diff options
Diffstat (limited to '527')
125 files changed, 2914 insertions, 0 deletions
diff --git a/527/CH1/EX1.1/1_1exam.sce b/527/CH1/EX1.1/1_1exam.sce new file mode 100755 index 000000000..4a3fa1b77 --- /dev/null +++ b/527/CH1/EX1.1/1_1exam.sce @@ -0,0 +1,20 @@ +//Engineering and Chemical Thermodynamics
+//Example 1.1
+//Page no :22
+
+clear ; clc
+// Given the quality of the system is ,x=0.2
+
+// V_l = Specific volume of pure liquid
+// V_v = Specific volume of pure vapour
+// V = Molar volume of liquid-vapour mixture
+disp(" Example: 1.1 Page no : 22") ;
+disp(" V = V_l + x*(V_v - V_l)");
+disp(" 0.2 = (V - V_l) / (V_v - V_l)");
+disp(" 0.8 = (V_v - V) / (V_v - V_l)");
+
+disp(" The tie line is devided into two parts according to the fraction of each phase to get the state of the mixture . ");
+
+// The line segment representing the liquid is four times greater than that of vapour
+
+disp(" As no numerical values are given for specific volumes , we can not get numerical answer .");
\ No newline at end of file diff --git a/527/CH1/EX1.2/1_2exam.sce b/527/CH1/EX1.2/1_2exam.sce new file mode 100755 index 000000000..c2a382788 --- /dev/null +++ b/527/CH1/EX1.2/1_2exam.sce @@ -0,0 +1,30 @@ +//Engineering and Chemical Thermodynamics
+//Example 1.2
+//Page no :25
+
+clear ; clc
+P = 1.4 ; // [MPa]
+T = 333 ; //[K]
+
+//Given values are
+T1 = 320 ; //[K]
+T2 = 360 ; //[K]
+P_low = 1 ; //[MPa]
+P_high = 1.5 ; //[MPa]
+V_cap_T1_P1 = 0.2678 ;
+V_cap_T2_P1 = 0.2873 ;
+V_cap_T1_P1_5 = 0.1765 ;
+V_cap_T2_P1_5 = 0.1899 ;
+
+//At P = 1 MPa
+V_cap_T333_P1 = V_cap_T1_P1 + (V_cap_T2_P1 - V_cap_T1_P1)*((T - T1)/(T2- T1)); //[m^3/kg]
+
+//Similarly at P=1.5 MPa
+V_cap_T333_P1_5 = V_cap_T1_P1_5 + (V_cap_T2_P1_5 - V_cap_T1_P1_5)*((T - T1)/(T2 - T1)); //[m^3/kg]
+
+//At T=333*C
+V_cap_P1_5 = V_cap_T333_P1_5 ;
+V_cap_P1 = V_cap_T333_P1 ;
+V_cap_P1_4 = V_cap_P1 + (V_cap_P1_5 - V_cap_P1)*((P - P_low)/(P_high - P_low)) ; //[m^3/kg]
+disp(" Example: 1.2 Page no : 25") ;
+printf('\n Required specific volume = %g m^3/kg',V_cap_P1_4);
\ No newline at end of file diff --git a/527/CH1/EX1.3/1_3exam.sce b/527/CH1/EX1.3/1_3exam.sce new file mode 100755 index 000000000..9a3bdce8b --- /dev/null +++ b/527/CH1/EX1.3/1_3exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 1.3
+//Page no :27
+
+clear ; clc
+
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.")
\ No newline at end of file diff --git a/527/CH1/EX1.4/1_4exam.sce b/527/CH1/EX1.4/1_4exam.sce new file mode 100755 index 000000000..5f6bc69c0 --- /dev/null +++ b/527/CH1/EX1.4/1_4exam.sce @@ -0,0 +1,22 @@ +//Engineering and Chemical Thermodynamics
+//Example 1.3
+//Page no :27
+
+clear ; clc
+//From Ideal gas law we have v=(R*T)/P
+
+//Given data
+P = 1.4 ; //[MPa]
+P_low = 1 ;//[MPa]
+P_high = 1.5;//[MPa]
+
+//At T=333*C from interpolation we have
+v_cap_P1_5 = 0.18086 ;//[m^3/kg]
+v_cap_P1 = 0.27414 ;//[m^3/kg]
+
+//Molar volume is inversely proportional to pressure
+v_cap_P1_4 = v_cap_P1 +(v_cap_P1_5 - v_cap_P1)*((1/P - 1/P_low)/(1/P_high - 1/P_low));
+x=(0.19951-0.19418)/0.19418*100 ;
+disp(" Example: 1.4 Page no : 28") ;
+printf('\n Specific volume (m^3/kg) = %g',v_cap_P1_4);
+printf('\n Percentage difference = %g',x);
\ No newline at end of file diff --git a/527/CH2/EX2.1/2_1exam.sce b/527/CH2/EX2.1/2_1exam.sce new file mode 100755 index 000000000..91b6f135f --- /dev/null +++ b/527/CH2/EX2.1/2_1exam.sce @@ -0,0 +1,15 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.1
+//Page no :33
+
+clear ; clc
+z1 = 10 ; //[m]
+z2 = 0 ; //[m],Taking ground as state 2,reference
+v1 = 0 ;
+
+//From conservation of total energy we get
+// (1/2*m*v2^2-1 / 2*m*v1^2)+(m*g*z2 - m*g*z1) = 0
+// 1/2*m*v2^2 - m*g*z1 = 0
+v2 = sqrt(2 * 9.8 * z1) ; //[m/s]
+disp(" Example: 2.1 Page no : 33") ;
+printf('\n Final velocity = %g (m/s) ',v2);
\ No newline at end of file diff --git a/527/CH2/EX2.10/2_10exam.sce b/527/CH2/EX2.10/2_10exam.sce new file mode 100755 index 000000000..c29e64f73 --- /dev/null +++ b/527/CH2/EX2.10/2_10exam.sce @@ -0,0 +1,25 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.10
+//Page no :68
+
+clear ; clc
+//Given data:
+n_dot = 10 ; //[mol/s]
+T1 = 298.2 ; //[K]
+T2 = 342 ; //[K]
+T3 = 373.2 ; //[K]
+Cp_298_342 = 216.3 ; //[J/molK]
+A = 3.025 ;
+B = 53.722 * 10^-3 ;
+C = -16.791 * 10^-6 ;
+del_h_vap = 28.88 ; //[kJ/mol]
+
+del_h_1 = Cp_298_342 * (T2 - T1) * 10^-3 ; //[kJ/mol]
+del_h_2 = del_h_vap ;
+function y=f(T),y=8.314*(A*T + (B/2)*(T^2) + (C/3)*(T^3))* 10^-3 ;
+endfunction
+del_h_3 = f(T3) - f(T2) ;
+
+Q = n_dot * (del_h_1 + del_h_2 + del_h_3) ;
+disp(" Example: 2.10 Page no : 68") ;
+printf('\n Rate of heat supplied = %d kJ/s',Q );
\ No newline at end of file diff --git a/527/CH2/EX2.11/2_11exam.sce b/527/CH2/EX2.11/2_11exam.sce new file mode 100755 index 000000000..37b5be127 --- /dev/null +++ b/527/CH2/EX2.11/2_11exam.sce @@ -0,0 +1,24 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.11
+//Page no :69
+
+clear ; clc ;
+//Given data:
+m_1_v = 4.3 ; //[kg]
+m_1_l = 50 ; //[kg]
+u_cap_1_v = 2437.9 ; //[kJ/kg],From steam table
+u_cap_1_l = 191.8 ; //[kJ/kg],From steam table
+v_cap_1_v = 14.67 ; //[m^3],From steam table
+v_cap_1_l = 0.001 ; //[m^3],From steam table
+
+V2 = m_1_l * v_cap_1_l + m_1_v * v_cap_1_v ;
+m_2_v = m_1_l + m_1_v ;
+v_cap_2_v = V2 / m_2_v ; //[m^3/kg]
+
+// From table this specific volume matches at
+P2= 0.15 ; //[MPa]
+//At this condition
+u_cap_2_v = 2519.6 ; //(kJ/kg)
+Q = ((m_2_v * u_cap_2_v) -(m_1_l * u_cap_1_l + m_1_v * u_cap_1_v))*1000;
+disp(" Example: 2.11 Page no : 69") ;
+printf('\n Minimum amount of heat required = %e J',Q);
\ No newline at end of file diff --git a/527/CH2/EX2.12/2_12exam.sce b/527/CH2/EX2.12/2_12exam.sce new file mode 100755 index 000000000..ed9d65972 --- /dev/null +++ b/527/CH2/EX2.12/2_12exam.sce @@ -0,0 +1,14 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.12
+//Page no :73
+
+clear ; clc ;
+// From table we have
+del_h0_f_CO2 = -393.51 ; // [kJ/mol]
+del_h0_f_H2 = 0 ; // [kJ/mol]
+del_h0_f_H2O = -241.82 ; // [kJ/mol]
+del_h0_f_CH3OH = -200.66 ; // [kJ/mol]
+
+del_h0 = del_h0_f_CO2 + 3 * del_h0_f_H2 - del_h0_f_H2O - del_h0_f_CH3OH ;
+disp(" Example: 2.12 Page no : 73") ;
+printf('\n Enthalpy of reaction = %g kJ/mol',del_h0);
\ No newline at end of file diff --git a/527/CH2/EX2.13/2_13exam.sce b/527/CH2/EX2.13/2_13exam.sce new file mode 100755 index 000000000..e27d00bfc --- /dev/null +++ b/527/CH2/EX2.13/2_13exam.sce @@ -0,0 +1,49 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.13
+//Page no :73
+
+clear ; clc ;
+//Given data :
+del_h0_f_CO2 = -393.51 ; //[kJ/mol], From Appendix A.3
+del_h0_f_CO = -110.53 ; //[kJ/mol], From Appendix A.3
+del_h0_f_H2O = -241.82 ; //[kJ/mol], From Appendix A.3
+del_h0_f_C3H8 = -103.85 ; //[kJ/mol], From Appendix A.3
+del_h0_f_O2 = 0 ; //[kJ/mol], From Appendix A.3
+A_CO2 = 5.457 ; // From table E2.13
+B_CO2 = 1.05 * 10^-3 ;
+D_CO2 = -1.16 * 10^5 ;
+A_CO = 3.379 ;
+B_CO = 5.57 * 10^-4;
+D_CO = -3.1 * 10^3 ;
+A_H2O = 3.470 ;
+B_H2O = 1.45 * 10^-3;
+D_H2O = 1.21 * 10^4 ;
+A_N2 = 3.280 ;
+B_N2 = 5.93 * 10^-4;
+D_N2 = 4.00 * 10^3 ;
+
+//Let
+n_C3H8 = 10 ; //[mol]
+n_N2 = (0.79/0.21) * (9.7/2) * n_C3H8 ; //[mol]
+n_CO2 = 2.7 * n_C3H8 ; //[mol]
+n_CO = 0.3 * n_C3H8 ; //[mol]
+n_H2O = 4 * n_C3H8 ; //[mol]
+n_O2 = (9.7 / 2)* n_C3H8 ; //[mol]
+T_reff = 298 ; //[K]
+del_H_rxn_298 = n_CO2 * del_h0_f_CO2 + n_CO * del_h0_f_CO + n_H2O * del_h0_f_H2O - n_C3H8 * del_h0_f_C3H8 - n_O2 * del_h0_f_O2 ; //[kJ]
+
+//The co-efficients of T2 in the equation of degree 3 are
+a = 8.314*(n_CO2 * (B_CO2/2) + n_CO * (B_CO/2) + n_H2O * (B_H2O/2) + n_N2 * (B_N2/2));
+b = 8.314*(n_CO2 * A_CO2 + n_CO * A_CO + n_H2O * A_H2O + n_N2 * A_N2) ;
+d =8.314*(- n_CO2 * D_CO2 - n_CO * D_CO - n_H2O * D_H2O -n_N2 * D_N2) ;
+c = (del_H_rxn_298 *1000) + 8.314 * (n_CO2 * (- T_reff * A_CO2 - B_CO2/2 * T_reff^2 + D_CO2/T_reff) + n_CO * (- T_reff * A_CO - B_CO/2 * T_reff^2 + D_CO/T_reff) + n_H2O * (- T_reff * A_H2O - B_H2O/2 * T_reff^2 + D_H2O/T_reff) + n_N2 * (-T_reff * A_N2 - B_N2/2 * T_reff^2 + D_N2/T_reff));
+
+T2=poly(0,'T2');
+P = d + c*T2 + b*T2^2 + a*T2^3 ;
+M = roots(P);
+
+disp(" Example: 2.13 Page no : 73") ;
+disp(" The roots of the equation containig T2 as variable are (K)-")
+disp(M);
+disp(" But T2 must be more than 298K . So we have to choose the most suitable solution .")
+// The answer in the textbook does not statisfy the equation while it is counter-checked .
diff --git a/527/CH2/EX2.14/2_14exam.sce b/527/CH2/EX2.14/2_14exam.sce new file mode 100755 index 000000000..a665ad4b6 --- /dev/null +++ b/527/CH2/EX2.14/2_14exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.14
+//Page no :75
+
+clear ; clc ;
+disp(" Example 2.14 Page no : 75")
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.")
\ No newline at end of file diff --git a/527/CH2/EX2.15/2_15exam.sce b/527/CH2/EX2.15/2_15exam.sce new file mode 100755 index 000000000..050c13d3f --- /dev/null +++ b/527/CH2/EX2.15/2_15exam.sce @@ -0,0 +1,30 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.15
+//Page no :80
+
+clear ; clc ;
+
+//Given data
+V1 = 350 ; //[m/s]
+A = 3.355 ;
+B = 0.575*10^-3 ;
+D = -0.016*10^5 ;
+Tin = 283 ; //[K]
+MW = 29 * 10^-3 ; //[kg/mol]
+
+ek = 1/2 * MW * V1**2 ;
+//The co-efficients of T2 in the equation of degree 3 are
+a = B/2 ;
+b = A ;
+c = -(Tin * A + Tin^2*B/2 - (D/Tin) + ek/8.314) ;
+d=-D ;
+
+T2=poly(0,'T2');
+P = d + c*T2 + b*T2^2 + a*T2^3 ;
+M = roots(P);
+disp(" Example: 2.15 Page no : 80") ;
+disp( " The solutions are ")
+disp(M);
+disp(" But the outlet temp should be more than 283K(inlet temperature) .So we have to choose the most suitable solution .")
