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| author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
| commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
| tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3681 | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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initial commit / add all books
Diffstat (limited to '3681')
160 files changed, 1707 insertions, 0 deletions
diff --git a/3681/CH10/EX10.13/Ans10_13.PNG b/3681/CH10/EX10.13/Ans10_13.PNG Binary files differnew file mode 100644 index 000000000..713882089 --- /dev/null +++ b/3681/CH10/EX10.13/Ans10_13.PNG diff --git a/3681/CH10/EX10.13/Ex10_13.sce b/3681/CH10/EX10.13/Ex10_13.sce new file mode 100644 index 000000000..2975cdc4c --- /dev/null +++ b/3681/CH10/EX10.13/Ex10_13.sce @@ -0,0 +1,17 @@ +// Calculating the number of stator and rotor turns and rotor voltage between slip rings at standstill
+clc;
+disp('Example 10.13, Page No. = 10.35')
+// Given Data
+// 3 phase induction motor
+Nss = 54;// Number of stator slots
+Nrs = 72;// Number of rotor slots
+V = 400;// Applied voltage across the stator terminals
+// Calculation of the number of stator and rotor turns and rotor voltage between slip rings at standstill
+Ts = Nss*8/6;// Stator turns per phase. Since 8 conductors per slot
+Tr = Nrs*4/6;// Rotor turns per phase. Since 4 conductors per slot
+Es = 400/3^(1/2);// Stator voltage per phase
+Er = Es*Tr/Ts;// Rotor voltage per phase at standstill
+disp(Ts,'Stator turns per phase =');
+disp(Tr,'Rotor turns per phase =');
+disp(3^(1/2)*Er,'Rotor voltage between slip rings at standstill (Volts)=');
+//in book answers are 72, 48 and 266.7 Volts respectively. The answers vary due to round off error
diff --git a/3681/CH10/EX10.15/Ans10_15.PNG b/3681/CH10/EX10.15/Ans10_15.PNG Binary files differnew file mode 100644 index 000000000..d6e22fa3d --- /dev/null +++ b/3681/CH10/EX10.15/Ans10_15.PNG diff --git a/3681/CH10/EX10.15/Ex10_15.sce b/3681/CH10/EX10.15/Ex10_15.sce new file mode 100644 index 000000000..fe94cc837 --- /dev/null +++ b/3681/CH10/EX10.15/Ex10_15.sce @@ -0,0 +1,19 @@ +// Calculating the number of stator turns per phase
+clc;
+disp('Example 10.15, Page No. = 10.44')
+// Given Data
+// 3 phase star connected induction motor
+P = 75;// Power rating (in kw)
+V = 3000;// Voltage rating
+f = 50;// Frequency (in Hz)
+p = 8;// Number of poles
+AT60 = 500;// mmf required for flux density at 30 degree from pole axis
+Kws = 0.95;// Winding factor
+e = 0.94;// Full load efficiency
+pf = 0.86;// Full load power factor
+// Calculation of the number of stator turns per phase
+I = P*10^(3)/(3^(1/2)*V*e*pf);// Full load current (in ampere)
+Im = 0.35*I;// Magnetizing current (in Ampere). Since magnetizing current is 35% of full load current
+Ts = 0.427*p*AT60/(Kws*Im);// Stator turns per phase
+disp(Ts,'Stator turns per phase =');
+//in book answer is 288. The answers vary due to round off error
diff --git a/3681/CH10/EX10.16/Ans10_16.PNG b/3681/CH10/EX10.16/Ans10_16.PNG Binary files differnew file mode 100644 index 000000000..2807a0d4c --- /dev/null +++ b/3681/CH10/EX10.16/Ans10_16.PNG diff --git a/3681/CH10/EX10.16/Ex10_16.sce b/3681/CH10/EX10.16/Ex10_16.sce new file mode 100644 index 000000000..8b0fa7ea7 --- /dev/null +++ b/3681/CH10/EX10.16/Ex10_16.sce @@ -0,0 +1,36 @@ +// Calculating the magnetizing current per phase
+clc;
+disp('Example 10.16, Page No. = 10.44')
+// Given Data
+// 3 phase delta connected induction motor
+P = 75;// Power rating (in kw)
+V = 400;// Voltage rating
+f = 50;// Frequency (in Hz)
+p = 6;// Number of poles
+D = 0.3;// Diameter of motor core (in meter)
+L = 0.12;// Length of motor core (in meter)
+Nss = 72;// Number of stator slots
+Nc = 20;// Number of conductors per slot
+lg = 0.55;// Length of air gap (in meter)
+Kg = 1.2// Gap constraction factor
+Coil_Span = 11;// Coil span (slots)
+// Calculation of the magnetizing current per phase
+q = Nss/(3*p);// Slots per pole per phase
+Kd = sin(60/2*%pi/180)/(q*sin(60/(2*4)*%pi/180));// Distribution factor
+Ns_pole = Nss/p;// Slots per pole
+alpha = 1/Ns_pole*180;// Angle of chording (in degree). Since the winding is chorded by 1 slot pitch
+Kp = cos(alpha/2*%pi/180);// Pitch factor
+Kws = Kd*Kp;// Stator winding factor
+Ns = Nss*Nc;// Total stator conductors
+Ts = Ns/(3*2);// Stator turns per phase
+Eb = V;// Stator voltage per phase. Since machine is delta connected
+Fm = Eb/(4.44*f*Ts*Kws);// Flux per pole (in Wb)
+A = %pi*D*L/p;// Area per pole (in meter square)
+Bav = Fm/A;// Average air gap density (in Wb per meter square)
+Bg60 = 1.36*Bav;// Gap flux density at 30 degree from pole axis
+ATg = 800000*Bg60*Kg*lg*10^(-3);// Mmf required for air gap (in A)
+ATi = 0.35*ATg;// Mmf for iron parts (in A). Since mmf required for iron parts is 35% of air gap mmf
+AT60 = ATg+ATi;// Total mmf (in A)
+Im = 0.427*p*AT60/(Kws*Ts);// Magnetizing current per phase (in ampere)
+disp(Im,'Magnetizing current per phase (Ampere) =');
+//in book answer is 4.56 Ampere. The answers vary due to round off error
diff --git a/3681/CH10/EX10.19/Ans10_19.PNG b/3681/CH10/EX10.19/Ans10_19.PNG Binary files differnew file mode 100644 index 000000000..3bf5d16f2 --- /dev/null +++ b/3681/CH10/EX10.19/Ans10_19.PNG diff --git a/3681/CH10/EX10.19/Ex10_19.sce b/3681/CH10/EX10.19/Ex10_19.sce new file mode 100644 index 000000000..fff460a55 --- /dev/null +++ b/3681/CH10/EX10.19/Ex10_19.sce @@ -0,0 +1,27 @@ +// Calculating the current in rotor bars and in end rings
+clc;
+disp('Example 10.19, Page No. = 10.50')
+// Given Data
+p = 6;// Number of poles
+ms = 3;// Number of phases of stator
+Nss = 72;// Number of stator slots
+Nc = 15;// Number of conductors per slot
+Sr = 55;// Number of stator slots
+Is = 24.1;// Stator current (in Ampere)
+Coil_Span = 11;// Coil span (slots)
+pf = 0.83;// Power factor
+// Calculation of the current in rotor bars and in end rings
+q = Nss/(ms*p);// Stator slots per pole per phase
+Kd = sin(60/2*%pi/180)/(q*sin(60/(2*4)*%pi/180));// Distribution factor
+Ns_pole = Nss/p;// Slots per pole
+alpha = 1/Ns_pole*180;// Angle of chording (in degree). Since the winding is chorded by 1 slot pitch
+Kp = cos(alpha/2*%pi/180);// Pitch factor
+Kws = Kd*Kp;// Stator winding factor
+Ir_ = Is*pf;// Stator current equivalent to rotor current (in Ampere)
+Ns = Nss*Nc;// Total stator conductors
+Ts = Ns/(ms*2);// Stator turns per phase
+Ib = 2*ms*Kws*Ts*Ir_/Sr;// Current in each rotor bar (in Ampere)
+Ie = Sr*Ib/(%pi*p);// Current in each end ring (in Ampere)
+disp(Ib,'Current in each rotor bar (Ampere) =');
+disp(Ie,'Current in each end ring (Ampere) =');
+//in book answers are 375.4 Ampere and 1095.3 Ampere respectively. The answers vary due to round off error
diff --git a/3681/CH10/EX10.2/Ans10_2.PNG b/3681/CH10/EX10.2/Ans10_2.PNG Binary files differnew file mode 100644 index 000000000..ae212a782 --- /dev/null +++ b/3681/CH10/EX10.2/Ans10_2.PNG diff --git a/3681/CH10/EX10.2/Ex10_2.sce b/3681/CH10/EX10.2/Ex10_2.sce new file mode 100644 index 000000000..3c3edc928 --- /dev/null +++ b/3681/CH10/EX10.2/Ex10_2.sce @@ -0,0 +1,25 @@ +// Calculating the main dimentions of squirrel cage induction motor
+clc;
+disp('Example 10.2, Page No. = 10.14')
+// Given Data
+P = 15;// Power rating (in kW)
+V = 400;// Voltage rating (in Volts)
+rpm = 2810;// r.p.m.
+f = 50;// Frequency (in Hz)
+e = 0.88;// Efficiency
+pf = 0.9;// Full load power factor
+ac = 25000;// Specific electrical loading (in A per meter)
+Bav = 0.5;// Specific magnetic loading (in Wb per meter square)
+Kw = 0.955;
+// the rotor peripheral speed is approximately 20 meter per second at synchronous speed
+// Calculation of the main dimentions of squirrel cage induction motor
+Q = P/(e*pf);// kVA input
+Co = 11*Kw*Bav*ac*10^(-3);// Output co-efficient
+ns = 3000/60;// Synchronous speed corresponding to 50 Hz (in r.p.s.)
+D2L = Q/(Co*ns);// Product of D^(2)*L
+D = 20/(%pi*ns);// Since the rotor diameter in an induction motor is almost equal to stator bore
+L = D2L/(D*D);
+disp('Main dimentions of squirrel cage induction motor')
+disp(D,'D (meter)=');
+disp(L,'L (meter)=');
+//in book answers are 0.1257 meter and 0.177 meter respectively. The answers vary due to round off error
diff --git a/3681/CH11/EX11.10/Ans11_10.PNG b/3681/CH11/EX11.10/Ans11_10.PNG Binary files differnew file mode 100644 index 000000000..3ad97acde --- /dev/null +++ b/3681/CH11/EX11.10/Ans11_10.PNG diff --git a/3681/CH11/EX11.10/Ex11_10.sce b/3681/CH11/EX11.10/Ex11_10.sce new file mode 100644 index 000000000..47b50e1b3 --- /dev/null +++ b/3681/CH11/EX11.10/Ex11_10.sce @@ -0,0 +1,31 @@ +// Calculating the size of armature wire and the a.c. resistance of each pahase
+clc;
+disp('Example 11.10, Page No. = 11.34')
+// Given Data
+// 3 phase star connected synchronous generator
+p = 8;// Number of poles
+f = 50;// Frequency (in Hz)
+ys = 0.3;// Pole pitch (in meter)
+Iz = 100;// Line current (in Ampere)
+L = 0.3;// Gross axial length (in meter)
+Spp =3;// Slots per pole per phase
+Cs = 6;// Conductors per slot
+Kc_av = 1.3;// Average eddy current loss factor
+// Calculation of the suitable number of slots and conductors per slot
+D = ys*p/%pi;// Armature diameter (in meter)
+ns = 2*f/p;// Synchronous speed (in r.p.s.)
+Va = %pi*D*ns;// Peripheral speed (in meter per second)
+S = Spp*3*p;// Total number of slots
+Z = S*Cs;// Total number of conductors
+Tph = Z/6;// Turns per phase
+ac = Iz*Z/(%pi*D);// (in Ampere per meter)
+J = (43000/ac)+(Va/16);// Current density (in Ampere per mm square)
+as = 100/J;// Area of armature conductor
+disp(as,'(a) Area of armature conductor (mm square)=');
+L_active = 2*L;// Active length of each turn (in meter)
+Lmt = 2*L_active;// Since Total length of a turn is twice the active length (in meter)
+resistivity = 0.021;// Resistivity of copper at 75 degree celsius (in ohm per meter)
+r_dc = resistivity*Tph*Lmt/as;// D.C. resistance of each phase at 75 degree celsius (in ohm)
+r_ac = Kc_av*r_dc;// A.C. resistance of each phase
+disp(r_ac,'(b) A.C. resistance of each phase (ohm)=');
+//in book answers are 23.8 mm square and 0.099 ohm respectively. The answers vary due to round off error
diff --git a/3681/CH11/EX11.11/Ans11_11.PNG b/3681/CH11/EX11.11/Ans11_11.PNG Binary files differnew file mode 100644 index 000000000..e0d58bd6a --- /dev/null +++ b/3681/CH11/EX11.11/Ans11_11.PNG diff --git a/3681/CH11/EX11.11/Ex11_11.sce b/3681/CH11/EX11.11/Ex11_11.sce new file mode 100644 index 000000000..58deef421 --- /dev/null +++ b/3681/CH11/EX11.11/Ex11_11.sce @@ -0,0 +1,26 @@ +// Calculating the length of air gap
+clc;
+disp('Example 11.11, Page No. = 11.35')
+// Given Data
+// 3 phase silient pole alternator
+kVA = 500;// kVA rating
+V = 3.3;// Voltage rating (in kV)
+f = 50;// Frequency (in Hz)
+rpm = 600;// R.p.m.
+Tph = 180;// Turns per phase
+Bav = 0.54;// Average flux density (in Wb per meter square)
+SCR = 1.2;// Short circuit ratio
+Kw = 0.955;// Winding factor
+Kg = 1.15;// Gap constraction factor
+Kf = 0.65;// Since field form factor is equal to the ratio of pole arc to pole pitch
+// Calculation of the length of air gap
+ns = rpm/60;// Synchronous speed (in r.p.s.)
+p = 2*f/ns;// Number of poles
+Iph = kVA*1000/(3^(1/2)*V*1000);// Armature diameter (in meter)
+ATa = 2.7*Iph*Tph*Kw/p;// Armature mmf per pole (in A)
+AT_f0 = SCR*ATa;// No load field mmf per pole
+Bg = Bav/Kf;// Maximum flux density in air gap (in Wb per meter square)
+lg = 0.8*AT_f0/(800000*Bg*Kg);// Length of air gap
+// Since mmf required for gap is 80% of no load field mmf
+disp(lg*1000,'Length of air gap (mm)=');
+//in book answer is 5.2 mm. The answers vary due to round off error
diff --git a/3681/CH11/EX11.13/Ans11_13.PNG b/3681/CH11/EX11.13/Ans11_13.PNG Binary files differnew file mode 100644 index 000000000..358dec20e --- /dev/null +++ b/3681/CH11/EX11.13/Ans11_13.PNG diff --git a/3681/CH11/EX11.13/Ex11_13.sce b/3681/CH11/EX11.13/Ex11_13.sce new file mode 100644 index 000000000..06f8f19c9 --- /dev/null +++ b/3681/CH11/EX11.13/Ex11_13.sce @@ -0,0 +1,41 @@ +// Calculating the stator bore and stator core length and turns per phase and armature mmf per pole and mmf for air gap and field current
+clc;
+disp('Example 11.13, Page No. = 11.37')
+// Given Data
+// 3 phase synchronous generator
+Q = 1250;// kVA rating
+E = 3300;// Voltage rating (in kV)
+f = 50;// Frequency (in Hz)
+rpm = 300;// R.p.m.
+Bav = 0.58;// Specific magnetic loading (in Wb per meter square)
+ac = 33000;// Specific electric loading (in Ampere per meter)
+lg = 5.5;// Gap length (in mm)
+T_field = 60;// Field turns per pole
+SCR = 1.2;// Short circuit ratio
+Kw = 0.955;// Winding factor
+Va = 30;// Peripheral speed (in meter per second)
+// Calculation of the stator bore and stator core length and turns per phase and armature mmf per pole and mmf for air gap and field current
+ns = rpm/60;// Synchronous speed (in r.p.s.)
