From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001 From: priyanka Date: Wed, 24 Jun 2015 15:03:17 +0530 Subject: initial commit / add all books --- 1910/CH2/EX2.12/Chapter212.sce | 70 ++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 70 insertions(+) create mode 100755 1910/CH2/EX2.12/Chapter212.sce (limited to '1910/CH2/EX2.12/Chapter212.sce') 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 + + + + + -- cgit