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+clear;

+clc;

+disp('Example 5.26');

+

+//  aim : To determine

+//  the volume of the pressure vessel and the volume of the gas before transfer

+

+//  Given values

+

+P1 = 1400;//  initial pressure,[kN/m^2]

+T1 = 273+85;//  initial temperature,[K]

+

+P2 = 700;//  final pressure,[kN/m^2]

+T2 = 273+60;//  final temperature,[K]

+

+m = 2.7;// mass of the gas passes,[kg]

+cp = .88;//  [kJ/kg]

+cv = .67;//  [kJ/kg]

+

+//  solution

+

+//  steady flow equation is, u1+P1*V1+C1^2/2+Q=u2+P2*V2+C2^2/2+W  [1], 

+//  given, there is no kinetic energy change and neglecting potential energy term

+W = 0;// no external work done

+//  so final equation is,u1+P1*v1+Q=u2   [2]

+// also u2-u1=cv*(T2-T1)

+// hence Q=cv*(T2-T1)-P1*v1    [3]

+//  and for unit mass P1*v1=R*T1=(cp-cv)*T1  [4]

+//  so finally

+Q = cv*(T2-T1)-(cp-cv)*T1;//  [kJ/kg]

+//  so total heat transferred is

+Q = m*Q;//  [kJ] 

+

+//  using eqn [4]

+v1 = (cp-cv)*T1/P1;//  [m^3/kg]

+//  Total volume is

+V1 = m*v1;//  [m^3]

+

+//  using ideal gas equation P1*V1/T1=P2*V2/T2

+V2 = P1*T2*V1/(P2*T1);//  final volume,[m^3]

+

+mprintf('\n The volume of gas before transfer is  =  %f  m^3\n',V1);

+mprintf('\n The volume of pressure vessel is  =  %f  m^3\n',V2);

+ 

+//  End