Some Summaries

Power and Torque in AC Machines

• Power and efficiency in AC machines

o Electrical Power (in or out)
o Mechanical Power (in or out)
o Losses
 Electrical
 Copper Losses
 Core Losses (including hysteresis and eddy current)
 Mechanical
 Windage
 Friction
 Vibration
 Stray
• Ways to increase efficiency in AC machines
o Lower resistance windings (larger cross-section or
superconducting)
o Laminated core (high quality magnetic steel)
o Non-salient Pole rotor, Hydrogen filled
o Magnetic bearings
o Precision balanced rotor

Synchronous Generators

• Synchronous Machine
o Stator construction provides a rotating magnetic field.
o Number of poles
o Synchronous speed
o Salient vs. non-Salient pole machines
• Synchronous Machine excitation
o Standard excitation (slip rings and brushes)
o Brushless excitation

• Synchronous Generators
o Stator construction
 Built with windings for phases offset by 120 electrical
degrees
o Rotor construction
 Salient pole vs. non-salient pole
 Windings for formation of magnetic field
 Slip rings
 Brushless excitation
o Relation between electrical frequency and mechanical rate of rotation.

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  Note that the denominator consists of a conversion
factor (because the rotational speed is in rpm we need to
convert to seconds in the denominator) and a factor of two
to change from poles to pole pairs.
o Flow of power in Synchronous Generator

Synchronous Generator Equivalent Circuit

• Synchronous Generators
o Armature Resistance
 Inherent characteristic of windings made of copper (as opposed to
superconductors)
o Synchronous Reactance
 Inductance of windings
 Reaction to stator field
o Induced voltage


o Equivalent Circuit:

Synchronous Generator Power and Torque

• Induced voltage in synchronous generator

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 o Relation between flux and voltage
o Effect of saturation
• Voltage drops in synchronous generator
o Winding resistance
o Armature reaction
o Self Inductance
o Phasor Diagram
• Connections of stator windings
o Wye
o Delta
• Power and Torque
o Torque Angle
o Power as a function of torque angle
o Steady-State Stability limit

Synchronous Generators Excitation and Operation

• Synchronous Generator Losses

o Generator Power Flow
 Stray Losses
 Friction and Windage Losses (mech losses)
 Core Losses
 Rotor Losses
 Copper Losses
o Efficiency
• Power and torque
o Power Flow
o Static stability limit
o Torque
o Torque angle (angle between Vφ and EA)
o
• Modes of excitation
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 o Brushes
o Brushless excitation
• Synchronous Generator Parameters
o Open Circuit characteristic
o Short circuit characteristic
o DC resistance test

• Stand-Alone operation
o EA constant
o δ changes
o Vφ changes with load
• Parallel operation
o Why?
 Supply larger load
 Increased reliability
 Maintenance opportunities
 Efficiency
o Required conditions
 Line voltages equal
 Same phase sequence
 Same phase angles
 On-coming generator slightly faster
o Vφ constant
o Power delivered dependent on input torque
o If adjustments change power factor
o Prime Movers
 Steam turbine
 Gas turbine
 Diesel engine
 Water turbine
 Wind turbine

Example 1 :

1. A 13.8 kV, 200 MVA, 3600 rpm synchronous generator having an excitation
characteristic as shown is connected to an infinite bus operating at rated voltage, and
delivering 80 MW at 85% power factor lagging. The resistance of the armature is .0126
Ω , and the synchronous reactance is .077 Ω . Windage losses are approximately 5 MW,
and core losses are about 4 MW. Note: the open circuit characteristic may be modeled
with the following equation:

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 Determine the following:

a. The induced voltage for these operating conditions. _______________

b. The field current. _________________
c. The power factor if the generator input power is doubled without adjusting the
field current. (estimate based on an appropriate change in torque angle)
________________
d. The power factor if the field current is doubled without adjusting the generator
output power._____________
e. The efficiency for the original case (80 MW, 85%pf)_____________
f. The efficiency for the case in part (c)_________________

2. The mine at which you are employed is operating with a number of induction motors
being used for various ore processing equipment. The total load on the bus is 3MW, at a
85% lagging power factor. Your boss has just received a notice from New Century
Energies (the local utility) that the power factor must be improved to 95% or significant
penalties will be added to the electric bill. The timing is actually quite good to be
reviewing the power system, because expansion in the mine will require a 900 hp motor
to drive a rock crusher. Your assignment is to determine whether a synchronous motor
should be used to drive the rock crusher and correct the power factor, or an induction
motor used to drive the rock crusher and capacitors used to correct power factor. Use the
following assumptions:

o Either type of motor will have an efficiency of 95%.

o The synchronous motor cannot operate below 70% power factor.
o The induction motor would have a power factor of 85% lagging.