+
+
\ No newline at end of file diff --git a/527/CH2/EX2.16/2_16exam.sce b/527/CH2/EX2.16/2_16exam.sce new file mode 100755 index 000000000..6b3e73c7f --- /dev/null +++ b/527/CH2/EX2.16/2_16exam.sce @@ -0,0 +1,16 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.16
+//Page no :81
+
+clear ; clc ;
+//Given data:
+V_dot_2 = 0.001 ; //[m^3/kg]
+v_cap_2 = 0.001 ; //[m^3/kg], Specific volume of water
+z2 = 250 ; //[m] ; Taking ground as the reference level
+e_cap_2 = 9.8 * z2 ; //[kg*m^2/s^2]
+
+m_dot_2 = V_dot_2 / v_cap_2 ; //[kg/s]
+//Neglecting the kinetic energy ,frictional losses ftrom energy balance equation we have
+W_dot_s = m_dot_2 * e_cap_2 * 10^-3 ;
+disp(" Example: 2.16 Page no : 81") ;
+printf('\n Minimum power required is = %g kW',W_dot_s);
diff --git a/527/CH2/EX2.17/2_17exam.sce b/527/CH2/EX2.17/2_17exam.sce new file mode 100755 index 000000000..afacfbe21 --- /dev/null +++ b/527/CH2/EX2.17/2_17exam.sce @@ -0,0 +1,32 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.17
+//Page no :82
+
+clear ; clc ;
+//Given data
+n_dot = 10 ; //[mol/min]
+del_h_vap_CO2 = 10400 ; //[J/mol]
+A_CO2 = 5.457 ; //From appendix A.3
+B_CO2 = 1.045 * 10^-3 ;
+D_CO2 = -1.157 * 10^5 ;
+A_air = 3.355 ;
+B_air = 0.575 * 10^-3 ;
+D_air = -0.016 * 10^5 ;
+T1 = 273 ; //[K]
+T2 = 283 ; //[K]
+T3 = 323 ; //[K]
+T4 = 293 ; //{k}
+
+function y=f1(T),y=8.314 * (A_CO2 * T + (B_CO2/2) * T^2 - D_CO2/T)
+endfunction
+
+sen_heat_CO2 = f1(T2) - f1(T1) ;
+Q_dot = n_dot * (del_h_vap_CO2 + sen_heat_CO2) ; //[J/min]
+
+function y=f2(T),y=8.314 * (A_air * T + B_air/2*T^2 - D_air /T)
+endfunction
+sen_heat_air = f2(T4) - f2(T3);
+n_dot_air = - Q_dot / sen_heat_air ;
+disp(" Example: 2.17 Page no : 82") ;
+printf('\n Air required = %g mol/min',n_dot_air);
+
\ No newline at end of file diff --git a/527/CH2/EX2.18/2_18exam.sce b/527/CH2/EX2.18/2_18exam.sce new file mode 100755 index 000000000..434933064 --- /dev/null +++ b/527/CH2/EX2.18/2_18exam.sce @@ -0,0 +1,13 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.18
+//Page no :84
+
+clear ; clc ;
+m_dot_1 = 10 ; //[kg/s]
+h_cap_1 = 3238.2 ;//[kJ/kg], Super heated steam at 500*C & 200bar
+h_cap_2 = 93.3 ;//[kL/kg], subcooled liquid at 20*C & 100bar
+h_cap_3 = 2724.7 ;//{kJ/kg}, Super heated vapour at 100bar
+
+m_dot_2 = m_dot_1 * (h_cap_1 - h_cap_3) / (h_cap_3 - h_cap_2);
+disp(" Example: 2.18 Page no : 84") ;
+printf('\n Flow of liquid stream = %.2f kg/s',m_dot_2);
\ No newline at end of file diff --git a/527/CH2/EX2.19/2_19exam.sce b/527/CH2/EX2.19/2_19exam.sce new file mode 100755 index 000000000..dedc24dce --- /dev/null +++ b/527/CH2/EX2.19/2_19exam.sce @@ -0,0 +1,16 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.19
+//Page no :85
+
+clear ; clc ;
+//From steam table
+h_cap_st_1 = 2923.4 ; // [kJ/kg]
+h_cap_200 = 2875.3 ; // {kJ/kg} , At 100kPa
+h_cap_250 = 2974.3 ; // {kJ/kg} , At 100 kPa
+del_T = 250-200 ;
+
+T1 = 200 ; //[K]
+h_cap_st_2 = h_cap_st_1 ;//Assumimg bulk kinetic energy of the stream and heat transfered is negligible
+T2 = T1 + del_T * (h_cap_st_2 - h_cap_200) / (h_cap_250 - h_cap_200) ;
+disp(" Example: 2.19 Page no : 85") ;
+printf('\n The exit temperature is = %d *C',T2) ;
\ No newline at end of file diff --git a/527/CH2/EX2.2/2_2exam.sce b/527/CH2/EX2.2/2_2exam.sce new file mode 100755 index 000000000..5cbaf474d --- /dev/null +++ b/527/CH2/EX2.2/2_2exam.sce @@ -0,0 +1,28 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.2
+//Page no :36
+
+clear ; clc
+
+//Given data
+V2 = 14 ; // [m/s]
+u_cap_l1 = 104.86 ; //[kJ/kg],at 25*C internal energy of saturated water
+u_cap_l_t25 = 104.86 ; //[kJ/kg], From steam table
+u_cap_l_t30 = 125.77 ; //[kJ/kg], From steam table
+T1 = 25 ; //[*C]
+T2 = 30 ; //[*C]
+
+//For unit mass change in kinetic energy
+Delta_e_cap_k = 1/2 * V2^2 * 10^-3 ; //[kJ/kg]
+
+Delta_u_cap = Delta_e_cap_k ;
+
+//For final state of water:
+u_cap_l2 = Delta_u_cap + u_cap_l1 ;
+
+//From table
+
+x = (u_cap_l2 - u_cap_l_t25) / (u_cap_l_t30 - u_cap_l_t25) ;
+T_unknown = T1 + x*(T2 - T1) ;
+disp(" Example: 2.2 Page no : 36") ;
+printf('\n Final temperature of water = %g *C',T_unknown);
\ No newline at end of file diff --git a/527/CH2/EX2.20/2_20exam.sce b/527/CH2/EX2.20/2_20exam.sce new file mode 100755 index 000000000..78508cc93 --- /dev/null +++ b/527/CH2/EX2.20/2_20exam.sce @@ -0,0 +1,79 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.20
+//Page no :89
+
+clear ; clc ;
+//solution (a)
+//Given data
+Cv = 3/2 * 8.314 ;
+Cp = 5/2 * 8.314 ;
+n = 1;
+R = 8.314 ;
+T1 = 1000 ; //[K]
+P1 = 10 ; //[bar]
+T2 = 1000 ; //[K]
+P2 = 0.1 ; //[bar]
+T3 = 300 ; //[K]
+T4 = 300 ; //[K]
+
+k = Cp / Cv ;
+P3 = P2 * (T3 / T2)^(k/(k-1)); //[bar]
+P4 = P1 * (T4 / T1)^(k/(k-1)) ; //[bar]
+
+// (1)
+ del_U_12 = 0 ; // As process 1-2 is isothermal
+ W_12 = n * R * T1 * log(P2 / P1);
+ Q_h_12 = W_12 ;
+ disp(" Example: 2.20 Page no : 89") ;
+printf('(a)\n (1)\n del_U = %d J',del_U_12) ;
+printf('\n Work = %d J',W_12) ;
+printf('\n Heat = %d J',Q_h_12) ;
+
+//(2)
+ Q_23 = 0 ; // As adiabatic process
+ del_U_23 = n * Cv *(T3 - T2) ;
+ W_23 = del_U_23 ;
+printf('\n (2)\n del_U = %g J',del_U_23) ;
+printf('\n Work (J) = %d J',W_23) ;
+printf('\n Heat (J) = %d J',Q_23) ;
+
+//(3)
+ del_U_34 = 0 ; // As isothermal process
+ W_34 = n * R * T3 * log(P4 / P3) ; // Eqn E2.20.A
+ Q_c_34 = del_U_34 - W_34 ;
+printf('\n (3)\n del_U = %g J',del_U_34) ;
+printf('\n Work = %d J',W_34) ;
+printf('\n Heat = %d J',Q_c_34) ;
+
+//(4)
+ Q_41 = 0 ; // As adiabatic process
+ del_U_41 = n * Cv * (T1 - T4) ;
+ W_41 = del_U_41 ;
+printf('\n (4)\n del_U = %g J',del_U_41) ;
+printf('\n Work = %d J',W_41) ;
+printf('\n Heat = %d J',Q_41) ;
+
+//Solution (b)
+//Users can refer figure E2.20
+
+//Solution (c)
+ W_total = W_12 + W_23 + W_34 + W_41 ;
+ Q_absor = Q_h_12 ;
+ effi = W_total / Q_absor ;
+printf('\n\n(c) efficiency = %g',effi)
+
+//Solution (d)
+x = 1 - T3 / T1 ;
+printf('\n\n(d) 1 - Tc/Th = %g',x);
+disp(" i.e Efficiency = 1 - Tc/Th");
+
+//Solution (e)
+disp("(e) The process can be made more efficient by raising Th or by lowering Tc .");
+disp("Table E2.20B") ;
+disp(" T(K) P(bar) v(m^3/mol)") ;
+P = [P1 , P2 , P3 , P4 ] ;
+T = [T1 , T2 , T3 , T4 ] ;
+for i = 1:4
+ v(i) = R * T(i) * 10^-5/ P(i) ;
+ printf("\n %d %.4f %f \n",T(i) ,P(i) ,v(i)) ;
+end
diff --git a/527/CH2/EX2.3/2_3exam.sce b/527/CH2/EX2.3/2_3exam.sce new file mode 100755 index 000000000..cd379d489 --- /dev/null +++ b/527/CH2/EX2.3/2_3exam.sce @@ -0,0 +1,15 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.3
+//Page no :38
+
+clear ; clc
+
+//External pressure is constant
+P_ex = 1*10^5 ; //[Pa]}
+
+//To calculte work done
+function y = f(x),y = 1,endfunction
+I = intg(10,15.2,f) ;
+W = -P_ex * I * 10^-3 ; //[J]
+disp(" Example: 2.1 Page no : 33") ;
+printf('\n Work done = %g J',W);
\ No newline at end of file diff --git a/527/CH2/EX2.4/2_4exam.sce b/527/CH2/EX2.4/2_4exam.sce new file mode 100755 index 000000000..aa5e76aa5 --- /dev/null +++ b/527/CH2/EX2.4/2_4exam.sce @@ -0,0 +1,17 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.4
+//Page no :55
+
+clear ; clc
+//From steam table specific enthalpy at state1 and state2 are
+h_cap_1 = 3373.6 ; //[kJ/kg]
+h_cap_2 = 2675.5 ; //[kJ/kg]
+
+m_dot1 = 10; //[kg/s],As we are dealing with steady state
+m_dot2 = 10; //[kg/s]
+
+//Neglecting heat dissipation compared to shaft work we have
+// m_dot1*h_cap_1 - m_dot2*h_cap_2 + Ws_dot = 0
+Ws_dot = m_dot1 * (h_cap_2 - h_cap_1) ; //[kW]
+disp(" Example: 2.4 Page no : 55") ;
+printf('\n Power generated = %g kW',Ws_dot);
diff --git a/527/CH2/EX2.5/2_5exam.sce b/527/CH2/EX2.5/2_5exam.sce new file mode 100755 index 000000000..ef0c89cd3 --- /dev/null +++ b/527/CH2/EX2.5/2_5exam.sce @@ -0,0 +1,25 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.5
+//Page no :55
+
+clear ; clc
+
+//Solution(a)
+//Unsteady state analysis
+h_cap_in = 3241 ; //[kJ/kg] , From steam table
+P_final = 10 ; //[MPa]
+
+//From Eqn. Eq2.5A , Eq2.5B , Eq2.5C we get
+u_cap_2 = h_cap_in ;
+//At codition of P = 10MPa , u_cap_2 = 3241 kJ/kg the final temperature of the system is
+T2 = 600 ; // From steam table .No calculation is involved .
+disp(" Example: 2.5 Page no : 55") ;
+printf('\n (a)\n The final temperature of the system = %g *C\n',T2);
+
+//Closed system analysis
+//From equation E2.5E , E2.5F , E2.6G we get
+u_cap_2 = h_cap_in ;
+// So temperature is T2 = 600*C (From table).
+
+//Solution(b)
+disp(" (b) The temperature of the fluid increases in the system due to the receipent of flow work .")
\ No newline at end of file diff --git a/527/CH2/EX2.6/2_6exam.sce b/527/CH2/EX2.6/2_6exam.sce new file mode 100755 index 000000000..a0514ef09 --- /dev/null +++ b/527/CH2/EX2.6/2_6exam.sce @@ -0,0 +1,31 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.6
+//Page no :62
+
+clear ; clc
+//Q=n*Delta_h
+//Given data
+n =2 ; //[mol]
+A = 3.470 ;
+B = 1.450*10^-3 ;
+D = 0.121*10^5 ;
+T1 = 473 ; //[K]
+T2 = 773 ; //[K]
+
+function y = f(T),y = 8.314*(A + B*T + D*T^-2),endfunction
+Delta_h = intg(T1,T2,f);
+
+Q = n * Delta_h ;
+disp(" Example: 2.6 Page no : 62") ;
+printf('\n (a)Heat required = %g J',Q);
+
+//Solution (b)
+
+//From steam table
+h_cap_1 = 2827.9 ; //[kJ/kg]
+h_cap_2 = 3478.4 ; //[kJ/kg]
+m = 2*0.018 ; //[kg]
+
+Delta_h_cap = (h_cap_2 - h_cap_1) * 10^3 ; //[J/kg]
+Q = m * Delta_h_cap;
+printf('\n\n (b)Heat required = %g J',Q);
\ No newline at end of file diff --git a/527/CH2/EX2.7/2_7exam.sce b/527/CH2/EX2.7/2_7exam.sce new file mode 100755 index 000000000..c8fd7ec68 --- /dev/null +++ b/527/CH2/EX2.7/2_7exam.sce @@ -0,0 +1,20 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.7
+//Page no :63
+
+clear ; clc
+//Given data
+T1 = 298;
+T2_start = 300;
+A = 3.355;
+B = 0.575*10^-3;
+D = -0.016*10^5;
+
+function y = f(T),y = 8.314*[A*T + B/2*T^2 - D/T]
+endfunction;
+disp(" Example: 2.7 Page no : 63") ;
+for T2_start = 300:100:1000;
+ del_h = f(T2_start) - f(T1);
+ Cp = del_h /(T2_start - 298);
+ mprintf('\n At temperature(K) %g, Molar heat capacity (J/molK) %g',T2_start,Cp);
+end
\ No newline at end of file diff --git a/527/CH2/EX2.8/2_8exam.sce b/527/CH2/EX2.8/2_8exam.sce new file mode 100755 index 000000000..64248f7f1 --- /dev/null +++ b/527/CH2/EX2.8/2_8exam.sce @@ -0,0 +1,17 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.8
+//Page no :64
+
+clear ; clc
+// Given data
+n_dot_air = 10 ; //[mol/min]
+C_bar_P_900 = 30.71 ; //[J/molK]
+C_bar_P_600 = 29.97 ; //[J/molK]
+T1 = 600 ; //[K]
+T2 = 900 ; //[K]
+T_ref = 298 ; //[K]
+
+// Q_dot = n_dot_air * (h_900 - h_600)...........Eqn E2.8A
+Q_dot = n_dot_air * (C_bar_P_900 * (T2 - T_ref) - C_bar_P_600 * (T1 - T_ref));
+disp(" Example: 2.8 Page no : 33") ;
+printf('\n Heat rate required = %g J/min',Q_dot);
\ No newline at end of file diff --git a/527/CH2/EX2.9/2_9eaxm.sce b/527/CH2/EX2.9/2_9eaxm.sce new file mode 100755 index 000000000..0b245e1a1 --- /dev/null +++ b/527/CH2/EX2.9/2_9eaxm.sce @@ -0,0 +1,45 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.9
+//Page no :65
+
+clear ; clc
+//solution(a)
+
+// Given data:
+P1 = 100000 ; // [N/m^2]
+T1 = 298 ; //[K]
+V1 = 0.1 * 0.1 ; // [m^3]
+T2 = 373 ; // [N]
+P_ext = 100000 ; //[N/m^2]
+k = 50000 ; //[N/m]
+A = 0.1 ; //[m^2]
+
+// Applying ideal gas law we getan quadritic eqn of the form :
+// a * V2^2 + b * V2 + c = 0 where
+a = k / (T2 * A^2) ;
+b = (P_ext / T2) - k * V1 / (A^2 * T2) ;
+c = -P1 * V1 / T1 ;
+V2 = (-b + sqrt ( b^2 - (4*a*c))) / (2 * a) ;
+W = -P_ext * (V2 - V1) - ( k * (V2 - V1)^2)/(2 * A**2);//From eqn E2.9C
+disp(" Example: 2.9 Page no : 65") ;
+printf('\n (a) Work required = %g J \n\n',W);
+
+
+//Solution(b):
+
+//Given data:
+A = 3.355 ;
+B = 0.575 * 10^-3 ;
+D = -0.016 * 10^5 ;
+P1 = 10^5 ; //[N/m^2]
+V1 = 0.01 ; //[m^3]
+R = 8.314 ;
+T1 = 298 ;
+
+n = (P1 * V1) / (R * T1) ;
+function y=f(T),y=R*((A - 1) * T + B/2 * T^2 -D/T)
+endfunction
+del_u = f(373) - f(298) ;
+del_U = n * del_u ;
+Q = del_U - W;
+printf('\n (b).Heat transfered = %.4f J',Q);
\ No newline at end of file diff --git a/527/CH3/EX3.10/3_10exam.sce b/527/CH3/EX3.10/3_10exam.sce new file mode 100755 index 000000000..3b521c99b --- /dev/null +++ b/527/CH3/EX3.10/3_10exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 3.10:
+//Page no:131
+
+clear ; clc ;
+disp(" Example: 3.10 Page no : 131") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.") ;
\ No newline at end of file diff --git a/527/CH3/EX3.11/3_11exam.sce b/527/CH3/EX3.11/3_11exam.sce new file mode 100755 index 000000000..58f6a19e9 --- /dev/null +++ b/527/CH3/EX3.11/3_11exam.sce @@ -0,0 +1,33 @@ +//Engineering and Chemical Thermodynamics
+//Example 3.11
+//Page no:131
+
+clear ; clc ;
+P_1 = 10 ; //[bar]
+T_1 = 298 ; // [K]
+P_2 = 1 ; //[bar]
+T_2 = 298 ; // [K]
+P_3 = 1 ; //[bar]
+R = 8.314 ; // [J/mol K]
+n = 4 ; //[mol]
+X = 0.01 ;
+
+//Step 1 :
+del_S_sys = - R * log(P_2 / P_1);
+del_S_surr = - R * (1 - P_2 / P_1) ;
+del_s_univ_1 = del_S_sys + del_S_surr ;
+Del_S_univ_1 = n * del_s_univ_1 ;
+
+//Step 2 :
+Del_S_univ_2 = 0 ;
+n_3 = n * P_3 / P_1 ;
+
+//Step 3 :
+n_out = n - n_3 ;
+del_S_sys_3 = - n_out * R * log(X) ;
+Del_S_univ_3 = del_S_sys_3 ; // Assuming the composition of air in the room does not noticeably change by the dilute addition of argon
+Del_S_univ = Del_S_univ_1 + Del_S_univ_2 + Del_S_univ_3 ;
+
+disp(" Example: 3.11 Page no : 131") ;
+printf("\n Total entropy change of universe = %.2f J/K \n\n",Del_S_univ) ;
+disp(" No matter how slow the leak , the driving force for the expansion is finite . So the process canot be reverssible .")
