0.19 ohm
RF  75 ohm
RS  0.02 ohm Radj = 100  to 400  ohm
ωrated  1800 RPM
rad
ωrated  188.496 
sec
RPM 
2π
Prated  30 hp
60
 Hz
Protational  3550 W
In problems 8.1 through 8.7, assume that the motor described above can be connected in shunt.
The equivalent circuit of the shunt motor is shown in Figure P8-2.
8.1 If the resistor Radj is adjusted to 175 ohms, what is the rotational speed of the motor at
no-load conditions?
Radj  175  ohm
ωfl  ωrated
No load means no armature current.
IAnl  0  A
VAnl  VT
VAnl  240 V
Find the generated voltage at no load. With no current, it's the same at the terminal voltage
EAnl  VAnl  RA IAnl
EAnl  240 V
Calculate the field current in the machine
VT
IF 
IF  0.96 A
RF  Radj
From Figure P8-1, at 1800 RPM, this field current gives a generated voltage of 241V.
EA0  241  V
ω0  1800 RPM
Calculate the speed that corresponds to the same field current (same flux) but 240V=EA.
EA0
K Φ  ω0
EAnl
rad
=
ωnl  ω0 
ωnl  187.713 
ωnl  1793 RPM
EAnl
K Φ  ωnl
EA0
sec
8.2 Assuming no armature reaction, what is the speed of themotor at full load? What is
the speed regulation of the motor?
We have given the terminal full load current. There is no indication that the field has changed.
ILrated  110 A
IF  0.96 A
By the current law, we can find the armature current.
IAfl  ILrated  IF
IAfl  109.04 A
Use a loop equation to find the induced voltage.
EAfl  VT  IAfl RA
EAfl  219.3 V
From Figure P8-1, at 1800 RPM, this field current gives a generated voltage of 271V.
EA0  271  V
ω0  1800 RPM
Calculate the speed that corresponds to the same field current (same flux) but 218.3V=EA.
EA0
K Φ  ω0
EAfl
rad
=
ωfl  ω0 
ωfl  152.523 
ωfl  1456 RPM
EAfl
K Φ  ωfl
EA0
sec
ωnl  ωfl
SpeedRegulation 
SpeedRegulation  23.1 %
ωfl


