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

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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

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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).

numerical methods

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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

numerical methods

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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 Ï•, where Ï• is the power angle
of the circuit. P is given in units of watts.
Q: The reactive power → Q = E × I × sin Ï•, 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
Ï•, 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
Ï•1
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
Ï•1
I1
(a ) Underexcited
Lagging
Leading
Lagging
Generator
operation
Motor
operation
No-load or condenser
operation
V1 E1
φS
φR
φF
λ
Ï•1
Ï•1
Ï•1
Ï•1
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 Ï•1 (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 (Ï•1) 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.

Friday, 17 February 2012

interview question for electrical engineers

                                         Interview questions
Q:Difference between megger test equipment and contact resistance meter test instruments?
A: Megger test equipment used to measure cable electric resistance, conductor continuity, phase identification where as contact resistance meter test instruments used to measure low resistance like relays ,contactors.

Q:When we connect the large capacitor bank in series ?
A :we connect large capacitor bank in series to improve the voltage power supply at the load end in balanced transmission line when there is considerable voltage drop along the balanced transmission line due to high impedance of the line.So in order to bring the voltage at the load terminals within its limits (i.e (+ or - %6 )of the rated high terminal voltage )the large capacitor bank is used in series.

Q:What is electrical diversity factor in electrical installations?
A :Electrical diversity factor is the ratio of the sum of the individual maximum demands of the various subdivisions of a system, or part of a system, to the maximum demand of the whole system, or part of the system, under consideration. Electrical diversity factor is usually more than one.

Q:Why field rheostat is kept in minimum position while armature rheostat at maximum position?
A :In motors at the time of starting the armature resistance is introduced to reduce the high starting current and the field resistance is kept minimum to have high starting
torque.

Q:Why computer humming sound occurred in HT transmission line?
A :This computer humming sound is coming due to ionization (breakdown of air into charged particles) of air around transmission conductor. This effect is called as Corona effect, and it is considered as power loss.

Q:What is rated speed?
A :At the time of motor taking normal current (rated current)the speed of the motor is called rated speed. It is a speed at which any system take small current and give maximum efficiency.

Q:What is different between resistance grounding system and resistance earthing system?
A :Resistance grounding system means connecting the neutral point of the load to the ground to carry the residual current in case of unbalanced conditions through the neutral to the ground whereas resistance earthing system is done in an electric equipment in order to protect he equipment in occurence of fault in the system.

Q:Why should be the frequency 50 Hz 60Hz only why not others like 45, 95 56 or anything , why should we maintain the frequency constant if so why it is only 50 Hz 60Hz?
A :We can have the frequency at any frequency you like, but than you must also make your own motors,high voltage transformers or any other equipment you want to use.We maintain the frequency at 50hz or 60hz because the world maintains a standard at 50 /60hz and the equipments are are made to operate at these frequency.
Q:What is electric traction?
A:Electric traction means using the electric power for traction system (i.e. for railways,trams, trolleys etc). Electric traction means use of the electricity for all the above machines. Now a days, magnetic traction is also used for bullet trains.and basically dc motors are used for electric traction systems.

Q:How can you start-up the 40w tube lite with 230v AC/DC without using any choke/Coil?
A:It's possible by means of Electronic chokes,otherwise it's not possible to ionize the particles in tube light with normal voltage.

Q:What is "pu" in electrical engineering?
A:Pu stands for per unit and this will be used in single line diagram of power distribution and it is like a huge electrical circuit with no of components (generators, transformers, loads) with different ratings (in MVA and KV). To bring all the ratings into common platform we use pu concept in which, in general largest MVA and KV ratings of the component is considered as base values, then all other component ratings will get back into this basis.Those values are called as pu values. (p.u=actual value/base value).

Q:Operation carried out in Thermal power stations?
A:The water is obtained in the boiler and the coal is burnt so that steam is obtained this steam is allowed to hit the turbine , the turbine which is coupled with the generator generates the electricity.

Q:Why link is provided in neutral of an ac circuit and fuse in phase of ac circuit?
A:Link is provided at a Neutral common point in the circuit from which various connection are taken for the individual control circuit and so it is given in a link form to withstand high Amps. But in the case of Fuse in the Phase of AC circuit it is designed such that the fuse rating is calculated for the particular circuit (i.e load) only.So if any malfunction happen the fuse connected in the particular control circuit alone will blow off.

