Monday, 31 December 2012

DoorBell circuit using UM 66 IC


DoorBell circuit using UM 66 IC

Doorbell Circuit

Description

Here is a simple and easy to build doorbell circuit using IC UM 66.The  details of UM 66  is given in the older post Melody Generator using UM 66″. This is a slight modification of that circuit.In the previous circuit you have to keep the switch pressed for making the IC play the full music.Here if once the push button is pressed C1 is charged and the transistor Q2 will keep the IC playing the music till it ends.The time for the IC to play depends on discharging time of C1  which can be set by R1.Set R1 to select your time ,whether full tone or a part in one press.Transistor Q2 drives the speaker.

Doorbell Circuit Diagram with Parts List

Door bell circuit

Notes

  • Do not give more than 4.5 V to UM 66.
  • POT R2 is not necessary and can be bye passed .But if included you can set the loudness.
We have many other Tone Generator circuits that you may like take a look:
1. Tone Generator
2. High Power Siren Circuit
3. Melody Generator
4. Doorbell Circuit using 555 Timer
5. UJT Organ Circuit

Low cost / Automatic Emergency Light

Low cost / Automatic Emergency Light
author: Izhar Fareed - izhargmx.us - extremecircuits.net


 

Description

Here is a white-LED-based emergency light that offers the following advantages:

1. It is highly bright due to the use of white LEDs.
2. The light turns on automatically when mains supply fails, and turns off when mains power resumes.
3. It has its own battery charger. When the battery is fully charged, charging stops automatically.

The circuit comprises two sections: charger power supply and LED driver.The charger power supply section is
built around 3-terminal adjustable regulator (IC1) LM317, while the LED driver section is built around transistor BD140(T2). In the charger power supply section, input AC mains is stepped down by transformer to deliver 9V, 500mA to the bridge rectifier, which comprises diodes (IN4007x4). Filter capacitor (25v/1000uf)eliminates ripples. Unregulated DC voltage is fed to input pin 3 of IC1 and provides charging current through diode IN4007(D5) and limiting resistor (16ohm)R16. By adjusting preset 2.2K(VR1), the output voltage can be adjusted to deliver the required charging current. When the battery gets charged to 6.8V, zener diode conducts and charging current from regulator (IC1) finds a path through transistor BC547(T1) to ground and it stops charging of the battery. The LED driver section uses a total of twelve 10mm white LEDs. All the LEDs are connected in parallel with a 100-ohm resistor in series with each. The common-anode junction of all the twelve LEDs is connected to the collector of pnp transistor T2 and the emitter of transistor T2 is directly connected to the positive terminal of 6V battery. The unregulated DC voltage, produced at the cathode junction of Bridge(Diodes), is fed to the base of transistor T2 through a 1k resistor. When mains power is available, the base of transistor T2 remains high and T2 does not conduct. Thus LEDs are off. On the other hand, when mains fails, the base of transistor T2 becomes low and it conducts. This makes all the LEDs (LED1 through LED12) glow. The mains power supply, when available, charges the battery and keeps the LEDs off as transistor T2 remains cut-off. During mains failure, the charging section stops working and the battery supply makes the LEDs glow. Assemble the circuit on a general-purpose PCB and enclose in a cabinet with enough space for battery and switches. Mount the LEDs on the cabinet such that they light up the room. A hole in the cabinet should be drilled to connect 230V AC input for the primary of the transformer. I have tested the circuit with twelve 10mm white LEDs.You can use more LEDs provided the total current consumption does not exceed 1.5A. Driver transistor T2 can deliver up to 1.5A with proper heat-sink arrangement.

 
Download


low-cost-water-pump-controller-circuit.

Here the circuit diagram of low cost water pump controller. The automatic pump controller minimizes the need for any manual switching of water pumps installed for the functionality of pumping water from a reservoir to an overhead tank . It instantly switches on the pump once the water level within the tank falls below a specific low level (L), provided the water level in the reservoir is above a specific level (R). Subsequently, because the water level in the tank increases to an upper level (M), the pump is turned off instantly. The pump is turned on again only if the water level once again falls beneath level L in the tank, provided the level inside the reservoir is above R. This automated action
continues.


The circuit is intended to ‘overlook’ the transient oscillations of the water level that would otherwise trigger the logic to modify its state rapidly and unnecessarily. The circuit works by using a single CMOS chip (CD4001) for logic processing.
No utilization of any moving electromechanical elements within the water-level sensor has been made. This assures fast reaction, no wear and tear, and no mechanical problems. The circuit diagram can be seen on above image. The unit performed satisfactorily on a test run in conjunction with a 0.5 HP motor and pump.
The sensors applied to the circuit could be any two conducting probes, preferably resistant to electrolytic corrosion. For example, in the easiest case, a appropriately sealed audio jack could be utilized to operate as the sensor.
Low Cost Water Pump Controller circuit
The circuit may also be utilized like a constant fluid level maintainer. For this objective, the probes M and L are brought pretty near to one another to make sure that the fluid level is maintained within the M and L levels.
The benefit of this system is that it could be applied to tanks/reservoirs of any volume whatsoever. Even so, the circuit can’t be applied for purely non-conducting fluids. For non-conducting fluids, some modifications have to be prepared in the fluidlevel sensors. The circuit can on the other hand be kept intact.

Low cost Mic Mixer


Low cost Mic Mixer

Description
This is the schematics of a low cost mic mixer that can be assembled from the components in your junk box.This two channel mike mixer is designed for handling high impedance dynamic microphones.Transistor Q1 can be any general purpose pnp transistor like BC177 or BC157B. Resistor R5 biases the transistor while R3 and R4 provides channel isolation. How ever better the transistor quality better the performance.  C1 and C2 performs the job of input coupling and DC isolation while C4 is the output DC decoupling capacitor.Resistors R1 and R2 can be used for the level control of input signals.
Circuit Diagram With Parts List.
low cost mic mixer circuit
Notes
  • All capacitors are of ceramic type.
  • Always use a battery two reduce noise .
  • Avoid using power transistors for Q1. This is because power transistors have high leakage current and it adds to the noise which is not suitable for our purpose here

Low cost AM radio


Low cost AM radio

Description.
Here is the circuit diagram of a simple and low cost AM radio. The working of this radio circuit is straight forward. Inductor L1 and capacitor C1 forms the necessary tank circuit. Diode D1 performs the job of de-modulation. Transistor T1 is wired as a preamplifier whose output is further amplified by the IC LM386 to drive the head phone Z1. The 10K POT R3 can be used as a volume controller.
Circuit diagram.
low cost am radio
Notes.
  • The circuit can be powered from a 9V PP3 battery.
  • Use a 100cm long metal wire as antenna.
  • Experiment with the value of L1 for catching other bands.
  • The circuit can be assembled on a Vero board.

Emergency light mini project

This is an LDR based Emergency Lamp that turns on a High watt White LED when there is darkness in the room. It can be used as a simple emergency lamp in the child’s room to avoid the panic situation in the event a sudden power failure.It gives ample light in the room.

The circuit is too simple so that it can be enclosed in a small box. A 12 volt miniature battery is used to power the circuit. Two transistors T1 and T2

 are used as electronic switches to turn on / off the white LED. When there is sufficient light in the room, LDR conducts so that the base of the PNP transistor T1becomes high and it remains off. T2 also remains off since its base is grounded. In this state, White LED remains off. When the light falling on the LDR decreases, it cease to conduct and T1 forward bias providing base current to T2. It then turns on and White LED switches on.White LED used in the circuit is 1 watt High bright Luxeon LED . Since 1 watt White LED consumes around 300 milli ampere current, it is better to switch off the lamp after few minutes to conserve battery power

Electronics For u free download

http://www.ziddu.com/download/14018038/deekshith_efy_ckts_bundle.zip.html

Sunday, 30 December 2012

Anna university results

ANNA UNIVERSITY NOV/DEC 2012 3RD,5TH,7TH SEMESTER EXAMS FOR BOTH CREDIT SYSTEM AND MARK SYSTEM
HI FRIENDS UNLIKE THE PREVIOUS SEMESTER THIS TIME ONLY 3 CHANGES AND ADDITIONS ARE MADE SO FAR TO THE TIME TABLE. SO THE STUDENTS ARE FEELING LITTLE BETTER THAN THEIR PREVIOUS SEMESTERS.

ANNA UNIVERSITY BE/BTECH NOVEMBER/DECEMBER 2012 3RD,5TH,7TH SEMESTER RESULT


WHEN WILL BE THE ANNA UNIVERSITY NOVEMBER/DECEMBER 2012 RESULTS ANNOUNCED?

WE ALL KNOW THE LAST SEMESTER RESULTS ARE THE MOST DELAYED RESULTS. IT HAS MANY REASONS LIKE TIME TABLE CORRECTIONS, QUESTION PAPER LEAK, ETC..,

THIS SEMESTER UP TO NOW 3 CHANGES IN THE TIME TABLE ARE MADE.

IF THE EXAMS GO SMOOTHLY AS ON SCHEDULE THE RESULTS WILL BE ANNOUNCED SOONER THAN THE LAST SEMESTER.

THIS TIME THERE IS A CHANCE FOR THE FIRST YEAR FIRST SEMESTER JAN 2013 RESULT TO BE ANNOUNCED SEPARATELY.


