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clear;
clc;
printf("\t Example 6.15\n");
t1=20; //ambient air temperature
t2=70; //exhaust air temperature
r1=150; //evaporation of water
r2=.25; //outlet solid moisture content
t3=15; //inlet solid temperature
t4=65; //outlet solid temperature
p=5; //power demand in KW
h=18; //heat loss in kj
h1=1; //mean specific heat of dry air in kj/kg*K
h2=1.25; //mean specific heat of dry material in kj/kg*K
h3=4.18; //mean specific heat of moisture in kj/kg*K
e=2626; //enthalpy of saturated water vapour in kj/kg
//basis is 1hr
a1=r1*h3*(t4-t3); //heat required for heating 150 kg of water from 15 to 65
a2=r1*e; //heat required for 150 kg water evaporation
a3=2000*h1*(t2-t1); //heat required for heating air from 20 to 70
a4=r2*h3*(t4-t3); //heat required for heating moisture from 15 to 65
a5=120*h2*(t4-t3); //heat required for heating dry solid from 15 to 65
hlost=h*3600; //heat lost in kj
total=(a2+a3+a4+a5+hlost)/3600; //total heat lost
printf("\n :%f kW of heat required for 2000kg/hr of dry air",total);
ans1=a2+a1; //heat needed for evaporation
printf("\n heat needed fro evaporation is :%f",ans1/3600);
ans2=(ans1/3600)/total; //fraction of this heat needed for evaporation
printf("\n fraction of this heat needed for evaporation:%f",ans2);
//end
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