\ No newline at end of file diff --git a/527/CH3/EX3.12/3_12exam.sce b/527/CH3/EX3.12/3_12exam.sce new file mode 100755 index 000000000..820df2c91 --- /dev/null +++ b/527/CH3/EX3.12/3_12exam.sce @@ -0,0 +1,16 @@ +//Engineering and Chemical Thermodynamics
+// Example 3.12
+//Page no:136
+
+clear ; clc ;
+//Given
+n_dot = 250 ; // [mol/s]
+P_1 = 125 * 10^5 ; // [N/m^2]
+V_cap_1 = 5 * 10^-4 ; // [m^3/mol]
+P_2 = 8 * 10^5 ; // [N/m^2]
+
+X = 3 * P_1^0.6667 * V_cap_1 * ( P_2^(1/3) - P_1^(1/3)) ;
+W_dot_s = n_dot * X * 10^-6 ;
+
+disp(" Example: 3.12 Page no : 136") ;
+printf('\n Power generated = %.1f MW',W_dot_s) ;
diff --git a/527/CH3/EX3.13/3_13exam.sce b/527/CH3/EX3.13/3_13exam.sce new file mode 100755 index 000000000..b6f2e6600 --- /dev/null +++ b/527/CH3/EX3.13/3_13exam.sce @@ -0,0 +1,11 @@ +//Engineering and Chemical Thermodynamics
+// Example 3.13
+//Page no:137
+
+clear ; clc ;
+//Given
+Ws_real = -2.1 ; //[MW]
+Ws_rev = -2.8 ; // [MW]
+n_tur = Ws_real / Ws_rev ;
+disp(" Example: 3.13 Page no : 137") ;
+printf("\n Isentropic efficiency of turbine = %.2f %%", n_tur * 100);
\ No newline at end of file diff --git a/527/CH3/EX3.14/3_14exam.sce b/527/CH3/EX3.14/3_14exam.sce new file mode 100755 index 000000000..f81f59ed6 --- /dev/null +++ b/527/CH3/EX3.14/3_14exam.sce @@ -0,0 +1,39 @@ +//Engineering and Chemical Thermodynamics
+//Example 3.14
+//Page no:140
+
+clear ; clc ;
+//Given
+P_1 = 10 * 10^6 ; // [N/m^2]
+T_1 = 600 + 273 ; //[K]
+T_H = T_1 ;
+T_C = 100 + 273 ; //[K]
+P_3 = 10 * 10^4 ; //[N/m^2]
+P_4 = P_1 ;
+h_cap_1 = 3625.3 ; // [kJ/kg],From steam table
+S_cap_1 = 6.9028 ; //[kJ/kgK],From steam table
+S_cap_2 = S_cap_1 ; //[kJ/kgK],From steam table
+S_cap_v = 7.3593 ; //[kJ/kgK],From steam table
+S_cap_l = 1.3025 ; //[kJ/kgK],From steam table
+h_cap_l = 417.44 ;// [kJ/kg],From steam table
+h_cap_v = 2675.5 ;// [kJ/kg],From steam table
+V_cap_l = 10^-3 ; // [m^3/kg],From steam table
+
+X = (S_cap_2 - S_cap_l) / (S_cap_v - S_cap_l);
+h_cap_2 = (1 - X) * h_cap_l + X * h_cap_v ;
+W_cap_s = h_cap_2 - h_cap_1 ;
+h_cap_3 = h_cap_l ;
+
+W_cap_c = V_cap_l * (P_4 - P_3) * 10^-3 ;
+h_cap_4 = h_cap_3 + W_cap_c ;
+W_net = W_cap_s + W_cap_c ; // [kJ/kg]
+
+n_turb = ( -W_cap_s - W_cap_c) / (h_cap_1 - h_cap_4) ;
+disp(" Example: 3.14 Page no : 140") ;
+printf("\n Efficiency of the Rankine cycle = %.3f %% \n\n",n_turb * 100 );
+
+n_carnot = 1 - T_C / T_H ;
+printf(" Efficiency of the Carnot cycle = %.3f %%\n\n",n_carnot * 100);
+
+disp(" The Rankine efficiecy is lower than Carnot efficiency .")
+
diff --git a/527/CH3/EX3.15/3_15exam.sce b/527/CH3/EX3.15/3_15exam.sce new file mode 100755 index 000000000..640d3cca3 --- /dev/null +++ b/527/CH3/EX3.15/3_15exam.sce @@ -0,0 +1,24 @@ +//Engineering and Chemical Thermodynamics
+//Example 3.15
+//Page no:141
+
+clear ; clc ;
+//Given
+n_turb = 0.85 ;
+n_comp = 0.85 ;
+W_cap_s_rev = -1120 ; //[kJ/kg]
+h_cap_1 = 3625.3 ; //[kJ/kg]
+h_cap_l = 417.44 ; //[kJ/kg]
+W_cap_c_rev = 9.9 ; //[kJ/kg]
+
+W_cap_s_act = n_turb * W_cap_s_rev ;
+h_cap_2_act = W_cap_s_act + h_cap_1 ;
+h_cap_3 = h_cap_l ;
+W_cap_c_act = W_cap_c_rev / n_comp ;
+h_cap_4_act = W_cap_c_act + h_cap_3 ;
+W_cap_net = W_cap_s_act + W_cap_c_act ;
+n_rank_act = (-W_cap_s_act - W_cap_c_act) / (h_cap_1 - h_cap_4_act) ;
+
+disp(" Example: 3.15 Page no : 141") ;
+printf("\n W_cap_net = %.1f kJ/kg",W_cap_net) ;
+printf("\n Efficiency of Rankine cycle = %.3f %%",n_rank_act*100) ;
diff --git a/527/CH3/EX3.16/3_16exam.sce b/527/CH3/EX3.16/3_16exam.sce new file mode 100755 index 000000000..e3788c40b --- /dev/null +++ b/527/CH3/EX3.16/3_16exam.sce @@ -0,0 +1,24 @@ +//Engineering and Chemical Thermodynamics
+// Example 3.16
+//Page no:144
+
+clear ; clc ;
+//Given
+P_1 = 120 * 10^3 ; //[N]
+P_2 = 900 * 10^3 ; //[N]
+h_4 = 25.486 ; //[kJ/mol], From table
+h_1 = h_4 ;
+h_2 = 39.295 ; //[kJ/mol], From table
+S_2 = 177.89 ; //[kJ/molK], From table
+S_3 = S_2 ; //[kJ/mol]
+h_3 = 43.578 ; //[kJ/mol] , Enthalpy corresponding to S3 value which equales to S2
+Q_dot_c_des = 10 ; //[kW]
+
+q_c = h_2 - h_1 ;
+Q_dot_c = h_2 - h_1 ;
+W_dot_c = h_3 - h_2 ;
+
+COP = Q_dot_c / W_dot_c ;
+n_dot = Q_dot_c_des / q_c ;
+disp(" Example: 3.16 Page no : 144") ;
+printf("\n COP of the refrigerator is = %.2f \n\n Mass flow rate needed = %.3f mol/s",COP,n_dot)
\ No newline at end of file diff --git a/527/CH3/EX3.17/3_17exam.sce b/527/CH3/EX3.17/3_17exam.sce new file mode 100755 index 000000000..9df477632 --- /dev/null +++ b/527/CH3/EX3.17/3_17exam.sce @@ -0,0 +1,12 @@ +//Engineering and Chemical Thermodynamics
+// Example 3.17
+//Page no :151
+
+clear ; clc ;
+disp(" Example: 3.17 Page no : 151") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved .")
+
+// Del_S_magnetization > 0 ;
+// Del_S_magnetization + Del_S_temperature = 0 ; therefore
+// Del_S_temperature < 0 ;
+// i.e. T2 < T1 ;
\ No newline at end of file diff --git a/527/CH3/EX3.2/3_2exam.sce b/527/CH3/EX3.2/3_2exam.sce new file mode 100755 index 000000000..978deb53b --- /dev/null +++ b/527/CH3/EX3.2/3_2exam.sce @@ -0,0 +1,22 @@ +//Engineering and Chemical Thermodynamics
+//Example 3.2
+//Page no : 119
+
+//Solution(a)
+clear ; clc ;
+//Given
+del_U = 0 ; // As no work or heat transfered across its boundaries during the process
+T_1 = 500 ; // [K]
+V1 = 1.6682 / 2 * 10^-3; // [m^3]
+V2 = 2 * V1 ;
+del_S_sur = 0 ; // As no heat transfered across its boundaries during the process
+disp(" Example 3.2 Page no : 119")
+disp("(a)");
+disp(" For an ideal gas u = u(T only)") ;
+printf('\n Final temperature = %g K \n\n',T_1);
+
+//Solution(b)
+q_rev = 8.314 * T_1 * log(V2/V1) ;
+del_S_sys = q_rev / T_1 ;
+del_S_univ = del_S_sys + del_S_sur ;
+printf('(b)\n\n Entropy change for universe = %.2f J/(molK)',del_S_univ);
diff --git a/527/CH3/EX3.3/3_3exam.sce b/527/CH3/EX3.3/3_3exam.sce new file mode 100755 index 000000000..51e70cd1a --- /dev/null +++ b/527/CH3/EX3.3/3_3exam.sce @@ -0,0 +1,15 @@ +//Engineering and Chemical Thermodynamics
+//Example 3.3
+//Page no:121
+
+clear ; clc ;
+// Given
+T_1_1 = 273 ;// {K}
+T_1_2 = 373 ; //[K]
+Cp = 24.5 ; // [J/molK]
+del_S_sur = 0 ; //Since the system is isolated
+T2 = (T_1_1 + T_1_2)/2 ;
+del_S = Cp / 2 * log(T2^2 / (T_1_1 * T_1_2)) ;
+
+disp(" Example 3.3 Page no : 121") ;
+printf("\n Entropy change for the system = %.2f J/(mol K)", del_S);
\ No newline at end of file diff --git a/527/CH3/EX3.4/3_4exam.sce b/527/CH3/EX3.4/3_4exam.sce new file mode 100755 index 000000000..a28c409df --- /dev/null +++ b/527/CH3/EX3.4/3_4exam.sce @@ -0,0 +1,12 @@ +//Engineering and Chemical Thermodynamics
+// Example 3.4
+//Page no : 122
+
+clear ; clc ;
+//Given
+del_h_vap = 38.56 * 10^3 ; //[J/mol] , From Table
+Tb = 78.2 + 273 ; //[K] ,From table
+
+del_S = - del_h_vap / Tb * 10^-3 ;
+disp(" Example 3.4 Page no : 122") ;
+printf("\n Change in entropy = %.4f kJ/mol K",del_S);
\ No newline at end of file diff --git a/527/CH3/EX3.5/3_5exam.sce b/527/CH3/EX3.5/3_5exam.sce new file mode 100755 index 000000000..af994106b --- /dev/null +++ b/527/CH3/EX3.5/3_5exam.sce @@ -0,0 +1,19 @@ +//Engineering and Chemical Thermodynamics
+//Example 3.5
+//Page no:124
+
+clear ; clc ;
+//Given
+P_1 = 300 * 10^3 ; //[N/m^2]
+T_1 = 700 ; // [*C]
+V_bar_1 = 20 ; //[m/s]
+P_2 = 200 * 10^3 ; // [N/m^2]
+h_cap_1 = 3927.1 * 10^3 ; // [J/kg] , From table
+S_cap_1 = 8.8319 ; // [kJ/kgK] , From table
+
+S_cap_2 = S_cap_1 ; // Reverssible adiabatic process
+T2 = 623 ; // [*C] ,From table by interpolation
+h_cap_2 = 3754.7 * 10^3 ; // [J/kgK] ,From table by interpolation
+V_bar_2 = sqrt(2 * (h_cap_1 - h_cap_2) + V_bar_1^2) ;
+disp(" Example: 3.5 Page no : 124") ;
+printf('\n The final temperature is %g C and the exit velocity is %g m/s',T2,V_bar_2);
diff --git a/527/CH3/EX3.6/3_6exam.sce b/527/CH3/EX3.6/3_6exam.sce new file mode 100755 index 000000000..010bd0d1e --- /dev/null +++ b/527/CH3/EX3.6/3_6exam.sce @@ -0,0 +1,21 @@ +//Engineering and Chemical Thermodynamics
+//Example 3.6
+//Page no:125
+
+clear ; clc ;
+//Given
+m_dot_1 = 10 ; // [kg/s]
+m_dot_2 = 1.95 ; // [kg/s]
+P_1 = 200 * 10^5 ; //[N/m^2]
+T_1 = 500 ; //[*C]
+P_2 = 100 * 10^5 ; // [N/m^2]
+T_2 = 20 ; //[*C]
+P_3 = 100 * 10^5 ; //[N/m^2]
+S_cap_1 = 6.14 * 10^3 ; //[J/kgK] , From table
+S_cap_2 = 0.2945 * 10^3 ; //[J/kgK] , From table
+S_cap_3 = 5.614 * 10^3 ; //[J/kgK] , From table
+
+m_dot = m_dot_1 + m_dot_2 ;
+dS_dt_univ = (m_dot * S_cap_3 -(m_dot_1 * S_cap_1 + m_dot_2 * S_cap_2)) * 10^-3;
+disp(" Example: 3.6 Page no : 125") ;
+printf('\n Entropy generated = %.2f kW/K ',dS_dt_univ);
\ No newline at end of file diff --git a/527/CH3/EX3.7/3_7exam.sce b/527/CH3/EX3.7/3_7exam.sce new file mode 100755 index 000000000..88fdaff75 --- /dev/null +++ b/527/CH3/EX3.7/3_7exam.sce @@ -0,0 +1,28 @@ +//Engineering and Chemical Thermodynamics
+//Example3.7
+//Page no:128
+
+//Solution:(a)
+clear ; clc ;
+//Given
+V_1 = 0.5 ; //[m^3]
+P_1 = 150 ; //[kPa]
+T_1 = 20 + 273 ; //[K]
+P_2 = 400 ; // [kPa]
+Cp = 2.5 * 8.314 ;
+Q = V_1 * (P_1 - P_2);
+disp(" Example: 3.7 Page no : 128") ;
+printf("\n (a)\n Heat transferd = %g kJ\n\n",Q);
+
+//Solution:(b)
+del_S_sys = (P_1 * V_1) / T_1 * -log(P_2 / P_1) ;
+printf(' (b)\n Entropy change of system = %.2f kJ/K \n',del_S_sys);
+Q_surr = - Q ;
+del_S_surr = Q_surr / T_1 ;
+printf(' Entropy change of surrounding = %.2f kJ/K \n',del_S_surr) ;
+del_S_univ = del_S_sys + del_S_surr ;
+printf(' Entropy change of universe =%.2f kJ/K \n',del_S_univ) ;
+
+//Solution:(c)
+disp(" (c)");
+disp(" Since entropy of the universe increases , the process is irreverssible .")
\ No newline at end of file diff --git a/527/CH3/EX3.8/3_8exam.sce b/527/CH3/EX3.8/3_8exam.sce new file mode 100755 index 000000000..21cc0435a --- /dev/null +++ b/527/CH3/EX3.8/3_8exam.sce @@ -0,0 +1,20 @@ +//Engineering and Chemical Thermodynamics
+//Example 3.8
+//Page no :129
+
+clear ; clc ;
+//Given
+A = 3.355 ; // from table
+B = 0.575 * 10^-3 ; // from table
+D = -0.016 * 10^5 ; // from table
+R = 8.314 ;
+P1 = 1 ; //[bar]
+P2 = 0.5 ; //[bar]
+function y=f(T),y = R * (A * log(T) + B * T + D / (2 * T^2)) ;
+endfunction ;
+S1 = f(373) - f(298) ;
+S2 = R * log(P1 / P2) ;
+del_S = S1 - S2 ;
+
+disp(" Example: 3.8 Page no : 129") ;
+printf('\n Entropy change = %.2f J/(mol K )',del_S);
\ No newline at end of file diff --git a/527/CH3/EX3.9/3_9exam.sce b/527/CH3/EX3.9/3_9exam.sce new file mode 100755 index 000000000..7587079a9 --- /dev/null +++ b/527/CH3/EX3.9/3_9exam.sce @@ -0,0 +1,19 @@ +//Engineering and Chemical Thermodynamics
+//Example 3.9
+//Page no:129
+
+clear ; clc ;
+//Given
+P = 1 ; //[bar]
+p_O2 = 0.5 ; //[bar]
+p_N2 = 0.5 ; // [bar]
+n_O2 = 1 ; //[mol]
+n_N2 = 1 ; //[mol]
+R = 8.314 ; // J/mol K
+del_S_1_O2 = -n_O2 * R * log(p_O2 / P) ;
+del_S_1_N2 = -n_N2 * R * log(p_N2 / P) ;
+del_S_2 = 0 ; // As both O2 and N2 behave idealy
+del_S = del_S_2 + del_S_1_O2 + del_S_1_N2 ;
+disp(" Example: 3.9 Page no : 129") ;
+printf("\n Entropy of mixing = %.2f J/K",del_S);
+
\ No newline at end of file diff --git a/527/CH4/EX4.1/4_1exam.sce b/527/CH4/EX4.1/4_1exam.sce new file mode 100755 index 000000000..5150cc756 --- /dev/null +++ b/527/CH4/EX4.1/4_1exam.sce @@ -0,0 +1,41 @@ +//Engineering and Chemical Thermodynamics
+//Example 4.1
+//Page no :175
+
+clear ; clc ;
+//Let
+H2O = 1 ;
+NH3 = 2 ;
+CH4 = 3 ;
+CH3Cl = 4 ;
+CCl4 = 5 ;
+
+M_11 = 1.85 ; alp_12 = 14.80 ; I_13 = 12.62 ;
+M_12 = 1.47 ; alp_22 = 22.20 ; I_23 = 10.07 ;
+M_31 = 0.00 ; alp_32 = 26.00 ; I_33 = 12.61 ;
+M_41 = 1.87 ; alp_42 = 45.30 ; I_43 = 11.26 ;
+M_51 = 0.00 ; alp_52 = 105.0 ; I_53 = 11.47 ;
+
+k =1.38 * 10^-16 ; //[ J/K]
+T = 298 ; //[K]
+A =[M_11 , alp_12 , I_13;
+M_12 , alp_22 , I_23 ;
+M_31 , alp_32 , I_33 ;
+M_41 , alp_42 , I_43 ;
+M_51 , alp_52 , I_53 ;] ;
+disp(" Example: 4.1 Page no : 175") ;
+disp(" Molecule M alp*10^25 I C*10^60 Cd_d Cind Cdis") ;
+for i=1:5
+ A(i,5) = ceil( 2/3 * A(i,1)^4 / (k * T) * 10^-12) ;
+ A(i,6) = ceil(2 * A(i,2) * A(i,1)^2 * 10^-1) ;
+ A(i,7) = ceil(3/4 * A(i,2)^2 * A(i,3) * 1.6 * 10^-2) ;
+ A(i,4) = ceil(A(i,5) + A(i,6) + A(i,7)) ; // ....E4.1D
+end ;
+
+printf(" H2O %.2f %.1f %.2f %d %d %d %d ",A(1,1),A(1,2),A(1,3),A(1,4),A(1,5),A(1,6),A(1,7)) ;
+printf("\n NH3 %.2f %.1f %.2f %d %d %d %d ",A(2,1),A(2,2),A(2,3),A(2,4),A(2,5),A(2,6),A(2,7)) ;
+printf("\n CH4 %.2f %.1f %.2f %d %d %d %d ",A(3,1),A(3,2),A(3,3),A(1,4),A(3,5),A(3,6),A(3,7)) ;
+printf("\n CH3Cl %.2f %.1f %.2f %d %d %d %d ",A(4,1),A(4,2),A(4,3),A(4,4),A(4,5),A(4,6),A(4,7)) ;
+printf("\n CCl4 %.2f %.1f %.2f %d %d %d %d \n",A(5,1),A(5,2),A(5,3),A(5,4),A(5,5),A(5,6),A(5,7)) ;
+
+disp(" Even though it is non polar , CCl4 exhibit the largest intermolecular forces . It is due to the large polarizability accociated with the four Cl atom in CCl4 .") ;
\ No newline at end of file diff --git a/527/CH4/EX4.10/4_10exam.sce b/527/CH4/EX4.10/4_10exam.sce new file mode 100755 index 000000000..c374a867a --- /dev/null +++ b/527/CH4/EX4.10/4_10exam.sce @@ -0,0 +1,51 @@ +//Engineering and Chemical Thermodynamics
+//Example 4.10
+//Page no :202
+
+//Solution(a)
+clear ; clc ;
+T = 100 + 273 ; //[K]
+P = 70 * 10^5 ; //[N/m^2]
+P_c = 42.2 * 10 ^ 5 ;
+T_c = 370 ; //[K]
+w = 0.153 ;// Interpolating from table C.1 and C.2
+z_0 = 0.2822 ;// Interpolating from table C.1 and C.2
+z_1 = - 0.0670 ;// Interpolating from table C.1 and C.2
+m = 20 * 10^3 ;//[g]
+MW = 44 ; //[g/mol]
+R = 8.314 ;
+
+P_r = P / P_c ;
+T_r = T / T_c ;
+z = z_0 + w * z_1 ;
+V = m / MW *z * R * T / P ;
+disp(" Example: 4.10 Page no : 202") ;
+printf("\n (1)\n Volume = %.4f m^3 \n\n", V )
+
+//Solution(b)
+T = 295 ;//[K]
+n = 50 ; // [mol]
+a = 0.42748 * R^2 * T_c^2.5 / P_c ;
+b = 0.08664 * R * T_c / P_c ;
+v = 0.1 ;
+P = R * T / (v - b) - a / (T^0.5 * v * (v + b)) ;
+x = P * n * 10^-6 ;
+printf("\n (2)\n Pressure = %d MPa \n\n", x )
+
+//Solution (c)
+y1 = 0.4 ;
+y2 = 1 - y1 ;
+n = 50 ;
+P_c = 48.7 * 10^5 ;//[N/m^2]
+T_c = 305.5 ; //[K]
+a1 = a ;
+b1 = b ;
+a2 = 0.42748 * R^2 * T_c^2.5 / P_c ;
+b2 = 0.08664 * R * T_c / P_c ;
+
+a_mix = y1^2 * a1 + 2 * y1 * y2 * sqrt(a1 * a2) + y2^2 * a2 ;
+b_mix = y1 * b1 + y2 * b2 ;
+P = R * T / (v - b_mix) - a_mix /(T^0.5 * v * (v + b_mix));
+x = P * n * 10^-6 ;
+
+printf("\n (3)\n Pressure = %.2f MPa \n\n", x )
\ No newline at end of file diff --git a/527/CH4/EX4.2/4_2exam.sce b/527/CH4/EX4.2/4_2exam.sce new file mode 100755 index 000000000..9a0df64ef --- /dev/null +++ b/527/CH4/EX4.2/4_2exam.sce @@ -0,0 +1,15 @@ +//Engineering and Chemical Thermodynamics
+//Example 4.2
+//Page no :176
+
+clear ; clc ;
+//Given //
+C6_Ar_HCl_tab = 76 * 10^-60 ;//From table E4.2
+C6_Ar_Ar_tab = 52 * 10^-60 ;//From table E4.2
+C6_HCl_HCl_tab = 134 * 10^-60 ;//From table E4.2
+
+C6_Ar_HCl_gmean = sqrt(C6_Ar_Ar_tab * C6_HCl_HCl_tab) ; //[erg/cm^6]
+x = (C6_Ar_HCl_gmean - C6_Ar_HCl_tab) / C6_Ar_HCl_tab * 100 ;
+
+disp(" Example: 4.2 Page no : 176") ;
+printf("\n The geometric mean is different from that in table E4.2 by %d %%",x)
\ No newline at end of file diff --git a/527/CH4/EX4.3/4_3exam.sce b/527/CH4/EX4.3/4_3exam.sce new file mode 100755 index 000000000..a6376bced --- /dev/null +++ b/527/CH4/EX4.3/4_3exam.sce @@ -0,0 +1,9 @@ +//Engineering and Chemical Thermodynamics
+//Example 4.3
+//Page no :177
+
+clear ; clc ;
+//The problem contains only theory . There is no numerical part involved. Users can go through the book to obtain the required expression.
