initial commit / add all books

20 files changed, 227 insertions, 0 deletions

diff --git a/3681/CH5/EX5.12/Ans5_12.PNG b/3681/CH5/EX5.12/Ans5_12.PNG
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diff --git a/3681/CH5/EX5.12/Ex5_12.sce b/3681/CH5/EX5.12/Ex5_12.sce
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+// Calculating the resistance of secondary winding

+clc;

+disp('Example 5.12, Page No. = 5.89')

+// Given Data

+sp = 2.2;// Current density of primary winding(in Ampere per mm square)

+ss = 2.1;// Current density of secondary winding(in Ampere per mm square)

+rp = 8;// Resistance of primary inding (in ohm)

+R1 = 1/1.1;// Since length of mean turn of primary is 10% than that of the secondary

+R2 = 1/10;// Since ratio of transformation is 10:1

+// Calculation of the resistance of secondary winding

+rs = R2*R2*(ss/sp)*R1*rp;// Resistance of secondary winding (ohm)

+disp(rs,'Resistance of secondary winding (ohm)=');

+//in book answer is 0.0694 ohm.  The answers vary due to round off error

diff --git a/3681/CH5/EX5.13/Ans5_13.PNG b/3681/CH5/EX5.13/Ans5_13.PNG
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index 000000000..9e6f37822
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+++ b/3681/CH5/EX5.13/Ans5_13.PNG
diff --git a/3681/CH5/EX5.13/Ex5_13.sce b/3681/CH5/EX5.13/Ex5_13.sce
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index 000000000..601315a10
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+// Calculating the leakage reactance of the transformer referred to the h.v. side

+clc;

+disp('Example 5.13, Page No. = 5.89')

+// Given Data

+// 6600/400 V, delta/star 3-phase core type transformer

+Q = 300;// kVA rating

+f = 50;// Frequency (in Hz)

+u0 = 4*%pi*10^(-7);

+Tp = 830;// h.v winding turns

+Lmt = 0.9;// Length of mean turn (in meter)

+Lc = 0.5;// Height of coils (in meter)

+a = 0.015;// Width of duct between h.v and l.v. windings (in meter)

+bp = 0.025;// Width of h.v. winding (in meter)

+bs = 0.016;// Width of l.v. winding (in meter)

+// Calculation of the leakage reactance of the transformer referred to the h.v. side

+Xp = 2*%pi*f*u0*Tp*Tp*Lmt/Lc*(a+(bp+bs)/3);// Leakage reactance referred to the primary side (ohm)

+disp(Xp,'(a) Leakage reactance referred to the primary side (ohm)=');

+// If the l.v. winding divided into two parts, one on each side of h.v. winding

+Xp = %pi*f*u0*Tp*Tp*Lmt/Lc*(a+(bp+bs)/6);// Leakage reactance referred to the primary side (ohm)

+disp(Xp,'(b) Leakage reactance referred to the primary side (ohm)=');

+//in book answers are 14 ohm and 5.36 ohm respectively.  The answers vary due to round off error

diff --git a/3681/CH5/EX5.14/Ans5_14.PNG b/3681/CH5/EX5.14/Ans5_14.PNG
new file mode 100644
index 000000000..e6e6ff8be
--- /dev/null
+++ b/3681/CH5/EX5.14/Ans5_14.PNG
diff --git a/3681/CH5/EX5.14/Ex5_14.sce b/3681/CH5/EX5.14/Ex5_14.sce
new file mode 100644
index 000000000..a5e47dcf7
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+++ b/3681/CH5/EX5.14/Ex5_14.sce
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+// Calculating the per unit leakage reactance

+clc;

+disp('Example 5.14, Page No. = 5.90')

+// Given Data

+// 2000/400 V, single phase shell type transformer

+Q = 100;// kVA rating

+f = 50;// Frequency (in Hz)

+u0 = 4*%pi*10^(-7);

+Tp = 200;// h.v winding turns

+Lmt = 1.5;// Length of mean turn (in meter)

+W = 0.12;// Width of winding (in meter)

+a = 0.016;// Width of duct between h.v and l.v. windings (in meter)

+bp = 0.04;// Width of h.v. winding (in meter)

+bs = 0.036;// Width of l.v. winding (in meter)

+// Calculation of the per unit leakage reactance

+Xp = %pi*f*u0*Tp*Tp/2*Lmt/W*(a+(bp+bs)/6);// Leakage reactance referred to the primary side (ohm)

+I_hv = Q*10^(3)/2000;// H.V. winding current at full load (in ampere)

+Xp_pu = Xp*I_hv/2000;// Per unit leakage reactance

+disp(Xp_pu,'Per unit leakage reactance=');

