diff options
| author | Siddharth Agarwal | 2019-09-03 18:27:40 +0530 |
|---|---|---|
| committer | Siddharth Agarwal | 2019-09-03 18:27:40 +0530 |
| commit | 8ac15bc5efafa2afc053c293152605b0e6ae60ff (patch) | |
| tree | e1bc17aae137922b1ee990f17aae4a6cb15b7d87 /Working_Examples/2777/CH3/EX3.14 | |
| parent | 52a477ec613900885e29c4a0b02806a415b4f83a (diff) | |
| download | Xcos_block_examples-master.tar.gz Xcos_block_examples-master.tar.bz2 Xcos_block_examples-master.zip | |
Diffstat (limited to 'Working_Examples/2777/CH3/EX3.14')
| -rwxr-xr-x | Working_Examples/2777/CH3/EX3.14/Ex3_14.sce | 75 |
1 files changed, 75 insertions, 0 deletions
diff --git a/Working_Examples/2777/CH3/EX3.14/Ex3_14.sce b/Working_Examples/2777/CH3/EX3.14/Ex3_14.sce new file mode 100755 index 0000000..e0b0ab9 --- /dev/null +++ b/Working_Examples/2777/CH3/EX3.14/Ex3_14.sce @@ -0,0 +1,75 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.14
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+S = 10 * 10 ^ 3; // Rating of the Single Transformer in VA
+f = 50; // Frequency in Hertz
+Pc = 110; // Required input no-load at normal voltage in Watts (Core loss)
+Psc = 120; // Required input Short-circuit at full-load current in Watts (copper loss or short circuit loss)
+
+
+// CALCUATIONS
+// case (a) for Unity power factor
+
+cos_theta1 = 1; // Unity Power factor
+K1 = 1.0; // Full load
+K2 = 0.5; // Half load
+eta_11 = 100 * (K1*S*cos_theta1)/((K1*S*cos_theta1)+Pc+( K1 ^ 2 )*Psc); // Efficiency at unity factor and full load ( beacuse taken k1 = 1 ) in percentage
+eta_12 = 100 * (K2*S*cos_theta1)/((K2*S*cos_theta1)+Pc+( K2 ^ 2 )*Psc); // Efficiency at unity factor and half load ( beacuse taken k2 = 0.5 ) in percentage
+
+// case (b) for 0.8 power factor lagging
+
+cos_theta2 = 0.8; // 0.8 power factor lagging
+eta_21 = 100 * (K1*S*cos_theta2)/((K1*S*cos_theta2)+Pc+( K1 ^ 2 )*Psc); // Efficiency at 0.8 power factor lagging and full load ( beacuse taken k1 = 1 ) in percentage
+eta_22 = 100 * (K2*S*cos_theta2)/((K2*S*cos_theta2)+Pc+( K2 ^ 2 )*Psc); // Efficiency at 0.8 power factor lagging and half load ( beacuse taken k2 = 0.5 ) in percentage
+
+// Case (c) for 0.8 poer factor leading
+
+eta_31 = eta_21; // Efficiency at 0.8 power factor leading and full load will be same as the Efficiency at 0.8 power factor lagging and full load in percentage
+eta_32 = eta_22; // Efficiency at 0.8 power factor leading and half load will be same as the Efficiency at 0.8 power factor lagging and half load in percentage
+
+// Case (d) Maximum Efficiency assumed that unity power factor
+// Psc = Pc At Maximum Efficiency
+
+eta_41 = 100 * (K1*S*cos_theta1)/((K1*S*cos_theta1)+Pc+Pc); // Maximum Efficiency at unity factor and full load ( beacuse taken k1 = 1 ) in percentage
+
+// Case (e) Maximum Efficiency assumed that 0.8 power factor lagging
+// Psc = Pc At Maximum Efficiency
+
+eta_51 = 100 * (K1*S*cos_theta2)/((K1*S*cos_theta2)+Pc+Pc); // Maximum Efficiency at unity factor and full load ( beacuse taken k1 = 1 ) in percentage
+
+// Case (f) Maximum Efficiency assumed that 0.8 power factor leading
+// Psc = Pc At Maximum Efficiency
+
+eta_61 = eta_51; // Maximum Efficiency at 0.8 power factor leading and full load will be same as the Maximum Efficiency at 0.8 power factor lagging and full load in percentage
+out1 = K1*S*cos_theta1; // Output at which maximum efficiency occurs at unity power factor at full load in Watts
+out2 = K1*S*cos_theta2; // Output at which maximum efficiency occurs at 0.8 power factor lagging at full load in Watts
+out3 = K1*S*cos_theta2; // Output at which maximum efficiency occurs at unity power factor leading at full load in Watts
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.14 : SOLUTION :-") ;
+printf("\n (a.1) Efficiency at unity power factor and full load , eta = %.2f Percent \n ",eta_11);
+printf("\n (a.2) Efficiency at unity power factor and half load , eta= % .2f Percent \n",eta_12);
+printf("\n (b.1) Efficiency at 0.8 power factor lagging and full load , eta = %.2f Percent \n ",eta_21);
+printf("\n (b.2) Efficiency at 0.8 power factor lagging and half load , eta= % .2f Percent \n",eta_22);
+printf("\n (c.1) Efficiency at 0.8 power factor leading and full load , eta = %.2f Percent \n ",eta_31);
+printf("\n (c.2) Efficiency at 0.8 power factor leading and half load , eta= % .2f Percent \n",eta_32);
+printf("\n (d) Maximum Efficiency at unity power factor and full load , eta = %.2f Percent \n ",eta_41);
+printf("\n (e) Maximum Efficiency at 0.8 power factor lagging and full load , eta = %.2f Percent \n ",eta_51);
+printf("\n (f) Maximum Efficiency at 0.8 power factor leading and full load , eta = %.2f Percent \n ",eta_61);
+printf("\n (g) Output at which maximum efficiency occurs at unity power factor at full load = %.f W \n ",out1);
+printf("\n (h) Output at which maximum efficiency occurs at 0.8 power factor lagging at full load = %.f W \n ",out2);
+printf("\n (i) Output at which maximum efficiency occurs at 0.8 power factor leading at full load = %.f W \n ",out3);
+printf("\n IN THE ABOVE PROBLEM MAXIMUM EFFICIENCY AND THE OUTPUT AT WHICH THE MAXIMUM EFFICIENCY OCCURS IS NOT CALCULATED IN THE TEXT BOOK \n")
|