+p = 2*f/ns;// Number of poles
+Co = 11*Kw*Bav*ac*10^(-3);// Output co-efficient
+D2L = Q/(Co*ns);// Product of D*D*L
+D = Va/(%pi*ns);// Stator bore (in meter)
+disp(D,'Stator bore (meter) =');
+L = D2L/D^(2);// Stator core length (in meter)
+disp(L,'Stator core length (meter)=');
+A_pole = %pi*D*L/p;// Area per pole
+F_pole = Bav*A_pole;// Flux per pole
+Eph = E/3^(1/2);// Voltage per phase
+Tph = int(Eph/(4.44*f*F_pole*Kw));// Turns per phase
+disp(Tph,'Turns per phase =');
+Iph = Q*1000/(3^(1/2)*E);// Current per phase
+ATa = 2.7*Iph*Tph*Kw/p;// Armature mmf per pole (in A)
+disp(ATa,'Armature mmf per pole (Ampere)=');
+A_effective = 0.6*A_pole;// Effective gap area is 0.6 times the actual area
+KgBg = F_pole/A_effective;// Effective gap density (in Wb per meter square)
+mmf_airgap = 800000*KgBg*lg*10^(-3);// Mmf for air gap (in A)
+disp(mmf_airgap,'Mmf for air gap (Ampere)=');
+AT_f0 = SCR*mmf_airgap;// No load field mmf per pole
+If = AT_f0 /T_field;// Field current at no load
+disp(If,'Field current at no load (Ampere)=');
+//in book answers are 1.9 meter, 0.345 meter, 150, 4240 ampere, 4250 ampere and 85 ampere respectively. The answers vary due to round off error
diff --git a/3681/CH11/EX11.14/Ans11_14.PNG b/3681/CH11/EX11.14/Ans11_14.PNG Binary files differnew file mode 100644 index 000000000..75fb2b4be --- /dev/null +++ b/3681/CH11/EX11.14/Ans11_14.PNG diff --git a/3681/CH11/EX11.14/Ex11_14.sce b/3681/CH11/EX11.14/Ex11_14.sce new file mode 100644 index 000000000..fe62a34c3 --- /dev/null +++ b/3681/CH11/EX11.14/Ex11_14.sce @@ -0,0 +1,44 @@ +// Calculating the flux per pole and length and width of pole and winding height and pole height
+clc;
+disp('Example 11.14, Page No. = 11.40')
+// Given Data
+// 3 phase star connected selient pole alternator
+Q = 2500;// kVA rating
+E = 2400;// Voltage rating (in kV)
+f = 60;// Frequency (in Hz)
+rpm = 225;// R.p.m.
+D = 2.5;// Stator bore (in meter)
+L = 0.44;// Core length (in meter)
+Nspp = 3;// Number of slot per pole per phase
+Ncs = 4;// Number of conductors per slot
+a = 2;// Circuits per phase
+Bp = 1.5;// Flux density in pole core (in Wb per meter square)
+df = 30;// Depth of winding (in mm)
+Sf = 0.84;// Field widind space factor
+Cl = 1.2;// Leakage factor
+Kw = 0.95;// Winding factor
+qf =1800;// Loss dissipated by field winding
+h_insulation = 30;// Height of insulation
+// Calculation of the flux per pole and length and width of pole and winding height and pole height
+ns = rpm/60;// Synchronous speed (in r.p.s.)
+p = 2*f/ns;// Number of poles
+S = 3*p*3.5;// Total number of slots
+Z = Ncs*S;// Total number of conductors
+Tph = int(Z/6);// Turns per phase
+Eph =E/3^(1/2);// Voltage per phase
+F_pole = Eph*a/(4.44*Tph*f*Kw);// Flux per pole (in Wb)
+disp(F_pole,'(a) Flux per pole (Wb) =');
+Fp = Cl*F_pole;// Flux in pole body (in Wb)
+Ap = Fp/Bp;// Area of pole body (in meter square)
+Lp = L;// Length of pole body = Length of armature core
+bp = Ap/Lp;// Width of pole body
+disp(Lp,'(b) Length of pole body (meter) =');
+disp(bp,' Width of pole body (meter) =');
+Iph = Q*1000/(3^(1/2)*E);// Current in each phase
+Iz = Iph/a;// Current in each conductor
+ATa = 2.7*Iz*Tph*Kw/p;// Armature mmf per pole (in A)
+AT_fl = 2*ATa;// Field mmf at full load (in A)
+hf = AT_fl/(10^(4)*(Sf*df*10^(-3)*qf)^(1/2));// Height of field winding (in meter)
+disp(hf,'(c) Height of field winding (meter) =');
+disp(hf+h_insulation*10^(-3),'(d) Height of pole (meter) =');
+//in book answers are 0.049 Wb, 0.44 meter, 0.089 meter, 0.16 meter and 0.19 meter respectively. The answers vary due to round off error
diff --git a/3681/CH11/EX11.18/Ans11_18.PNG b/3681/CH11/EX11.18/Ans11_18.PNG Binary files differnew file mode 100644 index 000000000..bf388dceb --- /dev/null +++ b/3681/CH11/EX11.18/Ans11_18.PNG diff --git a/3681/CH11/EX11.18/Ex11_18.sce b/3681/CH11/EX11.18/Ex11_18.sce new file mode 100644 index 000000000..a7e7622ab --- /dev/null +++ b/3681/CH11/EX11.18/Ex11_18.sce @@ -0,0 +1,34 @@ +// Calculating the direct and quadrature axis synchronous reactances
+clc;
+disp('Example 11.18, Page No. = 11.52')
+// Given Data
+// 3 phase star connected selient pole alternator
+Q = 2500;// kVA rating
+E = 2400;// Voltage rating (in kV)
+f = 60;// Frequency (in Hz)
+p = 32;// Number of poles
+D = 2.5;// Stator bore (in meter)
+L = 0.44;// Core length (in meter)
+Tph = 224;// Turns per phase
+lg = 10;// Air gap length (in meter)
+Kg = 1.11;// Air gap constraction factor
+Kw = 0.95;// Winding factor
+R = 0.69;// Ratio of pole arc to pole pitch
+A1 = 1.068;// Ratio of amplitude of fundamental of gap flux density to maximum gap density
+Xl = 0.14;// Per unit leakage reactance
+// Calculation of the direct and quadrature axis synchronous reactances
+xm = 7.54*f*Tph*Tph*Kw*Kw*D*L/(p*p*lg*10^(-3)*Kg)*10^(-6);// Magnetic reactance per phase (in ohm)
+Eph =E/3^(1/2);// Voltage per phase
+Iph = Q*1000/(3^(1/2)*E);// Current in each phase
+Xm = Iph*xm/Eph;// Per unit magnetising reactance
+a = R*%pi;// Angle embraced by pole arc (in rad)
+pd = (a+sin(a))/(4*sin(a/2));// Reduction factor for direct axis armature mmf
+Ad1 = pd*A1;// Flux distribution factor for direct axis
+Xad = Ad1*Xm;// Per unit direct axis armature reaction reactance
+Aq1 = ((4*R+1)/5)-(sin(R*%pi)/%pi);// Flux distribution co-efficient for quadrature axis
+Xaq = Aq1*Xm;// Per unit quadrature axis armature reaction reactance
+Xd = Xl+Xad;// Per unit direct axis synchronous reactance
+Xq = Xl+Xaq;// Per unit quadrature axis synchronous reactance
+disp(Xd,'Per unit direct axis synchronous reactance =');
+disp(Xq,'Per unit quadrature axis synchronous reactance =');
+//in book answers are 0.916 and 0.533 respectively. The answers vary due to round off error
diff --git a/3681/CH11/EX11.20/Ans11_20.PNG b/3681/CH11/EX11.20/Ans11_20.PNG Binary files differnew file mode 100644 index 000000000..b7f55017e --- /dev/null +++ b/3681/CH11/EX11.20/Ans11_20.PNG diff --git a/3681/CH11/EX11.20/Ex11_20.sce b/3681/CH11/EX11.20/Ex11_20.sce new file mode 100644 index 000000000..a56be267e --- /dev/null +++ b/3681/CH11/EX11.20/Ex11_20.sce @@ -0,0 +1,32 @@ +// Calculating the kVA output of the machine
+clc;
+disp('Example 11.20, Page No. = 11.56')
+// Given Data
+// 3 phase turbo-alternator
+rpm = 3000;// R.p.m.
+f = 50;// Frequency (in Hz)
+L = 0.94;// Core length (in meter)
+Bav = 0.45;// Average gap density (in Wb per meter sqaure)
+ac = 25000;// Ampere conductors per meter
+Va = 100;// Peripheral speed of rotor (in meter per second)
+lg = 20;// Length of air gap (in mm)
+Kw = 0.95;// Winding factor
+// Winding is infinitely distributed with a phase spread of 60 degree
+// Calculation of the kVA output of the machine
+ns = rpm/60;// R.p.s
+Dr = Va/(%pi*ns);// Diameter of rotor (in meter)
+D = Dr+(2*lg*10^(-3));// Stator bore (in meter)
+// for full pitch
+Kd = 0.955;// Distribution factor
+Kp = 1;// Pitch factor
+Kw = Kd*Kp;// Winding factor
+Q = 11*Kw*Bav*ac*D*D*L*ns*10^(-3);// kVA output
+disp(Q,'(a) kVA output of machine (kVA)=');
+// for chorded by 1/3 pole pitch
+alpha = 180/3;// Angle of chording
+Kp = cos(alpha*%pi/180/2);// Pitch factor
+Kd = 0.955;// Distribution factor
+Kw = Kd*Kp;// Winding factor
+Q = 11*Kw*Bav*ac*D*D*L*ns*10^(-3);// kVA output
+disp(Q,'(b) kVA output of machine (kVA)=');
+//in book answers are 2480 kVA and 2147 kVA respectively. The provided in the textbook is wrong
diff --git a/3681/CH11/EX11.32/Ans11_32.PNG b/3681/CH11/EX11.32/Ans11_32.PNG Binary files differnew file mode 100644 index 000000000..377caf3b5 --- /dev/null +++ b/3681/CH11/EX11.32/Ans11_32.PNG diff --git a/3681/CH11/EX11.32/Ex11_32.sce b/3681/CH11/EX11.32/Ex11_32.sce new file mode 100644 index 000000000..d560a36dd --- /dev/null +++ b/3681/CH11/EX11.32/Ex11_32.sce @@ -0,0 +1,32 @@ +// Calculating the number of stator slots and average flux density
+clc;
+disp('Example 11.32, Page No. = 11.58')
+// Given Data
+// 3 phase star connected direct watercooled generator
+Q = 588;// MVA rating
+E = 22000;// Voltage rating
+p =2;// Number of poles
+rpm = 2500;// R.p.m.
+f = 50;// Frequency (in Hz)
+D = 1.3;// Stator bore (in meter)
+L = 6;// Core length (in meter)
+Nc =2;// Number of conductors per slot
+a = 2;// Circuits per phase
+ac = 200000;// Ampere conductors per meter
+Kw = 0.92;// Winding factor
+// Winding is infinitely distributed with a phase spread of 60 degree
+// Calculation of the number of stator slots and average flux density
+ns = rpm/60;// Speed (r.p.s)
+Eph = E/3^(1/2);// Voltage per phase
+Iph = Q*10^(6)/(3^(1/2)*E);// Current per phase
+Is = Iph/a;// Current in each conductor (in ampere)
+Z = %pi*D*ac/Is;// Total number of armature conductors
+Tph = int(Z/6+1);// Turns per phase for a three phase machine
+Z = 6*Tph;// Actual number of conductors used
+S = Z/Nc;// Number of slots
+disp(S,'(a) Number of slots =');
+F_pole = a*Eph/(4.44*f*Tph*Kw);// Flux per pole (in Wb)
+pole_pitch = %pi*D/p;//Pole pitch (in meter)
+Bav = F_pole/(pole_pitch*L);// Average flux density (in Wb per meter square)
+disp(Bav,'(b) Average flux density (Wb per meter square) =');
+//in book answers are 54 and 0.565 Wb per meter square respectively. The answers vary due to round off error
diff --git a/3681/CH11/EX11.4/Ans11_4.PNG b/3681/CH11/EX11.4/Ans11_4.PNG Binary files differnew file mode 100644 index 000000000..de1172609 --- /dev/null +++ b/3681/CH11/EX11.4/Ans11_4.PNG diff --git a/3681/CH11/EX11.4/Ex11_4.sce b/3681/CH11/EX11.4/Ex11_4.sce new file mode 100644 index 000000000..1bdacecba --- /dev/null +++ b/3681/CH11/EX11.4/Ex11_4.sce @@ -0,0 +1,26 @@ +// Calculating the suitable number of slots and conductors per slot
+clc;
+disp('Example 11.4, Page No. = 11.28')
+// Given Data
+// 3 phase star connected alterator (Single layer winding)
+rpm = 300;// R.p.m.