THE RESULT FOR THE 2ND,3RD,4TH YEAR STUDENTS ARE EXPECTED TO BE ANNOUNCED IN THE MONTH OF JAN/FEB 



click here to read more: http://www.annaunivedu.in/2012/11/anna-university-novdec-2012-3rd5th7th.html#ixzz2GbjxisNH

Anna University even sem last working day

Anna University Academic Schedule for Even Semester Jan 2013 to May 2013

Anna University Academic Schedule for Even Semester 4th, 6th 8th Semester Reopening date, Last working Date, Date of Commencement of semester exam, Reopening date for Next semester


CENTRE FOR ACADEMIC COURSES ANNA UNIVERSITY: : CHENNAI — 600 025 ACADEMIC SCHEDULE for the January 2013 — May 2013 (EVEN SEMESTER) SESSION OF THE ACADEMIC YEAR 2012 — 2013 UG & PG DEGREE PROGRAMMES OFFERED IN AFFILIATED INSTITUTIONS  
RE - OPENING DAY FOR THE NEXT SEMESTER 01.07.2013  
NOTE: 1. The Theory and Practical Examination schedules which will be published in the due course by the Controller of Examinations, Anna University, Chennai should be followed. (Practical Examinations will be conducted before the commencement of theory examinations). 
If required Saturdays may also be made as working days. 

SI .
No.
Programme
Semester
Date of
Commencement
of Classes
Last working day
Date of
Commencement
of End Semester
Theory
Examinations
1.
B.E. B.Tech. (Full-time)
IV, VI, VIII
02.01.2013
03.05.2013
09.05.2013
2.
B.E. (Part-Time)
III,V,V11
02.01.2013
03.05.2013
09.05.2013
3.
B.Arch. (Full-Time)
IV,VI, VIII,X
02.01.2013
03.05.2013
09.05.2013
4.
M.E. M.Tech./ M. Arch. (FT/PT)
IV,VI
02.01.2013
03.05.2013
09.05.2013
5.
M.C.A. (Full-Time)
IV,V1
02.01.2013
03.05.2013
09.05.2013
6.
M.B.A. (FT/PT)
IV,V1
02.01.2013
03.05.2013
09.05.2013
7.
M.Sc ( 5 Yrs-Integrated)
IV,VI,VIII,X
02.01.2013
03.05.2013
09.05.2013
8.
M.Sc.(2 Yrs)
IV
02.01.2013
03.05.2013
09.05.2013
9.
B.Sc BCA
IV, VI
02.01.2013
03.05.2013
09.05.2013


click here to read more: http://www.annaunivedu.in/2012/12/anna-university-academic-schedule-for.html#ixzz2GbjTQ0KF
more question bank and prevoius year question paper 

Thursday, 27 December 2012

மு ஹம்மத்  ஆஷாருதீன்
மு ஹம்மத் அலி  தெரு 

Tuesday, 17 July 2012

inauguration function at Einstein college of engineering


                                           we welcome all the friends to attend the function.
This will be our function.EEE dept is the gethu dept,
go ZETA go ZETA
Date:21.07.2012
Time:3.00 to 5.00
place:Einstein college of engineering

Saturday, 18 February 2012

numerical methods

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NUMERICAL METHODS
QUESTION BANK
UNIT-5
PART-A
1. Define the local truncation error.
2. Write down the standard five point formula used in solving laplace equation U xx + U yy =
0 at the point ( i x, j y ).
3. Derive Crank-Niclson scheme.
4. State Bender Schmidt’s explicit formula for solving heat flow equations
5. Classify x 2 f xx + (1-y 2 ) f yy = 0
6. What is the truncation error of the central difference approximation of
y ' (x)?
7. What is the error for solving Laplace and Poissson’s equation by finite difference method.
8. Obtain the finite difference scheme fore the difference equations 2
2
2
dx
d y
+ y = 5.
9. Write dowm the implicit formula to solve the one dimensional heat equation.
10. Define the diagonal five point formula .
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1. Solve the equqtion Ut = U xx subject to condition u(x,0) = sin x ; 0 x 1,u(0,t) =
u(1,t) =0 using Crank- Nicholson method taking h = 1/3 k = 1/36(do on time step)
2. Solve U xx + U yy = 0 for the following square mesh with boundary values
1 2
1 4
2 5
4 5
3. Solve U xx = Utt with boundary condition u(0,t) = u(4,t) and the initial condition
ut (x,0) = 0 , u(x,0)=x(4-x) taking h =1, k = ½ (solve one period)
4. Solve xy II + y = 0 , y(1) =1,y(2) = 2, h = 0.25 by finite difference method.
5. Solve the boundary value problem xy II -2y + x = 0, subject to y(2) = 0 =y(3).Find
y(2.25),y(2.5),y(2.75).
6 . Solve the vibration problem
2
2
4
x
y
t
y
subject to the boundary conditions
y(0,t)=0,y(8,0)=0 and y(x,0)=
2
1
x(8-x).Find y at x=0,2,4,6.Choosing x = 2, t =
2
1
up
compute to 4 time steps.
7. Solve 2 u = -4(x + y) in the region given 0 x 4, 0 y 4. With all boundaries kept
at 0 0 and choosing x = y = 1.Start with zero vector and do 4 Gauss- Seidal iteration.
0 0 0 0 0 0 0 0 0 0
u1 u 2
u 3 u 4
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0 0
0 0
0 0
0 0 0 0 0 0 0 0 0 0
8. Solve u xx + u yy = 0 over the square mesh of sid e 4 units, satisfying the following
conditions .
u(x,0) =3x for 0 x 4
u(x, 4) = x 2 for 0 x 4
u(0,y) = 0, for 0 y 4
u(4,y) = 12+y for 0 y 4
9. Solve
t
u
x
u
2 2
2
= 0, given that u(0,t)=0,u(4.t)=0.u(x,0)=x(4-x).Assume h=1.Find
the values of u upto t =5.
10. Solve y tt = 4y xx subject to the condition y(0,t) =0, y(2,t)=o, y(x,o) = x(2-x),
(x,0) 0
t
y
. Do 4steps and find the values upto 2 decimal accuracy.

numerical methods

Powered By www.technoscriptz.com
NUMERICAL METHODS
QUESTION BANK
UNIT-5
PART-A
1. Define the local truncation error.
2. Write down the standard five point formula used in solving laplace equation U xx + U yy =
0 at the point ( i x, j y ).
3. Derive Crank-Niclson scheme.
4. State Bender Schmidt’s explicit formula for solving heat flow equations
5. Classify x 2 f xx + (1-y 2 ) f yy = 0
6. What is the truncation error of the central difference approximation of
y ' (x)?
7. What is the error for solving Laplace and Poissson’s equation by finite difference method.
8. Obtain the finite difference scheme fore the difference equations 2
2
2
dx
d y
+ y = 5.
9. Write dowm the implicit formula to solve the one dimensional heat equation.
10. Define the diagonal five point formula .
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1. Solve the equqtion Ut = U xx subject to condition u(x,0) = sin x ; 0 x 1,u(0,t) =
u(1,t) =0 using Crank- Nicholson method taking h = 1/3 k = 1/36(do on time step)
2. Solve U xx + U yy = 0 for the following square mesh with boundary values
1 2
1 4
2 5
4 5
3. Solve U xx = Utt with boundary condition u(0,t) = u(4,t) and the initial condition
ut (x,0) = 0 , u(x,0)=x(4-x) taking h =1, k = ½ (solve one period)
4. Solve xy II + y = 0 , y(1) =1,y(2) = 2, h = 0.25 by finite difference method.
5. Solve the boundary value problem xy II -2y + x = 0, subject to y(2) = 0 =y(3).Find
y(2.25),y(2.5),y(2.75).
6 . Solve the vibration problem
2
2
4
x
y
t
y
subject to the boundary conditions
y(0,t)=0,y(8,0)=0 and y(x,0)=
2
1
x(8-x).Find y at x=0,2,4,6.Choosing x = 2, t =
2
1
up
compute to 4 time steps.
7. Solve 2 u = -4(x + y) in the region given 0 x 4, 0 y 4. With all boundaries kept
at 0 0 and choosing x = y = 1.Start with zero vector and do 4 Gauss- Seidal iteration.
0 0 0 0 0 0 0 0 0 0
u1 u 2
u 3 u 4
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0 0
0 0
0 0
0 0 0 0 0 0 0 0 0 0
8. Solve u xx + u yy = 0 over the square mesh of sid e 4 units, satisfying the following
conditions .
u(x,0) =3x for 0 x 4
u(x, 4) = x 2 for 0 x 4
u(0,y) = 0, for 0 y 4
u(4,y) = 12+y for 0 y 4
9. Solve
t
u
x
u
2 2
2
= 0, given that u(0,t)=0,u(4.t)=0.u(x,0)=x(4-x).Assume h=1.Find
the values of u upto t =5.
10. Solve y tt = 4y xx subject to the condition y(0,t) =0, y(2,t)=o, y(x,o) = x(2-x),
(x,0) 0
t
y
. Do 4steps and find the values upto 2 decimal accuracy.