Q: What is the difference between electronic regulator and ordinary electrical rheostat  regulator for fans?
A:The difference between the electronic and ordinary electrical  regulator is that in electronic regulator power losses are less because as we decrease the speed the electronic regulator gives the power needed for that particular speed but in case of ordinary rheostat type regulator, the power wastage is same for every speed and no power is saved.In electronic regulator, triac is employed for speed control by varying the firing angle speed and it is controlled but in rheostatic ,control resistance is decreased by steps to achieve speed control.
Q:How tube light circuit is connected and how it works?
A:A choke is connected in one end of the tube light and a starter is in series with the circuit. When supply is provided ,the starter will interrupt the supply cycle of AC. Due to the sudden change of supply the chock will generate around 1000volts . This volt will capable of to break the electrons inside the tube to make electron flow. once the current passes through the tube the starter circuit will be out of part. now there is no change of supply causes choke voltage normalized and act as minimize the current.

Q:whats is MARX CIRCUIT?
A:It is used with generators for charging a number of capacitor in parallel and discharging them in series.It is used when voltage required for testing is higher than the available.

Q:What is encoder, how it function?
A:An encoder is a device used to change a signal (such as a bitstream) or data into a code. The code may serve any of a number of purposes such as compressing information for transmission or storage, encrypting or adding redundancies to the input code, or translating from one code to another. This is usually done by means of a programmed algorithm,especially if any part is digital, while most analog encoding is done with analog circuitry.

Q:What are the advantages of speed control using thyristor?
A:Advantages :1. Fast Switching Characterstics than Mosfet, BJT, IGBT 2. Low cost 3. Higher Accurate.

Q:Why Human body feel Electric shock ?? n in an Electric train during running , We did nt feel any Shock ? why?
A:Unfortunately our body is a pretty good conductor of electricity, The golden rule is Current takes the lowest resistant path if you have insulation to our feet as the circuit is not complete (wearing rubber footwear which doing some repairs is advisable as our footwear is a high resistance path not much current flows through our body).The electric train is well insulated from its electrical system.

Q:what is the principle of motor?
A:Whenever a current carrying conductor is placed in an magnetic field it produce turning or twisting movemnt is called as torque.
Q:Why, when birds sit on transmission lines or current wires doesn't get shock?
A:Its true that if birds touch the single one line (phase or neutral) they don't get electrical shock... if birds touch 2 lines than the circuit is closed and they get electrical shock.. so if a human touch single one line(phase) then he doesn't get shock if he is in the air (not touching - standing on the ground if he is standing on the ground then touching the line (phase) he will get a shock because the ground on what we standing is like line (ground bed - like neutral)। and in the most of electric lines the neutral is grounded..so that means that human who touch the line closes the circuit between phase and neutral.

Q:what is meant by armature reaction?
A:The effect of armature flu to main flux is called armature reaction. The armature flux may support main flux or opposes main flux.

Q:what happen if we give 220 volts dc supply to d bulb r tube light?
A:Bulbs [devices] for AC are designed to operate such that it offers high impedance to AC supply. Normally they have low resistance. When DC supply is applied, due to low resistance, the current through lamp would be so high that it may damage the bulb element.

Q:Which motor has high Starting Torque and Staring current DC motor, Induction motor or Synchronous motor?
A:DC Series motor has high starting torque. We can not start the Induction motor and Synchronous motors on load, but can not start the DC series motor without load.

Q:what is ACSR cable and where we use it?
A:ACSR means Aluminium conductor steel reinforced, this conductor is used in transmission & distribution.

Q:What is vaccum currcuit breaker.define with cause and where be use it Device?
A:A breaker is normally used to break a ciruit. while breaking the circuit, the contact terminals will be separated. At the time of seperation an air gap is formed in between the terminals. Due to existing current flow the air in the gap is ionised and results in the arc. various mediums are used to quench this arc in respective CB's. but in VCB the medium is vaccum gas. since the air in the CB is having vaccum pressure the arc formation is interrupted. VCB's can be used upto 11kv.
Q:What will happen when power factor is leading in distribution of power?
A:If their is high power factor, i.e if the power factor is close to one:
1.losses in form of heat will be reduced,
2.cable becomes less bulky and easy to carry, and very
cheap to afford, &
3. it also reduces over heating of tranformers.

