eee formula sheet

8/9/2019 eee formula sheet

1/147

One stop series for

GATE/IES/PSU-2014

(Formula Sheet for EEE Dept.,)


2/147

Published & Marketed byInstitute of Engineering Studies (IES,Bangalore),

D.No.600, Ist floor,

46thCross, Corporation colony,

South End Main, Jayanagar 9thBlock,Bangalore- 560 011

Ph: 080-3255 2008, (+91) 99003 99699

Website:www.gateiespsu.com

Email:[email protected]

Publisher:

No part of this book may be reproduced in a retrieval system or transmitted, inany form or by any means, electronics, mechanical photocopying, recordingand or without permission of Institute of Engineering Studies, Bangalore.
http://www.gateiespsu.com/http://www.gateiespsu.com/http://www.gateiespsu.com/mailto:[email protected]:[email protected]:[email protected]:[email protected]://www.gateiespsu.com/


3/147

Formula Sheet(EEE Department)

Index Page. No.

1. Mathematics --- --- 1 to 9

2. Electromagnetic Fields --- 10 to 16

3. Signals & Systems --- --- 17 to 22

4. Electrical Machines --- --- 23 to 58

5. Power Systems --- --- 59 to 89

6. Control Systems --- --- 90 to 92

7. Measurements --- --- 93 to 114

8. Analog Electronics --- --- 115 to 123

9. Digital Electronics --- --- 124 to 126

10.Microprocessors --- --- 127 to 130

11.Power Electronics --- --- 131 to 143

*For updates

Facebook:www.facebook.com/onlineies

Twitter:www.twitter.com/ies_bangalore

Contact:

[email protected]

Mob: (+91) 99003 99699
http://www.facebook.com/onlineieshttp://www.facebook.com/onlineieshttp://www.facebook.com/onlineieshttp://www.twitter.com/ies_bangalorehttp://www.twitter.com/ies_bangaloremailto:[email protected]:[email protected]://www.twitter.com/ies_bangalorehttp://www.facebook.com/onlineies


4/147

Online & Offline Services for Aspirants

1. Aspirants, who joined for Classroom Coaching in our Institute ofEngineering Studies, get Diamond user access in our portal along

with Classes.2. Aspirants who take Postal series get Diamond access along with

books of our Study materials, Question Banks, Practice problems

books.

3. Aspirants who take Practice problems books (Without StudyMaterials) get Shift access in our portal.

4. Aspirants who want to take Online Tests Conducted for GATE, IESObjective and for Different PSU/Govt sector jobs will get Gold access

in our portal.

5. Those who dont pay will get Silver access in our portal.

To Know more about these access levels visit our portal at

www.gateiespsu.com

Through Mobile appalso you can take Online Tests, Practice

Questions, Report Questions and Check Solution of any

question with the Question ID provided at the end of each

question in our books.

Silver Gold Shift Diamond

Few Online Tests Premium Online Tests

(for both GATE/PSU)

Premium Online Tests

(for both GATE/PSU)

Premium Online Tests

(for both GATE/PSU)

Practice Qns. In online

without Solutions

Premium Downloads Premium Downloads Premium Downloads

Answering to

reported Questions

with Least Priority

Practice Questions in

online with Solutions





-NA- Answering to

reported Qns with

Highest Priority

Answering to

reported Qns with

Highest Priority

Answering to

reported Qns with

Highest Priority

-NA- -NA- All Practice problems

books to your

doorstep

All Practice problems

& Study Materials

books to your

doorstep

-NA- Check Soln of any Qn

from Practice books &

Report

Check Soln of any Qn


Report

Check Soln of any Qn


Report
http://www.gateiespsu.com/http://www.gateiespsu.com/http://www.gateiespsu.com/


5/147

Institute of Engineering Studies (IES,Bangalore) Mathematics Formula SheetMatheMatics

Matrix :- If |A| = 0

Singular matrix ; |A|

0 Non singular matrix

Scalar Matrixis a Diagonal matrix with all diagonal elements are equal Unitary Matrixis a scalar matrix with Diagonal element as 1 (AQ= (A)T= A) If the product of 2 matrices are zero matrix then at least one of the matrix has det zero Orthogonal Matrix if AAT= AT.A = I AT= A A = AT Symmetric

A = - ATSkew symmetricProperties :- (if A & B are symmetrical )

A + B symmetric KA is symmetric AB + BA symmetric AB is symmetric iff AB = BA

For any A A + AT symmetric ; A - AT skew symmetric. Diagonal elements of skew symmetric matrix are zero If A skew symmetric Asymmetric matrix ; Askew symmetric If A is null matrix then Rank of A = 0.

Consistency of Equations :-

r(A, B) r(A) is consistent r(A, B) = r(A) consistent &

if r(A) = no. of unknowns then unique solutionr(A) < no. of unknowns then solutions .

Hermition , Skew Hermition , Unitary & Orthogonal Matrices :-

AT= A then Hermition AT= A then Hermition Diagonal elements of Skew Hermition Matrix must be purely imaginary or zero Diagonal elements of Hermition matrix always real . A real Hermition matrix is a symmetric matrix. |KA| = K|A|Eigen Values & Vectors :- Char. Equation |A I| = 0.Roots of characteristic equation are called eigen values . Each eigen value corresponds to non zero

solution X such that (A I)X = 0 . X is called Eigen vector . Sum of Eigen values is sum of Diagonal elements (trace) Product of Eigen values equal to Determinent of Matrix . Eigen values of AT& A are same is Eigen value of A then 1/ A & || is Eigen value of adj A. , are Eigen values of A then

Leading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/PracticeBranches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email:[email protected]

Site:www.GateIesPsu.com Google+:www.gplus.to/onlineiesFB:www.facebook.com/onlineiesPg.No. 1 of 143
mailto:[email protected]:[email protected]:[email protected]://www.gateiespsu.com/http://www.gateiespsu.com/http://www.gateiespsu.com/http://www.gplus.to/onlineieshttp://www.gplus.to/onlineieshttp://www.gplus.to/onlineieshttp://www.facebook.com/onlineieshttp://www.facebook.com/onlineieshttp://www.facebook.com/onlineieshttp://www.facebook.com/onlineieshttp://www.gplus.to/onlineieshttp://www.gateiespsu.com/mailto:[email protected]


6/147

Institute of Engineering Studies (IES,Bangalore) Mathematics Formula Sheet

KA K , K ..K A

,

..

.