+E = 3300;// Voltage rating (in volts)
+f = 50;// Frequency (in Hz)
+D = 2.3;// Diameter of core (in meter)
+L = 0.35;// Length of core (in meter)
+Bm = 0.9;// Maximum flux density in the air gap (in Wb per meter square)
+// Calculation of the suitable number of slots and conductors per slot
+ns = rpm/60;// Synchronous speed (r.p.s)
+p = 2*f/ns;// Number of poles
+Bav = 2/%pi*Bm;// Average flux density in the air gap (in Wb per meter square)
+Flux_pole = Bav*%pi*D*L/p;// Flux per pole (in Wb)
+Eph = E/3^(1/2);// Voltage per phase (in volts)
+ys = 40;// Slot pitch (in mm). The slot pitch should be nearly 40 mm for 3.3 kV machines
+Kw = 0.955;// Taking winding factor
+Tph = int(Eph/(4.44*f*Flux_pole*Kw));// Turns per phase
+q = int(%pi*D/(3*p*ys*10^(-3)));// Slots per pole per phase
+S = 3*p*q;// Total number of stator slots
+Tph6 = 6*Tph;// Total number of stator conductors
+Zs = int(Tph6/S);// Conductors per slot
+disp(Zs*S,'Total stator conductors used =');
+disp(Zs*S/6,'Turns per phase used=');
diff --git a/3681/CH15/EX15.1/Ans15_1.PNG b/3681/CH15/EX15.1/Ans15_1.PNG Binary files differnew file mode 100644 index 000000000..f0e12a844 --- /dev/null +++ b/3681/CH15/EX15.1/Ans15_1.PNG diff --git a/3681/CH15/EX15.1/Ex15_1.sce b/3681/CH15/EX15.1/Ex15_1.sce new file mode 100644 index 000000000..3bc44e275 --- /dev/null +++ b/3681/CH15/EX15.1/Ex15_1.sce @@ -0,0 +1,17 @@ +// Calculating the current in exciting coil
+clc;
+disp('Example 15.1, Page No. = 15.7')
+// Given Data
+F = 200;// Mass (in kg)
+lg = 5;// Distance (in mm)
+A = 5*10^(-3);// Area of pole face (in meter square)
+T = 3000;// Exciting coil turns
+u0 = 4*%pi*10^(-7);// Permeability of free space
+// Calculation of the current in exciting coil
+B = (F*u0/(0.051*A))^(1/2);// flux density in air gap (in Wb per meter square)
+mmf_air = 800000*B*lg*10^(-3);// Mmf required for air (in A)
+mmf_iron = 0.1*mmf_air;// Mmf required for iron parts (in A). Since mmf required for iron parts is 10% of air gap mmf
+AT = mmf_air+mmf_iron;// Total mmf
+I = AT/T;// Current in exciting coil (in Ampere)
+disp(I,'Current in exciting coil (Ampere) =');
+//in book answer is 1.456 Ampere. The answers vary due to round off error
diff --git a/3681/CH15/EX15.4/Ans15_4.PNG b/3681/CH15/EX15.4/Ans15_4.PNG Binary files differnew file mode 100644 index 000000000..83e23e183 --- /dev/null +++ b/3681/CH15/EX15.4/Ans15_4.PNG diff --git a/3681/CH15/EX15.4/Ex15_4.sce b/3681/CH15/EX15.4/Ex15_4.sce new file mode 100644 index 000000000..9d6fbb645 --- /dev/null +++ b/3681/CH15/EX15.4/Ex15_4.sce @@ -0,0 +1,26 @@ +// Calculating the winding depth and winding space and space factor and the number of turns
+clc;
+disp('Example 15.4, Page No. = 15.9')
+// Given Data
+hf = 80;// in between flanges (in mm)
+Do = 75;// in flange diameter (in mm)
+Di = 30;// in gross diameter tube (in mm)
+a = 0.0357;// Area of copper wire
+d = 0.213;// Diameter of bare conductor (in mm)
+d1 = 0.213+2*0.05;// Diameter of insulated conductor (in mm)
+// Calculation of the winding depth and winding space and space factor and the number of turns
+df = (Do-Di)/2;// Winding depth (in mm)
+Aw = hf*10^(-3)*df*10^(-3);// Winding space
+disp(df,'(a) Winding depth =');
+disp(Aw,' Winding space =');
+disp('(b) for conductors when they bed')
+Sf = 0.9*(d/d1)^(2);// Space factor
+T = Sf*Aw/a*10^(6);// Number of turns
+disp(Sf,' Space factor =');
+disp(T,' Number of turns =');
+disp(' for conductors when they do not bed')
+Sf = 0.78*(d/d1)^(2);// Space factor
+T = Sf*Aw/a*10^(6);// Number of turns
+disp(Sf,' Space factor =');
+disp(T,' Number of turns =');
+//in book answers are 22.5 mm, 0.0018 mm square, 0.417, 21025, 0.361 and 18200. The answers vary due to round off error
diff --git a/3681/CH16/EX16.2/Ans16_2.PNG b/3681/CH16/EX16.2/Ans16_2.PNG Binary files differnew file mode 100644 index 000000000..ce0a35415 --- /dev/null +++ b/3681/CH16/EX16.2/Ans16_2.PNG diff --git a/3681/CH16/EX16.2/Ex16_2.sce b/3681/CH16/EX16.2/Ex16_2.sce new file mode 100644 index 000000000..a37256045 --- /dev/null +++ b/3681/CH16/EX16.2/Ex16_2.sce @@ -0,0 +1,13 @@ +// Calculating the inductance
+clc;
+disp('Example 16.2, Page No. = 16.6')
+// Given Data
+N = 25;// Number of turns
+Ac = 1;// Cross sectional area of the core (in cm square)
+u0 = 4*%pi*10^(-7);// Permeability of free space
+ur = 200;// Relative permeability
+lc = 15;// (in cm)
+// Calculation of the inductance
+L = u0*ur*Ac*10^(-4)*N^(2)/(lc*10^(-2))*10^(6);// Inductance (in micro H)
+disp(L,'Inductance (micro H) =');
+//in book answer is 105 micro H. The answers vary due to round off error
diff --git a/3681/CH18/EX18.1/Ans18_1.pdf b/3681/CH18/EX18.1/Ans18_1.pdf Binary files differnew file mode 100644 index 000000000..d9540724e --- /dev/null +++ b/3681/CH18/EX18.1/Ans18_1.pdf diff --git a/3681/CH18/EX18.1/Ex18_1.sce b/3681/CH18/EX18.1/Ex18_1.sce new file mode 100644 index 000000000..5e89a4063 --- /dev/null +++ b/3681/CH18/EX18.1/Ex18_1.sce @@ -0,0 +1,38 @@ +// Calculating the upper and lower limits of current during starting and resistance of each section
+clc;
+disp('Example 18.1, Page No. = 18.3')
+// Given Data
+// d.c. shunt motor
+P = 37;// Power rating (in kW)
+V = 250;// Voltage rating (in Volts)
+e = 0.84;// Full load efficiency
+rm = 0.2;// Armature circuit resistance (in ohm)
+ns = 8;// Number of studs
+// Maximum torque is 150% of full load torque
+// Calculation of the upper and lower limits of current during starting
+Ifl = P*10^(3)/(V*e);// Full load current (in Ampere)
+I1 = 1.5*Ifl;// Maximum current (in Ampere). Since torque is proportional to current
+n = ns-1;// Number of sections
+alpha = (rm*I1/V)^(1/n);
+I2 = alpha*I1;// Lower limit of current (in Ampere)
+disp(I1,'Upper limit of current (Ampere) =');
+disp(I2,'Lower limit of current (Ampere) =');
+// Calculation of the resistance of each section
+R1 = V/I1;// Total resistance at starting (in ohm)
+r1 = (1-alpha)*R1;
+r2 = alpha*r1;
+r3 = alpha*r2;
+r4 = alpha*r3;
+r5 = alpha*r4;
+r6 = alpha*r5;
+r7 = alpha*r6;
+disp(R1,'Total resistance at starting (ohm) =');
+disp('Resistance of each section')
+disp(r1,'r1 (ohm) =');
+disp(r2,'r2 (ohm) =');
+disp(r3,'r3 (ohm) =');
+disp(r4,'r4 (ohm) =');
+disp(r5,'r5 (ohm) =');
+disp(r6,'r6 (ohm) =');
+disp(r7,'r7 (ohm) =');
+//in book answers are I1 = 264 ampere, I2 = 211 ampere, R1 = 0.947 ohm, r1 = 0.189 ohm,, r2 = 0.151 ohm, r3 = 0.121 ohm, r4 = 0.097 ohm, r5 = 0.077 ohm, r6 = 0.062 ohm, r7 = 0.050 ohm. The answers vary due to round off error
diff --git a/3681/CH3/EX3.1/Ans3_1.PNG b/3681/CH3/EX3.1/Ans3_1.PNG Binary files differnew file mode 100644 index 000000000..3c3bac712 --- /dev/null +++ b/3681/CH3/EX3.1/Ans3_1.PNG diff --git a/3681/CH3/EX3.1/Ex3_1.sce b/3681/CH3/EX3.1/Ex3_1.sce new file mode 100644 index 000000000..56cd7d9c8 --- /dev/null +++ b/3681/CH3/EX3.1/Ex3_1.sce @@ -0,0 +1,16 @@ +// Calculating effective length of air gap +clc; +disp('Example 3.1, Page No. = 3.12') +// Given Data +Ws = 12;// Slot width in mm +Wt = 12;// Tooth width in mm +lg = 2;// Length of air gap in mm +Kcs = 1/(1+(5*lg/Ws));//Carter's co-efficient for slots +// Calculation of effective length of air gap +ys=Ws+Wt;//Slot Pitch in mm +Kgs=ys/(ys-(Kcs*Ws));//Gap contraction for slots +Kgd=1;//Gap contracion factor for ducts//Since there are no ducts +Kg=Kgs*Kgd;//Total gap contracion factor +lgs=Kg*lg;//Effective gap length in mm +disp(lgs,'Effective gap length(mm)='); +//in book answer is 2.74 mm. The answers vary due to round off error diff --git a/3681/CH3/EX3.11/Ans3_11.PNG b/3681/CH3/EX3.11/Ans3_11.PNG Binary files differnew file mode 100644 index 000000000..988470971 --- /dev/null +++ b/3681/CH3/EX3.11/Ans3_11.PNG diff --git a/3681/CH3/EX3.11/Ex3_11.sce b/3681/CH3/EX3.11/Ex3_11.sce new file mode 100644 index 000000000..9f9d8a124 --- /dev/null +++ b/3681/CH3/EX3.11/Ex3_11.sce @@ -0,0 +1,16 @@ +// Calculating the specific iron loss
+clc;
+disp('Example 3.11, Page No. = 3.34')
+// Given Data
+Bm = 3.2;// Maximum flux density in Wb per meter square
+f = 50;// Frequency in Hz
+t = 0.5*10^(-3);// Thickness of sheet in mm
+p = .3*10^(-6);// Resistivity of alloy steel in ohm*meter
+D = 7.8*10^(3);// Density in kg per meter cube
+ph_each = 400;// Hysteresis loss in each cycle in Joule per meter cube
+// Calculation of total iron loss
+pe = %pi*%pi*f*f*Bm*Bm*t*t/(6*p*D);// Eddy current loss in W per Kg
+ph = ph_each*f/D;// Hysterseis loss in W per Kg
+Pi = pe+ph;// Total iron loss in W per Kg
+disp(Pi,'Specific iron loss(W per Kg)=');
+//in book answer is 3.2 W per Kg. The provided in the textbook is wrong
diff --git a/3681/CH3/EX3.12/Ans3_12.PNG b/3681/CH3/EX3.12/Ans3_12.PNG Binary files differnew file mode 100644 index 000000000..6e5250863 --- /dev/null +++ b/3681/CH3/EX3.12/Ans3_12.PNG diff --git a/3681/CH3/EX3.12/Ex3_12.sce b/3681/CH3/EX3.12/Ex3_12.sce new file mode 100644 index 000000000..8d64df9fa --- /dev/null +++ b/3681/CH3/EX3.12/Ex3_12.sce @@ -0,0 +1,21 @@ +// Calculating the specific iron loss
+clc;
+disp('Example 3.12, Page No. = 3.35')
+// Given Data
+Bm = 1.0;// Maximum flux density in Wb per meter square
+f = 100;// Frequency in Hz
+t = 0.3*10^(-3);// Thickness of sheet in mm
+p = .5*10^(-6);// Resistivity of alloy steel in ohm*meter
+D = 7650;// Density in kg per meter cube
+pi_quoted = 1.2;// Quoted iron loss in W per Kg
+// Calculation of total iron loss
+S1 = 2*12;// Sides of hysteresis loop in A/m
+S2 = 2*1;// Sides of hysteresis loop in Wb per meter square
+A = S1*S2;// Area of hysteresis loop in W-s per meter cube
+ph_each = A;// Hysteresis loss in each cycle in Joule per meter cube
+ph = ph_each*f/D;// Hysterseis loss in W per Kg
+pe = %pi*%pi*f*f*Bm*Bm*t*t/(6*p*D);// Eddy current loss in W per Kg
+pi = pe+ph;// Total iron loss in W per Kg
+disp(pi,'Specific iron loss(W per Kg)=');
+disp('The calculated iron loss is smaller than the quoted.')
+//in book answer is 1.014 W per Kg. The answers vary due to round off error
diff --git a/3681/CH3/EX3.13/Ans3_13.PNG b/3681/CH3/EX3.13/Ans3_13.PNG Binary files differnew file mode 100644 index 000000000..997c16eff --- /dev/null +++ b/3681/CH3/EX3.13/Ans3_13.PNG diff --git a/3681/CH3/EX3.13/Ex3_13.sce b/3681/CH3/EX3.13/Ex3_13.sce new file mode 100644 index 000000000..be0189b29 --- /dev/null +++ b/3681/CH3/EX3.13/Ex3_13.sce @@ -0,0 +1,19 @@ +// Calculating the hysteresis loss
+clc;
+disp('Example 3.13, Page No. = 3.35')
+// Given Data
+Bm = 1.0;// Maximum flux density in Wb per meter square
+f = 50;// Frequency in Hz
+SGi = 7.5;// Specific gravity of iron
+ph = 4.9;// Hysterseis loss in W per Kg
+// Calculation of co-efficient 'n'
+Di = 7500;// Density of iron
+n = ph/(Di*f*(Bm^(1.7)));//
+disp(n,'(a) co-efficient (n)=');
+//in book answer is 1307*10^(-6). The answers vary due to round off error
+// Calculation of hysteresis loss
+Bm = 1.8;// Maximum flux density in Wb per meter square
+f = 25;// Frequency in Hz
+ph = n*f*Di*Bm^(1.7);// Hysterseis loss in W per Kg
+disp(ph,'(b) Hysterseis loss(W per Kg)=');
+//in book answer is 6.66 W per Kg. The answers vary due to round off error
diff --git a/3681/CH3/EX3.15/Ans3_15.PNG b/3681/CH3/EX3.15/Ans3_15.PNG Binary files differnew file mode 100644 index 000000000..e30d422fe --- /dev/null +++ b/3681/CH3/EX3.15/Ans3_15.PNG diff --git a/3681/CH3/EX3.15/Ex3_15.sce b/3681/CH3/EX3.15/Ex3_15.sce new file mode 100644 index 000000000..8cbf2f11f --- /dev/null +++ b/3681/CH3/EX3.15/Ex3_15.sce @@ -0,0 +1,28 @@ +// Calculating the magnetic pull, unbalanced magnetic pull and ratio of unbalanced magnetic pull to useful force
+clc;
+disp('Example 3.15, Page No. = 3.71')
+// Given Data
+Power = 75000;// Power rating in W
+f = 50;// Frequency in Hz
+p = 2;// Number of poles
+D = 0.5;// Stator bore in meter
+L = 0.2;// Axial length of core in meter
+lg = 5;// Length of air gap
+ATm = 4500;// Peak magnetizing mmf per pole
+Bm = ATm*4*%pi*10^(-7)/(lg*10^(-3));// Peak value of flux density in Wb per meter square
+// Calculation of magnetic pull per pole
+MP = Bm*Bm*D*L/(3*4*%pi*10^(-7));// Magnetic pull per pole (Flux Distribution is sinusoidal)
+disp(MP,'(a) Magnetic pull per pole (N)=');
+//in book answer is 33.9 in kN The answers vary due to round off error
+// Calculation of unbalanced magnetic pull
+e = 1;// Displacement of rotor axis in mm
+Pp = %pi*D*L*Bm*Bm*e/(lg*4*4*%pi*10^(-7));// Unbalanced magnetic pull per pair of poles
+disp(Pp,'(b) Unbalanced magnetic pull per pair of poles (N)=');
+//in book answer is 16000 in N The answers vary due to round off error
+// Calculation of Ratio of unbalanced magnetic pull to useful force
+Speed = 2*f/p;// Speed in r.p.s.
