initial commit / add all books

1 files changed, 50 insertions, 0 deletions

diff --git a/2417/CH11/EX11.22/Ex11_22.sce b/2417/CH11/EX11.22/Ex11_22.sce
new file mode 100755
index 000000000..5bd77c11f
--- /dev/null
+++ b/2417/CH11/EX11.22/Ex11_22.sce
@@ -0,0 +1,50 @@
+//scilab 5.4.1

+clear;

+clc;

+printf("\t\t\tProblem Number 11.22\n\n\n");

+// Chapter 11 : Heat Transfer

+// Problem 11.22  (page no. 595) 

+// Solution

+

+//For brick,concrete,plaster,hot film and cold film,

+A=1; //area //Unit:ft^2

+//For a plane wall,the areas are all the same,and if we use 1 ft^2 of wall surface as the reference area,

+//For Brick,

+deltax=6/12; //6 inch = 6/12 feet //deltax=length //unit:ft

+k=0.40; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity //From the table

+brickResistance=deltax/(k*A); //Thermal resistance //Unit:(hr*f)/Btu

+printf("For brick,");

+printf("The resistance is %f (hr*F)/Btu\n",brickResistance);

+

+//For Concrete,

+deltax=(1/2)/12; //(1/2) inch = (1/2)/12 feet //deltax=length //unit:ft

+k=0.80; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity //From the table

+concreteResistance=deltax/(k*A); //Thermal resistance //Unit:(hr*f)/Btu

+printf("For Concrete,");

+printf("The resistance is %f (hr*F)/Btu\n",concreteResistance);

+

+//For plaster,

+deltax=(1/2)/12; // (1/2) inch = 6/12 feet //deltax=length //unit:ft

+k=0.30; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity //From the table

+plasterResistance=deltax/(k*A); //Thermal resistance //Unit:(hr*f)/Btu

+printf("For plaster,");

+printf("The resistance is %f (hr*F)/Btu\n",plasterResistance);

+

+//For "hot film",

+h=0.9; //Coefficient of heat transfer //Unit:Btu/(hr*ft^2*F)

+hotfilmResistance=inv(h*A); //Thermal resistance //Unit:(hr*f)/Btu

+printf("For hot film,");

+printf("The resistance is %f (hr*F)/Btu\n",hotfilmResistance);

+

+//For "cold film",

+h=1.5; //Coefficient of heat transfer //Unit:Btu/(hr*ft^2*F)

+coldfilmResistance=inv(h*A); //Thermal resistance //Unit:(hr*f)/Btu

+printf("For cold film,");

+printf("The resistance is %f (hr*F)/Btu\n\n",coldfilmResistance);

+

+totalResistance=brickResistance+concreteResistance+plasterResistance+hotfilmResistance+coldfilmResistance; //the overall resistance  //Unit:(hr*f)/Btu

+printf("The overall resistance is %f (hr*F)/Btu\n",totalResistance);

+

+U=inv(totalResistance); //Unit:Btu/(hr*ft^2) //The overall conductance(or overall heat-transfer coefficient)

+printf("The overall conductance(or overall heat-transfer coefficient) is %f Btu/(hr/ft^2)\n",U);

+//In problem 11.21,the solution is straightforward,because the heat-transfer area is constant for all series resistances.