8.3 If the motor is operating at full load and if its variable resistance R adj is increased to
250Ω what is the new speed of the motor? Compare the full load speed with Radj=175Ω to
the full load speed with Radj=250Ω Assume no armature reaction.
When we change Radj to 250Ω, the field current changes.
VT
IF3 
RF  Radj
Radj  250  ohm
IF3  0.738 A
Armature current remains the same at rated conditions because ratings are based upon the
heating that occurs in response to armature current and field current levels. Therefore, induced
voltage remains the same.
IA3  IAfl
IA3  109.04 A
EA3  VT  IA3 RA
EA3  219.3 V
From Figure P8-1, at 1800 RPM, this field current gives a generated voltage of 215V
EA0  215  V
ω0  1800 RPM
Calculate the speed that corresponds to the same field current (same flux) but 218.3V=EA.
EA0
EA3
=
K Φ  ω0
K Φ  ω3
EA3
ω3  ω0 
EA0
rad
ω3  192.25
sec
ω3  1836 RPM
This speed is somewhat higher than the speed under the same load at a lower field resistance. Field
weakening increases speed for the same load.
3. Problems 8.4 and 8.6 on page 553 of the textbook.
8.4 Assume that the motor is operating at full load and that the variable resistor R adj is
again 175Ω. If the armature reaction is 2000 A*turns at full load, what is the speed of the
motor? How does it compare to the results of Problem 8-2?
Radj  175  Ω
turns  1
AR  2000 A turns
We have the same field current as we began with, but this will be reduced by the armature
reaction. The armature reaction is related to the field current by the number of field turns:
VT
IF 
RF  Radj
IF  0.96 A
AR
IFeff  IF 
NF
IFeff  0.219 A
Looking up the nominal induced voltage from the magnetization curve in
Figure P8-1 on page 554:
EA0  84V
ω0  1800 RPM
The armature current is:
IA4  ILrated  IFeff
IA4  109.781 A
Calculating the induced voltage for the load and armature current specified:
EA4  VT  IA4 RA
EA4  219.142 V
Compare this to the nominal induced voltage at nominal speed and we can find the operating speed.
Armature reaction has the
effect of weakening the field.
EA4
K φ ω4
EA4 ω0
Field weakening speeds up
=
ω4 
ω4  4696 RPM
EA0
K φ ω0
EA0
the machine. In this case,
the numbers are big.
8.6 What is the starting current of thes machine if it is started by connecting it directly to the
power supply V T? How does this starting current compare to the full load current of the
motor?
The starting current for a line start is found by setting the induced voltage to zero and using the
remaining armature loop to find the current.
VT
3
IAstart 
IAstart  1.263  10 A
RA
To get the line current, add in the field current. The field current is negligible under these
circumstances.
ILstart  IAstart  IF
3
ILstart  1.264  10 A
This is more than ten times the 110A full load machine current. A current this large, particularly if
the motor is started under load (which makes it quite slow to come up to speed), is likely to damage
the motor. For this reason, there is a practical limit of a few horsepower for machines to be
line-started. Larger machines must have some form of starter circuit for safe starting.
4. Problem 8.7 on page 554 of the textbook.
8.7 Plot the torque-speed characteristic of this motor assuming no armature reaction and
again assuming an armature reaction of 1200 A-turns.
Restate the given.
VT  240  V RF  75 Ω Radj  175  Ω RA  0.40 Ω
IL0  110  A NF  2700
Describe the magnetization curve. Here, I just took a
set of points that give a reasonable piecewise
linearization of the curve.
ω0  1800 RPM
Far0  1200 A
RPM 
2 π
60
 Hz
T
EAx  ( 18 50 82 113 140 153 184 203 220 233 243 253 261 267 273 )  V
T
IFx  ( 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 )  A
Define the armature reaction as linear with terminal current,as the textbook does in its example.
 IL 
FAR IL Far0  Far0 

 IL0 


Set up the equations. Use circuit analysis to find I A, EA, and IF.
VT
IA IL  IL 
RF  Radj
 
 
 
EA IL  VT  IA IL  RA

VT
FAR IL Far0
IF IL Far0 

NF
RF  Radj



Enter the magnetization curve and use linear interpolation to get the nominal value of EA0.




EA0 IL Far0  linterp IFx EAx IF IL Far0
Calculate the speed for this case.
 
EA IL
ω87 IL Far0 
 ω0
EA0 IL Far0




Calculate the torque for this case.
   


EA IL  IA IL
τind87 IL Far0 
ω87 IL Far0



240
220


200
ω 87 IL 1200 A
ω 87 IL 0 A
180
160
0

50


τ ind87 IL 0 A τ ind87 IL 1200 A
The machine exhibits instability when armature reaction is considered.

100
5. Problem 8.8 on page 554 of the textbook.
8.8 For Problems 8-8 and 8-9, the shunt dc motor is reconnected separately excited,as
shown in Figure P8-3. It has a fixed field voltage V F of 240V and an armature voltage V A
that can be varied from 120V to 240V. What is the no load speed of this separately
excited motor when R adj=175ohms and (a) V A=120V, (b) VA=180V, and (c) VA=240V?
VF  240  V
 120 
VA   180   V
 
 240 
Radj  175  ohm
Find the field current.
VF
IF 
IF  0.96 A
RF  Radj
From Figure P8-1, at 1800 RPM, this field current gives a generated voltage of 241V.
EA0  241  V
ω0  1800 RPM
At no load, the armature voltage is equal to the generated voltage; no load means no armature current.
IA  0  A
 120 
EAnl   180  V
 
