1 files changed, 54 insertions, 54 deletions
diff --git a/839/CH15/EX15.4/Example_15_4.sce b/839/CH15/EX15.4/Example_15_4.sce
index 2a1724600..857829a79 100755
--- a/839/CH15/EX15.4/Example_15_4.sce
+++ b/839/CH15/EX15.4/Example_15_4.sce
@@ -1,54 +1,54 @@
-//clear//
-clear;
-clc;
-
-//Example 15.4
-//Given
-N = 28;
-xF = 0.5/12; // [ft]
-yF = 0.035/12; //[ft]
-km = 26; // [Btu/ft-h-F]
-AT = 2.830; //[ft^2/ft]
-Ab = 0.416; //[ft^2/ft]
-hi = 1500; //[Btu/ft^2-h-F]
-G = 5000; //[lb/h-ft^2]
-Tavg = 130; //[F]
-Tw = 250; //[F]
-mu = 0.046; //[lb/ft-h], from Appendix 8 
-Cp = 0.25; //[Btu/lb-F], from Appendix 15
-k = 0.0162; //[Btu/ft-h-F], from Appendix 12
-ID_shell = 3.068/12; //[ft], from Appendix 5
-OD_pipe = 1.9/12; //[ft], from Appendix 5
-//cross sectional area of shell space
-Ac = %pi/4*(ID_shell^2-OD_pipe^2)-N*xF*yF //[ft^2]
-//The perimeter of air space
-Ap = %pi*ID_shell+AT; //[ft]
-//hydraulic radius
-rh = Ac/Ap; //[ft]
-//equivalent diameter
-De = 4*rh; //[ft]
-//Reynolds Number
-Nre = De*h/mu
-//In computing mu_w the resistance of the wall and the steam film
-//are considered negligible, so
-mu_w = 0.0528; //[lb/ft-h]
-Npr = mu*Cp/k
-//Using Fig. 15.17, the heat transfer factor is
-jh = 0.0031;
-ho = jh*Cp*G*(mu/mu_w)^0.14/Npr^(2/3); //[Btu/ft^2-h-F]
-
-//For rectangular fins, disreagrding the contribution of the ends of the fins to 
-//the perimeter, Lp = 2L and S = Lyf, where yf is the fin thickness and L is the 
-//length of the fin. Then, from Eq.(15.11)
-aFxF = xF*sqrt(2*ho/(km*yF));
-//From Fig. 15.16
-netaF = 0.93;
-Dt = 1.610/12; //[ft], from Appendix 5
-DLbar = (OD_pipe-Dt)/log(OD_pipe/Dt); //[ft]
-Ai = %pi*Dt*1.0; //[ft^2]
-AF = AT-Ab; //[ft^2/ft] 
-xw = (OD_pipe-Dt)/2; //[ft]
-
-//Using Eq.(15.10), the overall coefficient 
-Ut =  1/(Ai/(ho*(netaF*AF+Ab))+(xw*Dt/(km*DLbar))+1/hi);//[Btu/ft^2-h-F]
-disp('Btu/ft^2-h-F',Ut,'The overall heat transfer coefficent is')
+//clear//
+clear;
+clc;
+
+//Example 15.4
+//Given
+N = 28;
+xF = 0.5/12; // [ft]
+yF = 0.035/12; //[ft]
+km = 26; // [Btu/ft-h-F]
+AT = 2.830; //[ft^2/ft]
+Ab = 0.416; //[ft^2/ft]
+hi = 1500; //[Btu/ft^2-h-F]
+G = 5000; //[lb/h-ft^2]
+Tavg = 130; //[F]
+Tw = 250; //[F]
+mu = 0.046; //[lb/ft-h], from Appendix 8 
+Cp = 0.25; //[Btu/lb-F], from Appendix 15
+k = 0.0162; //[Btu/ft-h-F], from Appendix 12
+ID_shell = 3.068/12; //[ft], from Appendix 5
+OD_pipe = 1.9/12; //[ft], from Appendix 5
+//cross sectional area of shell space
+Ac = %pi/4*(ID_shell^2-OD_pipe^2)-N*xF*yF //[ft^2]
+//The perimeter of air space
+Ap = %pi*ID_shell+AT; //[ft]
+//hydraulic radius
+rh = Ac/Ap; //[ft]
+//equivalent diameter
+De = 4*rh; //[ft]
+//Reynolds Number
+Nre = De*G/mu
+//In computing mu_w the resistance of the wall and the steam film
+//are considered negligible, so
+mu_w = 0.0528; //[lb/ft-h]
+Npr = mu*Cp/k
+//Using Fig. 15.17, the heat transfer factor is
+jh = 0.0031;
+ho = jh*Cp*G*(mu/mu_w)^0.14/Npr^(2/3); //[Btu/ft^2-h-F]
+
+//For rectangular fins, disregarding the contribution of the ends of the fins to 
+//the perimeter, Lp = 2L and S = Lyf, where yf is the fin thickness and L is the 
+//length of the fin. Then, from Eq.(15.11)
+aFxF = xF*sqrt(2*ho/(km*yF));
+//From Fig. 15.16
+netaF = 0.93;
+Dt = 1.610/12; //[ft], from Appendix 5
+DLbar = (OD_pipe-Dt)/log(OD_pipe/Dt); //[ft]
+Ai = %pi*Dt*1.0; //[ft^2]
+AF = AT-Ab; //[ft^2/ft] 
+xw = (OD_pipe-Dt)/2; //[ft]
+
+//Using Eq.(15.10), the overall coefficient 
+Ut =  1/(Ai/(ho*(netaF*AF+Ab))+(xw*Dt/(km*DLbar))+1/hi);//[Btu/ft^2-h-F]
+disp('Btu/ft^2-h-F',Ut,'The overall heat transfer coefficient is')
+\ No newline at end of file