numerical methods

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MA1251 – NUMERICAL METHODS
UNIT – IV : INITIAL VALUE PROBLEMS FOR ORDINARY DIFFERENTIAL
EQUATIONS
PART – A
1. By Taylor series, find y(1.1) given y = x + y, y(1) = 0.
2. Find the Taylor series upto x3 term satisfying 2y y x 1, y(0) 1.
3. Using Taylor series method find y at x = 0.1 if 1, (0) 1 2 x y y
dx
dy
.
4. State Adams – Bashforth predictor and corrector formula.
5. What is the condition to apply Adams – Bashforth method ?
6. Using modified Euler’s method, find y(0.1) if , (0) 1 2 2 y x y
dx
dy
.
7. Write down the formula to solve 2nd order differential equation using Runge-
Kutta method of 4th order.
8. In the derivation of fourth order Runge-Kutta formula, why is it called fourth
order.
9. Compare R.K. method and Predictor methods for the solution of Initial value
problems.
10. Using Euler’s method find the solution of the IVP log(x y), y(0) 2
dx
dy
at x 0.2 taking h 0.2.
PART-B
11. The differential equation
2 y x
dx
dy
is satisfied
by y(0) 1, y(0.2) 1.12186, y(0.4) 1.46820, y(0.6) 1.7379.Compute the value
of y(0.8) by Milne’s predictor - corrector formula.
12. By means of Taylor’s series expension, find y at x = 0.1,and x = 0.2 correct to
three decimals places, given x y e
dx
dy
2 3 , y(0) = 0.
13. Given y xy y 0, y(0) 1, y (0) 0, find the value of y(0.1) by using
R.K.method of fourth order.
14. Using Taylor;s series method find y at x = 0.1, if 1 2 x y
dx
dy
, y(0)=1.
15. Given (1 ) 2 x y
dx
dy
, y(1) = 1, y(1.1) = 1.233, y(1.2) = 1.548, y(1.3)=1.979,
evaluate y(1.4) by Adam’s- Bashforth method.
16. Using Runge-Kutta method of 4th order, solve
2 2
2 2
y x
y x
dx
dy
with y(0)=1 at
x=0.2.
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17. Using Milne’s method to find y(1.4) given that 5 2 0 2 xy y given that
y(4) 1, y(4.1) 1.0049, y(4.2) 1.0097, y(4.3) 1.0143.
18. Given
, (0) 2, (0.2) 2.443214, (0.4) 2.990578, (0.6) 3.823516 3 x y y y y y
dx
dy
find y(0.8) by Milne’s predictor-corrector method taking h = 0.2.
19. Using R.K.Method of order 4, find y for x = 0.1, 0.2, 0.3 given that
, (0) 1 2 xy y y
dx
dy
also find the solution at x = 0.4 using Milne’s method.
20. Solve 2 y x
dx
dy
, y(0) = 1.
Find y(0.1) and y(0.2) by R.K.Method of order 4.
Find y(0.3) by Euler’s method.
Find y(0.4) by Milne’s predictor-corrector method.
21. Solve 0.1(1 ) 0 2 y y y y subject to y(0) 0, y (0) 1 using fourth order
Runge-Kutta Method.
Find y(0.2) and y (0.2) . Using step size x 0.2.
22. Using 4th order RK Method compute y for x = 0.1 given
2 1 x
xy
y given y(0) =
1 taking h=0.1.
23. Determine the value of y(0.4) using Milne’s method given , (0) 1 2 xy y y
dx
dy
,
use Taylors series to get the value of y at x = 0.1, Euler’s method for y at x = 0.2
and RK 4th order method for y at x=0.3.
24. Consider the IVP 1, (0) 0.5 2 y x y
dx
dy
(i) Using the modified Euler method, find y(0.2).
(ii) Using R.K.Method of order 4, find y(0.4) and y(0.6).
(iii) Using Adam- Bashforth predictor corrector method, find y(0.8).
25. Consider the second order IVP 2 2 int, 2 y y y e S t with y(0) = -0.4 and
y’(0)=-0.6.
(i) Using Taylor series approximation, find y(0.1).
(ii) Using R.K.Method of order 4, find y(0.2).

numerical methods

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MA1251 – NUMERICAL METHODS
UNIT – IV : INITIAL VALUE PROBLEMS FOR ORDINARY DIFFERENTIAL
EQUATIONS
PART – A
1. By Taylor series, find y(1.1) given y = x + y, y(1) = 0.
2. Find the Taylor series upto x3 term satisfying 2y y x 1, y(0) 1.
3. Using Taylor series method find y at x = 0.1 if 1, (0) 1 2 x y y
dx
dy
.
4. State Adams – Bashforth predictor and corrector formula.
5. What is the condition to apply Adams – Bashforth method ?
6. Using modified Euler’s method, find y(0.1) if , (0) 1 2 2 y x y
dx
dy
.
7. Write down the formula to solve 2nd order differential equation using Runge-
Kutta method of 4th order.
8. In the derivation of fourth order Runge-Kutta formula, why is it called fourth
order.
9. Compare R.K. method and Predictor methods for the solution of Initial value
problems.
10. Using Euler’s method find the solution of the IVP log(x y), y(0) 2
dx
dy
at x 0.2 taking h 0.2.
PART-B
11. The differential equation
2 y x
dx
dy
is satisfied
by y(0) 1, y(0.2) 1.12186, y(0.4) 1.46820, y(0.6) 1.7379.Compute the value
of y(0.8) by Milne’s predictor - corrector formula.
12. By means of Taylor’s series expension, find y at x = 0.1,and x = 0.2 correct to
three decimals places, given x y e
dx
dy
2 3 , y(0) = 0.
13. Given y xy y 0, y(0) 1, y (0) 0, find the value of y(0.1) by using
R.K.method of fourth order.
14. Using Taylor;s series method find y at x = 0.1, if 1 2 x y
dx
dy
, y(0)=1.
15. Given (1 ) 2 x y
dx
dy
, y(1) = 1, y(1.1) = 1.233, y(1.2) = 1.548, y(1.3)=1.979,
evaluate y(1.4) by Adam’s- Bashforth method.
16. Using Runge-Kutta method of 4th order, solve
2 2
2 2
y x
y x
dx
dy
with y(0)=1 at
x=0.2.
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17. Using Milne’s method to find y(1.4) given that 5 2 0 2 xy y given that
y(4) 1, y(4.1) 1.0049, y(4.2) 1.0097, y(4.3) 1.0143.
18. Given
, (0) 2, (0.2) 2.443214, (0.4) 2.990578, (0.6) 3.823516 3 x y y y y y
dx
dy
find y(0.8) by Milne’s predictor-corrector method taking h = 0.2.
19. Using R.K.Method of order 4, find y for x = 0.1, 0.2, 0.3 given that
, (0) 1 2 xy y y
dx
dy
also find the solution at x = 0.4 using Milne’s method.
20. Solve 2 y x
dx
dy
, y(0) = 1.
Find y(0.1) and y(0.2) by R.K.Method of order 4.
Find y(0.3) by Euler’s method.
Find y(0.4) by Milne’s predictor-corrector method.
21. Solve 0.1(1 ) 0 2 y y y y subject to y(0) 0, y (0) 1 using fourth order
Runge-Kutta Method.
Find y(0.2) and y (0.2) . Using step size x 0.2.
22. Using 4th order RK Method compute y for x = 0.1 given
2 1 x
xy
y given y(0) =
1 taking h=0.1.
23. Determine the value of y(0.4) using Milne’s method given , (0) 1 2 xy y y
dx
dy
,
use Taylors series to get the value of y at x = 0.1, Euler’s method for y at x = 0.2
and RK 4th order method for y at x=0.3.
24. Consider the IVP 1, (0) 0.5 2 y x y
dx
dy
(i) Using the modified Euler method, find y(0.2).
(ii) Using R.K.Method of order 4, find y(0.4) and y(0.6).
(iii) Using Adam- Bashforth predictor corrector method, find y(0.8).
25. Consider the second order IVP 2 2 int, 2 y y y e S t with y(0) = -0.4 and
y’(0)=-0.6.
(i) Using Taylor series approximation, find y(0.1).
(ii) Using R.K.Method of order 4, find y(0.2).

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Unit III
Differentiation and Integration
Part A
1. What the errors in Trapezoidal and Simpson’s rule.
2. Write Simpson’s 3/8 rule assuming 3n intervals.
3. Evaluate
1
1
4 1 x
dx
using Gaussian quadrature with two points.
4. In Numerical integration what should be the number of intervals to apply
Trapezoidal, Simpson’s 1/3 and Simpson’s 3/8.
5. Evaluate
1
1
4
2
1 x
x dx
using Gaussian three point quadrature formula.
6. State two point Gaussian quadratue formula to evaluate
1
1
f (x)dx .
7. Using Newton backward difference write the formula for first and second order
derivatives at the end value x = x0 upto fourth order.
8. Write down the expression for
dx
dy
and
2
2
dx
d y
at x = x0 using Newtons forward
difference formula.
9. State Simpson’s 1/3 and Simpson’s 3/8 formula.
10. Using trapezoidal rule evaluate
0
sin xdx by dividing into six equal parts.
Part B
1. Using Newton’s backward difference formula construct an interpolating polynomial
of degree three and hence find f(-1/3) given f(-0.75) = - 0.07181250, f(-0.5) =
- 0.024750, f(-0.25) = 0.33493750, f(0) = 1.10100.
2. Evaluate
x y
dxdy
1
by Simpson’s 1/3 rule with x y = 0.5 where 0<x,y<1.
3. Evaluate I =
2
1
2
1 x y
dxdy
by using Trapezoidal rule, rule taking h= 0.5 and h=0.25.
Hence the value of the above integration by Romberg’s method.
4. From the following data find y’(6)
X : 0 2 3 4 7 9
Y: 4 26 58 112 466 922
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5. Evaluate
2
1
2
1
2 2 x y
dxdy
numerically with h= 0.2 along x-direction and k = 0.25 along y
direction.
6. Find the value of sec (31) from the following data
(degree) : 31 32 33 34
Tan : 0.6008 0.6249 0.6494 0.6745
7. Find the value of x for which f(x) is maxima in the range of x given the following
table, find also maximum value of f(x).
X: 9 10 11 12 13 14
Y : 1330 1340 1320 1250 1120 930
8. The following data gives the velocity of a particle for 20 seconds at an interval of
five seconds. Find initial acceleration using the data given below
Time(secs) : 0 5 10 15 20
Velocity(m/sec): 0 3 14 69 228
9. Evaluate
7
3
2 1 x
dx
using Gaussian quadrature with 3 points.
10. For a given data find
dx
dy
and
2
2
dx
d y
at x = 1.1
X : 1.0 1.1 1.2 1.3 1.4 1.5 1.6
Y: 7.989 8.403 8.781 9.129 9.451 9.750 10.031