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 o The induction motor would cost $75,000. The synchronous motor would cost
$100,000. Capacitors would cost $30/kVar.

To assist you in your analysis, answer the following questions:

a. How many kW electric would be drawn by either motor?

____________________
b. How many kVar should the synchronous motor supply if used? ______________
c. What is the maximum number of kVars that can be supplied by the synchronous
motor? _______________
d. If this is insufficient, how many kVars should be supplied by capacitors if the
synchronous motor is used?____
e. How many kVar should be supplied by capacitors if the induction motor is used?
______________
f. Which system would have the lower cost?_____________
g. What additional issues should be considered in selection of the system?
_______________
h. Make a selection and support your decision (a short paragraph of explanation)

1. Solution:

a. The induced voltage for these operating conditions. _______________

The equation for induced voltage is

so first the current IA must be calculated using the operating conditions specified:

The angle, of course, must be calculated using the arccos of the power factor
angle. This is then substituted into the first equation to get:

b. The field current. _________________

The field current is calculate from the open circuit voltage characteristic, which
is dependent on the induced voltage just calculated. An assumption must be made
about which part of the curve the open circuit voltage will be located in to
calculate the current. If that proves to be incorrect, then the other equation must
be used. Because the open circuit characteristic is given in terms of terminal
voltage, the induced voltage must be multiplied by the square root of three:

c. The power factor if the generator input power is doubled without adjusting the
field current. (estimate based on an appropriate change in torque angle)
________________
Using the equation for power based on torque angle:

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 And recognizing that EA cannot change in magnitude unless the field current is
changed, a ratio may be used to calculate the new torque angle:

This information must then be used with the first equation introduced in part (a)
to calculate IA

from which the power factor angle may be taken to calculate the power factor:

d. The power factor if the field current is doubled without adjusting the generator
output power._____________
The approach here is similar to the approach to part (c). The open circuit
characteristic must be used to determine the effect of doubling the field current
on the magnitude of EA, then the torque angle must be determine based on the
power/torque characteristic. These are then used to determine the new IA:

e. The efficiency for the original case (80 MW, 85%pf)_____________

Efficiency is calculated from the output power and the sum of output power and
losses. The losses are given for windage and core, while the copper losses must

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 be calculated from the current and resistance:

f. The efficiency for the case in part (c)_________________

The approach is the same for as for part (e)

2. Solution :

a. How many kW electric would be drawn by either motor?

____________________
This is simply calculated by converting from horse power and taking account for
efficiency:

b. How many kVar should the synchronous motor supply if used? ______________
This will be calculated from the final power factor and the total of all load:

c. What is the maximum number of kVars that can be supplied by the synchronous
motor? _______________
This is calculated from the power factor limit given and the output power, similar
to part (b):

d. If this is insufficient, how many kVars should be supplied by capacitors if the

synchronous motor is used?____
The synchronous motor has sufficient capacity to supply the necessary kVars
e. How many kVar should be supplied by capacitors if the induction motor is used?
______________
This is similar to part (b) except that the "original" kVars must include those
required by the added induction motor, so that

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 f. Which system would have the lower cost?_____________
Sync Motor cost: $100,000 <--- Lower Cost
Ind Motor cost: $75,000+30*1079=$107,370
g. What additional issues should be considered in selection of the system?
_______________
Synchronous motor would provide constant speed despite widely varying load
that may be seen by a rock crusher. It would probably require more maintenance
because of the DC excitation system.
h. Make a selection and support your decision (a short paragraph of explanation)
I would choose the synchronous motor because of the lower initial cost and
probably more importantly because of the constant speed characteristic.
However, the induction motor could also be a viable option if supported with a
strong argument.

Example 2 : Questions & Answers

1. Give two advantages and two disadvantages of synchronous motors as

compared to induction motors. If you were required to specify either an induction
motor or a synchronous motor to drive a compressor in an industrial facility,
which would you choose and why?

Advantages: Constant speed, Controllable Power Factor

Disadvantages: No Starting torque, more expensive, more complex, higher maintenance

cost, fixed speed.

