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

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 2705/CH10/EX10.3/Ex10_3.sce | 44 ++++++++++++++++++++++++++++++++++++++++++++
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(limited to '2705/CH10/EX10.3')

diff --git a/2705/CH10/EX10.3/Ex10_3.sce b/2705/CH10/EX10.3/Ex10_3.sce
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+clear;
+clc;
+disp('Example 10.3');
+
+//  aim : To determine
+//  (a) the thermal efficiency of the boiler
+//  (b) the equivalent evaporation of the boiler
+//  (c) the new coal consumption 
+
+//  given values
+ms_dot = 5400;// steam feed rate, [kg/h]
+P = 750;// steam pressure, [kN/m^2]
+x = .98;// steam dryness fraction
+Tf1 = 41.5;// feed water temperature, [C]
+CV = 31000;// calorific value of coal used in the boiler, [kJ/kg]
+mc1 = 670;// rate of burning of coal/h, [kg]
+Tf2 = 100;// increased water temperature, [C]
+
+// solution
+//   (a)
+SRC = ms_dot/mc1;// steam raised/kg coal, [kg]
+hf = 709.3;// [kJ/kg]
+hfg = 2055.5;// [kJ/kg]
+h1 = hf+x*hfg;// specific enthalpy of steam raised, [kJ/kg]
+//  from steam table 
+hfw = 173.9;// specific enthalpy of feed water, [kJ/kg]
+EOB = SRC*(h1-hfw)/CV;// efficiency of boiler
+mprintf('\n (a) The thermal efficiency of the boiler is  =  %f percent\n',EOB*100);
+
+// (b)
+he = 2256.9;// specific enthalpy of evaporation, [kJ/kg]
+Ee = SRC*(h1-hfw)/he;// equivalent evaporation[kg/kg coal]
+mprintf('\n (b) The equivalent evaporation of boiler is  =  %f  kg/kg coal\n',Ee);
+
+// (c)
+hw = 419.1;// specific enthalpy of feed water at 100 C, [kJ/kg]
+Eos = ms_dot*(h1-hw);// energy of steam under new condition, [kJ/h]
+neb = EOB+.05;// given condition new efficiency of boiler if 5%more than previous
+Ec = Eos/neb;// energy from coal, [kJ/h]
+mc2 = Ec/CV;// mass of coal used per hour in new condition, [kg]
+mprintf('\n (c) Mass of coal used in new condition is  =  %f kg\n',mc2);
+mprintf('\n      The saving in coal per hour is  =  %f  kg\n',mc1-mc2);
+
+//  End
-- 
cgit