+T = Power/(2*%pi*Speed);// Useful torque in Nm
+F = T/(D/2);// Useful force in N
+Ratio = Pp/F;// Ratio of unbalanced magnetic pull to useful force
+disp(Ratio,'(c) Ratio of unbalanced magnetic pull to useful force=');
+//in book answer is 16.8 The answers vary due to round off error
diff --git a/3681/CH3/EX3.2/Ans3_2.PNG b/3681/CH3/EX3.2/Ans3_2.PNG Binary files differnew file mode 100644 index 000000000..250a98a07 --- /dev/null +++ b/3681/CH3/EX3.2/Ans3_2.PNG diff --git a/3681/CH3/EX3.2/Ex3_2.sce b/3681/CH3/EX3.2/Ex3_2.sce new file mode 100644 index 000000000..7b462963a --- /dev/null +++ b/3681/CH3/EX3.2/Ex3_2.sce @@ -0,0 +1,22 @@ +// Calculating the mmf required for the air gap of a machine
+clc;
+disp('Example 3.2, Page No. = 3.12')
+// Given Data
+L = 0.32;// Core length in meter
+nd = 4;// Number of ducts
+Wd = 10;// Duct width in mm
+Pa = 0.19;// Pole arc in meter
+ys=65.4;//Slot Pitch in mm
+lg = 5;// Length of air gap in mm
+Wo = 5;// Slot opening in mm
+Fpp = 52;// Flux per pole in mWb
+Kcs = 0.18;//Carter's co-efficient for slots
+Kcd = 0.28;//Carter's co-efficient for ducts
+// Calculation of mmf required for the air gap
+Kgs=ys/(ys-(Kcs*Wo));//Gap contraction for slots
+Kgd=L/(L-(Kcd*nd*Wd*10^(-3)));//Gap contraction for ducts
+Kg=Kgs*Kgd;//Total gap contracion factor
+Bg=Fpp*10^(-3)/(Pa*L);//Flux density at the centre of pole in Wb per meter square
+ATg=800000*Kg*Bg*lg*10^(-3);//mmf required for air gap in A
+disp(ATg,'mmf required for air gap(A)=');
+//in book answer is 3587 A. The answers vary due to round off error
diff --git a/3681/CH3/EX3.3/Ans3_3.PNG b/3681/CH3/EX3.3/Ans3_3.PNG Binary files differnew file mode 100644 index 000000000..1a8131954 --- /dev/null +++ b/3681/CH3/EX3.3/Ans3_3.PNG diff --git a/3681/CH3/EX3.3/Ex3_3.sce b/3681/CH3/EX3.3/Ex3_3.sce new file mode 100644 index 000000000..0461e961a --- /dev/null +++ b/3681/CH3/EX3.3/Ex3_3.sce @@ -0,0 +1,29 @@ +// Estimating the effective air gap area per pole
+clc;
+disp('Example 3.3, Page No. = 3.13')
+// Given Data
+P = 10;// Number of pole
+Sb = 0.65;// Stator bore in meter
+L = 0.25;// Core length in meter
+Nss = 90;// Number of stator slots
+Wos = 3;// Stator slot opening in mm
+Nrs = 120;// Number of rotor slots
+Wor = 3;// Rotor slot opening in mm
+lg = 0.95;// Length of air gap in mm
+Kcs = 0.46;//Carter's co-efficient for slots
+Kcd = 0.68;//Carter's co-efficient for ducts
+nd = 3;// Number of ventilating ducts
+Wd = 10;// Width of each ventilating Duct in mm
+// Estimation of effective air gap area per pole
+ys = 3.141592654*Sb*10^(3)/Nss;// Stator slot pitch
+Kgss=ys/(ys-(Kcs*Wos));//Gap contraction factor for stator slots
+Rd = Sb-2*lg*10^(-3);// Rotor diameter in meter
+yr = 3.141592654*Rd*10^(3)/Nrs;// Rotor slot pitch
+Kgsr=yr/(yr-(Kcs*Wor));//Gap contraction factor for rotor slots
+Kgs=Kgss*Kgsr;//Gap contraction factor for slots
+Kgd=L*10^(3)/(L*10^(3)-(Kcd*nd*Wd));//Gap contraction for ducts
+Kg=Kgs*Kgd;//Total gap contracion factor
+Ag = 3.141592654*Sb*L/P;// Actual area of air gap per pole in meter square
+Age = Ag/Kg;// Effective air gap area per pole in meter square
+disp(Age,'Effective air gap area per pole(meter square)=');
+//in book answer is .04052 A. The answers vary due to round off error
diff --git a/3681/CH3/EX3.4/Ans3_4.PNG b/3681/CH3/EX3.4/Ans3_4.PNG Binary files differnew file mode 100644 index 000000000..885c591b7 --- /dev/null +++ b/3681/CH3/EX3.4/Ans3_4.PNG diff --git a/3681/CH3/EX3.4/Ex3_4.sce b/3681/CH3/EX3.4/Ex3_4.sce new file mode 100644 index 000000000..999571179 --- /dev/null +++ b/3681/CH3/EX3.4/Ex3_4.sce @@ -0,0 +1,28 @@ +// Estimating the average flux density in the air gap
+clc;
+disp('Example 3.4, Page No. = 3.14')
+// Given Data
+MVA = 172;// MVA rating
+P = 20;// Number of pole
+D = 6.5;// Diameter in meter
+L = 1.72;// Core length in meter
+ys = 64;//Slot Pitch in mm
+Ws = 22;// Stator slot (open) width in mm
+lg = 30;// Length of air gap in mm
+nd = 41;// Number of ventilating ducts
+Wd = 6;// Width of each ventilating Duct in mm
+mmf = 27000// Total mmf per pole in A
+Kf = 0.7;// Field form factor
+// Estimation of effective air gap area per pole
+y=Ws/(2*lg);//Ratio for slots
+Kcs= (2/%pi)*(atan(y)-log10(sqrt(1+y^2))/y);//Carter's co-efficient for slots
+Kgs=ys/(ys-(Kcs*Ws));//Gap contraction for slots
+y=Wd/(2*lg);//Ratio for ducts
+Kcd= (2/%pi)*(atan(y)-log10(sqrt(1+y^2))/y);//Carter's co-efficient for slots
+Kgd=L*10^(3)/(L*10^(3)-(Kcd*nd*Wd));//Gap contraction for ducts
+Kg=Kgs*Kgd;//Total gap expansion factor
+ATg = 0.87*mmf;// The required for the air gap is 87% of the total mmf per pole in A
+Bg = ATg/(800000*Kg*lg*10^(-3));// Maximum flux density in air gap in Wb per meter square
+Bav= Kf*Bg;// Average flux density in air gap in Wb per meter square
+disp(Bav,'Average flux density in air gap (Wb per meter square)=');
+//in book answer is .615 Wb per meter square. The provided in the textbook is wrong
diff --git a/3681/CH3/EX3.7/Ans3_7.PNG b/3681/CH3/EX3.7/Ans3_7.PNG Binary files differnew file mode 100644 index 000000000..b7866a133 --- /dev/null +++ b/3681/CH3/EX3.7/Ans3_7.PNG diff --git a/3681/CH3/EX3.7/Ex3_7.sce b/3681/CH3/EX3.7/Ex3_7.sce new file mode 100644 index 000000000..e16269b3d --- /dev/null +++ b/3681/CH3/EX3.7/Ex3_7.sce @@ -0,0 +1,20 @@ +// Calculating the apparent flux density
+clc;
+disp('Example 3.7, Page No. = 3.22')
+// Given Data
+Ws = 10;// Slot width in mm
+Wt = 12;// Tooth width in mm
+L = .32;// Grass core Length in meter
+nd = 4;// Number of ventilating ducts
+Wd = 10;// Width of each ventilating Duct in mm
+Breal = 2.2;// Real flux density in Wb per meter square
+p = 31.4*10^(-6);// Permeability of teeth corresponding to real flux density in henry per meter
+Ki = 0.9;// Stacking Factor
+// Calculation of apparent flux density
+at = Breal/p;// mmf per meter corresponding to real flux density and permeability
+Li = Ki*(L-nd*Wd*10^(-3));// Net iron length
+ys = Wt+Ws;// Slot pitch
+Ks = L*ys/(Li*Wt);
+Bapp = Breal+4*%pi*10^(-7)*at*(Ks-1);
+disp(Bapp,'Apparent flux density(Wb per meter square)=');
+//in book answer is 2.317 Wb per meter square. The answers vary due to round off error
diff --git a/3681/CH3/EX3.8/Ans3_8.PNG b/3681/CH3/EX3.8/Ans3_8.PNG Binary files differnew file mode 100644 index 000000000..b73707728 --- /dev/null +++ b/3681/CH3/EX3.8/Ans3_8.PNG diff --git a/3681/CH3/EX3.8/Ex3_8.sce b/3681/CH3/EX3.8/Ex3_8.sce new file mode 100644 index 000000000..f3b988b8e --- /dev/null +++ b/3681/CH3/EX3.8/Ex3_8.sce @@ -0,0 +1,19 @@ +// Calculating the apparent flux density
+clc;
+disp('Example 3.8, Page No. = 3.23')
+// Given Data
+Ws = 10;// Slot width in mm
+ys = 28;// Slot pitch in mm
+L = .35;// Grass core Length in meter
+nd = 4;// Number of ventilating ducts
+Wd = 10;// Width of each ventilating Duct in mm
+Breal = 2.15;// Real flux density in Wb per meter square
+at = 55000;// mmf per meter corresponding to real flux density and permeability
+Ki = 0.9;// Stacking Factor
+// Calculation of apparent flux density
+Li = Ki*(L-nd*Wd*10^(-3));// Net iron length
+Wt = ys-Ws;// Tooth width in mm
+Ks = L*ys/(Li*Wt);
+Bapp = Breal+4*%pi*10^(-7)*at*(Ks-1);
+disp(Bapp,'Apparent flux density(Wb per meter square)=');
+//in book answer is 2.2156 Wb per meter square. The answers vary due to round off error
diff --git a/3681/CH4/EX4.1/Ans4_1.PNG b/3681/CH4/EX4.1/Ans4_1.PNG Binary files differnew file mode 100644 index 000000000..05d1d0cd9 --- /dev/null +++ b/3681/CH4/EX4.1/Ans4_1.PNG diff --git a/3681/CH4/EX4.1/Ex4_1.sce b/3681/CH4/EX4.1/Ex4_1.sce new file mode 100644 index 000000000..46968a4c9 --- /dev/null +++ b/3681/CH4/EX4.1/Ex4_1.sce @@ -0,0 +1,15 @@ +// Calculating the loss that will pass through copper bar to iron
+clc;
+disp('Example 4.1, Page No. = 4.3')
+// Given Data
+D = 12;// Diameter of copper bar in mm
+t = 1.5;// Thickness of micanite tube in mm
+p = 8;// Resistivity of macanite tube in ohm*meter
+T = 25;// Temperature difference in degree celsius
+L = 0.2;// Length of copper bar
+// Calculation of loss.that will pass through copper bar to iron
+S = %pi*(D+t)*10^(-3)*L;//Area of insulation in the path of heat flow
+R =( p*t*10^(-3))/S;// Thermal resistance of micanite tube
+Q_con= T/R;// Heat Dissipated
+disp(Q_con,'Heat Dissipated(W)=');
+//in book answer is 17.67 W. The answers vary due to round off error
diff --git a/3681/CH4/EX4.11/Ans4_11.PNG b/3681/CH4/EX4.11/Ans4_11.PNG Binary files differnew file mode 100644 index 000000000..72bd97cd4 --- /dev/null +++ b/3681/CH4/EX4.11/Ans4_11.PNG diff --git a/3681/CH4/EX4.11/Ex4_11.sce b/3681/CH4/EX4.11/Ex4_11.sce new file mode 100644 index 000000000..ed6d148a2 --- /dev/null +++ b/3681/CH4/EX4.11/Ex4_11.sce @@ -0,0 +1,20 @@ +// Calculating the heat conducted across the former from winding to core
+clc;
+disp('Example 4.11, Page No. = 4.17')
+// Given Data
+t = 2.5;// Thickness of former (in mm)
+t_air = 1;// Thickness of air space (in mm)
+lw = 150*250;// The inner dimentions of the former of field coil (in mm square)
+h = 200;// Winding height (in mm)
+s_former = 0.166;// Thermal conductivity of former (in W per meter per degree celsius)
+s_air = 0.05;// Thermal conductivity of air (in W per meter per degree celsius)
+T = 40;// Temperature rise (in degree celsius)
+// Calculation of the heat conducted across the former from winding to core
+S = 2*(150+250)*h*10^(-6);// Area of path of heat flow (in meter square)
+R_former = t*10^(-3)/(S*s_former);// Thermal resistance of former (in ohm)
+R_air = t_air*10^(-3)/(S*s_air);// Thermal resistance of former (in ohm)
+R0 = R_former+R_air;// Since R_former and R_air are in series. Total thermal resistance to heat flow (in ohm)
+Q_con = T/R0;// Heat conducted (in Watts)
+disp(Q_con,'Heat conducted across the former from winding to core (in Watts)=');
+//in book answers is 182.6 Watts. The answers vary due to round off error
+
diff --git a/3681/CH4/EX4.12/Ans4_12.PNG b/3681/CH4/EX4.12/Ans4_12.PNG Binary files differnew file mode 100644 index 000000000..27758af4c --- /dev/null +++ b/3681/CH4/EX4.12/Ans4_12.PNG diff --git a/3681/CH4/EX4.12/Ex4_12.sce b/3681/CH4/EX4.12/Ex4_12.sce new file mode 100644 index 000000000..6ff5f1e2e --- /dev/null +++ b/3681/CH4/EX4.12/Ex4_12.sce @@ -0,0 +1,20 @@ +// Estimating the final steady temperature rise of coil and its time constant
+clc;
+disp('Example 4.12, Page No. = 4.21')
+// Given Data
+S = 0.15;// Heat dissipating surface (in meter square)
+l = 1;// Length of mean turn in meter
+Sf = 0.56;// Space Factor
+A = 100*50;// Area of cross-section (in mm square)
+Q = 150;// Dissipating loss (in Watts)
+emissivity = 34;// Emissivity (in Watt per degree celsius per meter square)
+h = 390;// Specific heat of copper (in J per kg per degree celsius)
+// Calculation of the final steady temperature rise of coil and its time constant
+V = l*A*Sf*10^(-6);// Volume of copper (in meter cube)
+G = V*8900;// Since copper weighes 8900 kg per meter cube. Weight of copper(in kg)
+Tm = Q/(S*emissivity);// Final steady temperature rise (in degree celsius)
+Th = G*h/(S*emissivity);// Heating time constant (in seconds)
+disp(Tm,'Final steady temperature rise (degree celsius))=');
+disp(Th,'Heating time constant (seconds)=');
+//in book final steady temperature rise (in degree celsius) is equal to 29.4 and heating time constant (in seconds) is equal to 1906. The answers vary due to round off error
+
diff --git a/3681/CH4/EX4.13/Ans4_13.PNG b/3681/CH4/EX4.13/Ans4_13.PNG Binary files differnew file mode 100644 index 000000000..0419719c7 --- /dev/null +++ b/3681/CH4/EX4.13/Ans4_13.PNG diff --git a/3681/CH4/EX4.13/Ex4_13.sce b/3681/CH4/EX4.13/Ex4_13.sce new file mode 100644 index 000000000..bd6468868 --- /dev/null +++ b/3681/CH4/EX4.13/Ex4_13.sce @@ -0,0 +1,20 @@ +// Calculating the final steady temperature rise of coil surface and hot spot temperature rise
+clc;
+disp('Example 4.13, Page No. = 4.21')
+// Given Data
+S = 0.125;// Cooling surface (in meter square)
+l = 0.8;// Length of mean turn in meter
+Sf = 0.56;// Space Factor
+A = 120*50;// Area of cross-section (in mm square)
+Q = 150;// Dissipating loss (in Watts)
+emissivity = 30;// Specific heat dissipation (in Watt per degree celsius per meter square)
+pi = 8;// Thermal resistivity of insulating material (in ohm*meter)
+// Calculation of the final steady temperature rise of coil surface and hot spot temperature rise
+Tm = Q/(S*emissivity);// Final steady temperature rise (in degree celsius)
+p0 = pi*(1-Sf^(1/2));// Effective thermal resistivity (in ohm*meter)
+q = Q/(l*A*10^(-6));// Loss (in Watts per meter cube)
+T0 = q*p0*(50*10^(-3))^(2)/8;// Temperature difference between coil surtface and hot spot (in degree celsius)
+disp(Tm,'Final steady temperature rise (degree celsius)=');
+disp(Tm+T0,'Temperature rise of hot spot (degree celsius)=');
+//in book final steady temperature rise (in degree celsius) is equal to 40 and hot spot temperature rise(in degree celsius) is equal to 59.5. The answers vary due to round off error
+
diff --git a/3681/CH4/EX4.15/Ans4_15.PNG b/3681/CH4/EX4.15/Ans4_15.PNG Binary files differnew file mode 100644 index 000000000..7361446c0 --- /dev/null +++ b/3681/CH4/EX4.15/Ans4_15.PNG diff --git a/3681/CH4/EX4.15/Ex4_15.sce b/3681/CH4/EX4.15/Ex4_15.sce new file mode 100644 index 000000000..eec08cfeb --- /dev/null +++ b/3681/CH4/EX4.15/Ex4_15.sce @@ -0,0 +1,25 @@ +// Calculating the temperature rise and thermal time constant and rating of the machine
+clc;
+disp('Example 4.15, Page No. = 4.23')
+// Given Data
+D = 0.6;// Diameter of induction motor (in meter)
+L = 0.9;// Length of induction motor (in meter)