A + KI + k , + k , .. + k(A KI) ( k), ( k) Eigen values of orthogonal matrix have absolute value of 1 . Eigen values of symmetric matrix also purely real . Eigen values of skew symmetric matrix are purely imaginary or zero . , , distinct eigen values of A then corresponding eigen vectors X, X, .. Xfor

linearly independent set .

adj (adj A) = |A| ; | adj (adj A) | = |A|()Complex Algebra :-

Cauchy Rieman equations

=; =Neccessary & Sufficient Conditions for f(z) to be analytic

f(z)/(Z a)+ dz = ! [ f(a) ] if f(z) is analytic in region C & Z =a is single point f(z) = f(z0) + f(z0) ()! + f(z0) ()! + + f(z0) ()! + . Taylor Series

if z0= 0 then it is called Mclauren Series f(z) = a(z z0)0 ; when a= (

)

! If f(z) analytic in closed curve C except @ finite no. of poles then f(z)dz = 2i (sum of Residues @ singular points within C )

Res f(a) = lim( ()= (a) / (a)=

lim

()! ((Z

a)f(z) )

Calculus :-

Rolles theorem :-

If f(x) is

(a) Continuous in [a, b]




7/147

Institute of Engineering Studies (IES,Bangalore) Mathematics Formula Sheet(b) Differentiable in (a, b)

(c) f(a) = f(b) then there exists at least one value C (a, b) such that f(c)= 0 .Langranges Mean Value Theorem :-

If f(x) is continuous in [a, b] and differentiable in (a, b) then there exists atleast one value C in (a, b)

such that f(c) = (b)()b Cauchys Mean value theorem :-

If f(x) & g(x) are two function such that

(a) f(x) & g(x) continuous in [a, b]

(b) f(x) & g(x) differentiable in (a, b)

(c) g(x) 0 x in (a, b)Then there exist atleast one value C in (a, b) such that

f(c)/ g(c) = (b)()(b)()Properties of Definite integrals :-

a < c < b

f(x)

. dxb =

f(x)

. dx +

f(x)

. dxb

f(x)dx0 = f(a x)dx0 f(x). dx = 2 f(x)dx0 f(x) is even

= 0 f(x) is odd

f(x). dx0 = 2 f(x)dx0 if f(x) = f(2a- x) = 0 if f(x) = - f(2a x)

f(x). dx0 = n f(x)dx0 if f(x) = f(x + a) f(x). dxb = f(a + b x). dxb




8/147


x f(x). dx0 = f(x). dx0 if f(a - x) = f(x)

sinx/

0 =

cosx/

0 =

()(3)(5)

(

)(

4)

.3 if n odd

=()(3)()(4).. if n even

sinx/0 . cosx. dx = {()(3).(5)()}{()(3).()}.(+)(+)(+4) Where K = / 2 when both m & n are even otherwise k = 1

Maxima & Minima :-

A function f(x) has maximum @ x = a if f(a) = 0 and f(a) < 0A function f(x) has minimum @ x = a if f(a) = 0 and f(a) > 0Constrained Maximum or Minimum :-

To find maximum or minimum of u = f(x, y, z) where x, y, z are connected by (x, y, z) = 0Working Rule :-

(i) Write F(x, y, z) = f(x, y, z) +

(x, y, z)

(ii) Obtain F= 0, F= 0 , F= 0(ii) Solve above equations along with = 0 to get stationary point .Laplace Transform :-

L ()= sf(s) - sf(0) - sf(0) f(0) L { tf(t) } = (1) f(s)

(

)

f(s) ds f(u)0 du f(s) / s .

Inverse Transforms :-




9/147


(+) = t sin at

(

+)

=

[ sin at + at cos at]

(+) = [ sin at - at cos at]

= Cos hat

= Sin hat

Laplace Transform of periodic function : L { f(t) } = ()

Numerical Methods :-

Bisection Method :-

(1) Take two values of x& xsuch that f(x) is +ve & f(x) is ve then x3= + find f(x3) if f(x3)+ve then root lies between x3& xotherwise it lies between x& x3.Regular falsi method :-

Same as bisection except x= x0- ()() f(x0)Newton Raphson Method :-

x+= x ()()Pi cards Method :-

y+= y0+ f(x,y ) = f(x, y)Taylor Series method :-

= f(x, y) y = y

0+ (x- x

0) (y

)

0 +

(

)

!

(y)

0+ .

(

)

!

(y)

0

Eulers method :-

y= y0+ h f(x0, y0) = f(x, yy()= y0+ [f(x0, y0) + f(x0 + h, y)




10/147


y()= y0+ [f(x0, y0) + f(x0+, y()) ]:

:

Calculate till two consecutive value of y agree

y= y+ h f(x0 + h, y)y()= y0+ [f(x0 + h, y) + f(x0 + 2h, y)

Runges Method :-k= h f(x0, y0)k= h f(x0 +, y0 +k ) finally compute K = 6(K+ 4K+ K3)k= h f(x0+h , y0+ k)k3= h ( f (x0+h , y0+ k))

Runge Kutta Method :-k= h f(x0, y0)k= h f(x0 +, y0 +k ) finally compute K = 6(K+ 2K+ 2K3 + K4)k3= h f(x0 +, y0 +k ) approximation vale y = y0+ K .k3= h f (x0+h , y0+ k3)

Trapezoidal Rule :-

f(x). dx+ = [ ( y0+ y) + 2 (y+ y+ . y)]f(x) takes values y0, y..@ x0, x, x..

Simpsons one third rule :-




11/147

Institute of Engineering Studies (IES,Bangalore) Mathematics Formula Sheet f(x). dx+ = 3[ ( y0+ y) + 4 (y+ y3+ . y) + 2 (y + y4 + . + y)]Simpson three eighth rule :-

f(x). dx+ = 38 [ ( y0+ y) + 3 (y+ y+ y4+ y5+ . y)+ 2 (y3 + y6 + . + y3) ]Differential Equations :-

Variable & Seperable :-

General form is f(y) dy = (x) dxSol: f(y) dy = (x) dx + C .

Homo generous equations :-

General form = (,)(,) f(x, y) & (x, y)Homogenous of same degree

Sol : Put y = Vx = V + x & solveReducible to Homogeneous :-

General form= +b++b+

(i) bbSol : Put x = X + h y = Y + k

= +b+(+bk+)+b+(+bk+) Choose h, k such that becomes homogenous then solve by Y = VX(ii)

= bbSol : Let = bb= = +b+(+b)+Put ax + by = t = /b




12/147

Institute of Engineering Studies (IES,Bangalore) Mathematics Formula SheetThen by variable & seperable solve the equation .

Libnetz Linear equation :-

General form +py = Q where P & Q are functions of xI.F = e p.

Sol : y(I.F) = Q. (I. F)dx + C .Exact Differential Equations :-

General form M dx + N dy = 0 M f (x, y)N f(x, y)

If My = Nx then

Sol : M. dx + (termsofN containing x ) dy = C( y constant )

Rules for finding Particular Integral :-

(

D)

e =

(

)

e

= x ()e if f (a) = 0= x ()e if f(a) = 0(b)sin (ax + b) = () sin (ax + b) f(- a) 0

= x() sin (ax + b) f(- a) = 0 Same applicable for cos (ax + b)

= x

(

) sin (ax + b)

(D)x= [f(D)]x(D) e f(x) = e (D+) f(x)Vector Calculus :-




13/147

Institute of Engineering Studies (IES,Bangalore) Mathematics Formula SheetGreens Theorem :-

(dx + dy)C = x ydx dyThis theorem converts a line integral around a closed curve into Double integral which is special case ofStokes theorem .

