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authorpriyanka2015-06-24 15:03:17 +0530
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+//Part A Chapter 7 Example 14
+clc;
+clear;
+close;
+rcv=71.5;//cm of Hg(Recorded condenser vaccum)
+br=76.8;//cm of Hg(Barometer reading)
+Tc=35;//degree C(Temperature of condensation)
+Tw=27.6;//degree C(Temperature of hot well)
+mc=1930;//kg(Mass of condensate/hour)
+mw=62000;//kg(Mass of cooling water/hour)
+T1=8.51;//degree C(Inlet temperature)
+T2=26.24;//degree C(Outlet temperature)
+pc=(br-rcv)/73.55*101.325;//kPa(condenser pressure)
+p_partial=5.628;//kPa(at 35 degree C)
+hf=146.68;//kJ/kg
+hfg=2418.6;//kJ/kg
+x=(mw*(T2-T1)*4.18/mc+4.18*Tw-hf)/hfg;//dryness fraction
+disp("State of steam(Dryness fraction) entering condenser is "+string(x));