Q:whats the one main difference between UPS & inverter ? And electrical engineering & electronics engineering ?
A:uninterrupt power supply is mainly use for short time . means according to ups VA it gives backup. ups is also two types : on line and offline . online ups having high volt and amp for long time backup with with high dc voltage.but ups start with 12v dc with 7 amp. but inverter is startwith 12v,24,dc to 36v dc and 120amp to 180amp battery with long time backup.

Q:What is 2 phase motor?
A:A two phase motor is a motor with the the starting winding and the running winding have a phase split. e.g;ac servo motor.where the auxiliary winding and the control winding have a phase split of 90 degree.

Q:Advantages of vvvf drives over non vvvf drives for EOT cranes?
A:1.smooth start and stop.
2.no jerking of load.
3.exact posiitoning
4.better protection for motor.
5.high/low speed selection.
6.reliability of break shoe.
7.programmable break control.
8.easy circutry
9.reduction in controls
10.increases motor life

Q:What is the significance of vector grouping in Power Transformers?
A:Every power transformer has a vector group listed by its manufacturer. Fundamentally it tells you the information about how the windings are connected (delta or wye) and the phace difference betweent the current and voltage. EG. DYN11 means Delta primary, Wye Secondry and the current is at 11 o clock reffered to the voltage.

Q:Which type of A.C motor is used in the fan (ceiling fan, exhaust fan, padestal fan, bracket fan etc) which are find in the houses ?
A:Its Single Phase induction motor which mostly squirrel cage rotor and are capacitor start capacitor run.
Q:Give two basic speed control scheme of DC shunt motor?
A:1. By using flux control method:in this method a rheostat is connected across the field winding to control the field current.so by changing the current the flux produced by the field winding can be changed, and since speed is inversely proportional to flux speed can be controlled 2.armature control method:in this method a rheostat is connected across armature winding by varying the resistance the value of resistive drop(IaRa)can be varied,and since speed is directly proportional to Eb-IaRa the speed can be controlled.

Q:what is the principle of motor?
A:Whenever a current carrying conductor is placed in an magnetic field it produce turning or twisting movement is called as torque.

Q:what is meant by armature reaction?
A:The effect of armature flu to main flux is called armature reaction. The armature flux may support main flux or opposes main flux.

Q:Give two basic speed control scheme of DC shunt motor?
A:1. By using flux control method:in this method a rheostat is connected across the field winding to control the field current.so by changing the current the flux produced by the field winding can be changed, and since speed is inversely proportional to flux speed can be controlled 2.armature control method:in this method a rheostat is connected across armature wdg.by varying the resistance the value of resistive drop(IaRa)can be varied,and since speed is directly proportional to Eb-IaRa the speed can be controlled.

Q:what is the difference between synchronous generator & asynchronous generator?
A:In simple, synchronous generator supply's both active and reactive power but asynchronous generator(induction generator) supply's only active power and observe reactive power for magnetizing.This type of generators are used in windmills.

Q:What is the Polarization index value ? (pi value)and simple definition of polarization index ?
A:Its ratio between insulation resistance(IR)i.e meggar value
for 10min to insulation resistance for 1 min. It ranges from 5-7 for new motors & normally for motor to be in good condition it should be Greater than 2.5 .
Q:Why syn. generators r used for the production of electricity?
A:synchronous machines have capability to work on different power factor(or say
different imaginary power varying the field emf. Hence syn. generators r used for the production of electricity.

Q:What is the difference between synchronous generator & asynchronous generator?
A:In simple, synchronous generator supply's both active and reactive power but asynchronous generator(induction generator) supply's only active power and observe reactive power for magnetizing.This type of generators are used in windmills.

Q:1 ton is equal to how many watts?
A:1 ton = 12000 BTU/hr and to convert BTU/hr to horsepower,
12,000 * 0.0003929 = 4.715 hp therefore 1 ton = 4.715*.746 = 3.5 KW.

Q:why syn. generators r used for the production of electricity?
A:synchronous machines have capability to work on differentpower factor(or say
different imaginary pow varying the field emf. Hence syn. generators r used for the production of electricity.