Series expansion :-

Taylor Series :-

f(x) = f(a) +()! (x-a) + ()! (x a)+ + ()! (x a)

f(x) = f(0) +(0)! x + (0)! x+ + (0)! x + . (mc lower series )

(1 + x)= 1+ nx + () x+ | nx| < 1e= 1 + x + !+ ..Sin x = x -

3!+ 5! - ..Cos x = 1 -

!+ 4! - ..




14/147


15/147

Institute of Engineering Studies (IES,Bangalore) Electro-Magnetic Theory Formula SheetFor monopole E ; Dipole E 3.V ; V

Electric lines of force/ flux /direction of E always normal to equipotential lines .

Energy Density W= QVN= = D. E dv= Edv Continuity Equation .J = - . = e/T where T= Relaxation / regeneration time = /(less for good conductor )

Boundary Conditions :- E = E Tangential component of E are continuous across dielectric-dielectric Boundary . Tangential Components of D are dis continues across Boundary .

E= E; = / . Normal components are of D are continues , where as E are dis continues.

D-

D=

; E

=

E

;

=

=

H= H B= BtB= B H= H

Maxwells Equations :-faraday law V= E.dI= - B.dsTransformer emf = E.dI= - ds E = -

s

Motional emf = E= (B).H = J + Electromagnetic wave propagation :- H = J + D = E E= E = - B = H H= . D J E.B 0

= -

=

/ ; E.H = 0 E

H in UPW

For loss less medium E- E = 0 = j( + j) = + j.= 1 + 1




16/147

Institute of Engineering Studies (IES,Bangalore) Electro-Magnetic Theory Formula Sheet

= 1 + + 1

E(z, t) = Eezcos(t z) ; H= E/ . = + || < || = /+ / tan 2= /. = + j attenuation constant Neper /m . | N|= 20 log =8.686 dB For loss less medium = 0; = 0. phase shift/length ; = / ; = 2/.

= = / = tan loss tanjent = 2 If tan

is very small (

< > ) good conductor Complex permittivity = 1 = - j . Tan = = .Plane wave in loss less dielectric :- ( 0)

= 0 ; = ; ; 2/ ; / 0. E & H are in phase in lossless dielectric

Free space :-(= 0, = , = )

= 0 ,

=

; u = 1/

,

2/ ; /

< 0 120 0

Here also E & H in phase .

Good Conductor :- > > / ; = = f; u = 2/ ; = 2/ ; = W 45 Skin depth 1/ 2e/4 + Skin resistance R

R= R.

R= .

Poynting Vector :-

(E H) ds = - [ E H] dv EdvS vLeading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/PracticeBranches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email:[email protected]



17/147


(z) = ||ezcos az Total time avge power crossing given area P= P (s)ds

S

Direction of propagation :- ()aa= aaa= aBoth E & H are normal to direction of propagationMeans they form EM wave that has no E or H component along direction of propagation .

Reflection of plane wave :-(a) Normal incidence

Reflection coefficient = = +Tcoefficient = = +Medium-I Dielectric , Medium-2 Conductor :-> :->0 , there is a standing wave in medium & Twave in medium 2.Max values of | E| occursZ - n/ ; n 0, 1, 2.Z (+) (+)4

< :- Eoccurs @ Z= (+) Z= (+) = (+)4 Z= nZ= =

Hmin occurs when there is |t|maxS =

|||| =

|||| =

+||||; | | = +Since || < 1 1 Transmission Lines :- Supports only TEM mode LC = ; G/C = /.

V

z -

rV= 0 ; I

z -

rI= 0

= (R + jL)( G + jC)= + j V(z, t) = V+ezcos (t- z) + Vezcos (t + z) Z= VI = R+ = G+= R+G+Lossless Line : (R = 0 =G; = 0)




18/147

Institute of Engineering Studies (IES,Bangalore) Electro-Magnetic Theory Formula Sheet= + j= jLC ; 0, w LC ; 1/ f LC , u 1/ LCZ L/C

Distortion less :(R/L = G/ )

RG ; LGR CRG LC Z RG ; 1/f LC ; u V; uz 1/C , u/z 1/Li/p impedance :-

Z= Z+Z +Z for lossless line = j tan hjl = j tan lZ= Z+Z +Z

VSWR =

=

ZZZ+Z

CSWR = - Transmission coefficient S = 1 + SWR =

VV =II =

+| ||| =ZZ =

ZZ(Z> Z) (Z< Z)

|Z| = VI = SZ |Z| = VI = Z/S

Shorted line :- = -1 , S = Z= Z= jZtan l = -1 , S = Z= Z= j Ztan l. Zmay be inductive or capacitive based on length 0

If l < / 4 inductive (Z+ve)4< l< /2 capacitive (Z-ve)Open circuited line :-

Z= Z= -jZcot l= 1 s = l< / 4 capacitive4< l + b Propegation mode j 0

k b f= + b u = phase velocity = is lossless dielectric medium = u/f= ( )+()

= 1 = / W = phase constant in dielectric medium. u= / = 2/ = u/f phase velocity & wave length in side wave guide T= = - = = 1

T= 1 impedance of UPW in mediumTE Modes :- (= 0)Hz= Hcos cos ez




20/147


T= = / 1

T> TTEDominant modeAntennas :-

Hertzian Dipole :- H= I4 sin e E= HHalf wave Dipole :-

H= I ; E= H




21/147


22/147

Institute of Engineering Studies (IES,Bangalore) Signals & Systems Formula SheetLaplace Transform :-

x(t) =1j X(s)e+jj ds

X(s) = x(t)e dsInitial & Final value Theorems :x(t) = 0 for t < 0 ; x(t) doesnt contain any impulses /higher order singularities @ t =0 then

x( 0+) = lim ()x() = lim ()

Properties of ROC :-

1. X(s) ROC has strips parallel to jaxis2. For rational laplace transform ROC has no poles

3. x(t) finite duration & absolutely integrable then ROC entire s-plane4. x(t) Right sided then ROC right side of right most pole excluding pole s = 5. x(t) left sided ROC left side of left most pole excluding s= - 6. x(t) two sided ROC is a strip7. if x(t) causal ROC is right side of right most pole including s = 8. if x(t) stable ROC includes j-axisZ-transform :-

x[n] =1j x(z)z1dz

X(z) = x[n]= zInitial Value theorem :

If x[n] = 0 for n < 0 then x[0] = lim()Final Value theorem :-

lim[]= lim1( 1)X(z)Properties of ROC :-




23/147

Institute of Engineering Studies (IES,Bangalore) Signals & Systems Formula Sheet

1.ROC is a ring or disc centered @ origin2. DTFT of x[n] converter if and only if ROC includes unit circle3. ROC cannot contain any poles

4. if x[n] is of finite duration then ROC is enter Z-plane except possibly 0 or 5. if x[n] right sided then ROC outside of outermost pole excluding z = 06. if x[n] left sided then ROC inside of innermost pole including z =07. if x[n] & sided then ROC is ring8. ROC must be connected region9.For causal LTI system ROC is outside of outer most pole including 10.For Anti Causal system ROC is inside of inner most pole including 011. System said to be stable if ROC includes unit circle .12. Stable & Causal if all poles inside unit circle13. Stable & Anti causal if all poles outside unit circle.

