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+clear;

+clc;

+disp('Example 5.14');

+

+//  aim : To determine the

+//  (a)heat transfer

+//  (b)change of internal energy

+//  (c)mass of gas

+

+//   Given values

+V1 = .4;//  initial volume, [m^3]

+P1 = 100;//  initial pressure, [kN/m^2]

+T1 = 273+20;//  temperature, [K]

+P2 = 450;//  final pressure,[kN/m^2]

+cp = 1.0;//  [kJ/kg K]

+Gamma = 1.4; //  heat capacity ratio

+

+//  solution

+

+//  (a)

+//  for the isothermal compression,P*V=constant,so

+V2 = V1*P1/P2;//  [m^3]

+W = P1*V1*log(P1/P2);//  formula of workdone for isothermal process,[kJ]

+

+//  for isothermal process, del_U=0;so

+Q = W;

+mprintf('\n (a) The heat transferred during compression is Q  =  %f kJ\n',Q);

+

+

+//  (b)

+V3 = V1;

+//  for adiabatic expansion

+//  also

+

+P3 = P2*(V2/V3)^Gamma;//  [kN/m^2]

+W = -(P3*V3-P2*V2)/(Gamma-1);//  work done formula for adiabatic process,[kJ]

+//  also, Q=0,so using Q=del_U+W

+del_U = -W;//  [kJ]

+mprintf('\n (b) The change of the internal energy during the expansion is,del_U  =  %f kJ\n',del_U);

+

+//  (c)

+//  for ideal gas

+//  cp-cv=R, and cp/cv=gamma, hence

+R = cp*(1-1/Gamma);//  [kj/kg K]

+

+//  now using ideal gas equation

+m = P1*V1/(R*T1);//  mass of the gas,[kg]

+mprintf('\n (c) The mass of the gas is,m  =  %f  kg\n',m);

+

+// There is calculation mistake in the book

+

+

+//  End

+