Q:Enlist types of dc generator?
A:D.C.Generators are classified into two types 1)separatly exicted d.c.generator 2)self exicted d.c.generator, which is further classified into;1)series 2)shunt and
3)compound(which is further classified into cumulative and differential).
Q:What is Automatic Voltage regulator(AVR)?
A:AVR is an abbreviation for Automatic Voltage Regulator.It is important part in Synchronous Generators, it controls theoutput voltage of the generator by controlling its excitation current. Thus it can control the output Reactive Power of the Generator.

Q:What is an exciter and how does it work?
A:There are two types of exciters, static exciter and rotory exciter.purpose of excitor is to supply the excitation dc voltage to the fixed poles of generator.Rotory excitor is an additional small generator mounted on the shaft of main generator. if it is dc generator, it will supply dc to the rotory poles through slip ring and brushes( conventional alternator). if it is an ac excitor, out put of ac excitor is rectified by rotating diodes and supply dc to main fixed poles.ac excitor is the ac generator whose field winding are stationary and armature rotates. initial voltage is built up by residual magnetism.It gives the starting torque to the generator.

Q:Difference between a four point starter and three point starter?
A:The shunt connection in four point stater is provided separately form the line where as in three point stater it is connected with line which is the drawback in three point stater

Q:Why use the VCB at High Transmission System ? Why can't use ACB?
A:Actually the thing is vacuum has high arc queching property compare to air because in VCB ,the die electric strengths equal to 8 times of air . That y always vaccum used as inHT breaker and air used as in LT .

Q:What is the difference between surge arrestor and lightning arrestor?
A:LA is installed outside and the effect of lightning is grounded,where as surge arrestor installed inside panels comprising of resistors which consumes the energy and nullify the effect of surge.
Q:What happens if i connect a capacitor to a generator load?
A:Connecting a capacitor across a generator always improves powerfactor,but it will help depends up on the engine capacity of the alternator,other wise the alternator will be over loaded due to the extra watts consumed due to the improvement on pf. Secondly, don't connect a capacitor across an alternator while it is picking up or without any other load.

Q:Why the capacitors works on ac only?
A:Generally capacitor gives infinite resistance to dc components(i.e., block the dc components). it allows the ac components to pass through.

Q:Explain the working principal of the circuit breaker?
A:Circuit Breaker is one which makes or breaks the circuit.It has two contacts namely fixed contact & moving contact.under normal condition the moving contact comes in contact with fixed contact thereby forming the closed contact for the flow of current. During abnormal & faulty conditions(when current exceeds the rated value) an arc is produced between the fixed & moving contacts & thereby it forms the open circuitArc is extinguished by the Arc Quenching media like air, oil, vaccum etc.

Q:How many types of colling system it transformers?
A:1. ONAN (oil natural,air natural)
2. ONAF (oil natural,air forced)
3. OFAF (oil forced,air forced)
4. ODWF (oil direct,water forced)
5. OFAN (oil forced,air forced)

Q:What is the function of anti-pumping in circuit breaker?
A:when breaker is close at one time by close push button,the anti pumping contactor prevent re close the breaker by close push button after if it already close.

Q:what is stepper motor.what is its uses?
A:Stepper motor is the electrical machine which act upon input pulse applied to it. it is one type of synchronous motor which runs in steps in either direction instead of running in complete cycle.so, in automation parts it is used.
Q:How to calculate capacitor bank value to maintain unity power factor with some suitable example?
A:KVAR= KW(TAN(COS(-1)#e)- TAN(COS(-1)#d) )
#e= EXISTING P.F.
#d= DESIRED P.F.

Q:Tell me in detail about c.t. and p.t. ?(Company:reliance)
A:The term C.T means current transformer,and the term P.T means potential transformer.In circuit where measurements of high voltage and high current is involved they are used there.Particularly when a measuring device like voltmeter or ammeter is not able to measure such high value of quantity because of large value of torque due to such high value it can damage the measuring device.so, CT and PT are introduced in the circuits. They work on the same principle of transformer, which is based on linkage of electromagneticflux produced by primary with secondary.They work on the ratio to they are designed.E.g if CTis of ratio 5000\5A and it has to measure secondary current of
8000A.then ANS=8000*5\5000=8Aand this result will be given to ammeter .and after measuring 8A we can calculate the primary current.same is
the operation of PT but measuring voltage.

Q:There are a Transformer and an induction machine. Those two have the same supply. For which device the load current will be maximum? And why?
A:The motor has max load current compare to that of transformer because the motor consumes real power.. and the transformer is only producing the working flux and its not consuming.. hence the load current in the transformer is because of core loss so it is minimum.