Phase Delay & Group Delay :-

When a modulated signal is fixed through a communication channel , there are two different delays to beconsidered.

(i) Phase delay:Signal fixed @ o/p lags the fixed signal by () phaseP= - () where () = K H(j)

Frequency response of channel

Group delay = d()d = for narrow Band signalSignal delay / Envelope delay

Probability & Random Process:-

P (A/B) = P(AB)P(B) Two events A & B said to be mutually exclusive /Disjoint if P(A B) =0Two events A & B said to be independent if P (A/B) = P(A) P(A B) = P(A) P(B)P(Ai / B) = P(AB)P(B) = P P(A) P P(A) CDF :-Cumulative Distribution function F

x(x) = P { X

x}

Properties of CDF :

Fx() = P { X } = 1 Fx(- ) = 0 Fx(x1X x) = Fx(x) - Fx(x1)




24/147

Institute of Engineering Studies (IES,Bangalore) Signals & Systems Formula Sheet Its Non decreasing function P{ X > x} = 1 P { X x} = 1- Fx(x)PDF :-

Pdf = fx(x) = ddxFx(x)Pmf = fx(x) = P{X = x= } (x = x)

Properties:- fx(x) 0 Fx(x) = fx(x) * u(x) = fxx (x) dx Fx() = fx (x) dx =1 so, area under PDF = 1

P { x1< X x} = fx(x)dxxx Mean & Variance :-

Mean x= E {x} = x fx (x) dxVariance = E { (X x)} = E {x} - xE{g(x)} = g(x)fx (x)dxUniform Random Variables :

Random variable X ~u(a,b) if its pdf of form as shown below

fx(x) = 1 ; ;


25/147

Institute of Engineering Studies (IES,Bangalore) Signals & Systems Formula Sheet

fx(x) = 1 e(x)/X ~N (1)Mean = x 1 e(x)/dx = Variance =

1 x e(x)/ dx = Exponential Distribution :-

fx(x) = exu(x)Fx(x) = ( 1- ex) u(x)Laplacian Distribution :-

fx(x) = e|x|Multiple Random Variables :-

FXY(x , y) = P { X x , Y y } FXY(x , ) = P { X x } = Fx(x) ; Fxy(, y) = P { Y < y } = FY(y) Fxy(-, y) = Fxy(x, - ) = Fxy(-, -) = 0 fx(x)= fxy (x, y)dy ; fY(y) = fxy (x, y) dx F

Y/

X

YX

x

=

P{Yy, Xx}P

{

X

x}

=F(x,y)

F(

x)

fY/X(y/x) = f(x,y)f(x)

Independence :- X & Y are said to be independent if FXY(x , y) = FX(x) FY(y)fXY(x, y) = fX(x) . fX(y) P { X x, Y y} = P { X x} . P{Y y}Correlation:

Corr{ XY} = E {XY} = fxy (x, y). xy. dx dyIf E { XY} = 0 then X & Y are orthogonal .

Uncorrelated :-Covariance = Cov {XY} = E { (X - x) (Y- y}

= E {xy} E {x} E{y}.If covariance = 0 E{xy} = E{x} E{y}




26/147

Institute of Engineering Studies (IES,Bangalore) Signals & Systems Formula Sheet Independence uncorrelated but converse is not true.Random Process:-Take 2 random process X(t) & Y(t) and sampled @ t

1, t

X(t1), X(t) , Y(t1) , Y (t) random variablesAuto correlation Rx(t1, t) = E {X(t1) X(t) }Auto covariance Cx(t1, t) = E { X(t1) - x(t1)) (X(t) - x(t) } = Rx(t1, t) - x(t1) x(t)cross correlation Rxy(t1, t) = E { X(t1) Y(t) }cross covariance Cxy(t1, t) = E{ X(t1) - x(t1)) (Y(t) - y(t) } = Rxy(t1, t) - x(t1) y(t)CXY(t1, t) = 0 Rxy(t1, t) = x(t1) y(t) Un correlated RXY(t1, t) = 0 Orthogonal cross correlation 0 FXY(x, y ! t1, t) = Fx(x! t1) Fy(y ! t) independentProperties of Auto correlation :-

Rx(0) = E { x} Rx() = Rx(-) even | Rx() | Rx(0)

Cross Correlation

Rxy() = Ryx(-) Rxy () Rx(0) . Ry(0) 2 | R

xy(

)|

R

x(0) + R

y(0)

Power spectral Density :-

P.S.D Sx(j) = Rx ()ejdRx() = 1 (j)ejd

Sy(j) = Sx(j) |H(j)| Power = Rx(0) = 1 (j)d Rx() = k () white processProperties : Sx(j) even Sx(j) 0




27/147

Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet

DC MACHINES : -

Lap Winding Wave Winding

(1)Coil Span : Y

cs=

SP Y

cs=

SP

(2)Back Pitch Yb= U Ycs Yb= U Ycs(3)Commutator Pitch Yc= 1

for Progressive winding

Yc = -1for Retrogressive winding

Yc= (c+)p for Progressivewinding

Yc= -(c+)p for Restrogressive winding

(YcMust be integer)(4)Front Pitch Y=Yb+2

for Progressive winding

Y

=Y

b-2

for Retrogressive winding

Y=2Yc - Yb

(5)Parallel Paths A = P A = 2(6)Conductor Current Ic=IA Ic=I (7)No of brushes No of brushes = A = P No of brushes = 2

S = No of commutator segments

P = No of poles

U = No of coil sides / No of poles = CS C = No of coils on the rotor A = No of armature parallel paths

Ia= Armature current Distribution factor (Kd) = phasorsumcoilemarthematicsumocoilem= chordarc = Pitch factor (Kp) =elecrricalangleocoil *100%

electrical = P

mechanical

Armature mmf/Pole (Peak) , ATa= ZIAP AT (Compensating Winding) =

ZIAP* polearcpolepitch AT(Inter pole) = ATa+ Blgi

Where Bi= Flux density in inter pole airgapLeading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/PracticeBranches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email:[email protected]



28/147


lgi= length of inter pole airgap , =4 107 No of turns in each interpole , Ninterpole= AT(Interpole)I The no of compensating conductor per pole, Ncw/pole = ZAP( polearcpolepitch) The Mechanical power that is converted is given by Pconv = Tindm

