diff options
Diffstat (limited to '3681/CH4')
46 files changed, 419 insertions, 0 deletions
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
+
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