 240 
EAnl  VA  IA RA
Calculate the speed that corresponds to the same field current (same flux).
EA0
EAnl
=
K Φ  ω0
K Φ  ωnl
EAnl
ωnl  ω0 
EA0
 93.857 
rad
ωnl   140.785  

 sec
 187.713 
 896 
ωnl   1344   RPM


 1793 
Check: the third operating point here is the same as the operating conditions in Problem 8.1.
6. Problem 8.20 on page 559 of the textbook.
8.20 An automatic starter circuit is to be designed for a shunt motor rated at 20hp, 240V, and 75A.
The armature resistance is 0.12Ω and the shunt field resistance is 40Ω. The motor is to start with
no more than 250% of its rated armature current and as soon as the current falls to rated value, a
starting resistor stage is to be cut out. How many stages of starting resistance are need and how
big should each one be?
PR  20 hp
VAR  240  V
ILR  75 A
RA  0.12 Ω
VAR
IFR 
RF
IFR  6 A
IAR  ILR  IFR
RF  40 Ω
IX  250%
IAR  69 A
To limit the starting current to 250% of this rated value,
IAPk  IX IAR
IAPk  172.5 A
Starting resistance needed to limit the current to the peak value specified is found as follows:
VAR
RAstart1 
RAstart1  1.391 Ω
IAPk
We calculate the expected number of steps and rounding up:
  RA  
log 

  RAstart1  
n  ceil
3
 log IAR  



 IAPk  
Subtracting the armature resistance gives us the necessary value of added start resistance.
Rstart1  RAstart1  RA
Rstart1  1.271 Ω
When the machine accelerates, it eventually reaches the rated current value, at an induced voltage
value of

EA1  VT  IAR RA  Rstart1

EA1  144 V
We now reduce the starting resistance. Again, the limit is the specified peak current.
RAstart2 
VAR  EA1
RAstart2  0.557 Ω
IAPk
Subtracting the armature resistance to get the necessary value of added resistance.
Rstart2  RAstart2  RA
Rstart2  0.437 Ω
When the machine accelerates, it eventually reaches the rated current value, at an induced voltage
value of

EA2  VT  IAR RA  Rstart2

EA2  201.6 V
We find the next value of starting resistance using this induced voltage and the specified peak
current.
RAstart3 
VAR  EA2
RAstart3  0.223 Ω
IAPk
Subtracting the armature resistance to get the necessary value of added resistance.
Rstart3  RAstart3  RA
Rstart3  0.103 Ω
When the machine accelerates, it eventually reaches the rated current value, at an induced voltage
value of

EA3  VT  IAR RA  Rstart3

EA3  224.64 V
We find the next value of starting resistance using this induced voltage and the specified peak
current.
RAstart4 
VAR  EA3
IAPk
RAstart4  0.089 Ω
This is less than the armature resistance, so no further additional resistance is needed to complete
the starting process. There are only three stages of armature resistance needed for starting.
Rstart1  1.271 Ω
Rstart2  0.437 Ω
Rstart3  0.103 Ω
Thd individual starting resistances are found by evaluating each successive stage of the starting
process.
Rstart1 = R1  R2  R3
Rstart2 = R2  R3
Solving these for the starting resistances,
R3  Rstart3
R3  0.103 Ω
R2  Rstart2  R3
R2  0.334 Ω
R1  Rstart1  R2  R3
R1  0.835 Ω
Rstart3 = R3
7. Problem 8.14a on page 556 of the textbook; do only the 100% case.
8.14a A 20-hp, 240V, 76A, 900RPM series motor has a field winding of 33 turns per pole. Its
armature resistance is 0.09Ω and its field resisitance is 0.06Ω. The magnetization curve
expressed in terms of magnetomotive force versus EA at 900 RPM is given by the following
table
EA (V)
F (A*turns)
95
500
150
1000
188
1500
212
2000
229
2500
243
3000
Armature reaction is negligible for this machine. Computer the motor's torque, speed,
and output power at 100% of full load armature current. Neglect rotational losses.
Pout  20 hp
VT  240  V
IA  76 A
ωfl  900  RPM
T
NF  33
RA  0.09 Ω
RF  0.06 Ω
T
EA  ( 95 150 188 212 229 234 )  V
F  ( 500 1000 1500 2000 2500 3000 )  A
Set up the magnetization curve for interpolation.