Where T = Induced torque

m= Angular speed of the machines rotor The resulting electric power produced P

conv = E

AI

A

The power balance equation of the DC Machine is Tindm= EAIA The induced emf in the armature is Ea= ZNP6A Torque developed in Dc machine , Te= PZAIa

Where = Flux\pole , Z = No of armature conductors , P = No of poles , N = Speed in rpm ,A = No of armature parallel paths, Ia = Armature current

The terminal voltage of the DC generator is given by V

t= E

a- I

aR

a

The terminal voltage of the DC motor is given by Vt= Ea+ IaRa Speed regulation of dc machine is given by ,SR = * 100 % = NN * 100 % Voltage regulation , VR =

VV * 100 %Shunt Generator:

For a shunt generator with armature induced voltage Ea, armature current Iaandarmature resistance Ra, the terminal voltage V is:

V = Ea- IaRa

The field current Iffor a field resistance Rfis:If= V / Rf

The armature induced voltage Eaand torque T with magnetic flux at angularspeed are:Ea= kf= kmT = kfIa= kmIawhere kfand kmare design coefficients of the machine.




29/147


Note that for a shunt generator:- induced voltage is proportional to speed,- torque is proportional to armature current.

The airgap power Pefor a shunt generator is:Pe= T = EaIa= kmIa

Series Generator:

For a series generator with armature induced voltage Ea, armature current Ia,armature resistance Raand field resistance Rf, the terminal voltage V is:V = Ea- ( IaRa+ IaRf)= Ea- Ia(Ra+ Rf)The field current is equal to the armature current.

The armature induced voltage Eaand torque T with magnetic flux at angular

speed are:

Ea= kfIa= kmIa

T = kfIa2= kmIa2where kfand kmare design coefficients of the machine.

Note that for a series generator:- induced voltage is proportional to both speed and armature current,- torque is proportional to the square of armature current,- armature current is inversely proportional to speed for a constant Ea

The airgap power Pefor a series generator is:Pe= T = EaIa= kmIa2

Cumulatively compounded DC generator : - ( long shunt)

(a) Ia= I+ IL(b)Vt= Ea- Ia(Ra+ Rs)(c) Is= VR= shunt field current(d)The equivalent effective shunt field current for this machine is given by

Is=Is + NN Ia- (ArmaturereactionMMFN )Where Nse= No of series field turnsN = = No of shunt field turns




30/147


Differentially compounded DC generator : - ( long shunt)

(a) Ia= I+ IL(b) V

t= E

a- I

a(R

a+ R

s)

(c) Is= VR= shunt field current(d) The equivalent effective shunt field current for this machine is given by

Is=Is - NN Ia- (ArmaturereactionMMFN )Where Nse= No of series field turnsN = = No of shunt field turns

Shunt Motor:

For a shunt motor with armature induced voltage Ea, armature current Iaandarmature resistance Ra, the terminal voltage V is:V = Ea+ IaRaThe field current Iffor a field resistance Rfis:If= V / Rf

The armature induced voltage Eaand torque T with magnetic flux at angularspeed are:Ea= kf= kmT = kfIa= kmIawhere kfand kmare design coefficients of the machine.

Note that for a shunt motor:- induced voltage is proportional to speed,- torque is proportional to armature current.

The airgap power Pefor a shunt motor is:Pe= T = EaIa= kmIa

The speed of the shunt motor , = V- R()TWhere K =

PZA

Series Motor:

For a series motor with armature induced voltage Ea, armature current Ia,armature resistance Raand field resistance Rf, the terminal voltage V is:




31/147


V = Ea+ IaRa+ IaRf= Ea+ Ia(Ra+ Rf)The field current is equal to the armature current.

The armature induced voltage Eaand torque T with magnetic flux at angular

speed are:

Ea= kfIa= kmIa

T = kfIa2= kmIa2where kfand kmare design coefficients of the machine.

Note that for a series motor:- induced voltage is proportional to both speed and armature current,- torque is proportional to the square of armature current,

- armature current is inversely proportional to speed for a constant Ea

The airgap power Pefor a series motor is:Pe= T = EaIa= kmIa2

Losses:

constant losses (P k) =Pw f+ Pi o

Where,Pio= No of load core loss Pw= Windage & friction loss Variable losses (P

v) = P

c+ P

s

t+ P

b

where Pc= Copper losses = IaRaPst = Stray load loss = IPb= Brush Contact drop = VbIa, Where Vb= Brush voltage drop

The total machine losses , PL= Pk+VbIa+ KvIaEfficiency

The per-unit efficiency of an electrical machine with input power Pin, outputpower Poutand power loss Plossis:

= Pout/ Pin= Pout/ (Pout+ Ploss) = (Pin- Ploss) / Pin

Rearranging the efficiency equations:




32/147


Pin= Pout+ Ploss= Pout/ = Ploss/ (1 - )

Pout= Pin- Ploss= Pin= Ploss/ (1 - )

Ploss= Pin- Pout= (1 - )Pin= (1 - )Pout/

Temperature Rise:

The resistance of copper and aluminium windings increases with temperature,and the relationship is quite linear over the normal range of operatingtemperatures. For a linear relationship, if the winding resistance is R1attemperature 1and R2at temperature 2, then:

R1/ (1- 0) = R2/ (2- 0) = (R2- R1) / (2- 1)where 0is the extrapolated temperature for zero resistance.

The ratio of resistances R2and R1is:R2/ R1= (2- 0) / (1- 0)

The average temperature rise of a winding under load may be estimated frommeasured values of the cold winding resistance R1at temperature 1(usuallyambient temperature) and the hot winding resistance R2at temperature 2, using:= 2- 1= (1- 0) (R2- R1) / R1

Rearranging for per-unit change in resistance Rpurelative to R1:Rpu= (R2- R1) / R1= (2- 1) / (1- 0) = / (1- 0)

.Copper Windings:

The value of 0for copper is - 234.5 C, so that:= 2- 1= (1+ 234.5) (R2- R1) / R1

If 1is 20 C and is 1 degC:Rpu= (R2- R1) / R1= / (1- 0) = 1 / 254.5 = 0.00393

The temperature coefficient of resistance of copper at 20 C is 0.00393 perdegC.

Aluminium Windings :




33/147


The value of 0for aluminium is - 228 C, so that:= 2- 1= (1+ 228) (R2- R1) / R1

If 1is 20 C and is 1 degC:Rpu= (R2- R1) / R1= / (1- 0) = 1 / 248 = 0.00403

The temperature coefficient of resistance of aluminium at 20 C is 0.00403 perdegC.

