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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /1910/CH2/EX2.1 | |
| download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip | |
initial commit / add all books
Diffstat (limited to '1910/CH2/EX2.1')
| -rwxr-xr-x | 1910/CH2/EX2.1/Chapter21.sce | 83 |
1 files changed, 83 insertions, 0 deletions
diff --git a/1910/CH2/EX2.1/Chapter21.sce b/1910/CH2/EX2.1/Chapter21.sce new file mode 100755 index 000000000..b6ef7a2ae --- /dev/null +++ b/1910/CH2/EX2.1/Chapter21.sce @@ -0,0 +1,83 @@ +// 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 1")
+//The length of steel sheet (Ls)=1.5 mm and thermal conductivity (ks)=25 W/(mK) at the outer surface.
+Ls=1.5;
+ks=25;
+//The length of plywood (Lp)=10 mm and thermal conductivity (kp)= .05 W/(mK) at the inner surface.
+Lp=10;
+kp=.05;
+//The length of glass wool (Lg)=20 mm and thermal conductivity (kg)= .01W/(mK) in between steel sheet and plywood.
+Lg=20;
+kg=.01;
+//The temprature of van inside cold Enviroment is (Ti)= -15°C while the outside surface is exposed to a surrounding ambient temprature (To)=24°C
+To=24;
+Ti=-15;
+//The average value of heat transfer coefficients at the inner and outside surfaces of the wall are hi=12 W/(m^2*K) and ho= 20 W/(m^2*K)
+hi=12;
+ho=20;
+//The surface area of wall (A)= .75 m^2
+A=.75;
+//The convective resistance is Ro= 1/(ho*A) at the outer surface
+disp("The convective resistance Ro= 1/(ho*A) at the outer surface in KW^-1 is")
+Ro=1/(ho*A)
+//The conduction resistance is Rs= Ls/(ks*A) of steel sheet
+disp("The conduction resistance Rs= Ls/(ks*A) of steel sheet in KW^-1 is")
+Rs=Ls*10^-3/(ks*A)
+//The conduction resistance is Rg= Lg/(kg*A) of glass wool
+disp("The conduction resistance Rg= Lg/(kg*A) of glass wool in KW^-1 is")
+Rg= Lg*10^-3/(kg*A)
+//The conduction resistance is Rp= Lp/(kp*A) of plywood
+disp("The conduction resistance Rp= Lp/(kp*A) of plywood in KW^-1 is")
+Rp= Lp*10^-3/(kp*A)
+//The convective resistance is Ri= 1/(hi*A) at the outer surface
+disp("The convective resistance Ri= 1/(hi*A) at the outer surface in KW^-1 is")
+Ri= 1/(hi*A)
+//The rate of heat flow is Q=(To-Ti)/(Ro+Rs+Rg+Rp+Ri)
+disp("The rate of heat flow Q=(To-Ti)/(Ro+Rs+Rg+Rp+Ri) in W is")
+Q=(To-Ti)/(Ro+Rs+Rg+Rp+Ri)
+//The tempraure at the outer surface of wall is T1.
+//The temprature at the interface b/w steel sheet and glass wool is T2.
+//The temprature at the interface b/w glass wool and plywood is T3.
+//The tempraure at the inner surface of wall is T4.
+disp("The temprature at the outer surface of wall is T1=To-(Q*Ro) in °C")
+T1=To-(Q*Ro)
+disp("The temprature at the interface b/w steel sheet and glass wool is T2=T1-(Q*Rs) in°C")
+T2=T1-(Q*Rs)
+disp("The temprature at the interface b/w glass wool and plywood is T3=T2-(Q*Rg) in°C")
+T3=T2-(Q*Rg)
+disp("The temprature at the inner surface of wall is T4=T3-(Q*Rp)in °C")
+T4=T3-(Q*Rp)
+//Check for Ti(Temprature inside the van)
+disp("Check for Ti(in °C)")
+Ti=T4-(Q*Ri)
+disp("The value is same as given in the problem")
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