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Unit II : Interpolation and Approxiamtion
Part A
1. Construct a linear interpolating polynomial given the points (x0,y0) and (x1,y1).
2. Obtain the interpolation quadratic polynomial for the given data by using
Newton’s forward difference formula.
X : 0 2 4 6
Y : -3 5 21 45
3. Obtain the divided difference table for the following data.
X : -1 0 2 3
Y : -8 3 1 12
4. Find the polynomial which takes the following values.
X : 0 1 2
Y : 1 2 1
5. Define forward, backward, central differences and divided differences.
6. Evaluate 10 (1-x) (1-2x) (1-3x)--------(1-10x), by taking h=1.
7. Show that the divided difference operator is linear.
8. State the order of convergence of cubic spline.
9. What are the natural or free conditions in cubic spline.
10. Find the cubic spline for the following data
X : 0 2 4 6
Y : 1 9 21 41
11. State the properties of divided differences.
12. Show that
bcd a abcd
1
)
1
( 3 .
13. Find the divided differences of f(x) = x3 + x + 2 for the arguments 1,3,6,11.
14. State Newton’s forward and backward interpolating formula.
15. Using Lagranges find y at x = 2 for the following
X : 0 1 3 4 5
Y : 0 1 81 256 625
Part B
1. Using Lagranges interpolation formula find y(10) given that y(5) = 12, y(6) = 13,
y(9) = 14 and y(11) = 16.
2. Find the missing term in the following table
x : 0 1 2 3 4
y : 1 3 9 - 81
3. From the data given below find the number of students whose weight is between
60 to 70.
Wt (x) : 0-40 40-60 60-80 80-100 100-120
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No of
students : 250 120 100 70 50
4. From the following table find y(1.5) and y’(1) using cubic spline.
X : 1 2 3
Y : -8 -1 18
5. Given sin 450 = 0.7071, sin 500 = 0.7660, sin 550 = 0.8192, sin 600 = 0.8660, find
sin 520 using Newton’s forward interpolating formula.
6. Given log 10 654 = 2.8156, log 10 658 = 2.8182, log 10 659 = 2.8189, log 10 661 =
2.8202, find using Lagrange’s formula the value of log 10 656.
7. Fit a Lagrangian interpolating polynomial y = f(x) and find f(5)
x : 1 3 4 6
y : -3 0 30 132
8. Find y(12) using Newton’ forward interpolation formula given
x : 10 20 30 40 50
y : 46 66 81 93 101
9. Obtain the root of f(x) = 0 by Lagrange’s inverse interpolation given that f(30) = -30,
f(34) = -13, f(38) = 3, f(42) = 18.
10. Fit a natural cubic spline for the following data
x : 0 1 2 3
y : 1 4 0 -2
11. Derive Newton’s divided difference formula.
12. The following data are taken from the steam table:
Temp0 c : 140 150 160 170 180
Pressure : 3.685 4.854 6.502 8.076 10.225
Find the pressure at temperature t = 1420 and at t = 1750
13. Find the sixth term of the sequence 8,12,19,29,42.
14. From the following table of half yearly premium for policies maturing at different ages, estimate the premium for policies maturing at the age of 46.
Age x : 45 50 55 60 65
Premium y : 114.84 96.16 83.32 74.48 68.48
15. Form the divided difference table for the following data
x : -2 0 3 5 7 8
y : -792 108 -72 48 -144 -252

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NUMERICAL METHIODS
Unit I : Solution of equations and eigen value problems
Part A
1. If g(x) is continuous in [a,b] then under what condition the iterative method x =
g(x) has unique solution in [a,b].
2. Find inverse of A =
2 7
1 3
by Gauss – Jordan method.
3. Why Gauss Seidel iteration is a method of successive corrections.
4. Compare Gauss Jacobi and Gauss Siedel methods for solving linear system of
the form AX = B.
5. State the conditions for convergence of Gauss Siedel method for solving a
system of equations.
6. Compare Gaussian elimination method and Gauss-Jordan method.
7. What type of eigen value can be obtained using power method.
8. Find the dominant eigen value of A =
3 4
1 2
by power method.
9. How is the numerically smallest eigen value of A obtained.
10. State two difference between direct and iterative methods for solving system of
equations.
Part B
1. Find all the eigen values of the matrix
0 1 2
1 2 1
2 1 0
by power method (Apply
only 3 iterations).
2. Use Newton’s backward difference formula to construct an interpolating
polynomial of degree 3 for the data:
f( - 0.75) = - 0.0718125, f( - 0.5) = - 0.02475, f( - 0.25) = - 0.3349375 and f(0) =
1.101. Hence find f (-
3
1
).
6. Solve the system of equations using Gauss Seidel iterative methods.
20x – y – 2z = 17, 3x + 20y – z = -18, 2x – 3y +20z = 25.
7.Find the largest eigen values and its corresponding vector of the matrix
1 4 10
3 2 4
1 3 1
by power method.
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8. Using Gauss- Jordan obtain the inverse of the matrix
1 3 5
2 1 1
2 2 3
9. Using Gauss Seidel method solve the system of equations starting with the values
x = 1 , y = -2 and z = 3,
x + 3y + 5z = 173.61, x – 27y + 2z = 71.31, 41x – 2y + 3z = 65.46
10. Solve the following equations by Jacobi’s iteration method
x + y + z = 9, 2x – 3y + 4z = 13, 3x + 4y + 5z = 40.