+
+disp(" Example: 4.3 Page no : 177") ;
+disp(" (C6)SiCl4 > (C6)CCl4 > (C6)CF4")
\ No newline at end of file diff --git a/527/CH4/EX4.4/4_4exam.sce b/527/CH4/EX4.4/4_4exam.sce new file mode 100755 index 000000000..e6065261a --- /dev/null +++ b/527/CH4/EX4.4/4_4exam.sce @@ -0,0 +1,29 @@ +//Engineering and Chemical Thermodynamics
+//Example 4.4
+//Page no :185
+
+clear ; clc ;
+//Given
+Psat_wat_25 = 3.169 * 10^3 ;// From steam table
+Psat_wat_50 = 1.235 * 10^4 ;// From steam table
+Psat_wat_100 = 1.014 * 10^5 ;// From steam table
+A =11.9673 ;
+B = 3626.55 ;
+C = -34.29 ;
+T1 = 25 ; //[*C]
+T2 = 50 ; //[*C]
+T3 = 100 ; //[*C]
+
+M = [T1 , Psat_wat_25 ; T2 , Psat_wat_50 ; T3 , Psat_wat_100];
+for i=1:3
+ M(i,3) = exp(A - B / (M(i,1) + 273 + C)) * 10^5 ;
+end
+disp(" Example: 4.4 Page no : 185") ;
+disp(" T(*C) Water(Pa) Methanol(Pa)") ;
+disp(M);
+
+//Solution(1) :
+printf("\n(1)\n Water can form two hydrogen bonds . While CH4Oh can form only one . Thus at a given temperature , water has stronger attractive forces in the liquid and a lower vapour pressure .\n\n")
+
+//Solution(2):
+printf("(2)\n Since the Maxwell-Boltzmann distribution depends exponentially on temperature , Psat also increses exponentially with temperature .")
\ No newline at end of file diff --git a/527/CH4/EX4.5/4_5exam.sce b/527/CH4/EX4.5/4_5exam.sce new file mode 100755 index 000000000..ed6ec4f87 --- /dev/null +++ b/527/CH4/EX4.5/4_5exam.sce @@ -0,0 +1,10 @@ +//Engineering and Chemical Thermodynamics
+//Example 4.5
+//Page no :189
+
+clear ; clc ;
+//The problem contains only theory . There is no numerical part involved. Users can go through the book to obtain the required expression.
+
+disp(" Example: 4.5 Page no : 189") ;
+disp(" (a) a_SiCl3H > a_SiCl4 > a_CCl4 > a_CF4 ");
+disp(" (b) b_SiCl4 > b_CCl4 > b_SiCl3H > b_CF4 ");
\ No newline at end of file diff --git a/527/CH4/EX4.6/4_6exam.sce b/527/CH4/EX4.6/4_6exam.sce new file mode 100755 index 000000000..61af41d2e --- /dev/null +++ b/527/CH4/EX4.6/4_6exam.sce @@ -0,0 +1,23 @@ +//Engineering and Chemical Thermodynamics
+//Example 4.6
+//Page no :190
+
+clear ; clc ;
+//Given
+Pc_B = 49.1 ; // [bar] , From table
+Pc_T = 42.0 ; // [bar] , From table
+Pc_C = 40.4 ; // [bar] , From table
+Tc_B = 562 ; // [K] , From table
+Tc_T = 594 ; // [K] , From table
+Tc_C = 553 ; // [K] , From table
+R = 8.314 ;
+
+A = [Pc_B , Tc_B ; Pc_T , Tc_T ; Pc_C , Tc_C];
+for i=1:3
+ A(i,3) = 27/64 * (R * A(i,2))^2 /( A(i,1) * 10^5) ;
+ A(i,4) = R * A(i,2) / (8 * A(i,1) * 10^5) ;
+end
+disp(" Example: 4.6 Page no : 190") ;
+disp(" P_c T_c a b ") ;
+disp(A) ;
+disp(" The attractive interactions of all three compounds are dominated by dispersion interactions ( parameter a) , while size affects parameter b .")
\ No newline at end of file diff --git a/527/CH4/EX4.7/4_7exam.sce b/527/CH4/EX4.7/4_7exam.sce new file mode 100755 index 000000000..733033637 --- /dev/null +++ b/527/CH4/EX4.7/4_7exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 4.7
+//Page no :191
+
+clear ; clc ;
+disp(" Example: 4.7 Page no : 191") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.")
\ No newline at end of file diff --git a/527/CH4/EX4.8/4_8exam.sce b/527/CH4/EX4.8/4_8exam.sce new file mode 100755 index 000000000..15700ac2c --- /dev/null +++ b/527/CH4/EX4.8/4_8exam.sce @@ -0,0 +1,14 @@ +//Engineering and Chemical Thermodynamics
+//Example 4.8
+//Page no :197
+
+clear ; clc ;
+//Given
+B = 0.0486 * 10^-3 ;
+T1 = 20 + 273 ; //[K]
+T2 = 500 + 273 ; //[K]
+v1 = 7.11 ; // [cm^3/mol]
+
+v2 = v1 * exp( B * (T2 - T1)) ;
+disp(" Example: 4.8 Page no : 197") ;
+printf("\n Molar volume of solid state 2 = %.2f cm^3/mol", v2);
\ No newline at end of file diff --git a/527/CH4/EX4.9/4_9exam.sce b/527/CH4/EX4.9/4_9exam.sce new file mode 100755 index 000000000..525533543 --- /dev/null +++ b/527/CH4/EX4.9/4_9exam.sce @@ -0,0 +1,43 @@ +//Engineering and Chemical Thermodynamics
+//Example 4.9
+//Page no :199
+
+clear ; clc ;
+//Given
+P_c = 37.9 * 10^5 ;//[N/m^2] , From compressibility chart
+T_c = 425.2 ;// [K , From compressibility chart
+P = 50 * 10^5 ; //N/m^2]
+T = 333.2 ;//[K]
+R = 8.314 ;
+z_0 = 0.2148 ; // Using interpolation from table C.1 and C.2
+z_1 = -0.0855 ; // Using interpolation from table C.1 and C.2
+w = 0.199 ;
+m = 10 ;
+MW = 0.05812 ;
+
+// Using Redlich Kwong equation
+a = (0.42748 * R^2 * T_c^2.5) / P_c ;
+b = 0.08664 * R * T_c / P_c ;
+A = P * T^(1/2) ;
+B = -R * T^(3/2) ;
+C = (a - P * T^(1/2) * b^2 - R * T^(3/2)*b) ;
+D = - a * b;
+
+mycoeff = [ D , C , B , A] ;
+p = poly(mycoeff , "v" , "coeff" );
+M = roots(p);
+
+disp(" Example: 4.9 Page no : 199") ;
+for i = 1:3
+ sign(M(i,1)) ;
+ if ans == 1 then
+ V = m / MW *(M(i,1)) ;
+ printf("\n Using Redlich Kwong equation the volume is = %.3f m^3\n\n",V)
+ end
+end
+
+// Using compressibility chart
+z = z_0 + w * z_1 ;
+v = z * R * T / P ;
+V = m / MW * v ;
+ printf(" Using compressibility chart the volume is = %.3f m^3\n\n",V)
diff --git a/527/CH5/EX5.1/5_1exam.sce b/527/CH5/EX5.1/5_1exam.sce new file mode 100755 index 000000000..c67926013 --- /dev/null +++ b/527/CH5/EX5.1/5_1exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 5.1
+//Page no :218
+
+clear ; clc ;
+disp(" Example: 5.1 Page no : 218") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.")
\ No newline at end of file diff --git a/527/CH5/EX5.2/5_2exam.sce b/527/CH5/EX5.2/5_2exam.sce new file mode 100755 index 000000000..b0142015d --- /dev/null +++ b/527/CH5/EX5.2/5_2exam.sce @@ -0,0 +1,20 @@ +//Engineering and Chemical Thermodynamics
+//Example 5.2
+//Page no :222
+
+clear ; clc ;
+//Given
+T_c = 370 ; //[K]
+P_c = 41.58 * 10^5 ;//[N/m^2]
+R = 8.314 ;
+V1 = 0.001 ;//[m^3]
+V2 = 0.04 ; //[m^3]
+q = 600 ; //[J]
+
+a = 27/64 * (R ^2)*(T_c)^2 / P_c ;
+//Using E5.2D , E5.2E in E5.2C
+del_U = -0.96 * (1 / V2 - 1 / V1) ;
+W = del_U - q ;
+
+disp(" Example: 5.2 Page no : 222") ;
+printf("\n Work done for the expansion = %g J/mol",W) ;
\ No newline at end of file diff --git a/527/CH5/EX5.3/5_3exam.sce b/527/CH5/EX5.3/5_3exam.sce new file mode 100755 index 000000000..c1f310120 --- /dev/null +++ b/527/CH5/EX5.3/5_3exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 5.3
+//Page no :223
+
+clear ; clc
+disp(" Example: 5.3 Page no : 223") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.") ;
\ No newline at end of file diff --git a/527/CH5/EX5.4/5_4exam.sce b/527/CH5/EX5.4/5_4exam.sce new file mode 100755 index 000000000..6ec359881 --- /dev/null +++ b/527/CH5/EX5.4/5_4exam.sce @@ -0,0 +1,44 @@ +//Engineering and Chemical Thermodynamics
+//Example 5.4
+//Page no :225
+
+clear ; clc ;
+//Given
+P_1 = 9.43 * 10^5 ; //[N/m^2]
+P_2 = 18.9 * 10^5 ; //[N/m^2]
+T_1 = 80 + 273 ; //[K]
+T_2 = 120 + 273 ; //[K]
+A = 1.935 ;
+B = 36.915 * 10^-3 ;
+C = -11.402 * 10^-6 ;
+T_c = 425.2 ; // [K]
+P_c = 37.9 * 10^5 ;//[N/m^2]
+R = 8.314 ;
+del_h_1 = 1368 ; //[J/mol]
+del_h_3 = -2542 ; //[J/mol]
+Ws = 2100 ; //[J/mol]
+
+a = 0.42748 * R^2 * T_c^2.5 / P_c ;
+b = 0.08664 * R * T_c / P_c ;
+
+
+ function y = f1 (v) , y = R * T_1 / (v - b) - a / (sqrt(T_1) * v *(v + b)) - P_1;
+ endfunction ;
+za= fsolve([0.001] , f1) ;
+
+ function y = f2 (v) , y = R * T_2 / (v - b) - a / (sqrt(T_2) * v *(v + b)) - P_2;
+ endfunction ;
+ zb= fsolve([0.001] , f2) ;
+
+ function y = f(T),
+ y = R * ( A * T + B/2 * T^2 + C/3 * T^3) ;
+ endfunction ;
+
+del_h_2 = f(T_2) - f(T_1) ;
+del_h_total = del_h_1 + del_h_2 + del_h_3 ;
+q = del_h_total - Ws ;
+
+
+disp(" Example: 5.4 Page no : 225") ;
+printf("\n v1 = %f m^3/mol\n v2 = %f m^3/mol",za ,zb ) ;
+printf("\n\n The heat input = %g J/mol",ceil(q));
\ No newline at end of file diff --git a/527/CH5/EX5.5/5_5exam.sce b/527/CH5/EX5.5/5_5exam.sce new file mode 100755 index 000000000..e07695dd9 --- /dev/null +++ b/527/CH5/EX5.5/5_5exam.sce @@ -0,0 +1,39 @@ +//Engineering and Chemical Thermodynamics
+//Example 5.5
+//Page no :234
+
+clear ; clc ;
+//Given
+T_c = 425.2 ; //[K] ,From Appendix A.1
+P_c = 37.9 * 10^5 ; //[N/m^2] ,From Appendix A.1
+w = 0.199 ;// From Appendix A.1
+A = 1.935 ;
+B = 36.915 * 10^-3 ;
+C = -11.402 * 10^-6 ;
+Ws = 2100 ; //[J/mol]
+T1 = 353.15 ;//[K]
+T2 = 393.15 ;//[K]
+P1 =7.47 * 10^5 ;//[N/m^2]
+P2 = 18.9 * 10^5 ; //[N/m^2]
+R = 8.314 ;
+enth_dep1_0 = -0.413 ;// Table C.3,C.4 in Appendix C
+enth_dep1_1 = -0.622 ;// Table C.3,C.4 in Appendix C
+enth_dep1 = enth_dep1_0 + w * enth_dep1_1 ;//....E5.5B
+enth_dep2_0 = -0.771 ;// Table C.3,C.4 in Appendix C
+enth_dep2_1 = -0.994 ;// Table C.3,C.4 in Appendix C
+enth_dep2 = enth_dep2_0 + w * enth_dep2_1 ;//....E5.5C
+
+T1_r = T1 / T_c ;
+P1_r = P1 / P_c ;
+T2_r = T2 / T_c ;
+P2_r = P2 / P_c ;
+
+function y=f(T) , y = R * (A * T + B/2 * T^2 + C/3 * T^3)
+endfunction
+del_h = f(T2) - f(T1) ;//.....E5.5D
+
+Del_h = -enth_dep1 * R * T_c + del_h + enth_dep2 * R * T_c ;
+q = Del_h - Ws ;
+
+disp(" Example: 5.5 Page no : 235") ;
+printf("\n Heat input = %d J/mol",q)
\ No newline at end of file diff --git a/527/CH5/EX5.6/5_6exam.sce b/527/CH5/EX5.6/5_6exam.sce new file mode 100755 index 000000000..9da61389a --- /dev/null +++ b/527/CH5/EX5.6/5_6exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 5.6
+//Page no :237
+
+clear ; clc ;
+disp(" Example: 5.6 Page no : 237") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.")
\ No newline at end of file diff --git a/527/CH5/EX5.7/5_7exam.sce b/527/CH5/EX5.7/5_7exam.sce new file mode 100755 index 000000000..50d3a5a1e --- /dev/null +++ b/527/CH5/EX5.7/5_7exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 5.7
+//Page no :239
+
+clear ; clc ;
+disp(" Example: 5.7 Page no :239") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.")
