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+// Example 5.6
+// Determine (a) New operating speed if a system disturbance causes a 10% drop
+// in voltage and 6% drop in frequency (b) New shaft horsepower.
+// Page No. 190
+
+clc;
+clear;
+close;
+
+// Given data
+etaV=0.90;               // Efficiency related to voltage
+V=230;                   // Voltage
+etaF=0.94;               // Efficiency related to voltage
+f=60;                    // Frequency
+N=6;                     // Number of poles
+nr1=1175;                // Speed of motor
+P=20;                    // Horsepower of motor
+
+// (a)  New operating speed if a system disturbance causes a 10% drop in 
+// voltage and 6% drop in frequency
+V2=etaV*V;              // New voltage after 10% drop
+f2=etaF*f;              // New frequency after 6% drop
+ns1=120*f/N;
+ns2=120*0.94*f/N;
+s1=(ns1-nr1)/ns1;       // Speed difference
+
+s2=s1*((V/V2)^2)*(f2/f);  
+nr2=ns2*(1-s2);         // New speed
+
+// (b) New shaft horsepower
+P2=P*(nr2/nr1);         // With a constant torque load T2=T1
+
+// Display result on command window
+printf("\n New operating speed in case of voltage and frequency drop = %0.0f r/min ",nr2);
+printf("\n New shaft horsepower = %0.1f hp ",P2);
+