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diff --git a/2417/CH11/EX11.18/Ex11_18.sce b/2417/CH11/EX11.18/Ex11_18.sce new file mode 100755 index 000000000..164f9344a --- /dev/null +++ b/2417/CH11/EX11.18/Ex11_18.sce @@ -0,0 +1,23 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.18\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.18 (page no. 579)
+// Solution
+
+//We first check the Reynolds number and note that G is same as for problem 11.17.So,
+//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=0.062; //the viscosity of air //unit:lbm/(ft*hr)
+D=0.87/12; //Inside diameter //1 in^2=144 ft^2
+//Reynolds number is DG/mu,therefore
+Re=(D*G)/mu; //Reynolds number
+printf("The Reynolds number is %f\n",Re);
+//which places the flow in the turbulent regime.Because W/1000(W=weight flow) is same as for problem 11.17 and equals 1.2,we now enter figure 11.19 at 1.2 and 400F to obtain h1=135.Because the inside tube diameter is same as before,F=1.25.Therefore,
+h1=135; //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 inside film coefficient is %f Btu/(hr*ft^2*F)\n",h);
+//It is interesting that for equal mass flow rates,water yields a heat-transfer coefficient almost five times greater than air
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