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-rw-r--r--3681/CH9/EX9.10/Ans9_10.PNGbin0 -> 6377 bytes
-rw-r--r--3681/CH9/EX9.10/Ex9_10.sce17
-rw-r--r--3681/CH9/EX9.12/Ans9_12.PNGbin0 -> 4289 bytes
-rw-r--r--3681/CH9/EX9.12/Ex9_12.sce17
-rw-r--r--3681/CH9/EX9.26/Ans9_26.PNGbin0 -> 4510 bytes
-rw-r--r--3681/CH9/EX9.26/Ex9_26.sce19
-rw-r--r--3681/CH9/EX9.27/Ans9_27.PNGbin0 -> 8485 bytes
-rw-r--r--3681/CH9/EX9.27/Ex9_27.sce15
-rw-r--r--3681/CH9/EX9.32/Ans9_32.PNGbin0 -> 4149 bytes
-rw-r--r--3681/CH9/EX9.32/Ex9_32.sce12
-rw-r--r--3681/CH9/EX9.33/Ans9_33.PNGbin0 -> 4776 bytes
-rw-r--r--3681/CH9/EX9.33/Ex9_33.sce12
-rw-r--r--3681/CH9/EX9.34/Ans9_34.PNGbin0 -> 4629 bytes
-rw-r--r--3681/CH9/EX9.34/Ex9_34.sce24
-rw-r--r--3681/CH9/EX9.7/Ans9_7.PNGbin0 -> 4874 bytes
-rw-r--r--3681/CH9/EX9.7/Ex9_7.sce13
-rw-r--r--3681/CH9/EX9.8/Ans9_8.PNGbin0 -> 4794 bytes
-rw-r--r--3681/CH9/EX9.8/Ex9_8.sce11
-rw-r--r--3681/CH9/EX9.9/Ans9_9.PNGbin0 -> 7473 bytes
-rw-r--r--3681/CH9/EX9.9/Ex9_9.sce26
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diff --git a/3681/CH9/EX9.10/Ans9_10.PNG b/3681/CH9/EX9.10/Ans9_10.PNG
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+// 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
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+// 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
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diff --git a/3681/CH9/EX9.26/Ex9_26.sce b/3681/CH9/EX9.26/Ex9_26.sce
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+// 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
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diff --git a/3681/CH9/EX9.27/Ex9_27.sce b/3681/CH9/EX9.27/Ex9_27.sce
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+// 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
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+// 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
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+// 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
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diff --git a/3681/CH9/EX9.34/Ex9_34.sce b/3681/CH9/EX9.34/Ex9_34.sce
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+// 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
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+// 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
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+// 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
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+// 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
generated by cgit (git 2.34.1) at 2024-12-22 17:01:04 +0000