Dielectric Dissipation Factor:

If an alternating voltage V of frequency f is applied acrossan insulation systemcomprising capacitance C and equivalent series loss resistance RS, then thevoltage VRacross RSand the voltage VCacross C due to the resultingcurrent I are:VR= IRSVC= IXCV = (VR2+ VC2)

The dielectric dissipation factor of the insulation system is the tangent of thedielectric loss angle between VCand V:tan= VR/ VC= RS/ XC= 2fCRSRS= XCtan= tan/ 2fC

The dielectric power loss P is related to the capacitive reactive power QCby:

P = I2

RS= I2

XCtan= QCtan

The power factor of the insulation system is the cosine of the phaseangle between VRand V:cos= VR/ Vso that and are related by:+ = 90

tanand cosare related by:tan= 1 / tan= cos/ sin= cos/ (1 - cos2)so that when cosis close to zero, tancos

TRANSFORMERS:

Gross cross sectional area = Area occupied by magnetic material + Insulationmaterial.




34/147


Net cross sectional area = Area occupied by only magnetic material excluding areaof insulation material.

Hence for all calculations, net cross sectional area is taken since (flux) majorly

flows in magnetic material. = BAn

Specific weight of t/f =WeightoVAratingo

Stacking/iron factor :- (ks) = NetCrossSectionalareaGrossCrossSectionalarea k

sis always less than 1

Gross c.s Area = AG= length breadth Net c.s Area = An= ks AGUtilization factor of transformer core =

EectiveC.S.AreaTotalC.SArea U.F of cruciform core = 0.8 to0.85

Flux =mmFReluctance= = m sin t

According to faradays second law e = Nddt= N ddt m sin t

Transformer emf equations :-

E= 4.44 NBmaxAnf (1)E= 4.44 NBmaxAnf (2)

Emf per turn in Iry= EN= 4.44 BmaxAnf Emf per turn in IIry = EN= 4.44 BmaxAnf

Emf per turn on both sides of the transformer is sameEN = EN EE = NN = kTransformation ratio = K =

EE =

Instantaneous value

of emf in primarye = Nm sint 2




35/147

Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula SheetNNTurns ratio =

= N

N

For an ideal two-winding transformer with primary voltage V1appliedacross N1primary turns and secondary voltage V2appearing across N2secondaryturns:V1/ V2= N1/ N2

The primary current I1and secondary current I2are related by:I1/ I2= N2/ N1= V2/ V1

For an ideal step-down auto-transformer with primary voltage V1appliedacross (N1+ N2) primary turns and secondary voltage V2appearingacross N2secondary turns:V1/ V2= (N1+ N2) / N2

The primary (input) current I1and secondary (output) current I2are related by:

I1/ I2= N2/ (N1+ N2) = V2/ V1.

For a single-phase transformer with rated primary voltage V1, rated primarycurrent I1, rated secondary voltage V2and rated secondary current I2, the voltampererating S is:S = V1I1= V2I2

For a balanced m-phase transformer with rated primary phase voltage V1, ratedprimary current I1, rated secondary phase voltage V2and rated secondary current I2,the voltampere rating S is:S = mV1I1= mV2I2

The primary circuit impedance Z1referred to the secondary circuit for an idealtransformer with N1primary turns and N2secondary turns is:Z12= Z1(N2/ N1)2

During operation of transformer :-

Bm E V B

max= constant

V

= constant

Equivalent ckt of t/f under N.L condition :-




36/147


No load current = I = I + Iw = IIw = I cos I

= I

sin

No load power = vI cos = vIw =Iron losses.R = vI ; X = vI Iw = NoloadpowerV Transferring from to :-

IR = IRR

= R

II

= R R = RFrom to :-

IR = I. RR = II . R

R = R. K Total resistance ref to primary = R

+ R

R = R + R/k Total resistance ref to secondary = R + RR = R + kR

Total Cu loss = IROr

IR

RR

No load /shunt branch.

N N

EEV

II

w

R XI




37/147


Per unit resistance drops :-

P.U primary resistance drop =IR

E

P.U secondary resistance drop = I

RE Total P.U resistance drop ref to Iry= IRE Total P.U resistance drop ref to IIry= IRE The P.U resistance drops on both sides of the t/f is sameIRE = IRE Losses present in transformer :-

1. Cu losses in t/f:

Total Cu loss = IR + IR= IR= I

R

Rated current on Iry = VAratingot/E Similarly current on IIry= VAratingot/E

Cu losses Ior I. Hence there are called as variable losses. P.U Full load Cu loss =

FLCulossinwattsVAratingot/ =

IREI If VA rating of t/f is taken as base then P.U Cu loss Ias remaining terms are constant. P.U Cu loss at x of FL = x

PUFL Cu loss

P. U resistance drop ref to Iry

or

P. U resistance ref to Iry = IRE II

=IREI

1. Copper losses

2. Iron losses

3. Stray load losses

4. Dielectric losses

major losses

minor losses

t/f windings

t/f corecu parts

Iron parts

insulating materials.




38/147

Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula SheetP.U Resistance drop = P.U FL cu loss% FL Cu loss = % R = % Resistance drop.

Iron (or) Core losses in t/f :-

1. Hysteresis loss :

Steinmetz formula :-

Where

= stienmetz coefficientBmax= max. flux density in transformer core.f = frequency of magnetic reversal = supply freq.v = volume of core material

x = Hysteresis coeff (or) stienmetz exponent

= 1.6 (Si or CRGo steel)

2. Eddycurrent loss:

Eddy current loss ,(We) Rce IeAs area decreases in laminatedcore resistance increases as a result conductivity decreases.

During operation of transformer :-

Bm V Case (i) :-

V = constant, Bmax= const.we fwe = B f

Const.

wi = wh + wewi = Af + Bf

When Bmax= const.

W

e= K. B

max f. t

ConstantSupply freq

thickness of laminations.

it is a function of

Area under one hysteresis loop. Bmaxx . f . vWh=




39/147


Case (ii) :-V constant, Bm const.

we V . fw

e

V

wi = wh + wewi = AV.. + BV

P.U iron loss :-

P.U iron loss =IronlossinwattsVAratingot/

As VA rating is choosen as base then the P.U iron loss are also constant at all load conditions.

To find out constant losses :-

W= Losses in t/f under no load condition= Iron losses + Dielectric loss + no load primary loss (IR)

Constant losses = W IRWhere , R= LV winding resistance.

To find out variable losses :-

Wsc= Loss in t/f under S.C condition= F.L Cu loss + stray load losses (Cu and Iron) + Iron losses in both wdgs

Variable losses = WSCIron losses corresponding to VCCO.C test :-

VratedWiS.C test :-

V

SC (W

i)

S.

C

Wi VW(W) = VratedV (Wi)S.C = Wi VV Variable losses= WSC (Wi)SC VV Leading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/PracticeBranches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email:[email protected]



40/147


Under the assumption that small amount of iron losses corresponds to VSC and stray loadlosses are neglected the wattmeter reading in S.C test can be approximately taken as F.L

Cu losses in the transformer.

