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+//Example 10.4//
+
+//(a)= 0.5 wt % C we must quenchfrom the austenite boundary (770degree C) to ~520 degree in ~0.6, giving
+a=770;//degree C //austenite boundary
+b=520;//Degree C //temprature
+t=0.6;//s //seconds // time
+dt1=(a-b)/t
+mprintf("dt1 = %i degree C/s",dt1)
+//(b)=0.77 wt % C steel, we quench from the eutectoid temperature(727degree C) to ~550degree C in 0.7s, giving
+a1=727;//degree C //eutectoid temperature
+b1=550;//degree C //temperature
+t1=0.7;//s//seconds //time
+dt2=(a1-b1)/t1
+mprintf("\ndt2 = %i degree C/s",dt2)
+//(c)= 1.13 wt %C steel we quench from the austenite boundary (880degree C) to ~550degree C in ~3.5
+a2=880;//degree C //eutectoid temperature
+t3=0.35;//s //seconds //time
+dt3=(a2-b1)/t3
+mprintf("\ndt3 = %i degree C/s",dt3)
+mprintf("\nThe calculated answer in the textbook is wrong")