From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

---
 405/CH7/EX7.3/7_3.sce | 28 ++++++++++++++++++++++++++++
 1 file changed, 28 insertions(+)
 create mode 100755 405/CH7/EX7.3/7_3.sce

(limited to '405/CH7/EX7.3')

diff --git a/405/CH7/EX7.3/7_3.sce b/405/CH7/EX7.3/7_3.sce
new file mode 100755
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+++ b/405/CH7/EX7.3/7_3.sce
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+clear;
+clc;
+printf("\t\t\tExample Number 7.3\n\n\n");
+// heat transfer from horizontal tube in water
+// Example 7.3 (page no.-333) 
+// solution
+
+d = 0.02;// [m] diameter of heater
+Ts = 38;// [degree celsius] surface temperature of heater
+Tw = 27;// [degree celsius] water temperature
+// the film temperature is 
+Tf = (Ts+Tw)/2;// [degree celsius]
+// from appendix A the properties of water are 
+k = 0.630;// [W/m degree celsius] thermal conductivity
+// and the following term is particularly useful in obtaining the product GrPr product when it is multiplied by d^(3)*DT
+// g*Beta*rho^(2)*Cp/(mu*k) = 2.48*10^(10) [1/m^(3) degree celsius]
+K = 2.48*10^(10);// [1/m^(3) degree celsius]
+Gr_into_Pr = K*(Ts-Tw)*d^(3);
+// using table 7-1 (page number -328), we get 
+C = 0.53;
+m = 1/4;
+// so that
+Nu = C*(Gr_into_Pr)^(1/4);
+h = Nu*k/d;// [W/square meter degree celsius] convection heat transfer coefficient
+// the heat transfer is thus
+q_by_L = h*%pi*d*(Ts-Tw);// [W/m]
+printf("free-convection heat loss per unit length of heater is %f W/m",q_by_L);
+
-- 
cgit