M cs  cspline F EA
Calculate the generated voltage and the MMF under the conditions given.


EAx  VT  IA RA  RF
EAx  228.6 V
F0  NF IA
F0  2508 A
Perform the interpolation to get the normalized generated voltage at the nominal speed of ω0  900  RPM


EA0  interp M cs F EA F0  229.197 V
Calculate the speed that corresponds to the same field current (same flux) but at the calculated E A. Then
initiate a collection of values for later plotting.
EA0
EAx
=
K Φ  ω0
K Φ  ωx
EAx
ωx  ω0 
EA0
ωx  897.656  RPM
rad
ωm  ωx  ( 94.002 ) 
sec
 
Output power is
Pconv  EAx IA
Pconv  17.374 kW
Pconv  23.298 hp
which is close to the
nominal value
Torque is
τind 
Pconv
ωx
Collect torque values for plotting:
τind  184.821  N m
Check
τind ωx  17.374 kW


τm  τind  ( 184.821 )  N m
IA  76 A 0.33  25.08 A
EAx  236.238 V
F0  NF IA
Repeat the calculations for 33% load current:


EAx  VT  IA RA  RF

F0  827.6 A

EA0  interp M cs F EA F0  133.018 V
EA0
EAx
=
K Φ  ω0
EAx
ωx  ω0 
EA0
K Φ  ωx
Pconv  EAx IA
τind 
ω0  900  RPM
ωx  1598.4 RPM
 94   rad
ωm  stack ωm ωx  

 167.4  sec
Pconv  7.945  hp

Pconv  5.925  kW
Pconv
τind  35.397 N m
ωx



τm  stack τm τind 
Check
Repeat the calculations for 67% load current:


EAx  VT  IA RA  RF

τind ωx  5.925 kW
IA  76 A 0.67  50.92 A
EAx  232.362 V
F0  NF IA

EA0  interp M cs F EA F0  197.842 V
EA0
EAx
=
K Φ  ω0
Pconv  EAx IA
τind 
ωx  1057 RPM
 94 
rad
ωm  stack ωm ωx   167.4  

 sec
Pconv  15.867 hp
 110.7 

Pconv  11.832 kW
Pconv
τind  106.89 N m
ωx
F0  1680.4 A
ω0  900  RPM
EAx
ωx  ω0 
EA0
K Φ  ωx
 184.821   N m


 35.397 

 184.821 
τm  stack τm τind    35.397   N m


 106.89 
τind ωx  11.832 kW
Check
Repeat the calculations for 133% load current: IA  76 A 1.33  101.08 A


EAx  VT  IA RA  RF

EAx  224.838 V
F0  NF IA

EA0  interp M cs F EA F0  226.396 V
EA0
EAx
=
K Φ  ω0
K Φ  ωx
Pconv  EAx IA
τind 
Pconv
ωx
EAx
ωx  ω0 
EA0
Pconv  22.727 kW
ω0  900  RPM
 94 
ωx  893.8  RPM
167.4  rad
ωm  stack ωm ωx  

 110.7  sec
Pconv  30.477 hp
 93.6 



τind  242.808  N m
Check
F0  3335.6 A
τind ωx  22.727 kW

 184.821 
35.397 
τm  stack τm τind   
 N m
 106.89 
 242.808 


Plot the torque vs speed curve.
300
200
τm
100
0
80
100
120
140
ωm
160
180