1 files changed, 31 insertions, 0 deletions
diff --git a/416/CH20/EX20.1/example20_1.sce b/416/CH20/EX20.1/example20_1.sce
new file mode 100755
index 000000000..de23afab7
--- /dev/null
+++ b/416/CH20/EX20.1/example20_1.sce
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+clc
+clear
+disp('example 20.1')
+lod=1 //industrial installation load
+pf=0.78  //power factor
+tf=200   //tariff 
+md=3.5   //extra maximum demand
+ic=500   //installation of capacitor
+id=0.15  //interest and depreciation
+lf=0.8  //load factor
+sinp=ic*id/tf
+ph2=asind(sinp)
+epf2=cosd(ph2)
+ph1=acosd(pf)
+ph1=round(ph1*10^2)/10^2
+ph2=round(ph2*10^2)/10^2
+q=lod*(tand(ph1)-tand(ph2))
+q=round(q*10^4)/10^4
+ikva=lod/pf
+ikv=round(ikva*(10^5))/10^2
+aeu=lod*lf*8760*10^6
+eb=ikv*tf+aeu*md
+printf("(a)\neconomic power factor %.3flagging \n(b) \ncapacitor kVAr to improve the power factor %.4f \n(c) \ninitial kVA %.2fKVA \nannual energy used %0.3ekWh \nelectrical bill Rs%e per year",epf2,q,ikv,aeu,eb)
+kvc=round((lod*10^3/(round(epf2*1000)/10^3))*10^2)/10^2
+ebc=kvc*tf+aeu*md
+aidc=q*10^3*ic*id
+te=ebc+aidc
+asc=eb-te
+printf("\n(d)\nKVA after installation of capacitors %.2fKVA \n",kvc)
+printf("energy bill after installation of capacitor Rs%e per year \n",ebc)
+printf("annual interest and depreciation of capacitor bank Rs%.1fper year \ntotal expendition after installation of capacitors Rs%e per year \n annual savings due to installation of capacitors Rs%d per year",aidc,te,asc)
+\ No newline at end of file