+out = 7500;// Output of induction motor (in W)
+e = 0.9;// Efficiency
+G = 375;// Weight of material (in kg)
+h = 725;// Specific heat (in J/kg degree celsius)
+Lem = 12;// Specific heat dissipation (in Watt per meter square degree celsius)
+// Calculation of the temperature rise and thermal time constant of the machine
+S = (%pi*D*L)+(2*%pi/4*D^(2));// Total heat dissipating surface (in meter square)
+Q = (out/e)-out;// Losses (in Watts)
+Tm = Q/(S*Lem);// Final temperature rise (in degree celsius)
+Th = G*h/(S*Lem);// Time constant (in seconds)
+disp(Tm,'(a) Final temperature rise (degree celsius) =');
+disp(Th,' Time constant (seconds) =');
+// Calculation of the rating of the machine
+Lem_new = 25;// Specific heat dissipation (in Watt per meter square degree celsius)
+Q = Tm*S*Lem_new;// Losses (in Watts)
+out = (e*Q)/(1-e);// Output of induction motor (in W)
+disp(out,'(b) Rating of the machine (Watt) =');
+//in book answers are 30.85 degree celsius, 10025 seconds and 15687 watts. The answers vary due to round off error
+
diff --git a/3681/CH4/EX4.17/Ans4_17.PNG b/3681/CH4/EX4.17/Ans4_17.PNG Binary files differnew file mode 100644 index 000000000..57e75dacf --- /dev/null +++ b/3681/CH4/EX4.17/Ans4_17.PNG diff --git a/3681/CH4/EX4.17/Ex4_17.sce b/3681/CH4/EX4.17/Ex4_17.sce new file mode 100644 index 000000000..9ccf3b01e --- /dev/null +++ b/3681/CH4/EX4.17/Ex4_17.sce @@ -0,0 +1,15 @@ +// Calculating the temperature of machine after one hour of its final steady temperature rise
+clc;
+disp('Example 4.17, Page No. = 4.24')
+// Given Data
+Ti = 40;// Initial temperature (in degree celsius)
+T_ambient = 30;// Ambient temperature (in degree celsius)
+Tm = 80;// Final steady temperature rise (in degree celsius)
+Th = 2;// Heating time constant (in hours)
+t = 1;// Since we have to calculate temperature of machine after one hour of its final steady temperture rise (in hours)
+// Calculation of the final steady temperature rise of coil surface and hot spot temperature rise
+Ti_rise = Ti-T_ambient;// Initial temperature rise (in degree celsius)
+T = Tm*(1-%e^(-t/Th))+(Ti_rise*%e^(-t/Th));// Temperature rise after one hour (in degree celsius)
+disp(T+T_ambient,'Temperature of machine after one hour (degree celsius)=');
+//in book answer is 67.54 (degree celsius). The answers vary due to round off error
+
diff --git a/3681/CH4/EX4.19/Ans4_19.PNG b/3681/CH4/EX4.19/Ans4_19.PNG Binary files differnew file mode 100644 index 000000000..100cc6dc1 --- /dev/null +++ b/3681/CH4/EX4.19/Ans4_19.PNG diff --git a/3681/CH4/EX4.19/Ex4_19.sce b/3681/CH4/EX4.19/Ex4_19.sce new file mode 100644 index 000000000..ca25ef1e3 --- /dev/null +++ b/3681/CH4/EX4.19/Ex4_19.sce @@ -0,0 +1,14 @@ +// Calculating the rate of change of temperature at t=0
+clc;
+disp('Example 4.19, Page No. = 4.27')
+// Given Data
+I = 2.5;// Current (in Amperes)
+V = 230;// Voltage (in volts)
+G = 60;// Weight of copper (in kg)
+h = 390;// Specific heat of copper (in J per kg per degree celsius)
+// Calculation of the rate of change of temperature at t=0
+Q = I*V;// Loss (in Watts)
+T_rate = Q/(G*h);// Rate of change of temperature at t=0 (in degree celsius per second)
+disp(T_rate,'Rate of change of temperature at t=0 (degree celsius per second)=');
+//in book answer is 0.0246 (in degree celsius per second). The answers vary due to round off error
+
diff --git a/3681/CH4/EX4.2/Ans4_2.PNG b/3681/CH4/EX4.2/Ans4_2.PNG Binary files differnew file mode 100644 index 000000000..d5839e5e9 --- /dev/null +++ b/3681/CH4/EX4.2/Ans4_2.PNG diff --git a/3681/CH4/EX4.2/Ex4_2.sce b/3681/CH4/EX4.2/Ex4_2.sce new file mode 100644 index 000000000..45f094c79 --- /dev/null +++ b/3681/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,17 @@ +// Calculating the loss that will be conducted across the the laminations
+clc;
+disp('Example 4.2, Page No. = 4.3')
+// Given Data
+Q_con_5 = 25;// Heat Dissipated
+t_5 = 20;// Thickness of laminations in mm
+S_5 = 2500;// Cross-section area of conduction in mm square
+T_5 = 5;// Temperature difference in degree celsius
+t_20 = 40;// Thickness of laminations in mm
+S_20 = 6000;// Cross-section area of conduction in mm square
+T_20 = 20;// Temperature difference in degree celsius
+// Calculation of heat conducted across the laminations
+p_along = (T_5*S_5*10^(-6))/(Q_con_5*t_5*10^(-3));// Thermal resistivity along the direction of laminations
+p_across = 20*p_along;// Thermal resistivity across the direction of laminations
+Q_con_20 = S_20*10^(-6)*T_20/(p_across*t_20*10^(-3));// Heat conducted across the the laminations
+disp(Q_con_20,'Heat conducted across the the laminations(W)=');
+//in book answer is 6 W. The answers vary due to round off error
diff --git a/3681/CH4/EX4.22/Ans4_22.PNG b/3681/CH4/EX4.22/Ans4_22.PNG Binary files differnew file mode 100644 index 000000000..05ac52ae2 --- /dev/null +++ b/3681/CH4/EX4.22/Ans4_22.PNG diff --git a/3681/CH4/EX4.22/Ex4_22.sce b/3681/CH4/EX4.22/Ex4_22.sce new file mode 100644 index 000000000..ee640b754 --- /dev/null +++ b/3681/CH4/EX4.22/Ex4_22.sce @@ -0,0 +1,21 @@ +// Calculating the volume of air required per second and fan power
+clc;
+disp('Example 4.22, Page No. = 4.50')
+// Given Data
+MVA = 50;// MVA rating of turbo-alternator
+Q = 1500;// Total loss (in kW)
+Ti = 25;// Inlet temperature of air (in degree celsius)
+T = 30;// Temperature limit (in degree celsius)
+H = 760;// Baromatric height (in mm of mercury)
+P = 2000;// Pressure (in N per meter square)
+nf = 0.4;// Fan efficiency
+// Assumption
+cp = 995;// Specific heat of air at constant pressure (in J per kg per degree celsius)
+V = 0.775;// Volume of 1 kg of air at N.T.P. (in meter cube)
+// Calculation of the volume of air required per second and fan power
+Va = (V*Q*10^(3)/(cp*T))*((Ti+273)/273)*(760/H);// Volume of air (in meter cube per second)
+Pf = (P*Va/nf)*10^(-3);// Fan power (in kW)
+disp(Va,'Volume of air (meter cube per second)=');
+disp(Pf,'Fan power (kW)=');
+//in book Va is equal to 42.6 (meter cube per second) and Pf is equal to 212.5 (kW). The answers vary due to round off error
+
diff --git a/3681/CH4/EX4.23/Ans4_23.PNG b/3681/CH4/EX4.23/Ans4_23.PNG Binary files differnew file mode 100644 index 000000000..9914c7dda --- /dev/null +++ b/3681/CH4/EX4.23/Ans4_23.PNG diff --git a/3681/CH4/EX4.23/Ex4_23.sce b/3681/CH4/EX4.23/Ex4_23.sce new file mode 100644 index 000000000..e66cd8ea8 --- /dev/null +++ b/3681/CH4/EX4.23/Ex4_23.sce @@ -0,0 +1,25 @@ +// Calculating the efficiency of machine and amount of cooling water
+clc;
+disp('Example 4.23, Page No. = 4.50')
+// Given Data
+MVA = 30;// MVA rating of turbo-alternator
+Ti = 15;// Inlet temperature of air (in degree celsius)
+To = 45;// Outlet temperature of air (in degree celsius)
+H = 750;// Baromatric height (in mm of mercury)
+Va = 30;// Volume of air (in meter cube per second)
+nf = 0.4;// Fan efficiency
+cp = 1000;// Specific heat of air at constant pressure (in J per kg per degree celsius)
+V = 0.78;// Volume of 1 kg of air at N.T.P. (in meter cube)
+pf = 0.8;// Power factor
+// Calculation of the efficiency of machine
+T = To-Ti;// Temperature rise limit (in degree celsius)
+Q = Va/((V*10^(3)/(cp*T))*((Ti+273)/273)*(760/H));// Total losses (in kW)
+P_out = 30*10^(3)*pf;// Output power (in kW)
+n = P_out/(P_out+Q)*100;// Fan power (in kW)
+disp(n,'(a) Efficiency of machine (in percentage)=');
+// Calculation of the amount of cooling water
+T = 8;// Temperature rise of water (in degree celsius)
+Vw = 0.24*Q/T;// Amount of cooling water (in litre per second)
+disp(Vw,'(b) Amount of cooling water (litre per second)=');
+//in book efficiency is equal to 95.7% and amount of cooling water 32.4 (litre per second). The answers vary due to round off error
+
diff --git a/3681/CH4/EX4.24/Ans4_24.PNG b/3681/CH4/EX4.24/Ans4_24.PNG Binary files differnew file mode 100644 index 000000000..c3a67b6e1 --- /dev/null +++ b/3681/CH4/EX4.24/Ans4_24.PNG diff --git a/3681/CH4/EX4.24/Ex4_24.sce b/3681/CH4/EX4.24/Ex4_24.sce new file mode 100644 index 000000000..420c89d19 --- /dev/null +++ b/3681/CH4/EX4.24/Ex4_24.sce @@ -0,0 +1,14 @@ +// Calculating the temperature rise of hydrogen
+clc;
+disp('Example 4.24, Page No. = 4.51')
+// Given Data
+Q = 750;// Losses (in kW)
+Ti = 25;// Inlet temperature of air (in degree celsius)
+H = (2000+760);// Baromatric height (in mm of mercury)
+VH = 10;// Volume of hydrogen leaving the coolers (in meter cube per second)
+cp = 12540;// Specific heat of air at constant pressure (in J per kg per degree celsius)
+V = 11.2;// Volume of 1 kg of air at N.T.P. (in meter cube)
+// Calculation of the temperature rise of hydrogen
+T = (V*Q*10^(3)/(cp*VH))*((Ti+273)/273)*(760/H);// Temperature rise of hydrogen(in degree celsius)
+disp(T,'Temperature rise of hydrogen (degree celsius)=');
+//in book ans is 20 (degree celsius). The answers vary due to round off error
diff --git a/3681/CH4/EX4.25/Ans4_25.PNG b/3681/CH4/EX4.25/Ans4_25.PNG Binary files differnew file mode 100644 index 000000000..5f3242dd7 --- /dev/null +++ b/3681/CH4/EX4.25/Ans4_25.PNG diff --git a/3681/CH4/EX4.25/Ex4_25.sce b/3681/CH4/EX4.25/Ex4_25.sce new file mode 100644 index 000000000..01d892fdf --- /dev/null +++ b/3681/CH4/EX4.25/Ex4_25.sce @@ -0,0 +1,17 @@ +// Calculating the amount of oil and amount of water
+clc;
+disp('Example 4.25, Page No. = 4.51')
+// Given Data
+MVA = 40;// MVA rating of transformer
+Q = 200;// Total losses (in kW)
+Q_oil = 0.8*Q;// Since 20% of losses are dissipated by tank walls Heat taken up by oil (in kW)
+// Calculation of the amount of oil
+T = 20;// Temperature rise of oil (in degree celsius)
+cp = 0.4;// by assuming
+Vo = 0.24*Q_oil/(cp*T);// Amount of oil (in litre per second)
+disp(Vo,'Amount of oil (litre per second)=');
+// Calculation of the amount of water
+T = 10;// Temperature rise of water (in degree celsius)
+Vw = 0.24*Q_oil/T;// Amount of water (in litre per second)
+disp(Vw,'Amount of water (litre per second)=');
+//in book Vo is equal to 4.8 (litre per second) and Vw is equal to 3.84 (litre per second). The answers vary due to round off error
diff --git a/3681/CH4/EX4.26/Ans4_26.PNG b/3681/CH4/EX4.26/Ans4_26.PNG Binary files differnew file mode 100644 index 000000000..26074580a --- /dev/null +++ b/3681/CH4/EX4.26/Ans4_26.PNG diff --git a/3681/CH4/EX4.26/Ex4_26.sce b/3681/CH4/EX4.26/Ex4_26.sce new file mode 100644 index 000000000..08bd149d3 --- /dev/null +++ b/3681/CH4/EX4.26/Ex4_26.sce @@ -0,0 +1,19 @@ +// Calculating the temperature rise of tank
+clc;
+disp('Example 4.26, Page No. = 4.52')
+// Given Data
+MVA = 15;// MVA rating of transformer
+Q_iron = 80;// Iron losses (in kW)
+Q_copper = 120;// Copper losses (in kW)
+T_water = 15;// Temperature rise of water (in degree celsius)
+Vw = 3;// Amount of water (in litre per second)
+Dimensions = 3.5*3.0*1.4;// Tank dimensions (in meter)
+l = 10;// Specific loss dissipation from tank walls (in Watt per degree celsius per meter square)
+// Calculation of the temperature rise of tank
+Q_total = Q_iron+Q_copper;// Total losses (in kW)
+Q = Vw*T_water/0.24;// Heat taken away by water (in kW)
+Q_walls = Q_total-Q;// Loss dissipated by walls (in kW)
+S = 2*3.5*(3+1.14);// Area of tank walls by neglecting top and bottom surfaces
+T = Q_walls*10^(3)/(S*l);// Temperature rise of tank (in degree celsius)
+disp(T,'Temperature rise of tank (degree celsius)=');
+//in book answer is 40.6 (degree celsius). The provided in the textbook is wrong
diff --git a/3681/CH4/EX4.27/Ans4_27.PNG b/3681/CH4/EX4.27/Ans4_27.PNG Binary files differnew file mode 100644 index 000000000..8943d08df --- /dev/null +++ b/3681/CH4/EX4.27/Ans4_27.PNG diff --git a/3681/CH4/EX4.27/Ex4_27.sce b/3681/CH4/EX4.27/Ex4_27.sce new file mode 100644 index 000000000..c9d752c53 --- /dev/null +++ b/3681/CH4/EX4.27/Ex4_27.sce @@ -0,0 +1,26 @@ +// Calculating the amount of water required per second, area of water duct and pumping power
+clc;
+disp('Example 4.27, Page No. = 4.52')
+// Given Data
+Q = 800;// Stator copper losses (in kW)
+Ti = 38;// Temperature of water inlet (in degree celsius)
+To = 68;// Temperature of water outlet (in degree celsius)
+Ns = 48;// Number of slots
+v = 1;// velocity (in meter per second)
+p = 300*10^(3);// Pumping pressure (in N per meter square)
+n = 0.6;// Efficiency
+// Calculation of the volume of water required per second
+T = To-Ti;// Temperature rise of water (in degree celsius)
+Vwl = 0.24*Q/T;// Amount of water (in litre per second)
+Vwm = Vwl*10^(-3);// Amount of water (in meter cube per second)
+N_cond = 2*Ns;// Since each slot has two conductors Total number of stator conductors
+N_sub_cond = 32*N_cond;// Since each conductor is subdivided into 32 sub-conductors
+Vw_sub_cond = Vwl/N_sub_cond;// Volume of water required for each sub-conductors (in litre per second)
+disp(Vw_sub_cond,'Volume of water required for each sub-conductors (litre per second)=');
+A = Vw_sub_cond*10^(-3)/v;// Area of each duct (in meter square)
+A = A*10^(6);// Area of each duct (in mm square)
+disp(A,'Area of each duct (mm square)=');
+Q = 800-500;// Since it ia a 500 KW direct cooled turbo-alternator (in kW)
+P = (Q*10^(3)*Vwm/n)*10^(-3);// Pumping power (in kW)
+disp(P,'Pumping power (kW)=');
+//in book Vwl is equal to 0.00208 (litre per second), A is 2 (mm square) and pumping power is 3.2 (kW). The answers vary due to round off error
diff --git a/3681/CH4/EX4.3/Ans4_3.PNG b/3681/CH4/EX4.3/Ans4_3.PNG Binary files differnew file mode 100644 index 000000000..5da87fba9 --- /dev/null +++ b/3681/CH4/EX4.3/Ans4_3.PNG diff --git a/3681/CH4/EX4.3/Ex4_3.sce b/3681/CH4/EX4.3/Ex4_3.sce new file mode 100644 index 000000000..a5e2de5b1 --- /dev/null +++ b/3681/CH4/EX4.3/Ex4_3.sce @@ -0,0 +1,11 @@ +// Calculating the heat radiated from the body