+//in book answer is 0.0353.  The answers vary due to round off error

diff --git a/3681/CH5/EX5.16/Ans5_16.PNG b/3681/CH5/EX5.16/Ans5_16.PNG
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index 000000000..5b3164df8
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+++ b/3681/CH5/EX5.16/Ans5_16.PNG
diff --git a/3681/CH5/EX5.16/Ex5_16.sce b/3681/CH5/EX5.16/Ex5_16.sce
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index 000000000..a25af31af
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+++ b/3681/CH5/EX5.16/Ex5_16.sce
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+// Calculating the instantaneous radial force on the h.v. winding if a short circuit occurs at the terminals of the l.v. winding with h.v. energised and the force at full load

+clc;

+disp('Example 5.16, Page No. = 5.97')

+// Given Data

+// 6600/400 V, delta/star 3-phase core type transformer

+Q = 1000;// kVA rating

+f = 50;// Frequency (in Hz)

+u0 = 4*%pi*10^(-7);

+T = 500;// h.v winding turns

+Lmt = 1.3;// Length of mean turn (in meter)

+Lc = 0.6;// Height of winding (in meter)

+m = 1.8;// Doubling effect multiplier

+// Calculation of the per unit leakage reactance

+I_fl = Q*1000/(3*6600);// Full load current per phase on h.v. side (in Ampere)

+i = m*2^(1/2)*(1/0.05)*I_fl;// Instantaneous peak value of short circuit current (in Ampere)

+Fr = u0/2*(i*T)^(2)*Lmt/Lc;// Total instantaneous radial force on the h.v. coil (in N)

+disp(Fr,'Total instantaneous radial force on the h.v. coil (N)=');

+Fr = u0/2*(I_fl*T)^(2)*Lmt/Lc;// Force at full load (in N)

+disp(Fr,'Force at full load (N)=');

+disp('This shows that the forces under short circuit conditions are considerably large as compared with forces at full load')

+//in book answers are 2330000 (N) and 866 (N).  The answers vary due to round off error

diff --git a/3681/CH5/EX5.17/Ans5_17.PNG b/3681/CH5/EX5.17/Ans5_17.PNG
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index 000000000..63a2db3f8
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+++ b/3681/CH5/EX5.17/Ans5_17.PNG
diff --git a/3681/CH5/EX5.17/Ex5_17.sce b/3681/CH5/EX5.17/Ex5_17.sce
new file mode 100644
index 000000000..e1421e0e1
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+++ b/3681/CH5/EX5.17/Ex5_17.sce
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+// Calculating the instantaneous radial force and instantaneous axial force on the h.v. winding under short circuit conditions

+clc;

+disp('Example 5.17, Page No. = 5.98')

+// Given Data

+// 7500/435 V, single phase core type transformer

+Q = 575;// kVA rating

+f = 50;// Frequency (in Hz)

+u0 = 4*%pi*10^(-7);

+Z_pu = 0.036;// Per unit impedance

+T = 190;// h.v winding turns

+Lmt = 1.25;// Length of mean turn (in meter)

+Lc = 0.35;// Height of coils (in meter)

+m = 1.8;// Doubling effect multiplier

+a = 0.015;// Width of duct (in meter)

+bp = 0.027;// Width of h.v. winding (in meter)

+bs = 0.023;// Width of l.v. winding (in meter)

+k = 0.05;// Since the h.v. winding is 5% shorter than the l.v. winding at one end

+// Calculation of the instantaneous radial force

+I_fl = Q*1000/7500;// Rms value of full load current (in Ampere)

+i = m*2^(1/2)*(1/Z_pu)*I_fl;// Instantaneous peak value of short circuit current (in Ampere)

+Fr = u0/2*(i*T)^(2)*Lmt/Lc;// Instantaneous radial force on the h.v. coil (in N)

+disp(Fr,'(a) Instantaneous radial force on the h.v. winding (N)=');//in book answer is 2380000 (N).  The answers vary due to round off error

+// Calculation of the instantaneous axial force

+Fa = u0/2*k*(i*T)^(2)*Lmt/(2*(a+bp+bs));// Total instantaneous radial force on the h.v. coil (in N)

+disp(Fa,'(b) Instantaneous axial force on the h.v. winding (N)=');

+//in book answer is 3200000 (N).  The provided in the textbook is wrong

+disp('This shows that there is a very large axial force, even though one of the winding is only 5% shorter than the other at one end')

diff --git a/3681/CH5/EX5.18/Ans5_18.PNG b/3681/CH5/EX5.18/Ans5_18.PNG
new file mode 100644
index 000000000..1a0613e2c
--- /dev/null
+++ b/3681/CH5/EX5.18/Ans5_18.PNG
diff --git a/3681/CH5/EX5.18/Ex5_18.sce b/3681/CH5/EX5.18/Ex5_18.sce
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index 000000000..486fd9017
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+++ b/3681/CH5/EX5.18/Ex5_18.sce
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+// Calculating the maximum flux and no load current of the transformer

