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+clear;

+clc;

+disp('Example 5.12');

+

+//  aim : T0 determine 

+//  (a) change in internal nergy of the air

+//  (b) work done

+//  (c) heat transfer

+

+//  Given values

+m = .25;//  mass, [kg]

+P1 = 140;//  initial pressure, [kN/m^2]

+V1 = .15;//  initial volume, [m^3]

+P2 = 1400;//  final volume, [m^3]

+cp = 1.005;//  [kJ/kg K]

+cv = .718;//  [kJ/kg K]

+

+//  solution

+

+//  (a)

+//  assuming ideal gas

+R = cp-cv;//  [kJ/kg K]

+//  also, P1*V1=m*R*T1,hence

+T1 = P1*V1/(m*R);//  [K]

+

+//  given that process is polytropic with 

+n = 1.25; //  polytropic index

+T2 = T1*(P2/P1)^((n-1)/n);//  [K]

+

+//  Hence, change in internal energy is,

+del_U = m*cv*(T2-T1);//  [kJ]

+mprintf('\n (a) The change in internal energy of the air is del_U  =  %f  kJ',del_U);

+if(del_U>0)

+    disp('since del_U>0, so it is gain of internal energy to the air')

+else

+    disp('since del_U<0, so it is gain of internal energy to the surrounding')

+end

+

+//  (b)

+W = m*R*(T1-T2)/(n-1);//  formula of work done for polytropic process,[kJ]

+mprintf('\n (b) The work done is W =  %f  kJ',W);

+if(W>0)

+    disp('since W>0, so the work is done by  the air')

+else

+    disp('since W<0, so the work is done on the air')

+end

+

+//  (c)

+Q = del_U+W;//  using 1st law of thermodynamics,[kJ]

+mprintf('\n (c) The heat transfer is Q  =  %f  kJ',Q);

+if(Q>0)

+    disp('since Q>0, so the heat is received by  the air')

+else

+    disp('since Q<0, so the heat is rejected by the air')

+end

+

+//  End