Electrical machines 1

CHAPTER 1
PRINCIPLES OF OPERATION
OF SYNCHRONOUS MACHINES
The synchronous electrical generator (also called alternator) belongs to the
family of electric rotating machines. Other members of the family are the directcurrent
(dc) motor or generator, the induction motor or generator, and a number
of derivatives of all these three. What is common to all the members of this family
is that the basic physical process involved in their operation is the conversion
of electromagnetic energy to mechanical energy, and vice versa. Therefore, to
comprehend the physical principles governing the operation of electric rotating
machines, one has to understand some rudiments of electrical and mechanical
engineering.
Chapter 1 is written for those who are involved in operating, maintaining and
trouble-shooting electrical generators, and who want to acquire a better understanding
of the principles governing the machine’s design and operation, but
who do not have an electrical engineering background. The chapter starts by
introducing the rudiments of electricity and magnetism, quickly building up to
a description of the basic laws of physics governing the operation of the synchronous
electric machine, which is the type of machine all turbogenerators
belong to.
1.1 INTRODUCTION TO BASIC NOTIONS ON ELECTRIC POWER
1.1.1 Magnetism and Electromagnetism
Certain materials found in nature exhibit a tendency to attract or repeal each
other. These materials, called magnets, are also called ferromagnetic because
they include the element iron as one of their constituting elements.
Operation and Maintenance of Large Turbo Generators, by Geoff Klempner and Isidor Kerszenbaum
ISBN 0-471-61447-5 Copyright  2004 John Wiley & Sons, Inc.
3
4 PRINCIPLES OF OPERATION OF SYNCHRONOUS MACHINES
Magnets always have two poles: one called north; the other called south. Two
north poles always repel each other, as do two south poles. However, north and
south poles always attract each other. A magnetic field is defined as a physical
field established between to poles. Its intensity and direction determine the forces
of attraction or repulsion existing between the two magnets.
Figures 1.1 and 1.2 are typical representations of two interacting magnetic
poles, and the magnetic field established between them.
Magnets are found in nature in all sorts of shapes and chemical constitution.
Magnets used in industry are artificially made. Magnets that sustain their magnetism
for long periods of time are denominated “permanent magnets.” These are
widely used in several types of electric rotating machines, including synchronous
machines. However, due to mechanical, as well as operational reasons, permanent
magnets in synchronous machines are restricted to those with ratings much
lower than large turbine-driven generators, which is the subject of this book.
Turbine-driven generators (for short: turbogenerators) take advantage of the fact
that magnetic fields can be created by the flow of electric currents in conductors.
See Figure 1.3.
N S
Lines of Force
Fig. 1.1 Schematic representation of two magnetic poles of opposite polarity, and the
magnetic field between them shown as “lines of force.”
N N
Lines of Force
Fig. 1.2 Schematic representation of two north poles, and the magnetic field between
them. South poles will create similar field patterns, but the lines of force will point toward
the poles.
INTRODUCTION TO BASIC NOTIONS ON ELECTRIC POWER 5
Electric
Current
Conductor
Lines of
Force
Fig. 1.3 Schematic representation of a magnetic field created by the flow of current in
a conductor. The direction of the lines of force is given by the “law of the screwdriver”:
mentally follow the movement of a screw as it is screwed in the same direction as that
of the current; the lines of force will then follow the circular direction of the head of the
screw. The magnetic lines of force are perpendicular to the direction of current.
A very useful phenomenon is that, forming the conductor into the shape of a
coil can augment the intensity of the magnetic field created by the flow of current
through the conductor. In this manner, as more turns are added to the coil, the
same current produces larger and larger magnetic fields. For practical reasons
all magnetic fields created by current in a machine are generated in coils. See
Figure 1.4.
1.1.2 Electricity
Electricity is the flow of positive or negative charges. Electricity can flow in
electrically conducting elements (called conductors), or it can flow as clouds of
Current
Flow
Lines of Force
Fig. 1.4 Schematic representation of a magnetic field produced by the flow of electric
current in a coil-shaped conductor.
6 PRINCIPLES OF OPERATION OF SYNCHRONOUS MACHINES
(a )
(b )
Fig. 1.5 Electricity. (a) Ionic clouds of positive and negative currents. The positive
clouds are normally atoms that lost one or more electrons; the negative clouds are normally
free electrons. (b) The flow of electrons inside a conductor material, for example, copper.
ions in space or within gases. As it will be shown in later chapters, both types
of electrical conduction are found in turbogenerators. See Figure 1.5.
1.2 ELECTRICAL—MECHANICAL EQUIVALENCE
There is an interesting equivalence between the various parameters describing
electrical and mechanical forms of energy. People with either electrical or
mechanical backgrounds find this equivalence useful to the understanding of the
physical process in either form of energy. Figure 1.6 describes the various forms
of electrical-mechanical equivalence.
1.3 ALTERNATED CIRCUITS (AC)
As it will be shown later, alternators operate with both alternating (ac) and
direct-current (dc) electric power. The dc can be considered a particular case of
the general ac, with frequency equal to zero.
The frequency of an alternated circuit is measured by the number of times the
currents and/or voltages change direction (polarity) in a unit of time. The Hertz is
the universally accepted unit of frequency, and measures cycles per second. One
ALTERNATED CIRCUITS (AC) 7
Fig. 1.6 Electrical-mechanical equivalence.
Hz equals one cycle per second. Alternated currents and voltages encountered in
the world of industrial electric power are for all practical purposes of constant
frequency. This is important because periodic systems, namely systems that have
constant frequency, allow the currents and voltages to be represented by phasors.
A phasor is a rotating vector. The benefit of using phasors in electrical engineering
analysis is that it greatly simplifies the calculations required to solve
circuit problems.
Figure 1.7 depicts a phasor of magnitude E, and its corresponding sinusoidal
trace representing the instantaneous value of the quantity e. The magnitude E
represents the maximum value of e.
When a sinusoidal voltage is applied to a closed circuit, a current will flow in
it. After a while the current will have a sinusoidal shape (this is called the steadystate
current component) and the same frequency as the voltage. An interesting
8 PRINCIPLES OF OPERATION OF SYNCHRONOUS MACHINES
Voltage (e)
α
α
E (phasor)
ω
Fig. 1.7 A phasor E, that can represent the voltage impressed on a circuit. The phasor
is made of a vector with magnitude proportional to the magnitude of E, rotating at a
constant rotational speed ω. The convention is that phasors rotate counterclockwise. The
vertical projection of the phasor results in a sinusoidal representing the instantaneous
voltage e existing at any time. In the graph, α = ω × t, where t is the time elapsed from
its zero crossing.
phenomenon in periodic circuits is that the resulting angle between the applied
voltage and the current depends on certain characteristics of the circuit. These
characteristics can be classified as being resistive, capacitive, and inductive. The
angle between the voltage and the current in the circuit is called the power angle.
The cosine of the same angle is called the power factor of the circuit, or for short,
the PF.
Note: As it will be shown latter, in synchronous machines the term power angle
is used to identify a different concept. To avoid confusion, in this book the angle
between the current and the voltage in the circuit will therefore be identified by
the “power factor.”
In the case of a circuit having only resistances, the voltages and currents are
in phase, meaning the angle between them equals zero. Figure 1.8 shows the
various parameters encountered in a resistive circuit. It is important to note that
resistances have the property of generating heat when a current flows through
them. The heat generated equals the square of the current times the value of the
resistance. When the current is measured in amperes and the resistance in ohms,
the resulting power dissipated as heat is given in watts. In electrical machines this
heat represents a loss of energy. It will be shown later that one of the fundamental
requirements in designing an electric machine is the efficient removal of these
resistive losses, with the purpose of limiting the undesirable temperature rise of
the internal components of the machine.
In resistive circuits the instantaneous power delivered by the source to the load
equals the product of the instantaneous values of the voltage and the current.
When the same sinusoidal voltage is applied across the terminals of a circuit
with capacitive or inductive characteristics, the steady-state current will exhibit
an angular (or time) displacement vis-`a-vis the driving voltage. The magnitude
ALTERNATED CIRCUITS (AC) 9
Fig. 1.8 Alternating circuits (resistive). Schematic representation of a sinusoidal voltage
of magnitude E applied on a circuit with a resistive load R. The schematics shows the
resultant current i in phase with the voltage v. It also shows the phasor representation of
the voltage and current.
of the angle (or power factor) depends on how capacitive or inductive the load
is. In a purely capacitive circuit, the current will lead the voltage by 90◦, while
in a purely inductive one, the current will lag the voltage by 90◦ (see Fig. 1.9).
A circuit that has capacitive or inductive characteristics is referred to as being
a reactive circuit. In such a circuit, the following parameters are defined:
S: The apparent power → S = E × I, given in units of volt-amperes or VA.
P: The active power → P = E × I × cos ϕ, where ϕ is the power angle
of the circuit. P is given in units of watts.
Q: The reactive power → Q = E × I × sin ϕ, given in units of
volt-amperes-reactive or VAR.
The active power P of a circuit indicates a real energy flow. This is power that
may be dissipated on a resistance as heat, or may be transformed into mechanical
energy, as it will be shown later. However, the use of the word “power” in the
name of S and Q has been an unfortunate choice that has resulted in confounding
10 PRINCIPLES OF OPERATION OF SYNCHRONOUS MACHINES
Fig. 1.9 Alternating circuits (resistive–Inductive–Capacitive). Here the sinusoidal voltage
E is applied to a circuit comprised of resistive, capacitive, and inductive elements.
The resulting angle between the current and the voltage depends on the value of the
resistance, capacitance, and inductance of the load.
most individuals without an electrical engineering background for many years.
The fact is that apparent power and reactive power does not represent any measure
of real energy. They do represent the reactive characteristic of a given load
or circuit, and the resulting angle (power factor) between the current and voltage.
This angle between voltage and current significantly affects the operation of an
electric machine, as it will be discussed later.
For the time being let us define another element of ac circuit analysis: the
power triangle. From the relationships shown above among S, P, Q, E, I, and
ϕ, it can be readily shown that S, P, and Q form a triangle. By convention, Q is
shown as positive (above the horizontal), when the circuit is inductive, and vice
versa when capacitive (see Fig. 1.10).
THREE-PHASE CIRCUITS 11
Fig. 1.10 Definition of the “power triangle” in a reactive circuit.
1.4 THREE-PHASE CIRCUITS
The two-wire ac circuits shown above (called single-phase circuits or systems),
are commonly used in residential, commercial, and low voltage—low power
industrial applications. However, all electric power systems to which industrial
generators are connected are three-phase systems. Therefore any discussion in
this book about the “power system” will refer to a three-phase system. Moreover
in industrial applications the voltage supplies are, for all practical reasons,
balanced, meaning all three-phase voltages are equal in magnitude and apart by
120 electrical degrees. In those rare events where the voltages are unbalanced, its
implication into the operation of the generator will be discussed in other chapters
of this book.
Three-phase electric systems may have a fourth wire, called “neutral.” The
“neutral” wire of a three-phase system will conduct electricity if the source and/or
the load are unbalanced. In three-phase systems two sets of voltages and currents
can be identified. These are the phase and line voltages and currents.
Figure 1.11 shows the main elements of a three-phase circuit. Three-phase
circuits can have their sources and/or loads connected in wye (star) or in delta.
(See Fig. 1.12 for a wye-connected source feeding a delta-connected load.)
Almost without exception, turbine-driven generators have their windings connected
in wye (star). Therefore in this book the source (or generator) will be
shown wye-connected. There are a number of important reasons why turbogenerators
are star-connected. They have to do with considerations about its effective
12 PRINCIPLES OF OPERATION OF SYNCHRONOUS MACHINES
Fig. 1.11 Three phase systems. Schematic depiction of a three-phase circuit and the
vector (phasor) diagram representing the currents, voltages, and angles between them.
Y-Connected
Source
Δ-Connected
Load
∼ ∼