+clc;
+disp('Example 4.3, Page No. = 4.5')
+// Given Data
+e = 0.8;// Co-efficient of emissivity
+T1 = 273+60;// Temperature of body in degree kelvin
+T0 = 273+20;// Temperature of walls in degree kelvin
+// Calculation of the heat radiated from the body
+q_rad = 5.7*10^(-8)*e*(T1^(4)-T0^(4));// Heat radiated from the body
+disp(q_rad,'Heat radiated from the body(Watt per square meter)=');
+//in book answer is 224.6 in Watt per square meter. The answers vary due to round off error
diff --git a/3681/CH4/EX4.35/Ans4_35.PNG b/3681/CH4/EX4.35/Ans4_35.PNG Binary files differnew file mode 100644 index 000000000..3fd926a08 --- /dev/null +++ b/3681/CH4/EX4.35/Ans4_35.PNG diff --git a/3681/CH4/EX4.35/Ex4_35.sce b/3681/CH4/EX4.35/Ex4_35.sce new file mode 100644 index 000000000..1e69d2ea2 --- /dev/null +++ b/3681/CH4/EX4.35/Ex4_35.sce @@ -0,0 +1,15 @@ +// Calculating the continuous rating of motor
+clc;
+disp('Example 4.35, Page No. = 4.67')
+// Given Data
+Psh = 37.5;// Power rating of motor (in kW)
+th = 30;// Time (in minuts)
+Th = 90;// Heating time constant (in minuts)
+// Calculation of the continuous rating of motor
+ph = 1/(1-%e^(-th/Th));// Heating overload ratio
+K = 0.7^(2);// Maximum efficiency occurs at 70% full load
+pm = ((K+1)*ph-K)^(1/2);// Mechanical overload ratio
+Pnom = Psh/pm;// Continuous rating of motor (in kW)
+disp(Pnom,'Continuous rating of motor (kW)=');
+//in book answer is 17.2 kW. The answers vary due to round off error
+
diff --git a/3681/CH4/EX4.37/Ans4_37.PNG b/3681/CH4/EX4.37/Ans4_37.PNG Binary files differnew file mode 100644 index 000000000..916ce4a66 --- /dev/null +++ b/3681/CH4/EX4.37/Ans4_37.PNG diff --git a/3681/CH4/EX4.37/Ex4_37.sce b/3681/CH4/EX4.37/Ex4_37.sce new file mode 100644 index 000000000..c55a21428 --- /dev/null +++ b/3681/CH4/EX4.37/Ex4_37.sce @@ -0,0 +1,19 @@ +// Calculating the mean temperature rise
+clc;
+disp('Example 4.37, Page No. = 4.73')
+// Given Data
+th = 20;// Heating time (in minuts)
+Th = 120;// Heating time constant (in minuts)
+tc = 15;// Cooling time (in minuts)
+Tc = 180;// Cooling time constant (in minuts)
+Tm = 50;// Final temperature rise on the continuous full load (in degree celsius)
+Loss_fl = 500;// Copper loss at full load (in Watt)
+Loss_nl = 300;// Copper loss at no load (in Watt)
+// Calculation of the mean temperature rise
+Total_Loss_fl = Loss_fl+Loss_nl;// Total loss at full load (in Watt)
+Total_Loss_nl = Loss_nl;// Total loss at no load (in Watt)
+Tn = Total_Loss_nl/Total_Loss_fl*Tm;// Final temperature rise when running on no load (in degree celsius)
+T = ((Tm*th/Th)+(Tn*tc/Tc))/(th/Th+tc/Tc);// Mean temperature rise (in degree celsius)
+disp(T,'Mean temperature rise (degree celsius)=');
+//in book answer is 39.58 degree celsius. The answers vary due to round off error
+
diff --git a/3681/CH4/EX4.4/Ans4_4.PNG b/3681/CH4/EX4.4/Ans4_4.PNG Binary files differnew file mode 100644 index 000000000..ce595e64b --- /dev/null +++ b/3681/CH4/EX4.4/Ans4_4.PNG diff --git a/3681/CH4/EX4.4/Ex4_4.sce b/3681/CH4/EX4.4/Ex4_4.sce new file mode 100644 index 000000000..2f3bff8b4 --- /dev/null +++ b/3681/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,23 @@ +// Calculating the length and width of strip
+clc;
+disp('Example 4.4, Page No. = 4.5')
+// Given Data
+e = 0.9;// Emissivity
+Radiating_efficiency = 0.75;// Radiating efficiency
+v = 250;// Voltage in volts
+P = 1000;// Power in Watts
+t = 0.2;// Thickness of nickel chrome strip
+T1 = 273+(300+30);// Temperature of strip in degree kelvin
+T0 = 273+30;// Temperature of ambient medium in degree kelvin
+p = 1*10^(-6);// Resistivity of nickel chrome
+// Calculation of length and width of strip
+e = e*Radiating_efficiency;// Effective co-efficient of emissivity
+q_rad = 5.7*10^(-8)*e*(T1^(4)-T0^(4));// Heat dissipated by radiation in Watt per meter square
+R = v*v/P;// Resistance of strip in ohm
+l_by_w = R*t*10^(-3)/p;// This is equal to l/w
+lw = 1000/(q_rad*2);// This is equal to l*w
+l = sqrt(lw*l_by_w);// Length of strip in meter
+w = (lw/l)*10^(3);// Width of strip in mm
+disp(l,'Length of strip (meter)=');
+disp(w,'Width of strip (mm)=');
+//in book Length is 36.2 meter and width is 2.9 mm. The answers vary due to round off error
diff --git a/3681/CH4/EX4.43/Ans4_43.PNG b/3681/CH4/EX4.43/Ans4_43.PNG Binary files differnew file mode 100644 index 000000000..1681bf2cf --- /dev/null +++ b/3681/CH4/EX4.43/Ans4_43.PNG diff --git a/3681/CH4/EX4.43/Ex4_43.sce b/3681/CH4/EX4.43/Ex4_43.sce new file mode 100644 index 000000000..668b97480 --- /dev/null +++ b/3681/CH4/EX4.43/Ex4_43.sce @@ -0,0 +1,15 @@ +// Calculating the temperature rise
+clc;
+disp('Example 4.43, Page No. = 4.77')
+// Given Data
+az = 30*10^(-6);// Cross-sectional area (in meter square)
+Iz = 20*10^(3);// Current (in Ampere)
+t = 50;// Time (in mili second)
+p = 0.021*10^(-6);// Resistivity of conductor (in ohm*meter)
+h = 418;// Specific heat (in J/kg degree celsius)
+g = 8900;// Density (in kg per meter cube)
+// Calculation of the temperature rise
+T = Iz^(2)*p*t*10^(-3)/(g*az^(2)*h);// Temperature rise (in degree celsius)
+disp(T,'Temperature rise (degree celsius)=');
+//in book answer is 125 degree celsius. The answers vary due to round off error
+
diff --git a/3681/CH4/EX4.6/Ans4_6.PNG b/3681/CH4/EX4.6/Ans4_6.PNG Binary files differnew file mode 100644 index 000000000..214a6b3a7 --- /dev/null +++ b/3681/CH4/EX4.6/Ans4_6.PNG diff --git a/3681/CH4/EX4.6/Ex4_6.sce b/3681/CH4/EX4.6/Ex4_6.sce new file mode 100644 index 000000000..207b066e2 --- /dev/null +++ b/3681/CH4/EX4.6/Ex4_6.sce @@ -0,0 +1,21 @@ +// Estimating the temperature of the hot spot
+clc;
+disp('Example 4.6, Page No. = 4.11')
+// Given Data
+t = 0.5;// Plate width of transformer core in meter
+Ki = 0.94;// Stacking Factor
+p_core = 3;// Core loss in Watt per kg
+thermal_conductivity = 150;// Thermal conductivity in Watt per degree celsius
+Ts = 40;// Surface temperature in degree celsius
+D = 7800;// Density of steel plate in kg per meter cube
+// Calculation of the temperature of the hot spot
+q = p_core*Ki*D;// Core loss per unit volume (Watt per meter cube)
+p = 1/thermal_conductivity;// thermal resistivity
+x =t;// Since heat is taken all to one end
+Tm = (q*p*x*x/2)+Ts;// Temperature of the hot spot, if heat is taken all to one end (degree celsius)
+disp(Tm,'(a) Temperature of the hot spot, if heat is taken all to one end (degree celsius)=');
+//in book answers is 58.3 degree celsius. The answers vary due to round off error
+x =t/2;// Since heat is taken to both the directions
+Tm = (q*p*x*x/2)+Ts;// Temperature of the hot spot, if heat is taken to both the directions (degree celsius)
+disp(Tm,'(b) Temperature of the hot spot, if heat is taken to both the directions (degree celsius)=');
+//in book answers is 44.6 degree celsius. The answers vary due to round off error
diff --git a/3681/CH4/EX4.7/Ans4_7.PNG b/3681/CH4/EX4.7/Ans4_7.PNG Binary files differnew file mode 100644 index 000000000..b23596bb1 --- /dev/null +++ b/3681/CH4/EX4.7/Ans4_7.PNG diff --git a/3681/CH4/EX4.7/Ex4_7.sce b/3681/CH4/EX4.7/Ex4_7.sce new file mode 100644 index 000000000..cd72d12fb --- /dev/null +++ b/3681/CH4/EX4.7/Ex4_7.sce @@ -0,0 +1,18 @@ +// Estimating the hot spot temperature
+clc;
+disp('Example 4.7, Page No. = 4.14')
+// Given Data
+l = 1;// Length of mean turn in meter
+Sf = 0.56;// Space Factor
+p = 120;// Total loss in the coil in Watt
+pi = 8;// Thermal resistivity in ohm*meter
+A = 100*50;// Area of cross-section in mm square
+t = 50*10^(-3);// Thickness of coil in meter
+// Calculation of the temperature of the hot spot
+pe = pi*(1-Sf^(1/2));// Effective thermal resistivity in ohm*meter
+V = A*l*10^(-6);// Volume of coil(in meter cube)
+q = p/V;// Heat dissipated in Watt per meter cube
+T0 =q*pe*t*t/8;// Assuming equal inward and outward heat flows
+disp(T0,'Temperature of the hot spot (degree celsius)=');
+//in book answers is 15 degree celsius. The answers vary due to round off error
+
diff --git a/3681/CH4/EX4.8/Ans4_8.PNG b/3681/CH4/EX4.8/Ans4_8.PNG Binary files differnew file mode 100644 index 000000000..d7fc0172f --- /dev/null +++ b/3681/CH4/EX4.8/Ans4_8.PNG diff --git a/3681/CH4/EX4.8/Ex4_8.sce b/3681/CH4/EX4.8/Ex4_8.sce new file mode 100644 index 000000000..dd3e3889b --- /dev/null +++ b/3681/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,16 @@ +// Calculating the maximum temperature difference between the coil surface and the winding
+clc;
+disp('Example 4.8, Page No. = 4.14')
+// Given Data
+t = 25*10^(-3);// Thickness of coil (in meter)
+Sf = 0.7;// Space Factor
+Loss_cu = 20;// Copper losses (in Watt per kg)
+pi = 8;// Thermal resistivity of paper insulation(in ohm*meter)
+D_cu = 8900;// Density of copper (in kg per meter cube)
+// Calculation of the maximum temperature difference between the coil surface and the winding
+pe = pi*(1-Sf^(1/2));// Effective thermal resistivity in (ohm*meter)
+q = Sf*Loss_cu*D_cu;// Losses(in Watt per meter cube)
+T =q*pe*t^(2)/2;// Maximum temperature difference between the coil surface and the winding (in degree celsius)
+disp(T,'Maximum temperature difference between the coil surface and the winding (degree celsius)=');
+//in book answer is 51 degree celsius. The answers vary due to round off error
+
diff --git a/3681/CH4/EX4.9/Ans4_9.PNG b/3681/CH4/EX4.9/Ans4_9.PNG Binary files differnew file mode 100644 index 000000000..392fcd357 --- /dev/null +++ b/3681/CH4/EX4.9/Ans4_9.PNG diff --git a/3681/CH4/EX4.9/Ex4_9.sce b/3681/CH4/EX4.9/Ex4_9.sce new file mode 100644 index 000000000..d198af966 --- /dev/null +++ b/3681/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,13 @@ +// Calculating the temperature difference beetween the centre of the embedded portion of a conductor and the overhang
+clc;
+disp('Example 4.9, Page No. = 4.16')
+// Given Data
+L = 0.5;// Length of the machine in meter
+pc = 0.0025;// Thermal resistivity of conductor in ohm*meter
+p = 0.021*10^(-6);// Electrical resistivity of conductor in ohm*meter
+s = 4;// Current density in the conductors (in A per mm square)
+// Calculation of the temperature difference beetween the centre of the embedded portion of a conductor and the overhang
+T = (s*10^(6))^(2)*(p*pc*L*L)/8;// Effective thermal resistivity in ohm*meter
+disp(T,'Temperature difference beetween the centre of the embedded portion of a conductor and the overhang (degree celsius)=');
+//in book answers is 26.3 degree celsius. The answers vary due to round off error
+
diff --git a/3681/CH5/EX5.12/Ans5_12.PNG b/3681/CH5/EX5.12/Ans5_12.PNG Binary files differnew file mode 100644 index 000000000..d36b75aaf --- /dev/null +++ b/3681/CH5/EX5.12/Ans5_12.PNG diff --git a/3681/CH5/EX5.12/Ex5_12.sce b/3681/CH5/EX5.12/Ex5_12.sce new file mode 100644 index 000000000..11d6ef3dd --- /dev/null +++ b/3681/CH5/EX5.12/Ex5_12.sce @@ -0,0 +1,13 @@ +// 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 Binary files differnew file mode 100644 index 000000000..9e6f37822 --- /dev/null +++ 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 new file mode 100644 index 000000000..601315a10 --- /dev/null +++ b/3681/CH5/EX5.13/Ex5_13.sce @@ -0,0 +1,21 @@ +// 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 Binary files differnew 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 --- /dev/null +++ b/3681/CH5/EX5.14/Ex5_14.sce @@ -0,0 +1,20 @@ +// 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 Binary files differnew file mode 100644 index 000000000..5b3164df8 --- /dev/null +++ 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 new file mode 100644 index 000000000..a25af31af --- /dev/null +++ b/3681/CH5/EX5.16/Ex5_16.sce @@ -0,0 +1,21 @@ +// 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 Binary files differnew file mode 100644 index 000000000..63a2db3f8 --- /dev/null +++ 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 --- /dev/null +++ b/3681/CH5/EX5.17/Ex5_17.sce @@ -0,0 +1,27 @@ +// 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 Binary files differnew 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 new file mode 100644 index 000000000..486fd9017 --- /dev/null +++ b/3681/CH5/EX5.18/Ex5_18.sce @@ -0,0 +1,29 @@ +// 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 Binary files differnew file mode 100644 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 --- /dev/null +++ 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 Binary files differnew file mode 100644 index 000000000..2a6909480 --- /dev/null +++ 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 --- /dev/null +++ b/3681/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,22 @@ +// 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 Binary files differnew file mode 100644 index 000000000..5c475d26a --- /dev/null +++ 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 --- /dev/null +++ b/3681/CH5/EX5.6/Ex5_6.sce @@ -0,0 +1,23 @@ +// 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 Binary files differnew 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 --- /dev/null +++ b/3681/CH5/EX5.9/Ex5_9.sce @@ -0,0 +1,21 @@ +// 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
diff --git a/3681/CH6/EX6.1/Ans6_1.PNG b/3681/CH6/EX6.1/Ans6_1.PNG Binary files differnew file mode 100644 index 000000000..094303470 --- /dev/null +++ b/3681/CH6/EX6.1/Ans6_1.PNG diff --git a/3681/CH6/EX6.1/Ex6_1.sce b/3681/CH6/EX6.1/Ex6_1.sce new file mode 100644 index 000000000..cd4a7fe5e --- /dev/null +++ b/3681/CH6/EX6.1/Ex6_1.sce @@ -0,0 +1,22 @@ +// Calculating the specific electric and specific magnetic loading