\ No newline at end of file diff --git a/527/CH5/EX5.8/5_8exam.sce b/527/CH5/EX5.8/5_8exam.sce new file mode 100755 index 000000000..68f6355c5 --- /dev/null +++ b/527/CH5/EX5.8/5_8exam.sce @@ -0,0 +1,33 @@ +//Engineering and Chemical Thermodynamics
+//Example 5.8
+//Page no :241
+
+clear ; clc ;
+//Given
+T_c = 126.2 ; //[K] , From appendix A.1
+P_c = 33.8 * 10^5 ; //[N/m^2] , From appendix A.1
+w = 0.039 ;// From appendix A.1
+enth_dep_1 = -2.81 ; // From table C.1 Appendix C
+A = 3.28 ;// From Appendix A.2
+B = 0.593 * 10^-3 ;// From Appendix A.2
+del_h_dep_l = -5.1 ;
+del_h_dep_v = -0.1 ;
+T1 = 151 ; //[K]
+P1 = 100 * 10^5 ; //[N/m^2]
+P2 = 1 * 10^5 ; //[N/m^2]
+T2_r = 0.61 ; // From figure 5.4
+T1_r = T1 / T_c ;
+P1_r = P1 / P_c ;
+P2_r = P2 / P_c ;
+
+T2 = T2_r * T_c ; //[K]
+function y=f(T),y = A * T + B/2 * T^2
+endfunction
+x = 1 / T_c *(f(T2) - f(T1)) ;
+
+y = enth_dep_1 - x ;
+
+disp(" Example: 5.8 Page no : 241") ;
+disp(y)
+X = ( y - del_h_dep_l) / (del_h_dep_v - del_h_dep_l);
+printf("\n Quality = %.2f",X) ;
\ No newline at end of file diff --git a/527/CH6/EX6.1/6_1exam.sce b/527/CH6/EX6.1/6_1exam.sce new file mode 100755 index 000000000..6adb5b8e7 --- /dev/null +++ b/527/CH6/EX6.1/6_1exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 6.1:
+//Page no :257
+
+clear ; clc ;
+disp(" Example: 6.1 Page no : 257") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.")
\ No newline at end of file diff --git a/527/CH6/EX6.10/6_10exam.sce b/527/CH6/EX6.10/6_10exam.sce new file mode 100755 index 000000000..6ab118d82 --- /dev/null +++ b/527/CH6/EX6.10/6_10exam.sce @@ -0,0 +1,28 @@ +//Engineering and Chemical Thermodynamics
+//Example 6.10
+//Page no :283
+
+clear ;clc ;
+//Given
+h_H2SO4 = 1.596 ; //[kJ/mol]
+h_H2O = 1.591 ; //[kJ/mol]
+C1 = -74.40 ;
+C2 = 0.561 ;
+A = [0 ,0.1 , 0.2 ,0.3 ,0.4 ,0.5 ,0.6 ,0.7 ,0.8 ,0.9 ,1] ;
+B = [1 ,0.9 ,0.8 ,0.7 ,0.6 ,0.5 ,0.4 ,0.3 ,0.2 ,0.1 ,0] ;
+
+disp(" Example: 6.10 Page no : 283") ;
+for i = 1:11
+ H_bar_H2SO4 = h_H2SO4 + C1 * B(1,i)^2 - 2 * C2 * C1 * A(1,i) * B(1,i)^2 ;
+ H_bar_H2O = h_H2O + C1 * A(1,i)^2 -C2 * C1 * A(1,i)^2 * (1 - 2 * B(1,i)) ;
+ y_data_1(1,i) = H_bar_H2SO4 ;
+ y_data_2(1,i) = H_bar_H2O ;
+ x_data(1,i) = A(1,i) ;
+end
+plot(x_data,y_data_1) ;
+plot(x_data,y_data_2) ;
+
+m = y_data_1(1,6) ;
+s = y_data_2(1,6) ;
+xtitle("Figue E6.10","x_H2SO4"," Partial molar enthalpy");
+printf("\n For equimolar mixture del_H_H2SO4 = %.1f kJ/mol del_H_H2O = %.1f kJ/mol",m,s);
\ No newline at end of file diff --git a/527/CH6/EX6.11/6_11exam.sce b/527/CH6/EX6.11/6_11exam.sce new file mode 100755 index 000000000..04f4b188f --- /dev/null +++ b/527/CH6/EX6.11/6_11exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 6.11
+//Page no :283
+
+clear ; clc ;
+disp(" Example: 6.11 Page no : 283") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.")
\ No newline at end of file diff --git a/527/CH6/EX6.12/6_12exam.sce b/527/CH6/EX6.12/6_12exam.sce new file mode 100755 index 000000000..3debb5f96 --- /dev/null +++ b/527/CH6/EX6.12/6_12exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 6.12
+//Page no :287
+
+clear ; clc ;
+disp(" Example: 6.12 Page no : 287") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.")
\ No newline at end of file diff --git a/527/CH6/EX6.13/6_13exam.sce b/527/CH6/EX6.13/6_13exam.sce new file mode 100755 index 000000000..2cc7c42d3 --- /dev/null +++ b/527/CH6/EX6.13/6_13exam.sce @@ -0,0 +1,36 @@ +//Engineering and Chemical Thermodynamics
+//Example 6.13
+//Page no :287
+
+clear ;clc ;
+//Given
+C1 = 1.596 ;
+C2 = 1.591 ;
+C3 = -74.40 ;
+C4 = -0.561 ;
+A = [ 0 ,0.1 ,0.2 ,0.3 ,0.4 ,0.5 ,0.6 ,0.7 ,0.8 ,0.9 ,1] ;
+m = (-C1 + C2 + C3 * ( C4 * 0.25)) * 1000 ;
+
+disp(" Example: 6.13 Page no : 287") ;
+for i = 1:11
+ x_H2O = A(1,i) ;
+ x_H2SO4 = 1- x_H2O ;
+ h = C1 * x_H2SO4 + C2 * x_H2O + C3 * x_H2SO4 * x_H2O *(1 + C4 * x_H2SO4) ;
+ C(1,i) = h * 10^3;
+end
+y1 = C(1,6) ;
+
+function y = f613(x) ,
+ y = -m * (x - 0.5 ) + y1 ;
+endfunction
+
+for i = 1:11
+ F(1,i) = f613(A(1,i)) ;
+end
+
+plot(A,C);
+plot(A,F)
+xtitle("Figure E6.13","x_H2O","h (J/mol)");
+
+printf("\n H_bar_H2SO4 = %d J/mol H_bar_H2O = %d J/mol\n ",F(1,1),F(1,11)) ;
+disp(" The partial molar property can be obtained by drawing tangent at mole fraction 0.5 .")
\ No newline at end of file diff --git a/527/CH6/EX6.14/6_14exam.sce b/527/CH6/EX6.14/6_14exam.sce new file mode 100755 index 000000000..24e6cf802 --- /dev/null +++ b/527/CH6/EX6.14/6_14exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 6.14:
+//Page 291
+
+clear ; clc ;
+disp(" Example: 6.14 Page no : 291") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.")
\ No newline at end of file diff --git a/527/CH6/EX6.2/6_2exam.sce b/527/CH6/EX6.2/6_2exam.sce new file mode 100755 index 000000000..6c9273bb0 --- /dev/null +++ b/527/CH6/EX6.2/6_2exam.sce @@ -0,0 +1,12 @@ +//Engineering and Chemical Thermodynamics
+//Example 6.2
+//Page no :261
+
+clear ; clc ;
+//Given
+slop = -4222.1 ;
+R = 8.314 ;
+del_h_vap = -R * slop * 10^-3 ;
+
+disp(" Example: 6.2 Page no : 261") ;
+printf("\n Enthalpy of vapourisation of Ga(CH3)3 = %.1f kJ/mol",del_h_vap) ;
\ No newline at end of file diff --git a/527/CH6/EX6.3/6_3exam.sce b/527/CH6/EX6.3/6_3exam.sce new file mode 100755 index 000000000..f590fe2c7 --- /dev/null +++ b/527/CH6/EX6.3/6_3exam.sce @@ -0,0 +1,15 @@ +//Engineering and Chemical Thermodynamics
+//Example 6.3
+//Page no :261
+
+clear ; clc ;
+//The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.
+
+
+disp(" Example: 6.3 Page no : 261") ;
+function y=f(x) , y = -4222.1 * x + 17.556
+endfunction
+xdata = linspace(0.0032,0.004,8) ;
+ydata = f(xdata) ;
+plot(xdata,ydata) ;
+xtitle("Figure E6.2","1/T","ln P_sat (kPa)")
\ No newline at end of file diff --git a/527/CH6/EX6.4/6_4exam.sce b/527/CH6/EX6.4/6_4exam.sce new file mode 100755 index 000000000..7b84cfaab --- /dev/null +++ b/527/CH6/EX6.4/6_4exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 6.4
+//Page no :268
+
+clear ; clc ;
+disp(" Example: 6.4 Page no : 268") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.")
\ No newline at end of file diff --git a/527/CH6/EX6.5/6_5exam.sce b/527/CH6/EX6.5/6_5exam.sce new file mode 100755 index 000000000..ff439bfad --- /dev/null +++ b/527/CH6/EX6.5/6_5exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 6.5
+//Page no :271
+
+clear ; clc ;
+disp(" Example 6.5 Page no:271")
+disp(" There is no numerical part involved in this problem . Users can refer Figure 6.5.")
\ No newline at end of file diff --git a/527/CH6/EX6.6/6_6exam.sce b/527/CH6/EX6.6/6_6exam.sce new file mode 100755 index 000000000..cbc8d2675 --- /dev/null +++ b/527/CH6/EX6.6/6_6exam.sce @@ -0,0 +1,28 @@ +//Engineering and Chemical Thermodynamics
+//Example 6.6
+//Page no :277
+
+clear ; clc ;
+//Given
+MW1 = 119.5 ;
+MW2 = 58 ;
+A = [0,4.77,9.83,14.31,19.38,23.27,25.53,25.07,21.55,13.56,0] ;
+B = [0,.1 , .2 , .3 ,.4 ,.5 ,.6 ,.7 ,.8 ,.9,1] ;
+
+disp(" Example: 6.6 Page no : 277") ;
+for i = 1:11
+
+ x1 = (B(1,i) / MW1) / (B(1,i) / MW1 + (1 - B(1,i)) / MW2) ;
+ x2 = 1 - x1 ;
+ MW = x1 * MW1 + x2 * MW2 ;
+ del_h_mix = - 1*(A(1,i)) * MW ;
+ C(1,i) = del_h_mix ;
+ D(1,i) = x1 ;
+
+ printf("\n For weight percent %.3f del_h_mix = %.1f J/mol\n",x1,del_h_mix)
+end
+
+xdata = D ;
+ydata = C ;
+plot(xdata ,ydata) ;
+xtitle("Figure E6.6B","x_CHCl3","Del_h_mix (J/mol)") ;
diff --git a/527/CH6/EX6.7/6_7exam.sce b/527/CH6/EX6.7/6_7exam.sce new file mode 100755 index 000000000..b3f64457e --- /dev/null +++ b/527/CH6/EX6.7/6_7exam.sce @@ -0,0 +1,16 @@ +//Engineering and Chemical Thermodynamics
+//Example 6.7
+//Page no :279
+
+clear ; clc ;
+//Given
+A = [-32669,-31840,-28727,-26978,-24301,-20083,-13113] ;
+B = [20 ,10 ,5 ,4 ,3 ,2 ,1] ;
+
+disp(" Example: 6.7 Page no : 279") ;
+for i = 1:7
+ del_h_mix = A(1,i) / (1 + B(1,i)) ;
+ C(1,i) = del_h_mix ;
+ D(1,i) = 1 / (1 + B(1,i)) ;
+ printf("\n For mole fraction %.3f the entropy of mixing is %d J/mol\n",D(1,i),C(1,i)) ;
+end
\ No newline at end of file diff --git a/527/CH6/EX6.8/6_8exam.sce b/527/CH6/EX6.8/6_8exam.sce new file mode 100755 index 000000000..486e5702f --- /dev/null +++ b/527/CH6/EX6.8/6_8exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 6.8
+//Page no :280
+
+clear ; clc ;
+disp(" Example: 6.8 Page no : 280") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.")
\ No newline at end of file diff --git a/527/CH6/EX6.9/6_9exam.sce b/527/CH6/EX6.9/6_9exam.sce new file mode 100755 index 000000000..9d31d6421 --- /dev/null +++ b/527/CH6/EX6.9/6_9exam.sce @@ -0,0 +1,22 @@ +//Engineering and Chemical Thermodynamics
+//Example 6.9
+//Page no :282
+
+clear ; clc ;
+//Given
+x1 = 0.1 ;
+x2 = 1- x1 ;
+B11 = -910 ;
+B22 = -1330 ;
+B12 = -2005 ;
+T = 333 ; //[K]
+P = 10 * 10^5 ;
+R = 8.314 ;
+v1 = R * T /P * 10^6 + B11 ;//....E6.9A
+
+disp(" Example: 6.9 Page no : 282") ;
+printf("\n v1 = %g cm^3/mol\n",v1)
+V_bar_1 = (R * T / P) * 10^6+ (x1^2 + 2 * x1 * x2) * B11 + 2 * x2^2 * B12 - x2^2 * B22 ;//.....E6.9B
+printf("\n V_bar_1 = %g cm^3/mol\n",V_bar_1) ;
+del_v_mix = x1 * x2 * (2 * B12 - B11 - B22) ;//.....E6.9C
+printf("\n del_v = %g cm^3/mol",del_v_mix);
\ No newline at end of file diff --git a/527/CH7/EX7.1/7_1exam.sce b/527/CH7/EX7.1/7_1exam.sce new file mode 100755 index 000000000..c0395d198 --- /dev/null +++ b/527/CH7/EX7.1/7_1exam.sce @@ -0,0 +1,25 @@ +//Engineering and Chemical Thermodynamics
+//Example 7.1
+//Page no :308
+
+clear ; clc ;
+//Given
+h_cap_H2O = 2676.0 ; //[kJ/kg],From steam table
+S_cap_H2O = 7.3548 ; //[kJ/kgK],From steam table
+h_cap_0_H2O = 2687.5 ; //[kJ/kg],From Appendix B
+S_cap_0_H2O = 8.4479 ; //[kJ/kgK],From Appendix B
+P_0_H2O = 10 ;//[kPa]
+T = 373.15 ;//[K]
+R = 8.314 / 18 ;
+P_sys = 101.35 ;//[kPa]
+
+g_cap_H2O = h_cap_H2O - T * S_cap_H2O ;
+g_cap_0_H2O = h_cap_0_H2O - T * S_cap_0_H2O ;
+
+f_H2O = P_0_H2O * exp((g_cap_H2O - g_cap_0_H2O ) / (R * T)) ;
+
+Sai_H2O = f_H2O / P_sys ;
+
+disp(" Example: 7.1 Page no : 308") ;
+
+printf("\n The fugacity = %.2f kPa \n\n The fugacity coefficient = %.3f",f_H2O ,Sai_H2O);
\ No newline at end of file diff --git a/527/CH7/EX7.10/7_10exam.sce b/527/CH7/EX7.10/7_10exam.sce new file mode 100755 index 000000000..cbe18ba62 --- /dev/null +++ b/527/CH7/EX7.10/7_10exam.sce @@ -0,0 +1,16 @@ +//Engineering and Chemical Thermodynamics
+//Example 7.10
+//Page no :343
+
+clear ; clc ;
+//Given
+gama_a_inf = 0.88 ;
+gama_b_inf = 0.86 ;
+R = 8.314 ;
+T = 39.33 + 273 ;
+
+A_1 = R * T * log(gama_a_inf) ;
+A_2 = R * T * log(gama_b_inf) ;
+A = (A_1 + A_2) / 2 ;
+disp(" Example: 7.10 Page no : 343") ;
+printf("\n The average value of two-suffix Margules parameter A = %g J/mol",A);
\ No newline at end of file diff --git a/527/CH7/EX7.11/7_11exam.sce b/527/CH7/EX7.11/7_11exam.sce new file mode 100755 index 000000000..07d0a7688 --- /dev/null +++ b/527/CH7/EX7.11/7_11exam.sce @@ -0,0 +1,8 @@ +//Engineering and Chemical Thermodynamics
+//Example 7.11
+//Page no :343
+
+
+clear ; clc ;
+disp(" Example: 7.11 Page no : 343") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.") ;
\ No newline at end of file diff --git a/527/CH7/EX7.12/7_12exam.sce b/527/CH7/EX7.12/7_12exam.sce new file mode 100755 index 000000000..e41c0d520 --- /dev/null +++ b/527/CH7/EX7.12/7_12exam.sce @@ -0,0 +1,19 @@ +//Engineering and Chemical Thermodynamics
+//Example 7.12
+//Page no :352
+
+clear ; clc ;
+//Given
+A_T1 = 1401 ; //[J/mol]
+T1 = 10 + 273 ;//[K]
+T2 = 60 + 273 ; //[K]
+C = 3250 ;
+A_T2_prev = 1143 ;//[J/mol]
+
+A_T2 = T2 * (C *(1/T2 - 1/T1) + A_T1 / T1);
+disp(" Example: 7.12 Page no : 352") ;
+printf("\n Value of A at 60*C = %f J/mol\n\n",A_T2) ;
+x = (A_T2_prev - A_T2) / A_T2_prev* 100 ;
+printf("\n The values differ by = %g %%",x)
+
+// The results given in the text book are wrong .
\ No newline at end of file diff --git a/527/CH7/EX7.2/7_2exam.sce b/527/CH7/EX7.2/7_2exam.sce new file mode 100755 index 000000000..84dc700c2 --- /dev/null +++ b/527/CH7/EX7.2/7_2exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 7.2
+//Page no :309
+
+clear ; clc ;
+disp(" Example: 7.2 Page no : 309") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.")