Fig. 1.12 A “Wye-connected” source feeding a “delta-connected” load.
protection as well as design (insulation, grounding, etc.). These will be discussed
in the chapters covering stator construction, and operations.
On the other hand, loads can be found connected in star, delta, or a combination
of the two. This book is not about circuit solutions; therefore the type of load
connection will not be brought up herein.
BASIC PRINCIPLES OF MACHINE OPERATION 13
1.5 BASIC PRINCIPLES OF MACHINE OPERATION
In Section 1.1, basic principles were presented showing how a current flowing
in a conductor produces a magnetic field. In this section three important laws of
electromagnetism will be presented. These laws, together with the law of energy
conservation, constitute the basic theoretical bricks on which the operation of
any electrical machine can be explained.
1.5.1 Faraday’s Law of Electromagnetic Induction
This basic law, due to the genius of the great English chemist and physicist
Michael Faraday (1791–1867), presents itself in two different forms:
1. A moving conductor cutting the lines of force (flux) of a constant magnetic
field has a voltage induced in it.
2. A changing magnetic flux inside a loop made from a conductor material
will induce a voltage in the loop.
In both instances the rate of change is the critical determinant of the resulting
differential of potential. Figure 1.13 illustrates both cases of electromagnetic
induction, and also provides the basic relationship between the changing flux and
the voltage induced in the loop, for the first case, and the relationship between
the induced voltage in a wire moving across a constant field, for the second case.
The figure also shows one of the simple rules that can be used to determine the
direction of the induced voltage in the moving conductor.
1.5.2 Ampere-Biot-Savart’s Law of Electromagnetic Induced Forces
This basic law is attributed to the French physicists Andre Marie Ampere (1775–
1836), Jean Baptiste Biot (1774–1862), and Victor Savart (1803–1862). In its
simplest form this law can be seen as the “reverse” of Faraday’s law. While
Faraday predicts a voltage induced in a conductor moving across a magnetic field,
the Ampere-Biot-Savart law establishes that a force is generated on a currentcarrying
conductor located in a magnetic field.
Figure 1.14 presents the basic elements of the Ampere-Biot-Savart’s law as
applicable to electric machines. The figure also shows the existing numerical relationships,
and a simple hand-rule to determine the direction of the resultant force.
1.5.3 Lenz’s Law of Action and Reaction
Both Faraday’s law and Ampere-Biot-Savart’s law neatly come together in Lenz’s
law written in 1835 by the Estonian-born physicist Heinrich Lenz (1804–1865).
Lenz’s law states that electromagnetic-induced currents and forces will try to
cancel the originating cause.
14 PRINCIPLES OF OPERATION OF SYNCHRONOUS MACHINES
Fig. 1.13 Both forms of Faraday’s basic law of electromagnetic induction. A simple
rule (the “right–hand” rule) is used to determine the direction of the induced voltage in
a conductor moving across a magnetic field at a given velocity.
For example, if a conductor is forced to move cutting lines of magnetic force,
a voltage is induced in it (Faraday’s law). Now, if the conductors’ ends are closed
together so that a current can flow, this induced current will produce (according
to Ampere-Biot-Savart’s law) a force acting upon the conductor. What Lenz’s
law states is that this force will act to oppose the movement of the conductor in
its original direction.
Here in a nutshell is the explanation for the generating and motoring modes
of operation of an electric rotating machine! This law explains why when a
generator is loaded (more current flows in its windings cutting the magnetic field
in the gap between rotor and stator), more force is required from the driving
turbine to counteract the induced larger forces and keep supplying the larger
load. Similarly Lenz’s law explains the increase in the supply current of a motor
as its load increases.
BASIC PRINCIPLES OF MACHINE OPERATION 15
Fig. 1.14 The Ampere-Biot-Savart law of electromagnetic induced forces as it applies
to electric rotating machines. Basic numerical relationships and a simple rule are used to
determine the direction of the induced force.
Figure 1.15 neatly captures the main elements of Lenz’s law as it applies to
electric rotating machines.
1.5.4 Electromechanical Energy Conversion
The fourth and final physical law that captures, together with the previous three,
all the physical processes occurring inside an electric machine, is the “principle
of energy conversion.” Within the domain of the electromechanical world of an
electric rotating machine, this principle states that:
All the electrical and mechanical energy flowing into the machine, less all the
electrical and mechanical energy flowing out the machine and stored in the machine,
equals the energy dissipated from the machine as heat.
It is important to recognize that while mechanical and electrical energy can
go in or out the machine, the heat generated within the machine always has a
16 PRINCIPLES OF OPERATION OF SYNCHRONOUS MACHINES
Fig. 1.15 The Lenz Law as it applies to electric rotating machines. Basic numerical
relationships and a simple rule are used to determine the direction of the induced forces
and currents.
Fig. 1.16 Principle of energy conversion, as applicable to electric rotating machines.
THE SYNCHRONOUS MACHINE 17
negative sign: namely heat generated in the machine is always released during
the conversion process. A plus sign indicates energy going in; a minus indicates
energy going out. In the case of the stored energy (electrical and mechanical), a
plus sign indicates an increase of stored energy, while a negative sign indicates
a reduction in stored energy.
The balance between the various forms of energy in the machine will determine
its efficiency and cooling requirements, both critical performance and construction
parameters in a large generator. Figure 1.16 depicts the principle of energy
conversion as applicable to electric rotating machines.
1.6 THE SYNCHRONOUS MACHINE
At this point the rudiments of electromagnetism have been presented, together
with the four basic laws of physics describing the inherent physical processes
coexisting in any electrical machine. Therefore it is the right time to introduce
the basic configuration of the synchronous machine, which, as mentioned before,
is the type of electric machine that all large turbine-driven generators belong to.
1.6.1 Background
The commercial birth of the alternator (synchronous generator) can be dated back
to August 24, 1891. On that day, the first large-scale demonstration of transmission
of ac power was carried out. The transmission extended from Lauffen,
Germany, to Frankfurt, about 110 miles away. The demonstration was carried out
during an international electrical exhibition in Frankfurt. This demonstration was
so convincing about the feasibility of transmitting ac power over long distances,
that the city of Frankfurt adopted it for their first power plant, commissioned in
1894. This happened about one hundred and eight years before the writing of
this book (see Fig. 1.17).
The Lauffen-Frankfurt demonstration—and the consequent decision by the
city of Frankfurt to use alternating power delivery—were instrumental in the
adoption by New York’s Niagara Falls power plant of the same technology. The
Niagara Falls power plant became operational in 1895. For all practical purposes
the great dc versus ac duel was over. Southern California Edison’s history book
reports that its Mill Creek hydro plant is the oldest active polyphase (three-phase)
plant in the United States. Located in San Bernardino County, California, its first
units went into operation on September 7, 1893, placing it almost two years
ahead of the Niagara Falls project. One of those earlier units is still preserved
and displayed at the plant.
It is interesting to note that although tremendous development in machine ratings,
insulation components, and design procedures has occurred now for over
one hundred years, the basic constituents of the machine have remained practically
unchanged.
18 PRINCIPLES OF OPERATION OF SYNCHRONOUS MACHINES
Fig. 1.17 The hydroelectric generator from Lauffen, now in the Deutches Museum,
Munich. (Reprinted with permission from The Evolution of the Synchronous Machine by
Gerhard Neidhofer, 1992, ABB)
Fig. 1.18 “Growth” graph, depicting the overall increase in size over the last century,
of turbine-driven generators.
The concept that a synchronous generator can be used as a motor followed
suit. Although Tesla’s induction motor replaced the synchronous motor as the
choice for the vast majority of electric motor applications, synchronous generators
remained the universal machines of choice for the generation of electric power.
The world today is divided between countries generating their power at 50 Hz
and others (e.g., the United States) at 60 Hz. Additional frequencies (e.g., 25 Hz)
can still be found in some locations, but they constitute the rare exception.
Synchronous generators have continuously grown in size over the years (see
Fig. 1.18). The justification is based on simple economies of scale: the output
THE SYNCHRONOUS MACHINE 19
rating of the machine per unit of weight increases as the size of the unit increases.
Thus it is not uncommon to see machines with ratings reaching up to 1500 MVA,
with the largest normally used in nuclear power stations. Interestingly enough, the
present ongoing shift from large steam turbines as prime movers to more efficient
gas turbines is resulting in a reverse of the trend toward larger and larger generators,
at least for the time being. Transmission system stability considerations
also place an upper limit on the rating of a single generator.
1.6.2 Principles of Construction
Chapter 2 includes a description of the design criteria leading to the construction
of a modern turbogenerator, as well as contains a detailed description of all
components most commonly found in such a machine. This section is limited
to the presentation of the basic components comprising a synchronous machine,
with the purpose of describing its basic operating theory.
Synchronous machines come in all sizes and shapes, from the miniature
permanent magnet synchronous motor in wall-clocks, to the largest steam-turbinedriven
generators of up to about 1500 MVA. Synchronous machines are one of
two types: the stationary field or the rotating dc magnetic field.
The stationary field synchronous machine has salient poles mounted on the
stator—the stationary member. The poles are magnetized either by permanent
magnets or by a dc current. The armature, normally containing a three-phase
winding, is mounted on the shaft. The armature winding is fed through three
sliprings (collectors) and a set of brushes sliding on them. This arrangement can
be found in machines up to about 5 kVA in rating. For larger machines—all those
covered in this book—the typical arrangement used is the rotating magnetic field.
The rotating magnetic field (also known as revolving-field) synchronous
machine has the field-winding wound on the rotating member (the rotor ), and the
armature wound on the stationary member (the stator). A dc current, creating a
magnetic field that must be rotated at synchronous speed, energizes the rotating
field-winding. The rotating field winding can be energized through a set of slip
rings and brushes (external excitation), or from a diode-bridge mounted on the
rotor (self-excited). The rectifier-bridge is fed from a shaft-mounted alternator,
which is itself excited by the pilot exciter. In externally fed fields, the source can