+clc;

+disp('Example 5.18, Page No. = 5.99')

+// Given Data

+Ep = 400;// Primary winding voltage (in volts)

+f = 50;// Frequency (in Hz)

+A = 2.5*10^(-3);// Area of cross section (in meter square)

+Sf = 0.9;// Stacking factor

+Tp = 800;// Primary winding turns

+li = 2.5;// Length of the flux path (in meter)

+u0 = 4*%pi*10^(-7);// Permeability of free space

+ur = 1000;// Relative ermeability

+D = 7.8*10^(3);// Density of iron (in kg per meter cube)

+FD_w = 2.6;// Working flux density (in W per kg)

+// Calculation of the maximum flux

+Ai = Sf*A;// Net iron area (in meter square)

+Bm = Ep/(4.44*f*Ai*Tp);// Maximum flux density of core (in Wb per meter square)

+Fm = Bm*Ai;// Maximum flux in the core (in Wb)

+disp(Fm,'Maximum flux in the core (Wb)=');

+// Calculation of the no load current

+AT0 = li/(ur*u0)*Bm;// Magnetic mmf (in A)

+Im = AT0/(2^(1/2)*Tp);// Magnetising current (in A)

+V = Ai*li;// Volume of the core (in meter cube)

+W = V*D;// Weight of core (in kg)

+Pi = W*FD_w;// Iron loss (in W)

+Il = Pi/Ep;// Loss component of no load current (in A)

+I0 =(Im*Im+Il*Il)^(1/2);// No load current (in A)

+disp(I0,'No load current (Ampere)=');

+//in book answers are 0.00225 (Wb) and 1.77 (Ampere) respectively.  The answers vary due to round off error

diff --git a/3681/CH5/EX5.20/Ans5_20.PNG b/3681/CH5/EX5.20/Ans5_20.PNG
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index 000000000..05a99d7d7
--- /dev/null
+++ b/3681/CH5/EX5.20/Ans5_20.PNG
diff --git a/3681/CH5/EX5.20/Ex5_20.sce b/3681/CH5/EX5.20/Ex5_20.sce
new file mode 100644
index 000000000..4f3410a03
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+++ b/3681/CH5/EX5.20/Ex5_20.sce
@@ -0,0 +1,30 @@
+// Calculating the number of turns and no load current

+clc;

+disp('Example 5.20, Page No. = 5.101')

+// Given Data

+E = 6600;// Primary winding voltage (in volts)

+f = 60;// Frequency (in Hz)

+Ai = 22.6*10^(-3);// Area of cross section (in meter square)

+Bm = 1.1;// Maximum flux density of core (in Wb per meter square)

+Af = 1.52;// Amplitude factor

+Tp = 800;// Primary winding turns

+l = 2.23;// Mean length (in meter)

+mmf =232;// mmf per meter (in A per meter)

+n = 4;// Number of lap joints

+Gs = 7.5*10^(3);// Specific gravity of plates

+Ls = 1.76;// Specific loss (in W per kg)

+// Calculation of the number of turns

+Tp = E/(4.44*f*Ai*Bm);// Number of turns

+disp(Tp,'(a) Number of turns=');

+// Calculation of the no load current

+mmf_iron = mmf*l;// Mmf required for iron parts

+mmf_joints = 4*(1/4)*mmf;// Mmf required for joints.  Since lap joints takes 1/4 times reactive mmf as required per meter of core

+AT0 = mmf_iron+mmf_joints;// Total magnetising mmf (in A)

+Kpk = Af*2^(1/2);// Peak factor

+Im = AT0/(Kpk*Tp);// Magnetising current (in A)

+W = Ai*l*Gs;// Weight of core (in kg)

+Pi = Ls*W;// Iron loss (in W)

+Il = Pi/E;// Loss component of no load current (in A)

+I0 =(Im*Im+Il*Il)^(1/2);// No load current (in A)

+disp(I0,'(b) No load current (Ampere)=');

+//in book answers are 1100 and 0.333 (A) respectively.  The provided in the textbook is wrong

diff --git a/3681/CH5/EX5.3/Ans5_3.PNG b/3681/CH5/EX5.3/Ans5_3.PNG
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index 000000000..2a6909480
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+++ b/3681/CH5/EX5.3/Ans5_3.PNG
diff --git a/3681/CH5/EX5.3/Ex5_3.sce b/3681/CH5/EX5.3/Ex5_3.sce
new file mode 100644
index 000000000..2c1bf0a53
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+++ b/3681/CH5/EX5.3/Ex5_3.sce
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+// Calculating the kVA output of a single phase transformer

+clc;

+disp('Example 5.3, Page No. = 5.78')

+// Given Data

+D = 0.4;// Distance between core centres (in meter)

+f = 50;// Frequency (in Hz)