W

se

F.L Cu loss

ISC . RR = WI Efficiency :-

Efficiency of transformer is given by =outputpowerinputpower

=outputpoweroutputpower+losses

=

E

I

cos

EI cos

+F.Lculosses+Ironlosses

F.L = EI cos EI cos +IR+W

Transformer efficiency = VA cosVA cos+w+Culosses Efficiency = =

x MVA Px MVA P+w x+w Total losses in transformer =

1output

Condition for maximum effieciency is, Cu losses = Iron losses Total losses at

max= 2W

i

%load at which maximum efficiency occurs % x = WIR*100 %= IronlossF.L.culoss*100 % KVA corresponding to max= F.L KVA IronlossF.Lculoss

x o F.L = x (EI) cos x (EI) cos + x(I R) + Wi

O.C testS.C test




41/147


allday =O hrs

Voltage drop in t/f at a Specific load p.f = I

R

cos

I

X

sin

% Voltage regulation = IR cos IX sinV 100= IRV cos IXV sin

P.U resistance P.U reactance

% Regulation = (P.U R) cos + (P.U X) sin 100Condition for max. regulation :-

% regulation = (% R) cos (% X) sin dregnd = 0Tan = % X% R = XR = Tan XRlagging

At maximum regulation = Tan XR= Tan XRXR = XR

Value of maximum regulation :-

% Regulation = (% R) cos + (% X) sin At max. regulation cos = % R% Z

Sin = % X% Zmax. % regulation = (% R)

% R% Z + (% X)(% X)% Z

=(%

R)

+(%

X)

% Z

=(% Z)

% Z max. % regn = % Z

= % of rated voltage required to produce rated short ckt current

.




42/147


Condition for zero regulation:-

If the voltage regulation in the t/f is zero, the t/f voltages are maintained at their nominalvalues even under load condition

% Regn = (% R) cos

(% X) sin

For zero regulation occurs at leading p.fs(% R) cos (% X)sin = 0

Tan = % R% X = Tan % R% X = Tan RXleading.

At zero regulation condition : = Tan XR = Tan RXXR

=RX

Regulation at x of FL = x [% R cos %X sin ]= x F.L regn

Regulation at U.P.F:-

Regulation at UPF = % R

= % F.L Cu loss

Scott Connection:

VAM= 0.866 VVAN =V3= 0.577 VVMN= 0.866 V0.577 V= 0.289 V

I

86.6%

0.289

2

:

1

M

N

0.866 V0.577 VA

B

C

V IBCV

b

i

b

ia

VaIa

IB

IC

.663

IA2




43/147


VAN VMN= 0.577 V0.289 V= 2 : 1

If a neutral pt is located on 3side, such that, voltage between any terminal to that neutralpoint is 0.577 V

then such neutral point divides the primary of teaser transformer in the ratio

of 2 : 1

Location of neutral point from top = 0.866 N 3 Location of neutral point from bottom = 0.866 N 3

Operation of Scott Connection with 2balanced load at UPF :-Teaser t/f :-

Ii = N.66NIa = N.66N iaLet N N = 1 1

IA = 1.15 iaMain t/fIi = NN

IBC = NN ibLet N: N = 1

1

I

BC= i

bIB = IBC IA IC = IBC IA Capacity of Scott Connection :-

(KVA)Scott = 3 VLILV = V IL = I

Vol. rating of 1 t/f Current rating of 1 t/f

(KVA)Scot = 3 VI(KVA)Scott = 3 (KVA)

Utilization factor =(

VA)

availableVA= 3VIVI = 0.866 Utilization factor of Scott connection with 2 identical 1 t/fs is 86.6%

AUTO TRANSFORMER:




44/147


Primary applied voltage, Vab= Secondary voltage Vreferred to primary + primary leakage impedancedrop + secondary leakage impedance drop ref. to primary.

V

ab=

NN

N V

+ I

(r

+ jx

) + (I

I

)(r

+ j x

)

NN

N

K of auto transformer = LVHV(KVA)induction = (V V)I

I/P KVA = VI(VA)i/pVA = (VV)IVI

= 1 LVHV

= 1 K

(KVA) induction = (1 K) i/p KVA(KVA) conduction = I/p KVA (KVA)ind

(KVA)conduction = K I/p KVA Wt. of conductor in section AB of auto t/f (N N)I Wt of conductor in section BC of auto t/f (I I)N

Total wt. of conductor in auto t/f is

I(N N) + (I I)N 2 (N N)I Total wt. of conductor in 2 wdg transformer IN + IN2 IN

wt.oconductorinanautot/wt.oconductorinwdgt/ = (NN)INI = 1

NN= 1 K

Wt. of conductor in auto t/f = (1 K) (wt. of conductor in 2 wdg t/f)

Thus saving of conductor material if auto t/f is used} = K {conductor wt in 2 wdg transformer. (% FL losses)Autot/ = (1 K)(% FL losses)wdgt/ (% Z)AT = (1 K)(% Z)wdgt/ (KVA)AT = (KVA)wdgt/.




45/147


SYNCHRONOUS MACHINES:

Principle of operation :-

Whenever a conductor cuts the magnetic flux, an emf is induced in that conductor

Faradays law of electromagnetic induction.

Coil span () :- It is the distance between two sides of the coil. It is expressed in terms ofdegrees, pole pitch, no. of slots / pole etc

Pole pitch :- It is the distance between two identical points on two adjacent poles.Pole pitch is always 180 e = slots / pole.

elec = Pmech Slot pitch or slot angle :- (T)Slot angle is the angle for each slot.

For a machine with P poles and s no. of slots, the slot angle = =

P(

)

d

Pitch factor or coil span factor or chording factor :- (KP)KP= Theeminduced| coilinshortpitchedwindingTheeminduced|coilinullpitchedwinding

=Thevectorsumoinducedem| coilArithmeticsumoinducedem| coil

KP = Ecos/

E

Pitch factor for nthharmonic i.e,

chording angle to eliminate nthharmonics ()= n coil spam to eliminate nthharmonics ,() = 180 nn Distribution factor | spread factor | belt factor | breadth factor(kd) :-

Kd= TheeminducedwhenthewindingisdistributedTheeminducedwhenthewindingisconcentratedKd= VectorsumoeminducedArithmeticsumoeminduced

=

Kp= cos /2Kp=cosn




46/147


Kd =sinmsin

The distribution factor for uniformly distributed winding is

For nthharmonic, kdn= sinmsin To eliminate nthharmonics ,phase spread (m) = 36n Generally, KVA rating, power output kd and Eph(induce emf) kd. Tph.

VA(

3)

VA(3)=

Pout(

3)

Pout(3)=

kdkd=

sinsin

mm=

sin3

sin6

6= 1.15

VA(3)VA() = Pout(3)Pout() = kdkd = sinsin 6= 1.06

VA(3)VA() = Pout(3)Pout() = kdkd = sinsin

6 = 1.5VA()VA() = PoutPout = kdkd = sin sin = 1.414

Speed of space harmonics of order (6k 1) is (6k) . Nswhere Ns= synchronous speed = p The order of slot harmonics is SP 1where S = no. of slots , P = no. of poles

Slot harmonics can be eliminated by skewing the armature slots and fractional slot winding.