\ No newline at end of file diff --git a/527/CH7/EX7.3/7_3exam.sce b/527/CH7/EX7.3/7_3exam.sce new file mode 100755 index 000000000..23124374c --- /dev/null +++ b/527/CH7/EX7.3/7_3exam.sce @@ -0,0 +1,21 @@ +//Engineering and Chemical Thermodynamics
+//Example 7.3
+//Page no :311
+
+clear ; clc ;
+//Given
+P = 50 ; //[bar]
+T = 25 + 273.2 ;//[K]
+P_c = 48.7 ; //[bar] , From Appendix A.1 Table C.7 & C.8
+T_c = 303.5 ; //[K] , From Appendix A.1 Table C.7 & C.8
+w = 0.099 ; // From Appendix A.1 Table C.7 & C.8
+log_w_0 = -0.216 ;// By interpolation
+log_w_1 = -0.060 ;// By interpolation
+
+X = log_w_0 + w * log_w_1 ;
+sai_eth = 10^(X) ;
+f_eth = sai_eth * P ;
+
+disp(" Example: 7.3 Page no : 311") ;
+printf("\n Fugacity = %g bar",f_eth);
+
\ No newline at end of file diff --git a/527/CH7/EX7.4/7_4exam.sce b/527/CH7/EX7.4/7_4exam.sce new file mode 100755 index 000000000..e07643720 --- /dev/null +++ b/527/CH7/EX7.4/7_4exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 7.4
+//Page no :316
+
+clear ; clc ;
+disp(" Example: 7.4 Page no : 316") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.") ;
\ No newline at end of file diff --git a/527/CH7/EX7.5/7_5exam.sce b/527/CH7/EX7.5/7_5exam.sce new file mode 100755 index 000000000..78e90d492 --- /dev/null +++ b/527/CH7/EX7.5/7_5exam.sce @@ -0,0 +1,31 @@ +//Engineering and Chemical Thermodynamics
+//Example 7.5
+//Page no :319
+
+clear ; clc ;
+//Given
+P = 50 ; // [bar]
+T = 25 + 273.2 ; //[K]
+y_eth = 0.2 ;
+y_pro = 0.8 ;
+T_c_eth = 305.5 ; //[K],From Appendix A.1
+T_c_pro = 370 ; //[K],From Appendix A.1
+P_c_eth = 48.7 ; //[bar],From Appendix A.1
+P_c_pro = 42.4 ; // [bar],From Appendix A.1
+w_eth = 0.099 ;//From Appendix A.1
+w_pro = 0.153 ;//From Appendix A.1
+log_w_0 = -0.579 ;// By double liner interpolation
+log_w_1 = -0.406 ;// By double liner interpolation
+T_pc = y_eth * T_c_eth + y_pro * T_c_pro ;
+P_pc = y_eth * P_c_eth + y_pro * P_c_pro ;
+w_mix = y_eth * w_eth + y_pro * w_pro ;
+
+Pr = P / P_pc ;
+Tr = T /T_pc ;
+X = log_w_0 + w_mix * log_w_1 ;
+
+sai = 10^(X) ;
+f = sai * P ;
+
+disp(" Example: 7.5 Page no : 319") ;
+printf("\n Fugacity co-efficient = %.2f\n\n Fugacity = %.1f bar",sai,f);
\ No newline at end of file diff --git a/527/CH7/EX7.6/7_6exam.sce b/527/CH7/EX7.6/7_6exam.sce new file mode 100755 index 000000000..13ee82260 --- /dev/null +++ b/527/CH7/EX7.6/7_6exam.sce @@ -0,0 +1,9 @@ +//Engineering and Chemical Thermodynamics
+//Example 7.6
+//Page no :324
+
+clear ; clc ;
+
+// The problem does not contain any numerical part . The readers can refer the text book to get the answer .
+disp(" Example: 7.6 Page no : 324") ;
+disp(" Like interactions are stronger than unlike interaction .")
\ No newline at end of file diff --git a/527/CH7/EX7.7/7_7exam.sce b/527/CH7/EX7.7/7_7exam.sce new file mode 100755 index 000000000..24814edaa --- /dev/null +++ b/527/CH7/EX7.7/7_7exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 7.7
+//Page no :331
+
+clear ; clc ;
+disp(" Example: 7.7 Page no : 331") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.") ;
\ No newline at end of file diff --git a/527/CH7/EX7.8/7_8eaxm.sce b/527/CH7/EX7.8/7_8eaxm.sce new file mode 100755 index 000000000..efb317ebf --- /dev/null +++ b/527/CH7/EX7.8/7_8eaxm.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 7.8
+//Page no :338
+
+clear ; clc ;
+disp(" Example: 7.8 Page no : 338") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.") ;
\ No newline at end of file diff --git a/527/CH7/EX7.9/7_9exam.sce b/527/CH7/EX7.9/7_9exam.sce new file mode 100755 index 000000000..382272365 --- /dev/null +++ b/527/CH7/EX7.9/7_9exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 7.9
+//Page no :339
+
+clear ; clc ;
+disp(" Example: 7.9 Page no : 339") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression.") ;
\ No newline at end of file diff --git a/527/CH8/EX8.1/8_1exam.sce b/527/CH8/EX8.1/8_1exam.sce new file mode 100755 index 000000000..43a95902c --- /dev/null +++ b/527/CH8/EX8.1/8_1exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 8.1
+//Page 369
+
+clear ; clc ;
+disp(" Example: 8.1 Page no : 369") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression .") ;
\ No newline at end of file diff --git a/527/CH8/EX8.10/8_10exam.sce b/527/CH8/EX8.10/8_10exam.sce new file mode 100755 index 000000000..0c4ad6cc0 --- /dev/null +++ b/527/CH8/EX8.10/8_10exam.sce @@ -0,0 +1,66 @@ +//Engineering and Chemical Thermodynamics
+//Example 8.10
+//Page No:390
+
+clear ; clc ;
+//Given
+R = 8.314 ;
+T = 10 + 273.15 ; //[K]
+A_B = 9.2806 ; // From Appendix A , Table A1.1
+B_B = 2788.5 ; // From Appendix A , Table A1.1
+C_B = -52.36 ; // From Appendix A , Table A1.1
+A_C = 9.1325 ; // From Appendix A , Table A1.1
+B_C = 2766.63 ; // From Appendix A , Table A1.1
+C_C = -50.50 ; // From Appendix A , Table A1.1
+
+x1 = [0,0.0610 ,0.2149 ,0.3187 ,0.4320 ,0.5246 ,0.6117 ,0.7265 ,0.8040 ,0.8830 ,0.8999 ,1] ; //From table E8.9A
+P_exp = [6344 ,6590 ,6980 ,7140 ,7171 ,7216 ,7140 ,6974 ,6845 ,6617 ,6557 ,6073] ; //From table E8.9A
+
+P_1_sat = 6073 ; //[Pa]
+P_2_sat = 6344 ; //[Pa]
+A = 1390:1410 ;
+B = 60:80 ;
+w = 1 / (R * T) ;
+for k = 1:21
+ y = A(k) ;
+ for i = 1:21
+ z = B(i) ;
+ for j = 1:12
+ P(1,j) = x1(1,j) * exp((y + 3 * z) * (1 - (x1(1,j)))^2 *w-4*z*(1-x1(1,j))^3* w )* P_1_sat + (1-x1(1,j))*exp((y -3*z)*(x1(1,j))^2 * w + 4 * z * (x1(1,j)^3) * w )*P_2_sat ;
+ R(1,j) =(P(1,j) - P_exp(1,j))^2 ;
+ end
+
+ m = 0 ;
+ for l = 1:12
+ m = m + R(1,l) ;
+ end
+ S(k,i) = m ;
+ end
+end
+for i = 1:21
+ k = S(i,1) ;
+ for l = 2:21
+ if S(i,l) < k then
+ k = S(i,l) ;
+ end
+ end
+ D(1,i) = k ;
+end
+
+a = D(1,1) ;
+for i = 2:21
+ if D(1,i) < a then
+ a = D(1,i) ;
+ end
+end
+disp(" Example: 8.10 Page no : 390") ;
+for i = 1:21
+ if D(1,i) == a then
+ for l = 1:21
+ if S(i,l) == a then
+ printf("\n A = %g J/mol",A(1,i));
+ printf("\n B = %g J/mol",B(1,l)) ;
+ end
+ end
+ end
+end
diff --git a/527/CH8/EX8.11/8_11exam.sce b/527/CH8/EX8.11/8_11exam.sce new file mode 100755 index 000000000..2dbea636d --- /dev/null +++ b/527/CH8/EX8.11/8_11exam.sce @@ -0,0 +1,45 @@ +//Engineering and Chemical Thermodynamics
+//Example 8.11
+//Page No:390
+
+clear ; clc ;
+//Given
+R = 8.314 ;
+T = 10 + 273.15 ; //[K]
+x1 = [0 ,0.0610 ,0.2149 ,0.3187 ,0.4320 ,0.5246 ,0.6117 ,0.7265 ,0.8040 ,0.8830 ,0.8999 ,1] ; //From table E8.9A
+P_exp = [6344 ,6590 ,6980 ,7140 ,7171 ,7216 ,7140 ,6974 ,6845 ,6617 ,6557 ,6073 ,6073] ; //From table E8.9A
+y1 = [ 1 ,0.0953 ,0.2710 ,0.3600 ,0.4453,0.5106 ,0.5735 ,0.6626 ,0.7312 ,0.8200 ,0.8382, 0 ] ;//From table E8.9A
+P_1_sat = 6073 ; //[Pa]
+P_2_sat = 6344 ; //[Pa]
+
+n = 0 ;
+for i = 2:11
+ x2(1,i) = 1 - x1(1,i) ;
+ y2(1,i) = 1 - y1(1,i) ;
+ g_E(1,i) = R * T *( x1(1,i) * log (( y1(1,i) * P_exp(1,i)) / (x1(1,i)* P_1_sat)) + x2(1,i) * log((y2(1,i) * P_exp(1,i)) / (x2(1,i) * P_2_sat)) ) ;
+ n = n + g_E(1,i) / ((x1(1,i) * x2(1,i)) * 10) ;
+ ydata(1,i-1) = (g_E(1,i)/(x1(1,i)*x2(1,i)));
+ xdata(1,i-1) = x1(1,i) - x2(1,i) ;
+end
+m= 0 ; n=0 ; o = 0 ; p= 0 ;N = 10 ;
+for i = 2:11
+ m = m + g_E(1,i) * (2 * x1(1,i) - 1) / ( x1(1,i) * x2(1,i)) ;
+ n = n + g_E(1,i) / ( x1(1,i) * x2(1,i)) ;
+ o = o + (2 * x1(1,i) - 1) ;
+ p = p + (2 * x1(1,i) - 1)^2 ;
+end
+x_bar = o / N ;
+y_bar = n / N ;
+a1 = (N * m - n * o)/(N * p - o^2) ;
+a0 = y_bar - a1 * x_bar ;
+
+for i = 1:10
+ ydata2(1,i) = a0 + a1*xdata(1,i) ;
+end
+plot(xdata,ydata,"+") ;
+plot(xdata,ydata2) ;
+xtitle("Figure E8.11","x1-x2","g_E/x1*x2") ;
+disp(" Example: 8.11 Page no : 390") ;
+printf("\n From average , the value of A = %d J/mol\n",n/10) ;
+printf("\n From linear regression best fit line the values of A and B are %.1f J/mol & %.1f J/mol respectively .",a0 , a1) ;
+//Readers can refer figure E8.11 .
\ No newline at end of file diff --git a/527/CH8/EX8.12/8_12exam.sce b/527/CH8/EX8.12/8_12exam.sce new file mode 100755 index 000000000..f2a0851d7 --- /dev/null +++ b/527/CH8/EX8.12/8_12exam.sce @@ -0,0 +1,15 @@ +//Engineering and Chemical Thermodynamics
+//Example 8.12
+//Page no :395
+
+clear ; clc ;
+//Given
+H_O2 = 44253.9 ;//[bar] , From table 8.1
+p_O2 = 0.21 ; //[bar]
+
+x_O2 = p_O2 / H_O2 ;
+v_H2O = 1/(1/0.001 * 1/0.018 * 0.001 );
+_O2_ = x_O2 / v_H2O ; //[M]
+disp(" Example: 8.12 Page no : 395") ;
+printf("\n Mole fraction of O2 = %g",x_O2 ) ;
+printf("\n Concentration of O2 = %g M ",_O2_) ;
diff --git a/527/CH8/EX8.13/8_13exam.sce b/527/CH8/EX8.13/8_13exam.sce new file mode 100755 index 000000000..bb7443fb6 --- /dev/null +++ b/527/CH8/EX8.13/8_13exam.sce @@ -0,0 +1,25 @@ +//Engineering and Chemical Thermodynamics
+//Example 8.13
+//Page no :396
+
+clear ; clc ;
+//Given
+P = 300 ; //[bar]
+V_bar_inf_N2 = 3.3 * 10^-5 ;
+R = 8.314 ;
+T = 298 ; //[K]
+y_N2 = 1 ; // At 25*C vapour pressure of water is small
+H_N2_1 = 87365 ; //[bar]
+P_c = 33.8 ; //[bar]
+T_c = 126.2 ;// [K]
+w = 0.039 ; // From Appendix A.1
+log_w_0 = 0.013 ;
+log_w_1 = 0.210 ;
+H_N2_300 = H_N2_1 * exp((V_bar_inf_N2 * (P -1) * 10^5 )/ (R * T)) ;
+
+k = log_w_0 + w * log_w_1 ;
+sai_N2 = 10^k ;
+x_N2 = y_N2 * sai_N2 * P / H_N2_300 ;
+
+disp(" Example: 8.13 Page no : 396")
+printf("\n Solubility of N2 in water = %.5f",x_N2) ;
\ No newline at end of file diff --git a/527/CH8/EX8.14/8_14exam.sce b/527/CH8/EX8.14/8_14exam.sce new file mode 100755 index 000000000..4a1db0861 --- /dev/null +++ b/527/CH8/EX8.14/8_14exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 8.14
+//Page no :400
+
+clear ; clc;
+disp(" Example: 8.14 Page no : 400") ;
+disp(" The problem does not contain any numerical calculation . The readers can go through the text book to get the required answer .")