Q:what is power factor? whether it should be high or low? why?
A:Power factor should be high in order to get smooth operation of the system.Low power factor means losses will be more.it is the ratio of true power to apperent power. it has to be ideally 1. if it is too low then cable over heating & equipment overloading will occur. if it is greater than 1 then load will act as capacitor and starts feeding the source and will cause tripping.(if pf is poor ex: 0.17 to meet actual power load has to draw more current(V constant),result in more lossesif pf is good ex: 0.95 to meet actual power load has to draw less current(V constant),result in less losses).
Q:What is the difference between Isolator and Circuit Breaker?
A:Isolator is a off load device which is used for isolating the downstream circuits from upstream circuits for the reason of any maintenance on downstream circuits. it is manually operated and does not contain any solenoid unlike circuit breaker. it should not be operated while it is having load. first the load on it must be made zero and then it can safely operated. its specification only rated current is given.But circuit breaker is onload automatic device used for breaking the circuit in case of abnormal conditions like short circuit, overload etc., it is having three specification 1 is rated current and 2 is short circuit breaking capacity and 3 is instantaneous tripping current.

Q:what is boucholz relay and the significance of it in to the transformer?
A:Boucholz relay is a device which is used for the protection of transformer from its internal faults, it is a gas based relay. whenever any internal fault occurs in a transformer, the boucholz relay at once gives a horn for some time, if the transformer is isolated from the circuit then it stop its sound itself other wise it trips the circuit by its own tripping mechanism.

Q:What is SF6 Circuit Breaker?
A:SF6 is Sulpher hexa Flouride gas.. if this gas is used as arc quenching medium in a Circuitbreaker means SF6 CB.

Q:what is ferrantic effect?
A:Output voltage is greater than the input voltage or receiving end voltage is greater than the sending end voltage.

Q: what is meant by insulation voltage in cables? explain it?
A:It is the property of a cable by virtue of it can withstand the applied voltage without rupturing it is known as insulation level of the cable.
Q: Why we do 2 types of earthing on transformer (ie:)body earthing & neutral earthing , what is function. i am going to install a 5oo kva transformer & 380 kva DG set what should the earthing value?
A:The two types of earthing are Familiar as Equipment earthing and system earthing. In Equipment earthing: body ( non conducting part)of the equipment shouldd be earthed to safegaurd the human beings.system Earthing : In this neutral of the supply source ( Transformer or Generator) should be grounded. With this,in case of unbalanced loading neutral will not be shifted.so that unbalanced voltages will not arise. We can protect the equipment also. With size of the equipment( transformer or alternator)and selection of relying system earthing will be further classified into directly earthed,Impedance earthing, resistive (NGRs) earthing.

Q:What is the difference between MCB & MCCB, Where it can be used?
A:MCB is miniature circuit breaker which is thermal operated and use for short circuit protection in small current rating circuit. MCCB moulded case circuit breaker and is thermal operated for over load current and magnetic operation for instant trip in short circuit condition.under voltage and under frequency may be inbuilt. Normally it is used where normal current is more than 100A.

Q:Where should the lighting arrestor be placed in distribution lines?
A:Near distribution transformers and out going feeders of 11kv and incomming feeder of 33kv and near power transformers in sub-stations.

Q:Define IDMT relay?
A:It is an inverse definite minimum time relay.In IDMT relay its operating is inversely proportional and also a characteristic of minimum time after which this relay operates.It is inverse in the sense ,the tripping time will decrease as the magnitude of fault current increase.

Q:What are the transformer losses?
A:TRANSFORMER LOSSES - Transformer losses have two sources-copper loss and magnetic loss. Copper losses are caused by the resistance of the wire (I2R). Magnetic losses are caused by eddy currents and hysteresis in the core. Copper loss is a constant after the coil has been wound and therefore a measurable loss. Hysteresis loss is constant for a particular voltage and current. Eddy-current loss, however, is different for each frequency passed through the transformer.
A:When the supply is cutt off for a running motor, it still continue running due to inertia. In order to stop it quickly we place a load(resitor) across the armature winding and the motor should have maintained continuous field supply. so that back e.m.f voltage is made to apply across the resistor and due to load the motor stops quickly.This type of breaking is called as "Regenerative Breaking".

