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+clear;

+clc;

+disp('Example 4.17');

+

+//  aim : To determine the

+//  (a) mass of steam in the vessel

+//  (b) final dryness of the steam

+//  (c) amount of heat transferrred during the cooling process

+

+//  Given values

+V1 = .8;//  [m^3]

+P1 = 360;//  [kN/m^2]

+P2 = 200;//  [kN/m^2]

+

+//  solution

+

+//  (a)

+// at 360 kN/m^2

+vg1 = .510;//  [m^3]

+m = V1/vg1;//  mass of steam,[kg]

+mprintf('\n (a) The mass of steam in the vessel is =  %f kg\n',m);

+

+//  (b)

+//  at 200 kN/m^2

+vg2 = .885;// [m^3/kg]

+//  the volume remains constant so

+x = vg1/vg2;// final dryness fraction

+mprintf('\n (b) The final dryness fraction of the steam is  =  %f \n',x);

+

+// (c)

+//  at 360 kN/m^2

+h1 = 2732.9;// [kJ/kg]

+//  hence

+u1 = h1-P1*vg1;//  [kJ/kg]

+

+//  at 200 kN/m^2

+hf = 504.7;// [kJ/kg]

+hfg=2201.6;//[kJ/kg]

+//  hence

+h2 = hf+x*hfg;// [kJ/kg]

+//  now

+u2 = h2-P2*vg1;//  [kJ/kg]

+//  so

+del_u = u2-u1;//  [kJ/kg]

+//  from the first law of thermodynamics del_U+W=Q, 

+W = 0;//  because volume is constant

+del_U = m*del_u;//  [kJ]

+//  hence

+Q = del_U;//  [kJ]

+mprintf('\n (c) The amount of heat transferred during cooling process is  =  %f kJ \n',Q);

+

+//  End