be a shaft-driven dc generator, a separately excited dc generator, or a solid-state
rectifier. Several variations to these arrangements exist.
The stator core is made of insulated steel laminations. The thickness of the laminations
and the type of steel are chosen to minimize eddy current and hysteresis
losses, while maintaining required effective core length and minimizing costs. The
core is mounted directly onto the frame or (in large two-pole machines) through
spring bars. The core is slotted (normally open slots), and the coils making the
winding are placed in the slots. There are several types of armature windings,
such as concentric windings of several types, cranked coils, split windings of
various types, wave windings, and lap windings of various types. Modern large
machines typically are wound with double-layer lap windings (more about these
winding types in Chapter 2).
20 PRINCIPLES OF OPERATION OF SYNCHRONOUS MACHINES
The rotor field is either of salient-pole (Fig. 1.19) or non-salient-pole construction,
also known as round rotor or cylindrical rotor (Fig. 1.20). Non-salient-pole
(cylindrical) rotors are utilized in two- or four-pole machines, and, very seldom,
in six-pole machines. These are typically driven by steam or combustion turbines.
The vast majority of salient-pole machines have six or more poles. They include
all synchronous hydrogenerators, almost every synchronous condenser, and the
overwhelming majority of synchronous motors.
Non-salient-pole rotors are typically machined out of a solid steel forging. The
winding is placed in slots machined out of the rotor body and retained against
the large centrifugal forces by metallic wedges, normally made of aluminum or
steel. The retaining rings restrain the end part of the windings (end-windings).
In the case of large machines, the retaining rings are made out of steel.
Large salient-pole rotors are made of laminated poles retaining the winding
under the pole head. The poles are keyed onto the shaft or spider-and-wheel
Fig. 1.19 Synchronous machine construction. Schematic cross section of a salient-pole
synchronous machine. In a large generator, the rotor is magnetized by a coil wrapped
around it. The figure shows a two-pole rotor. Salient-pole rotors normally have many more
than two poles. When designed as a generator, large salient-pole machines are driven by
water turbines. The bottom part of the figure shows the three-phase voltages obtained
at the terminals of the generator, and the equation relates the speed of the machine, its
number of poles, and the frequency of the resulting voltage.
THE SYNCHRONOUS MACHINE 21
Fig. 1.20 Schematic cross section of a synchronous machine with a cylindrical
round-rotor (turbogenerator). This is the typical design for all large turbogenerators. Here
both the stator and rotor windings are installed in slots, distributed around the periphery
of the machine. The lower part shows the resulting waveforms of a pair of conductors,
and that of a distributed winding. The formula giving the magneto-motive force (mmf)
created by the windings.
structure. Salient-pole machines have an additional winding in the rotating member.
This winding, made of copper bars short-circuited at both ends, is embedded
in the head of the pole, close to the face of the pole. The purpose of this winding
is to start the motor or condenser under its own power as an induction motor, and
take it unloaded to almost synchronous speed, when the rotor is “pulled in” by
the synchronous torque. The winding also serves to damp the oscillations of the
rotor around the synchronous speed, and is therefore named the damping-winding
(also known as amortisseurs or damper-windings).
This book focuses on large turbine-driven generators. These are always twoor
four-pole machines, having cylindrical rotors. The discussion of salient-pole
machines can be found in other books. (See the Additional Reading section at
the end of this chapter.)
22 PRINCIPLES OF OPERATION OF SYNCHRONOUS MACHINES
1.6.3 Rotor Windings
In turbogenerators, the winding producing the magnetic field is made of a number
of coils, single-circuit, energized with dc power fed via the shaft from the
collector rings riding on the shaft and positioned outside the main generator
bearings. In self-excited generators, shaft-mounted exciter and rectifier (diodes)
generate the required field current. The shaft-mounted exciter is itself excited
from a stationary winding. The fact that unlike the stator, the rotor field is fed
from a relatively low power, low voltage circuit has been the main reason why
these machines have the field mounted on the rotating member and not the other
way around. Moving high currents and high power through the collector rings and
brushes (with a rotating armature) would represent a serious technical challenge,
making the machine that much more complex and expensive.
Older generators have field supplies of 125 volts dc. Later ones have supplies
of 250 volts and higher. Excitation voltages of 500 volts or higher are common
in newer machines. A much more elaborated discussion of rotor winding design
and construction can be found in Chapter 2.
1.6.4 Stator Windings
The magnitude of the voltage induced in the stator winding is, as shown above,
a function of the magnetic field intensity, the rotating speed of the rotor, and the
number of turns in the stator winding. An actual description of individual coil
design and construction, as well as how the completed winding is distributed
around the stator, is meticulously described in Chapter 2. In this section a very
elementary description of the winding arrangement is presented to facilitate the
understanding of the basic operation of the machine.
As stated above, coils are distributed in the stator in a number of forms.
Each has its own advantages and disadvantages. The basic goal is to obtain three
balanced and sinusoidal voltages having very little harmonic content (harmonic
voltages and currents are detrimental to the machine and other equipment in a
number of ways). To achieve a desired voltage and MVA rating, the designer may
vary the number of slots, and the manner in which individual coils are connected,
producing different winding patterns. The most common winding arrangement is
the lap winding, and it shown in Figure 1.21.
A connection scheme that allows great freedom of choice in designing the
windings to accommodate a given terminal voltage is one that allows connecting
sections of the winding in parallel, series, and/or a combination of the two.
Figure 1.22 shows two typical winding arrangements for a four-pole generator.
1.7 BASIC OPERATION OF THE SYNCHRONOUS MACHINE
For a more in-depth discussion of the operation and control of large turbogenerators,
the reader is referred to Chapter 4. In this chapter the most elementary
principles of operation of synchronous machines will be presented. As it was
BASIC OPERATION OF THE SYNCHRONOUS MACHINE 23
1 N 2 S 3 N 4 S 1 N
Stator Slots
Fig. 1.21 “Developed” view of a four-pole stator, showing the slots, the poles, and a
section of the winding. The section shown is of one of the three phases. It can be readily
seen that the winding runs clockwise under a north pole, and counterclockwise under a
south pole. This pattern repeats itself until the winding covers the four poles. A similar
pattern is followed by the other two phases, but located at 120 electrical degrees apart.
Fig. 1.22 Schematic view of a two-pole generator with two possible winding configurations:
(1) A two parallel circuits winding, (2) A two series connected circuits per phase.
On the right, the three phases are indicated by different tones. Note that, some slots only
have coils belonging to the same phase, while in others, coils belonging to two phases
share the slot.
24 PRINCIPLES OF OPERATION OF SYNCHRONOUS MACHINES
mentioned above, all large turbogenerators are three-phase machines. Thus the
best place to start describing the operation of a three-phase synchronous machine
is a description of its magnetic field.
Earlier we described how a current flowing through a conductor produces a
magnetic field associated with that current. It was also shown that by coiling the
conductor, a larger field is obtained without increasing the current’s magnitude.
Recall that if the three phases of the winding are distributed at 120 electrical degrees
apart, three balanced voltages are generated, creating a three-phase system.
Now a new element can be brought into the picture. By a simple mathematical
analysis it can be shown that if three balanced currents (equal magnitudes and
120 electrical degrees apart) flow in a balanced three-phase winding, a magnetic
field of constant magnitude is produced in the airgap of the machine. This magnetic
field revolves around the machine at a frequency equal to the frequency of
the currents flowing through the winding (see Fig. 1.23). The importance of a
three-phase system creating a constant field cannot be stressed enough. The constant
magnitude flux allows hundred of megawatts of power to be transformed
Fig. 1.23 Production of stator rotating field. A constant magnitude and constant rotational
speed magnetic flux is created when three-phase balanced currents flow through a
three-phase symmetrical winding. In a two-pole winding, however, any the same result
applies for any number of pairs of poles.
BASIC OPERATION OF THE SYNCHRONOUS MACHINE 25
inside an electric machine from electrical to mechanical power, and vice versa,
without major mechanical limitations. It is important to remember that a constantmagnitude
flux produces a constant-magnitude torque. Now try to imagine the
same type of power being transformed under a pulsating flux (and therefore
pulsating torque), which is tremendously difficult to achieve.
It is convenient to introduce the fundamental principles describing the operation
of a synchronous machine in terms of an ideal cylindrical-rotor machine
connected to an infinite bus. The infinite bus represents a busbar of constant
voltage, which can deliver or absorb active and reactive power without any limitations.
The ideal machine has zero resistance and leakage reactance, infinite
permeability, and no saturation, as well as zero reluctance torque.
The production of torque in the synchronous machine results from the natural
tendency of two magnetic fields to align themselves. The magnetic field produced
by the stationary armature is denoted as φs. The magnetic field produced by the
rotating field is φf. The resultant magnetic field is
φr = φs + φf
The flux φr is established in the airgap (or gasgap) of the machine. (Bold symbols
indicate vector quantities.)
When the torque applied to the shaft equals zero, the magnetic fields of the
rotor and the stator become perfectly aligned. The instant torque is introduced
to the shaft, either in a generating mode or in a motoring mode, a small angle
is created between the stator and rotor fields. This angle (λ) is called the torque
angle of the machine.
1.7.1 No-Load Operation
When the ideal machine is connected to an infinite bus, a three-phase balanced
voltage (V1) is applied to the stator winding (within the context of this work,
three-phase systems and machines are assumed). As described above, it can
be shown that a three-phase balanced voltage applied to a three-phase winding
evenly distributed around the core of an armature will produce a rotating
(revolving) magneto-motive force (mmf) of constant magnitude (Fs). This mmf,
acting upon the reluctance encountered along its path, results in the magnetic
flux (φs) previously introduced. The speed at which this field revolves around
the center of the machine is related to the supply frequency and the number of
poles, by the following expression:
ns = 120