Q:Why is the starting current high in a DC motor?
A:In DC motors, Voltage equation is V=Eb-IaRa (V = Terminal voltage,Eb = Back emf in Motor,Ia = Armature current,Ra = Aramture resistance).At starting, Eb is zero. Therefore, V=IaRa, Ia = V/Ra ,where Ra is very less like 0.01ohm.i.e, Ia will become enormously increased.

Q:What are the advantages of star-delta starter with induction motor?
A:(1). The main advantage of using the star delta starter is reduction of current during the starting of the motor.Starting current is reduced to 3-4 times Of current of Direct online starting.(2). Hence the starting current is reduced , the voltage drops during the starting of motor in systems are reduced.

Q:Why Delta Star Transformers are used for Lighting Loads?
A:For lighting loads, neutral conductor is must and hence the secondary must be star winding. and this lighting load is always unbalanced in all three phases. To minimize the current unbalance in the primary we use delta winding in the primary. So delta / star transformer is used for lighting loads.

Q:Why in a three pin plug the earth pin is thicker and longer than the other pins?
A:It depends upon R=rho l/a where area(a) is inversely proportional to resistance (R), so if (a) increases, R decreases & if R is less the leakage current will take low resistance path so the earth pin should be thicker. It is longer because the The First to make the connection and Last to disconnnect should be earth Pin. This assures Safety for the person who uses the electrical instrument.

Q:Why series motor cannot be started on no-load?
A:Series motor cannot be started without load because of high starting torque. Series motor are used in Trains, Crane etc.

Q:Why ELCB can't work if N input of ELCB do not connect to ground?
A:ELCB is used to detect earth leakage fault. Once the phase and neutral are connected in an ELCB, the current will flow through phase and that much current will have to return neutral so resultant current is zero. Once there is a ground fault in the load side, current from phase will directly pass through earth and it will not return through neutral through ELCB. That means once side current is going and not returning and hence because of this difference in current ELCB wil trip and it will safe guard the other circuits from faulty loads. If the neutral is not grounded, fault current will definitely high and that full fault current will come back through ELCB, and there will be no difference in current.
Q:How electrical power is generated by an A.C Generator?
A:For the generation of elect power we need a prime mover which supplies mechanical power input to the alternator, can be steam turbines,or hydro turbines .When poles of the rotor moves under the armature conductors which are placed on the stator ,field flux cut the armature conductor ,therefore voltage is generated and is of sinusoidal in nature...due to polarity change of rotor poles(i,e) N-S-N-S.

Q:Why an ac solenoid valve attract the plunger even though we interchanges the terminal? Will the poles changes?
A:Yes because the poles changes for every half-cycle of ac voltage so the polarity of AC voltage is continuously changing for every half cycle. so, interchanging of terminals in ac system does not show any difference. That's why the ac solenoid attract the plunger even though it's terminals are interchanged. 

Q:What is derating?, why it is necessary, it is same for all means for drives, motors,and cables.
A:The current currying of cables will change depending upon the site temperature (location of site), type of run (it will run through duct, trench, buried etc.), number of tray, depth of trench, distance between cables. Considering this condition actual current currying capacity of cable reduce than current currying capacity (which given to cable Catalogue) this is called derating.

Q:Why temperature rise is conducted in bus bars and isolators? 
A:Bus bars and isolators are rated for continuous power flow, that means they carry heavy currents which rises their temperature. so it is necessary to test this devices for temperature rise.

Q:When voltage increases then current also increases then what is the need of over voltage relay and over current relay? Can we measure over voltage and over current by measuring current only?
A:No.We can't sense the over voltage by just measuring the current only because the current increases not only for over voltages but also for under voltage(As most of the loads are non-linear in nature).So,the over voltage protection & over current protection are completely different. Over voltage relay meant for sensing over voltages & protect the system from insulation break down and firing. Over current relay meant for sensing any internal short circuit, over load condition ,earth fault thereby reducing the system failure & risk of fire.So, for a better protection of the system.It should have both over voltage & over current relay.

Q:If one lamp connects between two phases it will glow or not?
A:If the voltage between the two phase is equal to the lamp voltage then the lamp will glow. When the voltage difference is big it will damage the lamp and when the difference is smaller the lamp will glow depending on the type of lamp.
Q:How do you select a cable size (Cu & Al) for a particular load?
A:At first calculate the electrical current of the load, after that derate the electrical current considering derating factor(depending on site condition and laying of cable) after choose the cable size from cable catalog considering derating electrical current.After that measure the length of cable required from supply point of load to load poin. Calculate the voltage drop which will max 3% (resistance and reactance of cable found from cable catalog of selecting cable) if voltage drop>3%
then choose next higher size of cable.