+clc;
+disp('Example 6.1, Page No. = 6.10')
+// Given Data
+P = 350;// Power rating (in kW)
+E = 500;// Voltage (in V)
+rpm = 450;
+p = 6;// Number of poles
+a = 6;// Since a=p for lap winding
+Z = 660;// Number of conductors
+L = 0.32;// Core length (in meter)
+D = 0.87;// Armature diameter (in meter)
+// Calculation of the specific electric loading
+Ia = P*1000/E;// Armature current (in A)
+Iz = Ia/a;// Current in each conductor (in A)
+ac = Iz*Z/(%pi*D);// Specific electric loading
+disp(ac,'Specific electric loading (ampere conductors per meter)=');
+// Calculation of the specific magnetic loading
+F = E*a/(Z*rpm/60*p);// Flux per pole (in Wb)
+Bac = p*F/(%pi*D*L);// specific magnetic loading
+disp(Bac,'Specific magnetic loading (Wb per meter square)=');
+//in book answers are 28200 (ampere conductors per meter) and 0.693 (Wb per meter square) respectively. The answers vary due to round off error
diff --git a/3681/CH6/EX6.5/Ans6_5.PNG b/3681/CH6/EX6.5/Ans6_5.PNG Binary files differnew file mode 100644 index 000000000..d126b5402 --- /dev/null +++ b/3681/CH6/EX6.5/Ans6_5.PNG diff --git a/3681/CH6/EX6.5/Ex6_5.sce b/3681/CH6/EX6.5/Ex6_5.sce new file mode 100644 index 000000000..722cac545 --- /dev/null +++ b/3681/CH6/EX6.5/Ex6_5.sce @@ -0,0 +1,13 @@ +// Calculating the power developed by the armature of motor
+clc;
+disp('Example 6.5, Page No. = 6.12')
+// Given Data
+P = 125;// Power rating (in W)
+E = 230;// Voltage (in V)
+rpm = 5000;
+// Calculation of the power developed by the armature
+Losses_total = P;// Total losses (in W)
+Losses_constant = P/3;// Constant losses (in W). Since the sum of iron, friction and windage losses is approximately 1/3 of total losses
+Pa = Losses_total+Losses_constant;// Power developed by the armature (in W)
+disp(Pa,'Power developed by the armature (W)=');
+//in book answer is 167 (W). The answers vary due to round off error
diff --git a/3681/CH6/EX6.6/Ans6_6.PNG b/3681/CH6/EX6.6/Ans6_6.PNG Binary files differnew file mode 100644 index 000000000..ed289c74f --- /dev/null +++ b/3681/CH6/EX6.6/Ans6_6.PNG diff --git a/3681/CH6/EX6.6/Ex6_6.sce b/3681/CH6/EX6.6/Ex6_6.sce new file mode 100644 index 000000000..df7228d5f --- /dev/null +++ b/3681/CH6/EX6.6/Ex6_6.sce @@ -0,0 +1,11 @@ +// Calculating the limiting value of specific magnetic loading
+clc;
+disp('Example 6.6, Page No. = 6.12')
+// Given Data
+Bt = 2.0;// Maximum flux density in the armature (in Wb per meter square)
+R = 0.7;// Ratio of pole arc to pole pitch
+Wt_ys = 0.4;// Ratio of minimum width of tooth to slot pitch
+// Calculation of the limiting value of specific magnetic loading
+Bav = R*Wt_ys*Bt;// Limiting value of specific magnetic loading (in W per meter square)
+disp(Bav,'Limiting value of specific magnetic loading (W per meter square)=');
+//in book answer is 0.56 (W per meter square). The answers vary due to round off error
diff --git a/3681/CH6/EX6.8/Ans6_8.PNG b/3681/CH6/EX6.8/Ans6_8.PNG Binary files differnew file mode 100644 index 000000000..609ec5582 --- /dev/null +++ b/3681/CH6/EX6.8/Ans6_8.PNG diff --git a/3681/CH6/EX6.8/Ex6_8.sce b/3681/CH6/EX6.8/Ex6_8.sce new file mode 100644 index 000000000..640a265ef --- /dev/null +++ b/3681/CH6/EX6.8/Ex6_8.sce @@ -0,0 +1,23 @@ +// Calculating the maximum permissible specific electric loading
+clc;
+disp('Example 6.8, Page No. = 6.13')
+// Given Data
+p_20 = 1.734*10^(-8);// Resistivity of copper at 20 degree celsius (in ohm*meter)
+alpha = 0.00393;// Resistance temperature co-efficient of copper at 20 degree celsius (in per degree celsius)
+s = 3.5;// Current density (in A per mm square)
+c = 0.03;// Cooling co-efficient
+Tm_ambient = 40;// Maximum ambient temperature (in degree celsius)
+Tm_rise_A = 50;// Maximum temperature rise for Class A insulation (in degree celsius)
+Tm_rise_E = 65;// Maximum temperature rise for Class E insulation (in degree celsius)
+// Calculation of the maximum permissible specific electric loading
+//for Class A insulation
+T_A = Tm_ambient+Tm_rise_A;// Operating temperature of copper conductors (in degree celsius)
+p = p_20*(1+alpha*(T_A-20));// Resistivity at operating temperature (in ohm*meter)
+ac = Tm_rise_A/(p*s*10^(6)*c);// Maximum permissible specific electric loading
+disp(ac,'Maximum allowable specific electric loading (ampere conductors per meter)=');
+T_E = Tm_ambient+Tm_rise_E;// Operating temperature of copper conductors (in degree celsius)
+//for Class E insulation
+p = p_20*(1+alpha*(T_E-20));// Resistivity at operating temperature (in ohm*meter)
+ac = Tm_rise_E/(p*s*10^(6)*c);// Maximum permissible specific electric loading
+disp(ac,'Maximum allowable specific electric loading (ampere conductors per meter)=');
+//in book answers are 21600 (ampere conductors per meter) and 26700 (ampere conductors per meter) respectively. The answers vary due to round off error
diff --git a/3681/CH6/EX6.9/Ans6_9.PNG b/3681/CH6/EX6.9/Ans6_9.PNG Binary files differnew file mode 100644 index 000000000..506337e08 --- /dev/null +++ b/3681/CH6/EX6.9/Ans6_9.PNG diff --git a/3681/CH6/EX6.9/Ex6_9.sce b/3681/CH6/EX6.9/Ex6_9.sce new file mode 100644 index 000000000..fef103093 --- /dev/null +++ b/3681/CH6/EX6.9/Ex6_9.sce @@ -0,0 +1,21 @@ +// Calculating the specific electric loading
+clc;
+disp('Example 6.9, Page No. = 6.13')
+// Given Data
+Pc = 1000;// Core loss (in W)
+R = 0.025;// Armature resistance (in ohm)
+l = 230;// Specific loss dissipation (in W per degree celsius per meter square)
+a = 2;// Since a=z for lap winding
+Z = 270;// Number of conductors
+L = 0.25;// Core length (in meter)
+D = 0.25;// Armature diameter (in meter)
+T = 40;// Temperature rise (degree celsius)
+// Calculation of the specific electric loading
+c = 1/l;// Cooling co-efficient
+S = %pi*D*L;// Dissipation surface (in meter square)
+Q = S*T/c;// Maximum allowable pwer dissipation from armature surface
+Ia = ((Q-Pc)/R)^(1/2);// Armature current (in Ampere)
+Iz = Ia/a;// Current in each conductor (in A)
+ac = Iz*Z/(%pi*D);// Specific electric loading
+disp(ac,'Specific electric loading (ampere conductors per meter)=');
+//in book answer is 31000 (ampere conductors per meter). The answers vary due to round off error
diff --git a/3681/CH7/EX7.33/Ans7_33.PNG b/3681/CH7/EX7.33/Ans7_33.PNG Binary files differnew file mode 100644 index 000000000..541a11334 --- /dev/null +++ b/3681/CH7/EX7.33/Ans7_33.PNG diff --git a/3681/CH7/EX7.33/Ex7_33.sce b/3681/CH7/EX7.33/Ex7_33.sce new file mode 100644 index 000000000..e889eda24 --- /dev/null +++ b/3681/CH7/EX7.33/Ex7_33.sce @@ -0,0 +1,16 @@ +// Calculating the rms line voltage and circulating current
+clc;
+disp('Example 7.33, Page No. = 7.75')
+// Given Data
+E = 1000;// Amplitude of fundamental emf (in V)
+R = 10;// Reactance per phase (in ohm)
+// Calculation of the rms line voltage and circulating current
+Eph1 = E/2^(1/2);// Rms value of fundamental emf per phase
+Eph3 = 0.2*Eph1;// Rms value of 3rd harmonic component of phase voltage (in V) Given 20%
+Eph5 = 0.1*Eph1;// Rms value of 5th harmonic component of phase voltage (in V) Given 10%
+Eph = (Eph1*Eph1+Eph5*Eph5)^(1/2);// Phase voltage considering no 3rd harmonic
+disp(3^(1/2)*Eph,'(a) rms line voltage when star connected (V)=');
+disp(Eph,'(b) rms line voltage when delta connected (V)=');
+I_circulating = 3*Eph3/(3*3*10);// Circulating current taking reactance corresponding to 3rd harmonic
+disp(I_circulating,'Circulating current (ampere)=');
+//in book answers are 1230.8 V, 710.6 v and 4.71 ampere respectively. The answers vary due to round off error
diff --git a/3681/CH7/EX7.41/Ans7_41.PNG b/3681/CH7/EX7.41/Ans7_41.PNG Binary files differnew file mode 100644 index 000000000..9ca2169b9 --- /dev/null +++ b/3681/CH7/EX7.41/Ans7_41.PNG diff --git a/3681/CH7/EX7.41/Ex7_41.sce b/3681/CH7/EX7.41/Ex7_41.sce new file mode 100644 index 000000000..98bc28ce9 --- /dev/null +++ b/3681/CH7/EX7.41/Ex7_41.sce @@ -0,0 +1,35 @@ +// Calculating the eddy current loss ratio and average loss ratio and critical depth for minimum loss
+clc;
+disp('Example 7.41, Page No. = 7.104')
+// Given Data
+Ws = 20;// Slot width (in mm)
+b = 14;// Width of copper conductors (in mm)
+h = 8;// Depth of copper conductors (in mm)
+f = 50// Frequency (in Hz)
+N = 5;// Number of layers
+// Calculation of eddy loss factor for different layers
+a = 100*(b/Ws)^(1/2);
+ah = a*h*10^(-3);
+ah4 = ah^(4);
+Ke1 = 1;// 1st layer
+Ke2 = 1+ah4*2*(2-1)/3;// 2nd layer
+Ke3 = 1+ah4*3*(3-1)/3;// 3rd layer
+Ke4 = 1+ah4*4*(4-1)/3;// 4th layer
+Ke5 = 1+ah4*5*(5-1)/3;// 5th layer
+disp(Ke1,'1st layer Ke1 =');
+disp(Ke2,'2nd layer Ke2 =');
+disp(Ke3,'3rd layer Ke3 =');
+disp(Ke4,'4th layer Ke4 =');
+disp(Ke5,'5th layer Ke5 =');
+// Calculation of average eddy current loss factor for all the five layers
+Ke_av = 1+ah4*N*N/9;
+disp(Ke_av,'Average eddy current loss factor for all the five layers =');
+// Calculation of critical depth for minimum loss
+hc = 1/(a*(3*N*N/9)^(1/4))*1000;// Critical depth (in mm)
+disp(hc,'Critical depth (mm)=');
+// Calculation of average eddy current loss factor for all the five layers for this critical depth
+ahc = a*hc*10^(-3);
+ahc4 = ahc^(4);
+Ke_av = 1+ahc4*N*N/9;
+disp(Ke_av,'Average eddy current loss factor for this critical depth=');
+//in book answers are 1, 1.13, 1.4, 1.8, 2.33, 1.55, 7 mm and 1.33 respectively. The answers vary due to round off error
diff --git a/3681/CH8/EX8.2/Ans8_2.PNG b/3681/CH8/EX8.2/Ans8_2.PNG Binary files differnew file mode 100644 index 000000000..7059c3185 --- /dev/null +++ b/3681/CH8/EX8.2/Ans8_2.PNG diff --git a/3681/CH8/EX8.2/Ex8_2.sce b/3681/CH8/EX8.2/Ex8_2.sce new file mode 100644 index 000000000..215ef0c87 --- /dev/null +++ b/3681/CH8/EX8.2/Ex8_2.sce @@ -0,0 +1,17 @@ +// Calculating the stress on the ring
+clc;
+disp('Example 8.2, Page No. = 8.8')
+// Given Data
+rpm = 3000;// Speed in r.p.m.
+Rm = 0.35;// Radius of overhang (in meter)
+Rmr = 0.49;// Radius of ring (in meter)
+G = 300;// Weight of copper winding (in kg)
+gr = 7800;// Density of ring material (in kg per meter cube)
+tb = 350*45*10^(-6);// Area of retaining ring
+// Calculation of the stress on the ring
+n = rpm/60;// Speed in r.p.s
+Dm = 2*Rm;// Diameter of overhang (in meter)
+Dmr = 2*Rmr;// Diameter of ring (in meter)
+ft = (%pi*n*n*G*Dm/tb)+(%pi*%pi*n*n*gr*Dmr*Dmr);// Stress on ring (in Newton per meter square)
+disp(ft,'Stress on ring (Newton per meter square)=');
+//in book answer is 289.5 (MN per meter square). The answers vary due to round off error
diff --git a/3681/CH8/EX8.4/Ans8_4.PNG b/3681/CH8/EX8.4/Ans8_4.PNG Binary files differnew file mode 100644 index 000000000..a5b51fa78 --- /dev/null +++ b/3681/CH8/EX8.4/Ans8_4.PNG diff --git a/3681/CH8/EX8.4/Ex8_4.sce b/3681/CH8/EX8.4/Ex8_4.sce new file mode 100644 index 000000000..8ae86eb05 --- /dev/null +++ b/3681/CH8/EX8.4/Ex8_4.sce @@ -0,0 +1,21 @@ +// Calculating the tensile stress and factor of safety
+clc;
+disp('Example 8.4, Page No. = 8.12')
+// Given Data
+rpm = 3000;// Speed in r.p.m.
+Dr1 = 1.15;// Outer diameter of rotor (in meter)
+Nrs = 39;// Number of rotor slot
+Drs = 140;// Depth of rotor slot (in mm)
+Wrs = 45;// Width of rotor slot (in mm)
+gs = 7800;// Density of steel (in kg per meter cube)
+yield_stress = 520*10^(6);// Yield stress of rotor steel (in Newton per meter square)
+// Calculation of the tensile stress and factor of safety
+n = rpm/60;// Speed in r.p.s
+Dr2 = Dr1-2*Drs*10^(-3);// Diameter of rotor at the bottom of slots (in meter)
+t = (%pi*Dr2*10^(3)/Nrs)-Wrs;// Width of tooth at the bottom of slot (in mm)
+alpha = 360/Nrs;// Angle subtended by each slot (in degree)
+f = %pi^(3)/(3*t*10^(-3))*gs*n*n*(alpha/360)*(Dr1^(3)-Dr2^(3));// Tensile stress (in Newton per meter square)
+disp(f,'Tensile stress at the root of the teeth at normal operating speed (Newton per meter square)=');
+f_20 = 1.2^(2)*f;// Tensile stress at 20% over speed. Since centrifugal force is propartional of square of speed
+disp(yield_stress/f_20,'Factor of safety at 20% over speed =');
+//in book answers are 178 (Mega Newton per meter square) and 2.03 respectively. The answers vary due to round off error
diff --git a/3681/CH8/EX8.5/Ans8_5.PNG b/3681/CH8/EX8.5/Ans8_5.PNG Binary files differnew file mode 100644 index 000000000..ee475d521 --- /dev/null +++ b/3681/CH8/EX8.5/Ans8_5.PNG diff --git a/3681/CH8/EX8.5/Ex8_5.sce b/3681/CH8/EX8.5/Ex8_5.sce new file mode 100644 index 000000000..6225abdb8 --- /dev/null +++ b/3681/CH8/EX8.5/Ex8_5.sce @@ -0,0 +1,14 @@ +// Calculating the inertia constant of the generator
+clc;
+disp('Example 8.5, Page No. = 8.14')
+// Given Data
+P = 500;// Power rating (in MW)
+f = 50;// Frequency (in Hz)
+J = 50*10^(3);// Moment of inertia (in kg-meter square)
+pf = 0.85;// Power factor
+// Calculation of the inertia constant of the generator
+w = 2*%pi*f;// Angular speed (in rad/s)
+Q = 500*10^(3)/pf;// kVA rating
+H = (1/2)*J*w*w/(Q*10^(3));// Inertia constant (in seconds)
+disp(H,'Inertia constant (seconds)=');
+//in book answer is 4.2 seconds. The answers vary due to round off error
diff --git a/3681/CH9/EX9.10/Ans9_10.PNG b/3681/CH9/EX9.10/Ans9_10.PNG Binary files differnew file mode 100644 index 000000000..9d48ce8f7 --- /dev/null +++ b/3681/CH9/EX9.10/Ans9_10.PNG diff --git a/3681/CH9/EX9.10/Ex9_10.sce b/3681/CH9/EX9.10/Ex9_10.sce new file mode 100644 index 000000000..3fa502598 --- /dev/null +++ b/3681/CH9/EX9.10/Ex9_10.sce @@ -0,0 +1,17 @@ +// Calculating the demagnetizing and cross magnetizing mmf per pole
+clc;
+disp('Example 9.10, Page No. = 9.38')
+// Given Data
+P = 500;// Power rating (in kW)
+rpm = 375;// Speed in r.p.m.
