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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