+Bm = 1.2;// Flux density of core (in Wb per meter square)

+Kw = 0.27;// Window space factor

+s = 2.3;// Current density (in Ampere per mm square)

+R1 = 2.8;// Ratio of core height and distance between core centres

+R2 = 0.56;// Ratio of circumscribing circle and distance between core centres

+R3 = 0.7;// Ratio of net iron area and area of circumscribing circle

+// Calculation of the kVA output of a single phase transformer

+Hw = R1*D;// Core heightor window height (in meter)

+d = R2*D;// Diameter of circumscribing circle (in meter)

+Ww = D-d;// Width of window (in meter)

+Aw = Hw*Ww;// Area of window (in meter square)

+A = (%pi/4)*d*d;// Area of circumscribing circle (in meter square)

+Ai = R3*A;// Net iron area (in meter square)

+Q = 2.22*f*Bm*Kw*s*10^(6)*Aw*Ai*10^(-3);// kVA output of a single phase transformer

+disp(Q,'kVA output of a single phase transformer (kVA)=');

+//in book answer is 450 kVA.  The answers vary due to round off error

diff --git a/3681/CH5/EX5.6/Ans5_6.PNG b/3681/CH5/EX5.6/Ans5_6.PNG
new file mode 100644
index 000000000..5c475d26a
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+++ b/3681/CH5/EX5.6/Ans5_6.PNG
diff --git a/3681/CH5/EX5.6/Ex5_6.sce b/3681/CH5/EX5.6/Ex5_6.sce
new file mode 100644
index 000000000..c602d176e
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+++ b/3681/CH5/EX5.6/Ex5_6.sce
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+// Calculating the net iron area and window area and full load mmf

+clc;

+disp('Example 5.6, Page No. = 5.80')

+// Given Data

+Q = 400;// kVA rating

+R = 2.4*10^(-6);// Ratio of flux to full load mmf

+f = 50;// Frequency (in Hz)

+Bm = 1.3;// Maximum flux density of core (in Wb per meter square)

+Kw = 0.26;// Window space factor

+s = 2.7;// Current density (in Ampere per mm square)

+// Calculation of the net iron area

+K = (4.44*f*R*10^(3))^(1/2);

+Et = K*Q^(1/2);// Voltage per turn (in Volts)

+Flux = Et/(4.44*f);// Flux (in Wb)

+Ai = Flux/Bm;// Net iron area (in meter square)

+disp(Ai,'Net iron area (meter square)=');

+// Calculation of the net window area

+Aw = Q/(2.22*f*Bm*Kw*s*10^(6)*Ai*10^(-3));// Window area (in meter square)

+disp(Aw,'Window area (meter square)=');

+// Calculation of the full load mmf

+AT = Flux/R;// Full load mmf (in A)

+disp(AT,'Full load mmf (A)=');

+//in book answers are 0.0507 (meter square), 0.0777 (meter square) and 27500 (A) respectively.  The answers vary due to round off error

diff --git a/3681/CH5/EX5.9/Ans5_9.PNG b/3681/CH5/EX5.9/Ans5_9.PNG
new file mode 100644
index 000000000..419561861
--- /dev/null
+++ b/3681/CH5/EX5.9/Ans5_9.PNG
diff --git a/3681/CH5/EX5.9/Ex5_9.sce b/3681/CH5/EX5.9/Ex5_9.sce
new file mode 100644
index 000000000..a54096047
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+++ b/3681/CH5/EX5.9/Ex5_9.sce
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+// Calculating the net iron area and window area

+clc;

+disp('Example 5.9, Page No. = 5.82')

+// Given Data

+Q = 400;// kVA rating

+f = 50;// Frequency (in Hz)

+Bm = 1.5;// Maximum flux density of core (in Wb per meter square)

+Kw = 0.12;// Copper space factor

+s = 2.2;// Current density (in Ampere per mm square)

+gc = 8.9*10^(3);// Density of copper (in kg per meter cube)

+gi = 7.8*10^(3);// Density of iron (in kg per meter cube)

+R1 = 0.5;// Ratio of length of mean turn of copper to length of mean flux path

+R2 = 4;// Ratio of weight of iron to weight of copper

+// Calculation of the net iron area

+C = (1/2.22*R1*gc/gi*10^(3))^(1/2);// Flux (in Wb)

+Ai = C*(Q*R2/(f*Bm*s*10^(6)))^(1/2);// Net iron area (in meter square)

+disp(Ai,'Net iron area (meter square)=');

+// Calculation of the net window area

+Aw = Q/(2.22*f*Bm*Kw*s*10^(6)*Ai*10^(-3));// Window area (in meter square)

+disp(Aw,'Window area (meter square)=');

+//in book answers are 0.0478 (meter square) and 0.183 (meter square) respectively.  The answers vary due to round off error