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+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 2, Example 12")
+//A spherical tank of internal diameter, Di=5m ,made of t=25mm thick stainless steel(thermal conductivity,k=15W/(m*K)),is used to store water at temprature(Ti)=0°C.Do is outer diameter
+Di=5;
+t=25;
+Do=5+2*(t/1000);//in metre
+k=15;
+Ti=0;
+//The tank is located in a room whose temprature is (To)=20°C.
+To=20;
+//Emmisivity is 1.
+//The convection heat transfer coefficients at the inner and outer surfaces of the tank are hi=80W/(m^2*K) and ho=10W/(m^2*K)
+hi=80;
+ho=10;
+//The heat of fusion of ice at atmospheric pressure is deltahf=334kJ/kg.The stefan-boltzman constant(sigma) is 5.67*10^-8W/m^2.
+sigma=5.67*10^-8;
+deltahf=334;
+//The inner surface area is (A1) and outer surface area is (A2)of the tank
+disp("The inner(A1) and outer surfaces(A2) areas of the tank in m^2 are") 
+A1=%pi*Di^2
+A2=%pi*Do^2
+//The individual thermal resistances can be determined as
+//The convective resistance is (Ri)
+disp("The convective resistance(Ri) at the inner surface in K/W is ")
+Ri=1/(hi*A1)
+//The conduction resistance is(Rs)
+disp("The conduction resistance(Rs)of the tank in K/W is")
+Rs=(Do-Di)/(2*k*%pi*Di*Do)
+//The convective resistance is(Roc)
+disp("The convective resistance(Roc) at the outer surface in K/W is")
+Roc=1/(ho*A2)
+//The radiative resistance(Ror) at the outer surface is Ror=1/(A2*hr) 
+//The radiative heat transfer coefficient hr is determined by hr=sigma*(T2^2+293.15^2)*(T2+293.15)
+//But we do not know the outer surface temprature T2 of the tank and hence we cannot determine the value of hr.
+//Therfore we adopt an iterative procedure.We assume T2=4°C=277.15K.Putting the value in hr=sigma*(T2^2+293.15^2)*(T2+293.15) we get
+T2=277.15;
+disp("The radiative heat transfer coefficient hr in W/(m^2*K) is")
+hr=sigma*(T2^2+293.15^2)*(T2+293.15)
+disp("Therefore the radiative resistance(Ror) at the outer surface in K/W is")
+Ror=1/(A2*hr)
+//The two parallel resistances Roc and Ror can be replaced by an equivalent resistance Ro as X=(1/Ro)=(1/Roc)+(1/Ror)
+disp("The equivalent resistance in K/W is")
+X=(1/Roc)+(1/Ror);
+Ro=1/X
+//Now the resistance Ri,Rs and Ro are in series an hence the total resistance becomes Rtotal=Ri+Rs+Ro 
+disp("The total resistance in K/W is")
+Rtotal=Ri+Rs+Ro 
+//The rate of heat transfer is given by Q=(To-Ti)/Rtotal
+disp("The rate of heat transfer,Q in W is")
+Q=(To-Ti)/Rtotal
+//The outer surface(T2) is calculated as T2=To-Q*Ro
+disp("The outer surface temprature in °C is")
+T2=To-Q*Ro
+disp("which is sufficiently close to the assumption.So there is no need of further iteration")
+//The total heat transfer is (Qt),during a 24-hour period
+disp("The total heat transfer(Qt) during a 24-hour period in KJ is")
+Qt=Q*24*3600/1000
+//the amount of ice in kG which melts during a 24 hour period is (mice)
+disp("Therefore,the amount of ice(mice)in kG which melts during a 24 hour period is")
+mice=Qt/deltahf
+
+
+
+
+