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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
commit7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch)
treedbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /1319/CH5/EX5.11/5_11.sce
parentb1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff)
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+//Calcualte efficiencies at various loads
+
+clc;
+clear;
+
+P=100*(10^3);// Power Input
+Pc=1000;// Copper Loss
+Pil=1000;// Iron Loss
+pf=0.8;
+
+deff('y=unity(x)','y=(P*100*x)/((P*x)+Pil+(Pc*(x^2)))')// Unit Power Factor
+deff('y=pfactor(x)','y=(P*100*x*pf)/((P*pf*x)+Pil+(Pc*(x^2)))')// 0.8 p.f
+
+printf('a) Unity power factor efficiencies at \n \n')
+printf('i) Half of full load = %f percent \n',unity(1/2))
+printf('ii) Full load = %f percent \n',unity(1))
+printf('iii) (5/4) of full load = %f percent \n \n',unity(5/4))
+
+
+printf('b) 0.8 power factor efficiencies at \n \n')
+printf('i) Half of full load = %f percent \n',pfactor(1/2))
+printf('ii) Full load = %f percent \n',pfactor(1))
+printf('iii) (5/4) of full load = %f percent \n',pfactor(5/4))