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clear;
clc;
printf("\t Example 10.1\n");
T1=2273; //temp. of liquid air,K
T2=303; //temp. of room,K
T3=973; //temp. of shield,K
D1=0.003; //diameter of crucible,m
D2=0.05; //diameter of shield,m
theta1=330; //surrounding angle of jet,degree
theta2=30 // angle of slit,degree
Fjr=theta2/360; //fraction of energy of view of jet occupied by room
Fjs=theta1/360 ; //fraction of energy of view of jet occupied by shield
Qnjr=%pi*D1*Fjr*5.67*10^-8*(T1^4-T2^4); //net heat transfer from jet to room,W/m
Qnjs=%pi*D1*Fjs*5.67*10^-8*(T1^4-T3^4); //net heat transfer from jet to shield,W/m
//to find the radiation from the inside of the shield to the room, we need Fshield-room.since any radiation passing out of the slit goes to the room,we can find this view factor equating view factors to the room with view factors to the slit.
Fsj=%pi*D1/0.01309*Fjr; //fraction of energy of view of slit occupied by jet
Fss=1-Fsj; //fraction of energy of view of slit occupied by shield.
Fsr=0.01309*Fss/(%pi*D2*Fjs); //fraction of energy of view of shield occupied by room
Qnsr=%pi*D2*Fjs*5.67*10^-8*Fsr*(T3^4-T2^4); //net heat transfer from shield to room, W/m
printf("\t heat transfer from jet to room through the slit is :%.0f W/m\n",Qnjr);
printf("\t heat transfer from the jet to shield is :%.0f W/m\n",Qnjs);
printf("\t heat transfer from inside of shield to the room is :%.0f W/m\n",Qnsr);
printf("\t both the jet and the inside of the shield have relatively small view factors to the room, so that comparatively little heat is lost through the silt.");
//end
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