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diff --git a/2417/CH11/EX11.17/Ex11_17.sce b/2417/CH11/EX11.17/Ex11_17.sce new file mode 100755 index 000000000..b453bf07b --- /dev/null +++ b/2417/CH11/EX11.17/Ex11_17.sce @@ -0,0 +1,24 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.17\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.17 (page no. 578)
+// Solution
+
+//The first step is to check Reynolds number.It will be recalled that the Reynolds number is given by (D*V*rho)/mu and is dimensionless.Therefore,we can use D, diameter in feet;V velocity in ft/hr;rho density in lbm/ft^3 and mu viscosity in lbm/(ft*hr).
+//Alternatively,the Reynolds number is given by (D*G)/mu,where G is the mass flow rate per unit area (lbm/(hr*ft^2)).
+G=((20*60)*(4*144)/(%pi*0.87^2)); //Unit:lbm/(hr*ft^2) //Inside diameter=0.87 inch ////1 in.^2=144 ft^2 //20 lbm/min of water(min converted to second)
+//the viscosity of air at these conditions is obtained from figure 11.17 as 0.062 lbm/(ft*hr).So,
+mu=0.33; //the viscosity of air //unit:lbm/(ft*hr)
+D=0.87/12; //Inside diameter //1 in^2=144 ft^2
+//Therefore Reynolds number is
+Re=(D*G)/mu; //Reynolds number
+//which is well into the turbulent flow regime.
+printf("The Reynolds number is %f\n",Re);
+//The next step is to enter Figure 11.18 at W/1000 of 20*(60/1000)=1.2 and 400F to obtain h1=630.
+//From the figure 11.20,we obtain F=1.25 for an inside diameter of 0.87 inch.So,
+h1=630; //basic heat transfer coefficient //unit:Btu/(hr*ft^2*F)
+F=1.25; //correction factor
+h=h1*F; //heat transfer coefficient //the inside film coefficient //unit:Btu/(hr*ft^2*F)
+printf("The heat-transfer coefficient is %f Btu/(hr*ft^2*F)\n",h);
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