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+//Problem 19.18: Three similar coils, each having a resistance of 8 ohm  and an inductive reactance of 8 ohm  are connected (a) in star and (b) in delta, across a 415 V, 3-phase supply. Calculate for each connection the readings on each of two wattmeters connected to measure the power by the two-wattmeter method.
+
+//initializing the variables:
+R = 8; // in ohms
+XL = 8; // in ohms
+VL = 415; // in Volts
+
+//calculation:
+//For a star connection:
+//IL = Ip
+//VL = Vp*(3^0.5)
+VLs = VL
+Vps = VLs/(3^0.5)
+//Impedance per phase,
+Zp = (R*R + XL*XL)^0.5
+Ips = Vps/Zp
+ILs = Ips
+//Power dissipated, P = VL*IL*(3^0.5)*cos(phi)  or  P = 3*Ip*Ip*Rp)
+pf = R/Zp
+Ps = VLs*ILs*(3^0.5)*pf
+//If wattmeter readings are P1 and P2 then P1 + P2 = Pst
+Pst = Ps
+// Pid = Pi1 - Pi2
+phi = acos(pf)
+Psd = Pst*tan(phi)/(3^0.5)
+//Hence wattmeter 1 reads
+Ps1 = (Psd + Pst)/2
+//wattmeter 2 reads
+Ps2 = Pst - Ps1
+
+//For a delta connection:
+//VL = Vp
+//IL = Ip*(3^0.5)
+VLd = VL
+Vpd = VLd
+Ipd = Vpd/Zp
+ILd = Ipd*(3^0.5)
+//Power dissipated, P = VL*IL*(3^0.5)*cos(phi)  or  P = 3*Ip*Ip*Rp)
+Pd = VLd*ILd*(3^0.5)*pf
+//If wattmeter readings are P1 and P2 then P1 + P2 = Pdt
+Pdt = Pd
+// Pid = Pi1 - Pi2
+Pdd = Pdt*tan(phi)/(3^0.5)
+//Hence wattmeter 1 reads
+Pd1 = (Pdd + Pdt)/2
+//wattmeter 2 reads
+Pd2 = Pdt - Pd1
+
+printf("\n\n Result \n\n")
+printf("\n (a)When the coils are star-connected the wattmeter readings are %.3E W and %.3E W",Ps1,Ps2)
+printf("\n (b)When the coils are delta-connected the wattmeter readings are are %.3E W and %.3E W",Pd1,Pd2)
+\ No newline at end of file