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

---
 2417/CH11/EX11.21/Ex11_21.sce | 25 +++++++++++++++++++++++++
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 create mode 100755 2417/CH11/EX11.21/Ex11_21.sce

(limited to '2417/CH11/EX11.21')

diff --git a/2417/CH11/EX11.21/Ex11_21.sce b/2417/CH11/EX11.21/Ex11_21.sce
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+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.21\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.21  (page no. 589) 
+// Solution
+
+//Because the conditions of illustrative problem 11.15 are the same as for problem 11.19 and 11.20,we can solve this problem in two ways to obtain a check.
+//Thus,adding the results of these problems yields,
+printf("Adding the results of the problems yields,\n")
+Qtotal=206.2+214.5; //Unit:Btu/hr //total heat loss
+printf("The heat loss due to convection is %f Btu/hr\n",Qtotal);
+
+//We can also approach this solution by obtaining radiation and convection heat-transfer co-efficcient.Thus,
+hcombined=0.9+0.94; //Coefficient of heat transfer //Unit:Btu/(hr*ft^2*F)
+D=3.5/12; //3.5 inch = 3.5/12 feet //Unit:ft //Outside diameter
+Ti=120; //Inside temperature //unit:fahrenheit
+To=70; //Outside temperature //unit:fahrenheit
+deltaT=Ti-To; //unit:fahrenheit //Change in temperature
+L=5; //Length //Unit:ft //From problem 11.10
+A=(%pi*D)*L;  //Area //Unit:ft^2 
+Qtotal=hcombined*A*deltaT; //Unit:Btu/hr //total heat loss due to convection //Newton's law of cooling
+printf("By obtaining radiation and convection heat-transfer co-efficcient,\n")
+printf("The heat loss due to convection is %f Btu/hr",Qtotal);
-- 
cgit