Q:What are HRC fuses and where it is used?
A:HRC stand for "high rupturing capacity" fuse and it is used in distribution system for electrical transformers.

Q:Which power plant has high load factor?
A:All base load power plants have a high load factor. If we use high efficiency power plants to supply the base load,we can reduce the cost of generation.Hydel power plants have a higher efficiency than thermal & nuclear power plants.

Q:Mention the methods for starting an induction motor?
A:The different methods of starting an induction motor
DOL:direct online starter
Star delta starter
Auto transformer starter
Resistance starter
Series reactor starter

Q:What is the difference between earth resistance and earth electrode resistance?
A:Only one of the terminals is evident in the earth resistance. In order to find the second terminal we should recourse to its definition: Earth Resistance is the resistance existing between the electrically accessible part of a buried electrode and another point of the earth, which is far away.
The resistance of the electrode has the following components:
(A) the resistance of the metal and that of the connection to it.
(B) the contact resistance of the surrounding earth to the electrode.

Q:What is use of lockout relay in ht voltage?
A:A lock-out relay is generally placed in line before or after the e-stop switch so the power can be shut off at one central location. This relay is powered by the same electrical source as the control power and is operated by a key lock switch. The relay itself may have up to 24 contact points within the unit itself. This allows the control power for multiple machines to be locked out by the turn of a single key switch.

Q:What is the power factor of an alternator at no load?
A:At no load Synchronous Impedance of the alternator is responsible for creating angle difference. So it should be zero lagging like inductor.
Q:Why most of analog o/p devices having o/p range 4 to 20 mA and not 0 to 20 mA ?
A:4-20 mA is a standard range used to indicate measured values for any process. The reason that 4ma is chosen instead of 0 mA is for fail safe operation .For example- a pressure instrument gives output 4mA to indicate 0 psi, up to 20 mA to indicate 100 psi, or full scale. Due to any problem in instrument (i.e) broken wire, its output reduces to 0 mA. So if range is 0-20 mA then we can differentiate whether it is due to broken wire or due to 0 psi.

Q:Two bulbs of 100w and 40w respectively connected in series across a 230v supply which bulb will glow bright and why?
A:Since two bulbs are in series they will get equal amount of  electrical current but as the supply voltage is constant across the bulb(P=V^2/R).So the resistance of 40W bulb is greater and voltage across 40W is more (V=IR) so 40W bulb will glow brighter.

Q:What is meant by knee point voltage?
A:Knee point voltage is calculated for electrical Current transformers and is very important factor to choose a CT. It is the voltage at which a CT gets saturated.(CT-current transformer).

Q:What is reverse power relay?
A:Reverse Power flow relay are used in generating stations's protection. A generating stations is supposed to fed power to the grid and in case generating units are off,there is no generation in the plant then plant may take power from grid. To stop the flow of power from grid to generator we use reverse power relay.

Q:What will happen if DC supply is given on the primary of a transformer?
A:Mainly transformer has high inductance and low resistance.In case of DC supply there is no inductance ,only resistance will act in the electrical circuit. So high  electrical current will flow through primary side of the transformer.So for this reason coil and insulation will burn out.

Q:What is the difference between isolators and  electrical circuit breakers? What is bus-bar?
A:Isolators are mainly for switching purpose under normal conditions but they cannot operate in fault conditions .Actually they used for isolating the CBs for maintenance. Whereas CB gets activated under fault conditions according to the fault detected.Bus bar is nothing but a junction where the power is getting distributed for independent loads.