f
p

where
f = electrical frequency in Hz
p = number of poles of the machine
ns = speed of the revolving field in revolutions per minute (rpm)
26 PRINCIPLES OF OPERATION OF SYNCHRONOUS MACHINES
S N φS
φF
φF
φR
φR
φF
φR
φS
φF
φF
φR
φS
ϕ1
V1 E1
V1 E1 V1 E1 E1
S
N
φS
φF
δ S
N
φS
φF
δ
φF
φR
φS
φS
φS
φS
δ
λ
ϕ1
I1
(a ) Underexcited
Lagging
Leading
Lagging
Generator
operation
Motor
operation
No-load or condenser
operation
V1 E1
φS
φR
φF
λ
ϕ1
ϕ1
ϕ1
ϕ1
Leading
−V1 E1
−V1
δ
δ
δ
I1
I1
I1
I1
V1
I1
λ
λ
(b ) Overexcited
Fig. 1.24 Phasor diagrams for a synchronous cylindrical-rotor ideal machine.
If no current is supplied to the dc field winding, no torque is generated, and
the resultant flux (φr), which in this case equals the stator flux (φs), magnetizes
the core to the extent the applied voltage (V1) is exactly opposed by a counterelectromotive
force (cemf) (E1). If the rotor’s excitation is slightly increased,
and no torque is applied to the shaft, the rotor provides some of the excitation
required to produce (E1), causing an equivalent reduction of (φs). This situation
represents the underexcited condition shown in condition no load (a) in
Figure 1.24. When operating under this condition, the machine is said to behave
as a lagging condenser, meanings it absorbs reactive power from the network.
If the field excitation is increased over the value required to produce (E1), the
stator currents generate a flux that counteracts the field-generated flux. Under
BASIC OPERATION OF THE SYNCHRONOUS MACHINE 27
this condition, the machine is said to be overexcited, shown as condition no load
(b) in Figure 1.24. The machine is behaving as a leading condenser; that is, it is
delivering reactive power to the network.
Under no-load condition both the torque angle (λ) and the load angle (δ) are
zero. The load angle is defined as the angle between the rotor’s mmf (Ff) or
flux (φf) and the resultant mmf (Fr) or flux (φr). The load angle (δ) is the most
commonly used because it establishes the torque limits the machine can attain
in a simple manner (discussed later). One must be aware that in many texts the
name torque angle is used to indicate the load angle. The name torque angle
is also sometimes given to indicate the angle between the terminal voltage (V1)
and the excitation voltage (E1). This happens because the leakage reactance is
generally very much smaller than the magnetizing reactance, and therefore the
load angle (δ) and the angle between (V1) and (E1) are very similar. In this
book the more common name power angle is used for the angle between (V1)
and (E1). In Figure 1.24, the power angle is always shown as zero because the
leakage impedance has been neglected in the ideal machine.
It is important at this stage to introduce the distinction between electrical and
mechanical angles. In studying the performance of the synchronous machine,
all the electromagnetic calculations are carried out based on electric quantities;
that is, all angles are electrical angles. To convert the electrical angles used in
the calculations to the physical mechanical angles, we observe the following
relationship:
Mechanical angle =

2
p

Electrical angle
1.7.2 Motor Operation
The subject of this book is turbogenerators. These units seldom operate as a
motor. (One such example is when the main generator is used for a short period
of time as a motor fed from a variable speed converter. The purpose of this
operation is for starting its own prime-mover combustion turbine). However, this
section presents an introductory discussion of the synchronous machine, and thus
the motor mode of operation is also covered. If a breaking torque is applied to
the shaft, the rotor starts falling behind the revolving-armature-induced magnetomotive
force (mmf) (Fs). In order to maintain the required magnetizing mmf (Fr)
the armature current changes. If the machine is in the underexcited mode, the
condition motor in Figure 1.24a represents the new phasor diagram.
On the other hand, if the machine is overexcited, the new phasor diagram
is represented by motor in Figure 1.24b. The active power consumed from the
network under these conditions is given by
Active power = V1 × I1 × cos ϕ1 (per phase)
If the breaking torque is increased, a limit is reached in which the rotor
cannot keep up with the revolving field. The machine then stalls. This is known
28 PRINCIPLES OF OPERATION OF SYNCHRONOUS MACHINES
as “falling out of step,” “pulling out of step,” or “slipping poles.” The maximum
torque limit is reached when the angle δ equals π/2 electrical. The convention is to
define δ as negative for motor operation and positive for generator operation. The
torque is also a function of the magnitude of φr and φf. When overexcited, the
value of φf is larger than in the underexcited condition. Therefore synchronous
motors are capable of greater mechanical output when overexcited. Likewise,
underexcited operation is more prone to result in an “out-of-step” situation.
1.7.3 Generator Operation
Let’s assume that the machine is running at no load and a positive torque is
applied to the shaft; that is, the rotor flux angle is advanced ahead of the stator
flux angle. As in the case of motor operation, the stator currents will change to
create the new conditions of equilibrium shown in Figure 1.24, under generator.
If the machine is initially underexcited, condition (a) in Figure 1.24 obtains. On
the other hand, if the machine is overexcited, condition (b) in Figure 1.24 results.
It is important to note that when “seen” from the terminals, with the machine
operating in the underexcited mode, the power factor angle (ϕ1) is leading (i.e., I1
leads V1). This means the machine is absorbing reactive power from the system.
The opposite occurs when the machine is in the overexcited mode. As for the
motor operation, an overexcited condition in the generating mode also allows for
greater power deliveries.
As generators are normally called to provide VARs together with watts, they
are almost always operated in the overexcited condition.
1.7.4 Equivalent Circuit
When dealing with three-phase balanced circuits, electrical engineers use the
one-line or single-line representation. This simplification is allowed because in
three-phase balanced circuits, all currents and voltages, as well as circuit elements
are symmetrical. Thus, “showing” only one phase, it is possible to represent the
three-phase system, as long as care is taken in using the proper factors. For
instance, the three-phase balanced system of Figure 1.11 or Figure 1.12 can be
represented as shown in Figure 1.25. Hereinafter, when describing a three-phase
generator by an electrical diagram, the one-line method will be applied.
The most convenient way to determine the performance characteristics of synchronous
machines is by means of equivalent circuits. These equivalent circuits
Generator
Line
Load

Fig. 1.25 One-line representation of circuit shown in Figure 1.10 and 1.11.
BASIC OPERATION OF THE SYNCHRONOUS MACHINE 29
Machine
terminals
E1 V1(Vt)
Zs
Xa Ra Ia
X
Em
+

Fig. 1.26 Steady-state equivalent circuit of a synchronous machine. X = leakage reactance,
Xa = armature reaction reactance, Xs = Xa + X = synchronous reactance, Ra =
armature resistance, Zs = synchronous impedance, V1(Vt ) = terminal voltages, and Em =
magnetizing voltage.
can become very elaborate when saturation, armature reaction, harmonic reactance,
and other nonlinear effects are introduced. However, the simplified circuit
in Figure 1.26 is conducive to obtaining the basic performance characteristics of
the machine under steady-state conditions.
In Figure 1.26 the reactance Xa represents the magnetizing or demagnetizing
effect of the stator windings on the rotor. It is also called the magnetizing
reactance. Ra represents the effective resistance of the stator. The reactance X
represents the stator leakage reactance. The sum of Xa and X is used to represent
the total reactance of the machine, and is called the synchronous reactance (Xs).
Zs is the synchronous impedance of the machine. It is important to remember
that the equivalent circuit described in Figure 1.26 represents the machine only
under steady-state condition.
The simple equivalent circuit of Figure 1.27 (a) suffices to determine the
steady-state performance parameters of the synchronous machine connected to
a power grid. These parameters include voltages, currents, power factor, and
load angle (see Fig. 1.27b). The regulation of the machine can be easily found
from the equivalent circuit for different load conditions by using the regulation
formula:
(%) = 100 × Vno−load − Vload/Vload
For a detailed review of the performance characteristics of the synchronous
machine, in particular the turbogenerator, the reader is referred to Chapter 4.
Note: Regulation in a generator indicates how the terminal voltage of the machine
varies with changes in load. When the generator is connected to an infinite bus
(i.e., a bus that does not allow the terminal voltage to change), a change in
30 PRINCIPLES OF OPERATION OF SYNCHRONOUS MACHINES
I
E
E
E I E
E
Lagging power factor (overexcited)
(a) Generator operation
Leading power factor (underexcited)
Leading power factor (overexcited)
(b) Motor operation
Lagging power factor (underexcited)
IZ
IZ
IZ
I
I
I
δ
δ
δ
φ
φ
φ
φ
V V
V
IZ
V
V E − IZ = V
Z
I
E V
Z
Fig. 1.27 Steady-state equivalent circuit and vector diagram.
load will affect the machine’s output in a number of ways. (See Chapter 4 for a
discussion of this topic.)
1.7.5 Machine Losses
In Item 1.5.4 above the balance of energy in an electric machine was discussed.
As part of the discussion reference was made to the fact that the current that flow
through the machine’s conductors generate heating (a loss). However, there are
a number of other sources within a working alternator that produced heat and,
thus, losses. The following is a list of those sources of losses. In the following
chapters these losses, their origin, control, and consequences to the machine’s
design and operation will be covered in detail.
ADDITIONAL READING 31
Machine Losses.
Winding Losses (Copper Losses).
ž I 2R stator loss
ž I 2R rotor loss
ž Eddy and circulating current loss in winding (parasitic currents induced in
the windings)
Iron Losses.
ž Mainly stator losses due to hysteresis loss and eddy current loss in stator
laminations
Parasitic Eddy Losses.
ž Induced currents in all metallic component (bolts, frame, etc.)
ž Friction and windage loss
ž Losses in fans, rotor and stator cooling vents
ž Losses in bearings
Exogenous Losses.
ž Losses in auxiliary equipment
Excitation
Lubrication oil pumps
H2 seal oil pumps
H2 and water cooling pumps
And so on . . .
ž Iso-phase or lead losses
ADDITIONAL READING
A wealth of literature exists for the reader interested in a more in-depth understanding
of synchronous machine theory. The following is only but a very short list of classic
textbooks readily available describing the operation and design of synchronous machines
in a manner accessible to the uninitiated.
1. D. Zorbas, Electric Machines-Principles, Applications, and Control Schematics. West,
1989.
2. M. G. Say, Alternating Current Machines. Pitman Publishing, 1978.
3. T. Wildi, Electrical Machines, Drives and Power Systems. Prentice Hall.
4. V. del Toro, Electric Machines and Power Systems. Prentice Hall, 1985.
5. M. Liwschitz-Garik and C. C. Whipple, Electric Machinery, Vols. 1–2. Van Nostrand.
32 PRINCIPLES OF OPERATION OF SYNCHRONOUS MACHINES
6. A. E. Fitzgerald and C. Kingsley, Electric Machinery. McGraw-Hill, 1971.
7. For a text describing the practical issues related to operation and maintenance of
both turbogenerators and hydrogenerators, see Isidor Kerszenbaum, Inspection of Large
Synchronous Machines. IEEE-Press, 1996.