I would choose and induction motor, typically the compressor would be cycling on and
off, and have relatively high starting torque, so the synchronous motor would not be the
best.

2. A three-phase synchronous generator is to be specified for a small hydroelectric

generating station. The existing water turbine will provide up to 1.5 MW input to
the generator, when turning at its rated 300 rpm. The generator will be Y
connected, with its terminals connected to a 60 Hz system at 4160 V. The
generator will be intended to operate at 90% power factor. Make the assumption
that friction and windage losses are 38 kW and core losses are 43 kW.
a. How many poles will be required for this synchronous generator?

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b. Determine the armature resistance allowable for an efficiency of 90%, if
the terminal voltage is 4160 V and the machine is operating at 0.9 power
factor lagging.

This question takes a little consideration about the nature of efficiency. Based on
the assumption of 90% efficiency, the output power, and hence the output current
may be calculated. Also, based on the knowledge of the total losses in the
machine (based on efficiency), the stator copper losses will be the difference
between the known losses and the actual losses for that level of efficiency. Based
on this and the current magnitude, the armature resistance may be calculated.

c. If the synchronous reactance is 2.5 Ω , determine the induced voltage for

the same operating conditions.

d. What is the torque angle for these operating conditions?

The torque angle is the angle of the induced voltage, which is the angle calculated
in part c: 8.76o.

e. What is the stator current if the excitation current has been adjusted for
operation at unity power factor?

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3. A 208 V, 6 pole, 60 Hz, 15 hp, Y connected squirrel cage motor is being used to
drive a pump. It has the following equivalent circuit components:

R1 = 0.220 Ω R2 = 0.127 Ω jXM = j15.0 Ω

jX1 = j0.430 Ω jX2 = j0.430 Ω

4. For a slip of 0.05 (rated load) calculate:

a. Motor speed.

Speed is determined from slip, from line frequency, and from the number of
poles:

b. Line current.

This is calculated from the equivalent circuit of the induction machine with the
parameters given:

c. Stator copper losses.

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 d. Input power.

e. Machine efficiency.

Examples 3 : Questions & Answers.

1. Name 3 methods (features) for improved efficiency in rotating machinery design.

Any of the following are acceptable answers:

o Lower resistance windings (larger cross-section or superconducting)

o Laminated core (high quality magnetic steel)Reduces eddy and hysteresis losses
o Non-salient Pole rotor (reduces windage losses)
o Hydrogen filled (reduces windage losses)
o Magnetic bearings (reduces friction losses)
o Precision balanced rotor (reduces losses due to vibration)
2. A three-phase synchronous motor is to be specified for a pump at a water
treatment facility. The water turbine will require 975 hp, when turning at its rated
300 rpm. The motor will be Y connected, with its terminals connected to a 60 Hz
system at 4.16 kV. The motor is intended to operate at 90% power factor. Make
the assumption that friction and windage losses are 19 kW and core losses are
23 kW. If the synchronous reactance is 4.5 Ω and the armature resistance is 0.6
Ω,

a. Determine the efficiency, if the motor is operating at rated load and

voltage, and .9 power factor leading.

As always,
The output power is given in the problem statement and only needs to be

converted to SI units:
The calculation of the input power requires the knowledge of the input current to
calculate the copper losses, but the input power must be used to calculate the
input current, so either iteration or a quadratic equation must be used to calculate.
By Iteration:

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 By closed form quadratic solution:

b. Determine the induced voltage for the same operating conditions.

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 c. If other loads at the same facility amount to 2 MW at .85 pf lagging, what
is the total apparent power supplied to the entire facility?

2. A three phase 13.8 kV., 10 MVA, 12 pole, 60 Hz., Y connected three-phase

synchronous generator is supplying a load of 7.5 MW at 0.8 power factor lagging
and rated voltage. The stator impedance is (.25+j4.1)Ω . The machine is
operating at an efficiency of 94%.
a. What is the induced voltage?

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b. What is the rotational speed of the generator?

c. What is the magnitude of the field current?

From the graph, based on the open circuit terminal voltage, equivalent to the
induced phase voltage converted to a terminal value ( ) the field
current is read from the graph to be approximately 7 A.
d. What is the torque supplied by the prime mover?

e. What is the torque angle?

The torque angle δ is the angle of the induced voltage, EA: 7.76o.

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