Q:What are the advantage of free wheeling diode in a Full Wave rectifier?
A:It reduces the harmonics and it also reduces sparking and arching across the mechanical switch so that it reduces the voltage spike seen in a inductive load
Q:What are Motor Generator Sets and explain the different ways the motor generator set can be used ?
A:Motor Generator Sets are a combination of an electrical generator and an engine mounted together to form a single piece of equipment. Motor generator set is also referred to as a genset, or more commonly, a generatorThe motor generator set can used in the following different ways:
              1.Alternating current (AC) to direct current (DC)
              2.DC to AC
              3.DC at one voltage to DC at another voltage
              4.AC at one frequency to AC at another harmonically-related frequency

Q:What is power quality meter ?
A:Power Quality meters are common in many industrial environment. Small units are now available for home use as well. They give operators the ability to monitor the both perturbations on the power supply, as well as power used within a building, or by a single machine or appliance. In some situations, equipment function and operation is monitored and controlled from a remote location where communication is via modem, or high-speed communication lines.So we can understand the importance of power measurement through power quality meters.

Q:What is the different between digital phase converter and ordinary phase converter?
A:Digital phase converter are a recent development in phase converter technology that utilizes proprietary software in a powerful microprocessor to control solid state power switching components. This microprocessor, called a digital signal processor (DSP), monitors the phase conversion process, continually adjusting the input and output modules of the converter to maintain perfectly balanced three-phase power under all load conditions.

Q:Explain the operation of variable frequency transformer?
A:A variable frequency transformer is used to transmit electricity between two asynchronous alternating current domains. A variable frequency transformer is a doubly-fed electric machine resembling a vertical shaft hydroelectric generator with a three-phase wound rotor, connected by slip rings to one external ac power circuit. A direct-current torque motor is mounted on the same shaft. Changing the direction of torque applied to the shaft changes the direction of power flow; with no applied torque, the shaft rotates due to the difference in frequency between the networks connected to the rotor and stator.The variable frequency transformer behaves as a continuously adjustable phase-shifting transformer. It allows control of the power flow between two networks .

Q:What is the main use of rotary phase converter ?
A:Rotary phase converter will be converting single phase power into true balanced 3 phase power,so it is often called as single phase to three phase converter .Often the advantages of 3 phase motors, and other 3 phase equipment, make it worthwhile to convert single phase to 3 phase so that small and large consumers need not want to pay for the extra cost of a 3 phase service but may still wish to use 3 phase equipment.
Q:Which type of oil is used as a transformer oil?
A:Transformer oil, or insulating oil, is usually a highly-refined mineral oil that is stable at high temperatures and has excellent electrical insulating properties. It is used in oil filled transformers, some types of high voltage capacitors, fluorescent lamp ballasts, and some types of high voltage switches and circuit breakers. Its functions are to insulate, suppress corona and arcing, and to serve as a coolant.

Well into the 1970s, polychlorinated biphenyls (PCB)s were often used as a dielectric fluid since they are not flammable. They are toxic, and under incomplete combustion, can form highly toxic products such as furan. Starting in the early 1970s, concerns about the toxicity of PCBs have led to their banning in many countries.

Today, non-toxic, stable silicon-based or fluoridated hydrocarbons are used, where the added expense of a fire-resistant liquid offsets additional building cost for a transformer vault. Combustion-resistant vegetable oil-based dielectric coolants and synthetic pentaerythritol tetra fatty acid (C7, C8) esters are also becoming increasingly common as alternatives to naphthenic mineral oil. Esters are non-toxic to aquatic life, readily biodegradable, and have a lower volatility and higher flash points than mineral oil.

Q: If we give 2334 A, 540V on Primary side of 1.125 MVA step up transformer, then what will be the Secondary Current, If Secondary Voltage=11 KV?
A:As we know the Voltage & current relation for transformer-V1/V2 = I2/I1
We Know, VI= 540 V; V2=11KV or 11000 V; I1= 2334 Amps.
By putting these value on Relation-
540/11000= I2/2334
So,I2 = 114.5 Amps

Q:what are the points to be consider for MCB(miniature circuit breaker selection?
A:I(L)*1.25=I(MAX) maximum current. Mcb specification are done on maximum current flow in circuit.

Q:what is the full form of KVAR?
A:We know there are three types of power in Electricals as Active, apparent & reactive. So KVAR is stand for ``Kilo Volt Amps with Reactive component.

Q:What is excitation?
A:Excitation is applying an external voltage to DC shunt coil
in DC motors.

Q:In three pin plug 6 Amp. 220v AC rating. why earth pin diameter is higher than othter two pin? what its purpose ?
A:Because Current flow in the conductor is inversely proportional to the conductor diameter. So if any short circuits occur in the system first high currents bypassed in the Earthling terminal.( R=Pl/a area of the conductor increases resistance
value decreases)