\ No newline at end of file diff --git a/527/CH8/EX8.15/8_15exam.sce b/527/CH8/EX8.15/8_15exam.sce new file mode 100755 index 000000000..85c6dfb8b --- /dev/null +++ b/527/CH8/EX8.15/8_15exam.sce @@ -0,0 +1,47 @@ +//Engineering and Chemical Thermodynamics
+//Example 8.15
+//Page no :402
+
+clear ; clc;
+//Given
+R = 8.314 ;
+T = 20 + 273 ;//[K]
+A = 6000 ; //[J/mol]
+B = -384 ; //[J/mol]
+x_a = [0.001 ,0.03 ,0.05 ,0.06 ,0.075 ,0.1 ,0.12 , 0.13 ,0.15 ,0.2 ,0.25 ,0.3 ,0.35 ,0.4 ,0.45,0.475 ,0.5 ,0.55 ,0.6 ,0.65 ,0.7 ,0.75 ,0.8 ,0.8475 ,0.85 ,0.9 ,0.925 ,0.95 ,0.975 ,0.999] ;
+
+for i = 1:30
+ y_data(1,i) = R * T * ( x_a(1,i) * log(x_a(1,i)) + (1 - x_a(1,i)) * log(1- x_a(1,i))) + x_a(1,i) * (1 - x_a(1,i)) * (A + B * (2*x_a(1,i) - 1 )) ;
+ y_data2(1,i) = - 82 * x_a(1,i)- 185.6 ;
+end
+
+m = min(y_data) ;
+for i = 1:30
+ if y_data(1,i) == m then
+ a = x_a(1,i) ;
+ end
+end
+
+
+for i = 1: 30
+ y_data2(1,i) = -(R * T *( log(a) - log(1 - a)) + A * (1 - 2*a) + B * (6 * a - 1 - 6 * a^2)) * (x_a(1,i) - a) + m ;
+end
+
+for i = 1:20
+ y_data3(1,i) = y_data(1,i) - y_data2(1,i) ;
+end
+n = min(y_data3) ;
+
+for i = 1:20
+ if y_data3(1,i) == n then
+ b = x_a(1,i) ;
+ end
+end
+
+
+disp(" Example: 8.15 Page no : 402") ;
+plot(x_a ,y_data) ;
+plot(x_a ,y_data2) ;
+xtitle(" Figure E8.15","x_a","g - x_a * g_a - x_b * g_b") ;
+
+printf("\n\n The equilibrium composition can be found by drawing a line tangent to the minima .\n\n In this case the answer is %.2f and %.1f ." , a ,b)
\ No newline at end of file diff --git a/527/CH8/EX8.16/8_16exam.sce b/527/CH8/EX8.16/8_16exam.sce new file mode 100755 index 000000000..697bc3782 --- /dev/null +++ b/527/CH8/EX8.16/8_16exam.sce @@ -0,0 +1,19 @@ +//Engineering and Chemical Thermodynamics
+//Example 8.16
+//Page no :403
+
+clear ; clc;
+//Given
+A = 6349 ; //[J/mol]
+B = -384 ; //[J/mol]
+R = 8.314 ;
+T = 20 + 273 ; //[K]
+
+k = 0.000001 ;
+disp(" Example: 8.16 Page no : 403") ;
+function y816 = f816(x_a) , y816 = R * T * (1/x_a + 1/(1 - x_a)) - 2 * A +6 * B * (1 - 2 * x_a) + k
+endfunction
+ans1 = fsolve([0.1],f816) ;
+ans2 = fsolve([0.5],f816) ;
+
+printf("\n %.3f < x_a < %.3f ",ans1,ans2)
diff --git a/527/CH8/EX8.17/8_17exam.sce b/527/CH8/EX8.17/8_17exam.sce new file mode 100755 index 000000000..dbfe93906 --- /dev/null +++ b/527/CH8/EX8.17/8_17exam.sce @@ -0,0 +1,27 @@ +//Engineering and Chemical Thermodynamics
+//Example 8.17
+//Page no :406
+clear ; clc;
+//Given
+
+T = 300 ; //[K]
+A = 6235 ; //[J/mol]
+P_a_sat = 100 * 10^3 ; //[Pa]
+P_b_sat = 50 * 10^3 ; //{Pa}
+R = 8.314 ;
+w = 1/(R * T) ;
+function Z817 = f817(R)
+ x_a_a = R(1) ;
+ x_a_b = R(2) ;
+ Z817(1) = x_a_b * exp(A * (1 - x_a_b) ^ 2 * w) - x_a_a * exp(A * (1 - x_a_a) ^ 2 * w) ; // E8.17A
+ Z817(2) = (1 - x_a_b) * exp(A * ( x_a_b) ^ 2 * w) - (1 - x_a_a) * exp(A * (x_a_a) ^ 2 * w ) ; // E8.17B
+endfunction
+x0 = [0.75 ; 0.1] ;
+[z,fxs,m] = fsolve(x0,f817) ;
+disp(" Example: 8.17 Page no : 406") ;
+printf("\n The compositions are : x_a_a = %.3f and x_a_b = %.3f",z(1,1), z(2,1)) ;
+
+P = z(1,1) * exp(A * z(2,1) ^ 2 * w) * P_a_sat + z(2,1) * exp(A * z(1,1) ^ 2 * w) * P_b_sat ;
+printf("\n Total pressure = %d kPa",P * 10^-3) ;
+y_a = z(1,1) * exp(A * z(2,1) ^ 2 * w) * P_a_sat / P ;
+printf("\n y_a = %.3f" , y_a ) ;
\ No newline at end of file diff --git a/527/CH8/EX8.18/8_18exam.sce b/527/CH8/EX8.18/8_18exam.sce new file mode 100755 index 000000000..7d7d4d281 --- /dev/null +++ b/527/CH8/EX8.18/8_18exam.sce @@ -0,0 +1,20 @@ +//Engineering and Chemical Thermodynamics
+//Example 8.18
+//Page no :418
+
+clear ; clc ;
+//Given
+T_b = 373.15 ; //[K]
+del_h_vap = 2257 ; //[J/g]
+MW_salt = 58.5 ; //[g/mol]
+MW_water = 18 ; //[g/mol]
+w_salt = 3.5 ;
+w_water = 100 - w_salt ;
+R = 8.314 ;
+
+x_salt = (w_salt / MW_salt) / (w_salt / MW_salt + w_water / MW_water) ;
+x_b = 2 * x_salt ;// We assume NaCl completely dissociates into Na+ & Cl- ions
+
+del_T = R * T_b^2 / (del_h_vap * MW_water)* x_b ;
+disp(" Example: 8.18 Page no : 418")
+printf("\n The temperature that sea water boils is = %.2f degreeC",100 + del_T);
\ No newline at end of file diff --git a/527/CH8/EX8.19/8_19exam.sce b/527/CH8/EX8.19/8_19exam.sce new file mode 100755 index 000000000..b2abe96e0 --- /dev/null +++ b/527/CH8/EX8.19/8_19exam.sce @@ -0,0 +1,20 @@ +//Engineering and Chemical Thermodynamics
+//Example 8.19
+//Page no :418
+
+clear ; clc ;
+//Given
+rho_w = 1000 ; // [kg/m^3]
+g = 9.8 ; // [m/s^2]
+h = 0.0071 ;//[m]
+m_b = 1.93 * 10^-3 ; // [kg]
+V = 520 * 10^-6 ; //[m^3]
+R = 8.314 ;
+T = 298 ;
+
+PI = rho_w * g * h ;
+C_b = m_b / V ;
+MW_b = R * T * C_b / PI ;
+
+disp(" Example: 8.19 Page no : 418")
+printf("\n The molecular weight of the protein = %d kg/mol", MW_b );
diff --git a/527/CH8/EX8.2/8_2exam.sce b/527/CH8/EX8.2/8_2exam.sce new file mode 100755 index 000000000..12cba0773 --- /dev/null +++ b/527/CH8/EX8.2/8_2exam.sce @@ -0,0 +1,29 @@ +//Engineering and Chemical Thermodynamics
+//Example 8.2
+//Page no :369
+
+clear ; clc ;
+//Given
+A_C5H12 = 9.2131 ; //From table E8.2A
+B_C5H12 = 2477.07 ; //From table E8.2A
+C_C5H12 = -39.94 ; //From table E8.2A
+A_C6H12 = 9.1325 ; //From table E8.2A
+B_C6H12 = 2766.63 ; //From table E8.2A
+C_C6H12 = -50.50 ; //From table E8.2A
+A_C6H14 = 9.2164 ; //From table E8.2A
+B_C6H14 = 2697.55 ; //From table E8.2A
+C_C6H14 = -48.78 ; //From table E8.2A
+A_C7H16 = 9.2535 ; //From table E8.2A
+B_C7H16 = 2911.32 ; //From table E8.2A
+C_C7H16 = -56.51 ; //From table E8.2A
+
+x_C5H12 = 0.3 ;
+x_C6H12 = 0.3 ;
+x_C6H14 = 0.2 ;
+x_C7H16 = 0.2 ;
+
+function y82 = f82(T), y82 = -1 + (x_C5H12 * exp(A_C5H12 - B_C5H12 / (T + C_C5H12)) + x_C6H12 * exp(A_C6H12 - B_C6H12 / (T + C_C6H12)) + x_C6H14 * exp(A_C6H14 - B_C6H14 / (T + C_C6H14)) + x_C5H12 * exp(A_C5H12 - B_C5H12 / (T + C_C5H12)) + x_C7H16 * exp(A_C7H16 - B_C7H16 / (T + C_C7H16)));
+endfunction ;
+y =fsolve([300],f82) ;
+disp(" Example: 8.2 Page no : 369") ;
+printf("\n The temperature at which the liquid develops the first bubble of vapour = %d K",y);
\ No newline at end of file diff --git a/527/CH8/EX8.3/8_3exam.sce b/527/CH8/EX8.3/8_3exam.sce new file mode 100755 index 000000000..fe13850aa --- /dev/null +++ b/527/CH8/EX8.3/8_3exam.sce @@ -0,0 +1,43 @@ +//Engineering and Chemical Thermodynamics
+//Example 8.3
+//Page no :370
+
+clear ; clc ;
+//Given
+A_C5H12 = 9.2131 ; //From table E8.2A
+B_C5H12 = 2477.07 ; //From table E8.2A
+C_C5H12 = -39.94 ; //From table E8.2A
+A_C6H12 = 9.1325 ; //From table E8.2A
+B_C6H12 = 2766.63 ; //From table E8.2A
+C_C6H12 = -50.50 ; //From table E8.2A
+A_C6H14 = 9.2164 ; //From table E8.2A
+B_C6H14 = 2697.55 ; //From table E8.2A
+C_C6H14 = -48.78 ; //From table E8.2A
+A_C7H16 = 9.2535 ; //From table E8.2A
+B_C7H16 = 2911.32 ; //From table E8.2A
+C_C7H16 = -56.51 ; //From table E8.2A
+
+y_C5H12 = 0.3 ;
+y_C6H12 = 0.3 ;
+y_C6H14 = 0.2 ;
+y_C7H16 = 0.2 ;
+P = 1 ; //[bar]
+
+function y83 = f83(T), y83 = -1 + P * ( y_C5H12 / exp(A_C5H12 - B_C5H12 / (T + C_C5H12)) + y_C6H12 / exp(A_C6H12 - B_C6H12 / (T + C_C6H12)) + y_C6H14 / exp(A_C6H14 - B_C6H14 / (T + C_C6H14)) + y_C7H16 / exp(A_C7H16 - B_C7H16 / (T + C_C7H16)));
+endfunction ;
+y =fsolve([300],f83) ;
+disp(" Example: 8.3 Page no : 370") ;
+printf("\n\n The temperature at which vapour develops the first drop of liquid = %.2f K",y) ;
+
+T = y ;
+P_sat_C5H12 = exp(A_C5H12 - B_C5H12 / (T + C_C5H12)) ;
+p_sat_C6H12 = exp(A_C6H12 - B_C6H12 / (T + C_C6H12)) ;
+P_sat_C6H14 = exp(A_C6H14 - B_C6H14 / (T + C_C6H14)) ;
+P_sat_C7H16 = exp(A_C7H16 - B_C7H16 / (T + C_C7H16)) ;
+
+x_C5H12 = y_C5H12 * P / P_sat_C5H12 ;
+x_C6H12 = y_C6H12 * P / p_sat_C6H12 ;
+x_C6H14 = y_C6H14 * P / P_sat_C6H14 ;
+x_C7H16 = y_C7H16 * P / P_sat_C7H16 ;
+
+printf("\n\n x_C5H12 = %f x_C6H12 = %f\n\n x_C6H14 = %f x_C7H16 = %f",x_C5H12,x_C6H12 ,x_C6H14,x_C7H16) ;
diff --git a/527/CH8/EX8.4/8_4exam.sce b/527/CH8/EX8.4/8_4exam.sce new file mode 100755 index 000000000..1b607c2ae --- /dev/null +++ b/527/CH8/EX8.4/8_4exam.sce @@ -0,0 +1,30 @@ +//Engineering and Chemical Thermodynamics
+//Example 8.4
+//Page no :371
+
+clear ; clc ;
+//Given
+P_a_sat = 0.53 ; //[bar]
+P_b_sat = 0.16 ; //[bar]
+X = 1/3 ;
+Y = 1- X ;
+x_a_feed = 0.5 ;
+x_b_feed = 0.5 ;
+a = Y * -(x_a_feed + x_b_feed) + Y^2 ;
+b = X * Y *(P_a_sat + P_b_sat) - (x_a_feed * P_b_sat + x_b_feed * P_a_sat)*X ;
+c = P_a_sat * P_b_sat * X^2;
+
+k=poly(0,'k');
+P = c + b*k^1 + a*k^2 ;
+M = roots(P);
+
+disp(" Example: 8.4 Page no : 371") ;
+for i = 1:2
+ sign(M(i,1)) ;
+ if ans == 1 then
+ printf("\n\n Pressure = %.2f bar",M(i,1)) ;
+ Xa = x_a_feed / (P_a_sat / M(i,1) * X + Y) ;//....E8.4D
+ Ya = Xa * P_a_sat / M(i,1) ;//.....E8.4B
+ printf("\n\n Xa = %.2f \n Ya = %.2f\n",Xa,Ya);
+ end
+end
diff --git a/527/CH8/EX8.5/8_5exam.sce b/527/CH8/EX8.5/8_5exam.sce new file mode 100755 index 000000000..9dc1facd6 --- /dev/null +++ b/527/CH8/EX8.5/8_5exam.sce @@ -0,0 +1,20 @@ +//Engineering and Chemical Thermodynamics
+//Example 8.5
+//Page no :378
+
+clear ; clc;
+//Given
+P1_sat = 0.72 ; //[bar]
+P2_sat = 0.31 ; //[bar]
+A = 3590 ;
+B = -1180 ;
+R = 8.314 ;
+T = 70 + 273 ;//[K]
+function y85 = f85(x1) , y85 = -.48 + ( x1 * exp((A + 3*B) * (1 - x1)^2 / (R * T) - 4 * B * (1 - x1)^3 / (R * T)) * P1_sat) / ( x1 * exp((A + 3*B) * (1 - x1)^2 / (R * T) -4 * B * (1 - x1)^3 / (R * T)) * P1_sat +(1 - x1) * exp((A - 3*B) * x1^2 / (R * T) -4 * B * x1^3 / (R * T)) * P2_sat ) ;
+endfunction
+y = fsolve([0.1],f85);
+x1 = y ;
+P = ( x1 * exp((A + 3*B) * (1 - x1)^2 / (R * T) - 4 * B * (1 - x1)^3 / (R * T)) * P1_sat) + (1 - x1) * exp((A - 3*B) * x1^2 / (R * T) -4 * B * x1^3 / (R * T)) * P2_sat ;
+disp(" Example: 8.5 Page no : 378") ;
+printf("\n The value of x1 = %.3f\n\n",y) ;
+printf(" Pressure = %.2f bar",P) ;
\ No newline at end of file diff --git a/527/CH8/EX8.6/8_6exam.sce b/527/CH8/EX8.6/8_6exam.sce new file mode 100755 index 000000000..396b31aeb --- /dev/null +++ b/527/CH8/EX8.6/8_6exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 8.6
+//Page no :378
+
+clear ; clc ;
+disp(" Example: 8.6 Page no : 378") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression .") ;
\ No newline at end of file diff --git a/527/CH8/EX8.7/8_7exam.sce b/527/CH8/EX8.7/8_7exam.sce new file mode 100755 index 000000000..a1774c4c8 --- /dev/null +++ b/527/CH8/EX8.7/8_7exam.sce @@ -0,0 +1,22 @@ +//Engineering and Chemical Thermodynamics
+//Example 8.7
+//Page no :385
+
+clear ; clc;
+//Given
+P = 0.223 ; //[bar]
+P_a_sat = 0.156 ; // [bar]
+P_b_sat = 0.124 ; //[bar]
+R = 8.314 ;
+T = 50 + 273 ;
+Xa = 0.554 ;
+Xb = 1 - Xa ;
+
+gama_a = P / P_a_sat ;
+A1 = R * T * log(gama_a) / (Xb^2) * 10^-3 ;
+gama_b = P / P_b_sat ;
+A2 = R * T * log(gama_b) / (Xa^2) * 10^-3 ;
+
+A = ceil((A1 + A2) / 2) ;
+disp(" Example: 8.7 Page no : 385") ;
+printf("\n Value of two suffix Marguels parameter = %.1f kJ/mol",A);
\ No newline at end of file diff --git a/527/CH8/EX8.8/8_8exam.sce b/527/CH8/EX8.8/8_8exam.sce new file mode 100755 index 000000000..b2983472b --- /dev/null +++ b/527/CH8/EX8.8/8_8exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 8.8
+//Page no :385
+
+clear ; clc ;
+disp(" Example: 8.8 Page no : 385") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression .") ;
\ No newline at end of file diff --git a/527/CH8/EX8.9/8_9exam.sce b/527/CH8/EX8.9/8_9exam.sce new file mode 100755 index 000000000..4deed432f --- /dev/null +++ b/527/CH8/EX8.9/8_9exam.sce @@ -0,0 +1,52 @@ +//Engineering and Chemical Thermodynamics
+//Example 8.9
+//Page No:388
+
+clear ; clc ;
+//Given
+R = 8.314 ;
+T = 10 + 273 ; //[K]
+A_B = 9.2806 ; // From Appendix A , Table A1.1
+B_B = 2788.5 ; // From Appendix A , Table A1.1
+C_B = -52.36 ; // From Appendix A , Table A1.1
+A_C = 9.1325 ; // From Appendix A , Table A1.1
+B_C = 2766.63 ; // From Appendix A , Table A1.1
+C_C = -50.50 ; // From Appendix A , Table A1.1
+
+x1 = [0 ,0.0610 ,0.2149 ,0.3187 ,0.4320 ,0.5246 ,0.6117 ,0.7265 ,0.8040 ,0.8830 ,0.8999 ,1] ; //From table E8.9A
+P_exp = [6344 ,6590 ,6980 ,7140 ,7171 ,7216 ,7140 ,6974 ,6845 ,6617 ,6557 ,6073] ; //From table E8.9A
+
+P_1_sat = 6072.15 ; //[Pa]
+P_2_sat = 6344 ; //[Pa]
+
+A = [1390 ,1391 ,1392 ,1393 ,1394 ,1395 ,1396 ,1397 ,1398 ,1399 ,1400 ,1401 ,1402 ,1403 ,1404 ,1405 ,1406 ,1407 ,1408 ,1409 ,1410 ] ;
+
+for k = 1:21
+ y = A(1,k) ;
+ for i = 1:12
+ P(1,i) = x1(1,i) * exp( y / (R * T ) * (1 - x1(1,i))^2) * P_1_sat+(1 - x1(1,i)) * exp(y / (R * T ) * x1(1,i)^2) * P_2_sat ;
+ C(k,i) = (P(1,i) - P_exp(1,i))^2 ;
+ end
+end
+
+for k = 1:21
+ y = 0 ;
+ for i = 1:12
+ y = y + C(k,i) ;
+ end
+ R(1,k) = y ;
+end
+
+k = 100000 ;
+for i = 1:21
+ K = R(1,i) ;
+ if K < k then
+ k = K ;
+ end
+end
+disp(" Example: 8.9 Page no : 388") ;
+for i = 1:21
+ if R(1,i) == k then
+ printf("\n The two suffix Margules co-efficient is = %g J/mol" ,A(1,i)) ;
+ end
+end
diff --git a/527/CH9/EX9.1/9_1exam.sce b/527/CH9/EX9.1/9_1exam.sce new file mode 100755 index 000000000..6e61c7309 --- /dev/null +++ b/527/CH9/EX9.1/9_1exam.sce @@ -0,0 +1,18 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.1
+//Page no :440
+
+clear ; clc ;
+n_o_CH3OH = 1 ; //[mol]
+n_o_H2O = 3 ; //[mol]
+S = 0.87 ;
+n_CH3OH = 1 - S ;
+n_H2O = 2 - S ;
+n_CO2 = S ;
+n_H2 = 3 * S ;
+n_v = n_CH3OH + n_CO2 + n_H2O + n_H2 ;
+
+y_H2 = n_H2 / n_v ;
+disp(" Example: 9.1 Page no : 440") ;
+printf("\n No of moles of H2 produced for 1mol of CH3OH = %.3f mol" , n_H2)
+printf("\n Mole fraction of H2 = %.2f" , y_H2) ;
diff --git a/527/CH9/EX9.10/9_10exam.sce b/527/CH9/EX9.10/9_10exam.sce new file mode 100755 index 000000000..95fa185b2 --- /dev/null +++ b/527/CH9/EX9.10/9_10exam.sce @@ -0,0 +1,16 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.10
+//Page no :456
+
+clear ; clc ;
+//Given
+del_g0_f_1 = 31.72 ; //[kJ/mol]
+del_g0_f_2 = 26.89 ; //[kJ/mol]
+R = 8.314 ;
+T = 298 ;//[K]
+del_g0_rxn = del_g0_f_2 - del_g0_f_1 ;
+K = exp( - del_g0_rxn * 10^3 / (R * T) ) ;
+x = K / (1 + K) ;
+
+disp(" Example: 9.10 Page no : 456") ;
+printf("\n x = %.3f \n\n At equillibrium %.1f %% of the liquid exists as cyclohexane.",x ,x * 100) ;
diff --git a/527/CH9/EX9.11/9_11exam.sce b/527/CH9/EX9.11/9_11exam.sce new file mode 100755 index 000000000..b5c2fe3bc --- /dev/null +++ b/527/CH9/EX9.11/9_11exam.sce @@ -0,0 +1,16 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.11
+//Page no :457
+
+clear ; clc;
+// Given
+del_g0_f_CaCO3 = -951.25 ;
+del_g0_f_CaO = -531.09 ;
+del_g0_f_CO2 = -395.81 ;
+R = 8.314 ;
+T = 1000 ;// [K]
+del_g0_rxn = del_g0_f_CaO + del_g0_f_CO2 - del_g0_f_CaCO3 ;
+K = exp (-del_g0_rxn * 10^3 / (R * T)) ;
+p_CO2 = K ;
+disp(" Example: 9.11 Page no : 457") ;
+printf("\n Equilibrium pressure = %.3f bar ",p_CO2) ;
\ No newline at end of file diff --git a/527/CH9/EX9.12/9_12exam.sce b/527/CH9/EX9.12/9_12exam.sce new file mode 100755 index 000000000..f19dc7437 --- /dev/null +++ b/527/CH9/EX9.12/9_12exam.sce @@ -0,0 +1,23 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.12
+//Page no :458
+
+clear ; clc ;
+//Given
+del_g0_f_B = 124.3 ; //[kJ/mol] , From Appendix A.3
+del_g0_f_Ac = 209.2 ; //[kJ/mol] , From Appendix A.3
+R = 8.314 ;
+T = 298 ; // [K]
+A = 9.2806 ;
+B = 2788.51 ;
+C = -52.36 ;
+ del_g0_rxn = del_g0_f_B - 3 * del_g0_f_Ac ;
+K = exp( - del_g0_rxn * 10^3 / (R * T)) ;
+
+//We assume no acetylene condenses and no Benzene is volatile .
