initial commit / add all books

1 files changed, 70 insertions, 0 deletions

diff --git a/1910/CH2/EX2.12/Chapter212.sce b/1910/CH2/EX2.12/Chapter212.sce
new file mode 100755
index 000000000..bcd831977
--- /dev/null
+++ b/1910/CH2/EX2.12/Chapter212.sce
@@ -0,0 +1,70 @@
+// 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

+

+

+

+

+