The angle of skew =

s= (slot angle)

= 2 harmonic pole pitches

= 1 slot pitch.

Distribution factor for slot harmonics, kdsp 1

kd= sim




47/147


Is kd = sinmsin i.e., same that of fundamental Pith factor for slot harmonics, k

p sp

1

= k

p = cos 2

The synchronous speed Nsand synchronous angular speedsof a machine with p polepairs running on a supply of frequency fsare:

s= 2fs/ p

Slip S =NNN

Where N

S=

p

= synchronous speed

The magnitude of voltage induced in a given stator phase is Ea= 2 Nc f= KWhere K = constant The output power Pmfor a load torque Tmis:

Pm= sTm

The rated load torque TMfor a rated output power PMis:

TM= PM/ s= PMp/ 2fs= 120PM/ 2Ns

Synchronous Generator:

For a synchronous generator with stator induced voltage Es, stator current Isandsynchronous impedance Zs, the terminal voltage V is:

V = E - IsZs= Es- Is(Rs+ jXs)

where Rsis the stator resistance and Xsis the synchronous reactance

E = (Vcos + IaRa) + (Vsin IaXs)+ lag p.fleading p.f.




48/147


Synchronous Motor:

For a synchronous motor with stator induced voltage Es, stator current Isand synchronousimpedance Zs, the terminal voltage V is:

V = Es+ IsZs= Es+ Is(Rs+ jXs)where Rsis the stator resistance and Xsis the synchronous reactance

Voltage regulation :

% regulation =|E||V|

|V| 100E V = IaZs% regulation = EVV

=

I

ZV 100

regulation ZsAs Zsincreases, voltages regulation increases. Condition for zero | min. voltage regulation is,Cos ( + ) = IZV Condition for max. Voltage regulation is, = Short circuit ratio (SCR) =

II = Z(adjusted)| = X(adjusted)|SCR

X

Armature

reaction

Voltage regulation Armature reaction SCR VoltageregulationSmall value of SCR represent poor regulation.a = armaturemmreluctance But reluctance

Air gap

a = armaturemmairgap a Airgaplength

Armature reaction a AirgaplengthLeading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/PracticeBranches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email:[email protected]



49/147


SCR Armaturereaction Airgap length

machine size SCR.Cost SCR

Power =EVX sin

P X SCR

Large value of SCR represent more power output. Synchronizing power coefficient or stability factor Psyisgiven asPsy= dpd = dd EVX sin

=EVX cos

Psyis a measure of stabilitystability PsyBut Psy X SCR

Stability SCR Air gap length

When the stator mmf is aligned with the d axis of field poles then flux dperpole is set upand the effective reactance offered by the alternator is Xd.Xd = maximumVoltageminimumcurrent = (V)line(at min. I )3I(min ) = Direct axis reactance

When the stator mmf is aligned with the q axis of field poles then flux qper pole is set upand the effective reactance offered by the alternator is Xq.Xq = minimumvoltgemaximumvoltage = Vline(atmaximumI )3I(max ) = Quadrature axis reactance

Air gap length SCR

Power SCR

Stability

SCR

Stability Air gap length




50/147


Cylindrical rotor Synchronous machine ,

The per phase power delivered to the infinite bus is given by P =EVX sin

Salient pole synchronous machine ,

The per phase power delivered to the infinite bus is given by

P =EVX sin + V X X sin2

Condition for max. power:-

For cylindrical rotor machine :-

At constant Vtand E, the condition for max. power is obtained by putting dpd= 0 dpd = EVX cos = 0Cos = 0

= 90

Hence maximum power occurs at = 90

For salient pole synchronous machine :-dpd= 0

VE

Xcos

+ Vt

X

Xcos

2= 0

Cos = EXVXX + EXVXXThe value of load angle is seed to be less than 90. max. power occurs at < 90

Synchronizing power = Psy. .=

EVX cos . . Synchronizing torque =

Synchronizingpower

.

Power flow in Alternator : -

Complex power = S = P + jQ = VIaWhere Active power flow (P) =

EVZ cos( ) VZ cos ;Leading coaching center for GATE/IES/PSU in Bangalore & All over India for Online Tests/PracticeBranches: Jayanagar & Malleshwaram Ph: 0 99003 99699 Email:[email protected]



51/147


Reactive power flow (Q) =EVZ sin( ) VZ sin ;

Condition for max. power output :-

P =

EVZcos(

)

V

Zcos

dpd= 0 for max power conditionie = 0

If Ra= 0; = = 90 ; then max power is given by

SYNCHRONOUS MOTORS:

Speed regulation =N.N.N. 100

=NNN 100= 0%

Slip S =NNN = NNN = 0%

NS =

p

The speed can be controlled by varying the frequencyVvFratio control is preferred for rated torque operation

Power flow in synchronous motor is given by

complex power i/p s = p + jQ = V Iawhere P = Resl power flow , Q = Reactive power flow

:

:If Ra= 0 ; ZS = XS; 90

Condition for max power :-

=

Pmax= EVZ VZcos

Pin= EVX sin Q = VX [V Ecos ]

P =VZ cos EVZ cos( + ) Q = VZ sin EVZ sin( + )




52/147


Pin = VZ cos EVZ cos( + )dPd = 0 0 + EVZ sin( + ) = 0Sin ( + ) = 0 = sin 180

Expression for mechanical power developed :-

Mechanical power developed = Pm= active component + [E Ia]

Condition for max. mechanical power developed :-dpd = EVZ sin( )= 0Sin ( ) = 0 = sin 0

=

This is the expression for the mechanical power developed interms of load angle and the

internal machine angle , for constant voltage Vphand constant E i.e., excitation Gross Torque =

Pmw =

Pm Ns = synchronous speed in r.p.mTg = 6 . PmN

Condition for excitation when motor develops Pmax:-For max power developed isdPmdE = 0ddt EVZ cos( ) EZ cos = 0 Condition for excitation when motor develops Pmaxis , Eb = VZR

= 180 0Pmax = VZ cos + EVZ

Pm = EVZ cos( ) + EZ cos

Pmmax = EVZ EZ cos

Tg =.55PmN




53/147


The corresponding value of max. power is

Pmax= VR VRPower flow in synchronous motors :-

For leading p.f

tan =

I

X cos +

R sinV+IX sinR cos

The mechanical power developed per phase is given by,s

Pm = EVX sin + V X X sin2

INDUCTION MACHINES:

The power flow diagram of 3 induction motor is

Pin

3VLIL cos

(input)

Stator

copper loss

3 I

aph

R

a

Mechanical power

developed in armature

P

m= 3E

bph. I

aphcos

Documents

eee formula sheet