+p = 8;// Number of poles
+flux = 0.0885;// Flux per pole (in Wb per meter)
+// Calculation of the demagnetizing and cross magnetizing mmf per pole
+n = rpm/60;// Speed in r.p.s.
+alpha = 5/100*180;// Brush shift (in electrical degree). Since the brushes are given a lead by of 5% of pole pitch
+ATa = P/(2*flux*n*p*p*10^(-3));// Armature mmf per pole (A)
+ATad = ATa*2*alpha/180;;// Demagnetizing mmf per pole (A)
+ATaq = ATa-ATad;// Cross magnetizing mmf per pole (A)
+disp(ATad,'Demagnetizing mmf per pole (A) =');
+disp(ATaq,'Cross magnetizing mmf per pole (A) =');
+//in book answers are 706 (A) and 6354 (A) respectively. The answers vary due to round off error
diff --git a/3681/CH9/EX9.12/Ans9_12.PNG b/3681/CH9/EX9.12/Ans9_12.PNG Binary files differnew file mode 100644 index 000000000..6fe2f2b02 --- /dev/null +++ b/3681/CH9/EX9.12/Ans9_12.PNG diff --git a/3681/CH9/EX9.12/Ex9_12.sce b/3681/CH9/EX9.12/Ex9_12.sce new file mode 100644 index 000000000..9e02a54cb --- /dev/null +++ b/3681/CH9/EX9.12/Ex9_12.sce @@ -0,0 +1,17 @@ +// Calculating the armature voltage drop
+clc;
+disp('Example 9.12, Page No. = 9.49')
+// Given Data
+P = 300;// Power rating (in kW)
+V = 500;// Voltage rating (in volts)
+a = 6;// Number of parallel paths (Since lap winding)
+p = 0.021;// resistivity (in ohm mm square)
+Ns = 150;// Number of slots
+Lmt = 2.5;// Length of mean turn (in meter)
+az = 25;// Area of each conductror (in mm square)
+// Calculation of the armature voltage drop
+Z = Ns*8;// Number of armature conductors. Since 8 conductors per slot
+ra = Z*p*Lmt/(2*a*a*az);// Resistance of armature (in ohm)
+Ia = P*10^(3)/V;// Armature current
+disp(Ia*ra,'Armature voltage drop (Volts) =');
+//in book answer is 21 (Volt). The answers vary due to round off error
diff --git a/3681/CH9/EX9.26/Ans9_26.PNG b/3681/CH9/EX9.26/Ans9_26.PNG Binary files differnew file mode 100644 index 000000000..d6a290a89 --- /dev/null +++ b/3681/CH9/EX9.26/Ans9_26.PNG diff --git a/3681/CH9/EX9.26/Ex9_26.sce b/3681/CH9/EX9.26/Ex9_26.sce new file mode 100644 index 000000000..7a14d511a --- /dev/null +++ b/3681/CH9/EX9.26/Ex9_26.sce @@ -0,0 +1,19 @@ +// Calculating the number of turns on each commutating pole
+clc;
+disp('Example 9.26, Page No. = 9.85')
+// Given Data
+p = 6;// Number of poles
+Bgi = 0.5;// Flux density (in Wb per meter square)
+Ia = 500;// Armature full load current (in ampere)
+Z = 540;// Number of conductors
+Kgi = 1;// Inerpole interaction factor
+lgi = 4;// Effective length of air gap
+// Calculation of the number of turns on each commutating pole
+a = p;// Number of parallel paths. Since armature is lap wound
+ATa = Ia/a*Z/(2*p);// Armature mmf per pole
+mmf_airgap = 800000*Bgi*Kgi*lgi*10^(-3);// Mmf required for air gap (in A)
+mmf_iron = 0.1*mmf_airgap;// Mmf required for iron parts (in A). Since mmf required is one-tenth that for air gap
+ATi = ATa+mmf_airgap+mmf_iron;// Total mmf per pole on each interpole (in A)
+Ti = ATi/Ia;// Number of turns on each interpole
+disp(Ti,'Number of turns on each interpole =');
+//in book answer is 11. The answers vary due to round off error
diff --git a/3681/CH9/EX9.27/Ans9_27.PNG b/3681/CH9/EX9.27/Ans9_27.PNG Binary files differnew file mode 100644 index 000000000..561ca6b7b --- /dev/null +++ b/3681/CH9/EX9.27/Ans9_27.PNG diff --git a/3681/CH9/EX9.27/Ex9_27.sce b/3681/CH9/EX9.27/Ex9_27.sce new file mode 100644 index 000000000..ecdb0aa81 --- /dev/null +++ b/3681/CH9/EX9.27/Ex9_27.sce @@ -0,0 +1,15 @@ +// Calculating the reactance voltage for a machine with straight line and sinusoidal commutation
+clc;
+disp('Example 9.27, Page No. = 9.86')
+// Given Data
+Ns = 60;// Number of segments
+rev = 10;// Number of revolution per second
+W = 1.5;// Brush width in segments
+L = 0.2;// Co-efficient of self-induction (in mH)
+I = 20;// Current per coil
+// Calculation of the reactance voltage for a machine with straight line and sinusoidal commutation
+Tc = W/(Ns*rev);// Time of commutation
+Erav = L*10^(-3)*2*I/Tc;// Average reactance voltage
+disp(Erav,'Reactance voltage with straight line commutation (Volts)=');
+disp(%pi/2*Erav,'Reactance voltage with sinusoidal commutation (Volts)=');
+//in book answers are 3.2 Volts and 5 Volts respectively. The answers vary due to round off error
diff --git a/3681/CH9/EX9.32/Ans9_32.PNG b/3681/CH9/EX9.32/Ans9_32.PNG Binary files differnew file mode 100644 index 000000000..c79d322dd --- /dev/null +++ b/3681/CH9/EX9.32/Ans9_32.PNG diff --git a/3681/CH9/EX9.32/Ex9_32.sce b/3681/CH9/EX9.32/Ex9_32.sce new file mode 100644 index 000000000..bde0c0ef8 --- /dev/null +++ b/3681/CH9/EX9.32/Ex9_32.sce @@ -0,0 +1,12 @@ +// Calculating the minimum number of poles
+clc;
+disp('Example 9.32, Page No. = 9.92')
+// Given Data
+P = 1200;// Power rating (in kW)
+Ec = 15;// Average voltage between commutator segments (in Volts)
+ATa = 10000;// Armature mmf per pole
+// Calculation of the minimum number of poles
+a = P*10^(3)/(ATa*Ec);// Minimum number of parallel paths
+p = a;// Minimum number of poles. Since these parallel paths can be obtained by using a simplex winding
+disp(p,'Minimum number of poles =');
+//in book answer is 8 poles. The answers vary due to round off error
diff --git a/3681/CH9/EX9.33/Ans9_33.PNG b/3681/CH9/EX9.33/Ans9_33.PNG Binary files differnew file mode 100644 index 000000000..a4256757e --- /dev/null +++ b/3681/CH9/EX9.33/Ans9_33.PNG diff --git a/3681/CH9/EX9.33/Ex9_33.sce b/3681/CH9/EX9.33/Ex9_33.sce new file mode 100644 index 000000000..1e393130a --- /dev/null +++ b/3681/CH9/EX9.33/Ex9_33.sce @@ -0,0 +1,12 @@ +// Calculating the maximum armature voltage
+clc;
+disp('Example 9.33, Page No. = 9.92')
+// Given Data
+Vc = 40;// Peripheral speed of commutator (in meter per second)
+Ec = 20;// Average emf between adjacent segments (in Volts)
+Bc = 4;// Minimum pitch of commutator segments (in mm)
+f = 40;// Frequency (in Hz)
+// Calculation of the maximum armature voltage
+E = Vc*Ec/(2*f*Bc*10^(-3));// Maximum armature voltage (in Volts)
+disp(E,'Maximum armature voltage (Volts)=');
+//in book answer is 2500 Volts. The answers vary due to round off error
diff --git a/3681/CH9/EX9.34/Ans9_34.PNG b/3681/CH9/EX9.34/Ans9_34.PNG Binary files differnew file mode 100644 index 000000000..29bc29fad --- /dev/null +++ b/3681/CH9/EX9.34/Ans9_34.PNG diff --git a/3681/CH9/EX9.34/Ex9_34.sce b/3681/CH9/EX9.34/Ex9_34.sce new file mode 100644 index 000000000..2982e6dc1 --- /dev/null +++ b/3681/CH9/EX9.34/Ex9_34.sce @@ -0,0 +1,24 @@ +// Calculating the total commutator losses
+clc;
+disp('Example 9.34, Page No. = 9.92')
+// Given Data
+P = 800;// Power rating (in kW)
+V = 400;// Voltage rating (in Volts)
+rpm = 300;// r.p.m.
+p = 10;// Number of poles
+Dc = 1;// Commutator diameter (in meter). Since 100 cm = 1 meter
+u = 0.23;// Co-efficient of friction
+Pb = 14.7;// Brush pressure (in kN per meter square)
+J = 0.075;// Current density in brushes (in A per mm square)
+Vcb = 2.2;// Total brush contact drop (in Volts)
+// Calculation of the total commutator losses
+n = rpm/60;// r.p.s.
+Ia = P*10^(3)/V;// Armature current (in Ampere)
+Ib = 2*Ia/p;// Current per brush arm (in Ampere)
+Ab = Ib/J;// Brush area per brush arm (in mm square)
+AB = p*Ab*10^(-6);// Total brush area on the commutator (in meter square)
+Vc = %pi*Dc*n;// Peripheral speed (in meter per second)
+Wcf = u*Pb*10^(3)*AB*Vc;// Brush friction loss (in Watts)
+Wcb = Ia*Vcb;// Brush contact loss (in Watts)
+disp(Wcf+Wcb,'Total commutator losses (Watts)=');
+//in book answer is 7230 Watts. The answers vary due to round off error
diff --git a/3681/CH9/EX9.7/Ans9_7.PNG b/3681/CH9/EX9.7/Ans9_7.PNG Binary files differnew file mode 100644 index 000000000..ae3b63f71 --- /dev/null +++ b/3681/CH9/EX9.7/Ans9_7.PNG diff --git a/3681/CH9/EX9.7/Ex9_7.sce b/3681/CH9/EX9.7/Ex9_7.sce new file mode 100644 index 000000000..805f6ce1e --- /dev/null +++ b/3681/CH9/EX9.7/Ex9_7.sce @@ -0,0 +1,13 @@ +// Calculating the maximum permissible core length for the machine
+clc;
+disp('Example 9.7, Page No. = 9.32')
+// Given Data
+Kf = 0.67;// Form factor
+Bg = 1;// Maximum gap density (in Wb per meter square)
+Va = 40;// Armature peripheral speed (in meter)
+E = 7;// Maximum permissible value of emf induced in a conductor at no load (in Volts)
+// Calculation of the maximum permissible core length for the machine
+Bav = Kf*Bg;// Average gap density (in Wb per meter square)
+L = E/(Bav*Va);// Maximum permissible core length (in meter)
+disp(L,'Maximum permissible core length (meter)=');
+//in book answer is 0.26 (meter). The answers vary due to round off error
diff --git a/3681/CH9/EX9.8/Ans9_8.PNG b/3681/CH9/EX9.8/Ans9_8.PNG Binary files differnew file mode 100644 index 000000000..dfd2cc856 --- /dev/null +++ b/3681/CH9/EX9.8/Ans9_8.PNG diff --git a/3681/CH9/EX9.8/Ex9_8.sce b/3681/CH9/EX9.8/Ex9_8.sce new file mode 100644 index 000000000..b10213d48 --- /dev/null +++ b/3681/CH9/EX9.8/Ex9_8.sce @@ -0,0 +1,11 @@ +// Calculating the maximum permissible output from a machine
+clc;
+disp('Example 9.8, Page No. = 9.33')
+// Given Data
+D = 2;// Diameter (in meter)
+ac = 50000;// Specific electric loading
+ez = 7.5;// emf generated in a conductor at no load (in Volts)
+// Calculation of the maximum permissible output from a machine
+P = %pi*D*ac*ez*10^(-3);// Maximum permissible output (in kW)
+disp(P,'Maximum permissible output (kW)=');
+//in book answer is 2350 (kW). The answers vary due to round off error
diff --git a/3681/CH9/EX9.9/Ans9_9.PNG b/3681/CH9/EX9.9/Ans9_9.PNG Binary files differnew file mode 100644 index 000000000..69419fcd2 --- /dev/null +++ b/3681/CH9/EX9.9/Ans9_9.PNG diff --git a/3681/CH9/EX9.9/Ex9_9.sce b/3681/CH9/EX9.9/Ex9_9.sce new file mode 100644 index 000000000..1a58ea40f --- /dev/null +++ b/3681/CH9/EX9.9/Ex9_9.sce @@ -0,0 +1,26 @@ +// Calculating the number of extra shunt field turns to neutralize the demagnetization
+clc;
+disp('Example 9.9, Page No. = 9.38')
+// Given Data
+p = 4;// Number of poles
+Is = 140;// Current supplied by generator (in ampere)
+Z = 480;// Number of armature conductors
+mech_degree = 10;// Since brushes are given an actual lead of 10 degree
+// Calculation of the extra shunt field turns to neutralize the demagnetization
+Ia = Is+10;// Armature current (A). Since field winding is shunt connected and takes a current of 10 ampere
+alpha = p/2*mech_degree;// Angle of lead (in electrical degree)
+disp('(a) Wave connected')
+a= 2 // With wave winding number of parallel paths
+ATa = Ia*Z/(a*2*p);// Armature mmf per pole (A)
+ATad = ATa*2*alpha/180;;// Demagnetizing mmf per pole (A)
+ATaq = ATa-ATad;// Cross magnetizing mmf per pole (A)
+Extra_turns = ATad/10;// Extra turns required on the shunt field. Since field winding is shunt connected and takes a current of 10 ampere
+disp(Extra_turns,'Extra turns required on the shunt field =');
+disp('(b) Lap connected')
+a= p // With lap winding number of parallel paths
+ATa = Ia*Z/(a*2*p);// Armature mmf per pole (A)
+ATad = ATa*2*alpha/180;;// Demagnetizing mmf per pole (A)
+ATaq = ATa-ATad;// Cross magnetizing mmf per pole (A)
+Extra_turns = ATad/10;// Extra turns required on the shunt field. Since field winding is shunt connected and takes a current of 10 ampere
+disp(Extra_turns,'Extra turns required on the shunt field =');
+//in book answers are 100 and 50 respectively. The answers vary due to round off error
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