+P = 1 / K^(1/3) ;
+X = A - B / (T + C) ;
+P_b = exp(X) ;
+disp(" Example: 9.12 Page no : 458") ;
+disp(" At equilibrium , the cylinder is almost completely filled with Benzene .")
+printf("\n System pressure = %.3f bar ",P_b)
\ No newline at end of file diff --git a/527/CH9/EX9.13/9_13exam.sce b/527/CH9/EX9.13/9_13exam.sce new file mode 100755 index 000000000..6cc0d9886 --- /dev/null +++ b/527/CH9/EX9.13/9_13exam.sce @@ -0,0 +1,20 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.13
+//Page no :466
+
+clear ; clc ;
+//Given
+E_0_c = 0.153 ; //[V]
+E_0_a = -0.521 ; // [v]
+T = 298 ; //[K]
+z = 1 ;
+F =96485 ; //[C/mol e-]
+R =8.314 ; //[J/mol K ]
+
+E_0_rxn = E_0_c + E_0_a ;
+del_g_0_rxn = - z * F * E_0_rxn ;
+
+K = exp( - del_g_0_rxn / ( R * T )) ;
+disp(" Example: 9.13 Page no : 466") ;
+printf("\n The equilibrium constant = %.3g \n",K)
+disp(" The equilibrium constant is small . So the etching will not proceed spontaneously . However if we apply work through application of an electrical potential , we can etch the copper .")
diff --git a/527/CH9/EX9.14/9_14exam.sce b/527/CH9/EX9.14/9_14exam.sce new file mode 100755 index 000000000..b6a56b205 --- /dev/null +++ b/527/CH9/EX9.14/9_14exam.sce @@ -0,0 +1,20 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.14
+//Page no :466
+
+clear ; clc ;
+//Given
+E_0_c = 0.34 ; //[V]
+E_0_a = -1.23 ; //[V]
+T = 298 ; // [K]
+pH = 1 ;
+z = 2 ;
+Cu2 = 0.07 ;
+F = 96485 ; //[C/mol e-]
+R = 8.314 ;
+
+E_0_rxn = E_0_c + E_0_a ;
+E = E_0_rxn + 2.303 * R * T * 2 * pH / (z * F) + R * T * log(Cu2) / (z * F) ;
+disp(" Example: 9.14 Page no : 466") ;
+printf("\n Del_E_0_rxn = %.2f ",E_0_rxn ) ;
+printf("\n\n We have to apply potential greater than %.2f V",-E) ;
\ No newline at end of file diff --git a/527/CH9/EX9.15/9_15exam.sce b/527/CH9/EX9.15/9_15exam.sce new file mode 100755 index 000000000..eabf75052 --- /dev/null +++ b/527/CH9/EX9.15/9_15exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.15
+//Page no :468
+
+clear ; clc ;
+disp(" Example: 9.15 Page no : 468") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression .") ;
\ No newline at end of file diff --git a/527/CH9/EX9.16/9_16exam.sce b/527/CH9/EX9.16/9_16exam.sce new file mode 100755 index 000000000..f4707be07 --- /dev/null +++ b/527/CH9/EX9.16/9_16exam.sce @@ -0,0 +1,14 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.16
+//Page no :469
+
+clear ; clc ;
+//Given
+m = 4 ;
+T = 2 ;
+Pai = 1 ;
+S = 1 ;
+
+R = m - T + 2 - Pai - S ;
+disp(" Example: 9.16 Page no : 469") ;
+printf("\n We must specify %g independent equations .",R)
\ No newline at end of file diff --git a/527/CH9/EX9.17/9_17exam.sce b/527/CH9/EX9.17/9_17exam.sce new file mode 100755 index 000000000..a9ef64465 --- /dev/null +++ b/527/CH9/EX9.17/9_17exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.17
+//Page no :470
+
+clear ; clc ;
+disp(" Example: 9.17 Page no : 470") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression .") ;
\ No newline at end of file diff --git a/527/CH9/EX9.18/9_18exam.sce b/527/CH9/EX9.18/9_18exam.sce new file mode 100755 index 000000000..7b8986b94 --- /dev/null +++ b/527/CH9/EX9.18/9_18exam.sce @@ -0,0 +1,134 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.18
+//Page no :470
+
+clear ; clc
+//Given
+del_g_f_CH4 = -50.72 ;
+del_g_f_H2 = 0 ;
+del_g_f_H2O = -228.57 ;
+del_g_f_CO = -137.17 ;
+del_g_f_CO2 = -394.36 ;
+del_h_f_CH4 = -74.81 ;
+del_h_f_H2 = 0 ;
+del_h_f_H2O = -241.82 ;
+del_h_f_CO = -110.53 ;
+del_h_f_CO2 = -393.51 ;
+
+v1_CH4 = -1 ;
+v1_H2 = 3 ;
+v1_H2O = -1 ;
+v1_CO = 1 ;
+v1_CO2 = 0 ;
+v2_CH4 = -1 ;
+v2_H2 = 4 ;
+v2_H2O = -2 ;
+v2_CO = 0 ;
+v2_CO2 = 1 ;
+
+A_CH4 = 1.702 ;
+B_CH4 = 9.08 * 10^-3 ;
+C_CH4 = -2.16 * 10^-6 ;
+D_CH4 = 0 ;
+A_H2 = 3.249 ;
+B_H2 = 4.22 * 10^-4 ;
+C_H2 = 0 ;
+D_H2 = 8.30 * 10^3 ;
+A_H2O = 3.47 ;
+B_H2O = 1.45 * 10^-3 ;
+C_H2O = 0 ;
+D_H2O = 1.21 * 10^4 ;
+A_CO = 3.376 ;
+B_CO = 5.57 * 10^-4 ;
+C_CO = 0 ;
+D_CO = -3.10 * 10^3 ;
+A_CO2 = 5.457 ;
+B_CO2 = 1.05 * 10^-3 ;
+C_CO2 = 0 ;
+D_CO2 = -1.16 * 10^5 ;
+
+M(:,1) = 600:50:1150 ;
+R = 8.314 ;
+P = 1 ; //[bar]
+T_ref = 298.15 ; //[K]
+
+del_g_f_1 = (v1_CO * del_g_f_CO + v1_H2 *del_g_f_H2 + v1_CH4 * del_g_f_CH4 + v1_H2O * del_g_f_H2O) * 1000 ;
+del_h_f_1 = (v1_CO * del_h_f_CO + v1_H2 *del_h_f_H2 + v1_CH4 * del_h_f_CH4 + v1_H2O * del_h_f_H2O) * 1000 ;
+del_g_f_2 = (v2_CO2 * del_g_f_CO2 + v2_H2 *del_g_f_H2 + v2_CH4 * del_g_f_CH4 + v2_H2O * del_g_f_H2O) * 1000 ;
+del_h_f_2 = (v2_CO2 * del_h_f_CO2 + v2_H2 *del_h_f_H2 + v2_CH4 * del_h_f_CH4 + v2_H2O * del_h_f_H2O) * 1000;
+Del_A_1 = v1_CO * A_CO + v1_H2 * A_H2 + v1_CH4 * A_CH4 + v1_H2O * A_H2O ;
+Del_B_1 = v1_CO * B_CO + v1_H2 * B_H2 + v1_CH4 * B_CH4 + v1_H2O * B_H2O ;
+Del_C_1 = v1_CO * C_CO + v1_H2 * C_H2 + v1_CH4 * C_CH4 + v1_H2O * C_H2O ;
+Del_D_1 = v1_CO * D_CO + v1_H2 * D_H2 + v1_CH4 * D_CH4 + v1_H2O * D_H2O ;
+Del_A_2 = v2_CO2 * A_CO2 + v2_H2 * A_H2 + v2_CH4 * A_CH4 + v2_H2O * A_H2O ;
+Del_B_2 = v2_CO2 * B_CO2 + v2_H2 * B_H2 + v2_CH4 * B_CH4 + v2_H2O * B_H2O ;
+Del_C_2 = v2_CO2 * C_CO2 + v2_H2 * C_H2 + v2_CH4 * C_CH4 + v2_H2O * C_H2O ;
+Del_D_2 = v2_CO2 * D_CO2 + v2_H2 * D_H2 + v2_CH4 * D_CH4 + v2_H2O * D_H2O ;
+
+
+K_298_1 = exp( - del_g_f_1 / (R * T_ref)) ;
+K_298_2 = exp( - del_g_f_2 / (R * T_ref)) ;
+disp(" Example: 9.18 Page no : 470") ;
+
+for i = 1:12
+ X = (-del_h_f_1 / R + Del_A_1 * T_ref + Del_B_1 / 2 * T_ref^2 + Del_C_1 /3* T_ref^3- Del_D_1 / T_ref) * (1/M(i,1) - 1/T_ref) + Del_A_1*log(M(i,1) / T_ref)+ Del_B_1 / 2 * (M(i,1) - T_ref) + Del_C_1 / 6 *(M(i,1)^2 - T_ref^2) + Del_D_1 / 2* (1/(M(i,1)^2) - 1/(T_ref^2));
+
+ M(i,2) = K_298_1 * exp(X) ;
+
+ Y = (-del_h_f_2 / R + Del_A_2 * T_ref + Del_B_2 / 2 * T_ref^2 + Del_C_2/3* T_ref^3- Del_D_2 / T_ref) * (1/M(i,1) - 1/T_ref) + Del_A_2 * log(M(i,1) / T_ref)+ Del_B_2 / 2 * (M(i,1) - T_ref) + Del_C_2 / 6 *(M(i,1)^2 - T_ref^2) + Del_D_2 / 2* (1/(M(i,1)^2) - 1/(T_ref^2));
+
+ M(i,3) = K_298_2 * exp(Y) ;
+ function y = f918(R),
+ s1 = R(1) ;
+ s2 = R(2) ;
+ y(1) = (s1 * (3 * s1 + 4 * s2)^3) / ((5 + 2 * s1 + 2 * s2)^2 * (1 - s1 -s2) * (4 - s1 - 2 * s2)) * P^2 - M(i,2) ;
+ y(2) = (s2 * (3 * s1 + 4 * s2)^4) / ((5 + 2 * s1 + 2 * s2)^2 * (1 - s1 -s2) * (4 - s1 - 2 * s2)^2) * P^2 - M(i,3) ;
+endfunction
+ z = fsolve([0.0001;0.0001],f918) ;
+ M(i,4) = z(1) ;
+ M(i,5) = z(2) ;
+ M(i,6) = (1 - M(i,4) - M(i,5)) / (5 + 2 * M(i,4) + 2 * M(i,5)) ;
+ M(i,7) = (4 - M(i,4) - 2 * M(i,5)) / (5 + 2 * M(i,4) + 2 * M(i,5)) ;
+ M(i,8) = (3 * M(i,4) + 4 * M(i,5)) / (5 + 2 * M(i,4) + 2 * M(i,5)) ;
+ M(i,9) = M(i,4) / (5 + 2 * M(i,4) + 2 * M(i,5)) ;
+ M(i,10) = M(i,5) / (5 + 2 * M(i,4) + 2 * M(i,5)) ;
+
+end
+
+disp(" T K1 K2 S1 S2 y_CH4 y_H2")
+
+for i = 1:10 // For convenient display of solution .
+ for j = 1:7
+ n1(i,j) = M(i,j) ;
+ end
+end
+for i = 1:10 // For convenient display of solution .
+ for j = 1:3
+ n2(i,j) = M(i,j+7) ;
+ end
+end
+disp(n1) ;
+disp(" y_H20 y_CO y_CO2 ") ;
+disp(n2) ;
+for i = 1:10
+ for j = 1:10
+ N(i,j) = M(i,j) ;
+ end
+end
+
+plot(N(:,1) , N(:,4),"+") ;
+plot(N(:,1), N(:,5),".") ;
+xtitle("Figure E9.18 Extent of reaxn vs temp","Temperature(K)","S") ;
+legend("S1","S2") ;
+
+h = figure(1) ;
+clf() ;
+set(h,"background",35) ;
+plot(N(:,1) , N(:,6), "o-") ;
+plot(N(:,1) , N(:,7), "s-");
+plot(N(:,1) , N(:,8), "^-") ;
+plot(N(:,1) , N(:,9), "x-") ;
+plot(N(:,1) , N(:,10), ".-") ;
+legend("y_CH4 ","y_H2 ","y_H2O ","y_CO ","y_CO2") ;
+
+xtitle("Figure E9.18 mole fractn vs temp","Temp","mole fraction") ;
\ No newline at end of file diff --git a/527/CH9/EX9.19/9_19exam.sce b/527/CH9/EX9.19/9_19exam.sce new file mode 100755 index 000000000..be4797061 --- /dev/null +++ b/527/CH9/EX9.19/9_19exam.sce @@ -0,0 +1,42 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.19
+//Page no :472
+
+clear ; clc
+//Given
+del_g_0_f_SiCl2 = - 216012 ;
+del_g_0_f_SiCl4 = - 492536 ;
+del_g_0_f_SiCl3H = -356537 ;
+del_g_0_f_SiCl2H2 = -199368 ;
+del_g_0_f_SiClH3 = -28482 ;
+del_g_0_f_SiH4 = -176152 ;
+del_g_0_f_HCl = -102644 ;
+del_g_0_f_H2 = 0 ;
+del_g_0_f_Si = 0 ;
+R = 8.314 ;
+T = 1300 ; //[K]
+Del_g_rxn_1 = del_g_0_f_SiCl2 + 2 * del_g_0_f_HCl - del_g_0_f_SiCl4 - del_g_0_f_H2 ;
+Del_g_rxn_2 = del_g_0_f_SiCl3H + del_g_0_f_HCl - del_g_0_f_SiCl4 - del_g_0_f_H2 ;
+Del_g_rxn_3 = del_g_0_f_SiCl2H2 + del_g_0_f_HCl - del_g_0_f_SiCl3H - del_g_0_f_H2 ;
+Del_g_rxn_4 = del_g_0_f_SiClH3 + del_g_0_f_HCl - del_g_0_f_SiCl2H2 - del_g_0_f_H2 ;
+Del_g_rxn_5 = del_g_0_f_SiH4 + del_g_0_f_HCl - del_g_0_f_SiCl3H - del_g_0_f_H2 ;
+Del_g_rxn_6 = del_g_0_f_Si + 4 * del_g_0_f_HCl - del_g_0_f_SiCl4 - 2 * del_g_0_f_H2 ;
+
+M(1,1) = exp( - Del_g_rxn_1 / (R * T)) ;
+M(2,1) = exp( - Del_g_rxn_2 / (R * T)) ;
+M(3,1) = exp( - Del_g_rxn_3 / (R * T)) ;
+M(4,1) = exp( - Del_g_rxn_4 / (R * T)) ;
+M(5,1) = exp( - Del_g_rxn_5 / (R * T)) ;
+M(6,1) = exp( - Del_g_rxn_6 / (R * T)) ;
+
+S = [0.0763 ;0.1979 ;0.0067 ;0.0001 ;0.0000 ;-0.0512] ;
+K_cal = [.00137 ;0.0457 ;0.00644 ;0.00181 ;0.000752 ;0.000509] ;
+disp(" Example: 9.19 Page no : 472") ;
+disp(" K_i S K_i_cal K_i - K_i_cal") ;
+for i = 1:6
+ M(i,2) = S(i,1) ;
+ M(i,3) = K_cal(i,1) ;
+ M(i,4) = M(i,1) - M(i,3) ;
+end
+disp(M)
+// Readers can refer figure E9.19 .
\ No newline at end of file diff --git a/527/CH9/EX9.2/9_2exam.sce b/527/CH9/EX9.2/9_2exam.sce new file mode 100755 index 000000000..2ae792351 --- /dev/null +++ b/527/CH9/EX9.2/9_2exam.sce @@ -0,0 +1,21 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.2
+//Page no :444
+
+clear ; clc ;
+//Given
+del_gf_0_CO2 = -394.36 ; //[kJ/mol],From Appendix A.3
+del_gf_0_H2 = 0 ; //[kJ/mol],From Appendix A.3
+del_gf_0_H2O = -228.57 ; //[kJ/mol],From Appendix A.3
+del_gf_0_CH3OH = -161.96 ; //[kJ/mol],From Appendix A.3
+n_CO2 = 1 ;
+n_H2 = 3 ;
+n_CH3OH = 1 ;
+n_H2O = 1 ;
+T = 298.15 ;//[K]
+R = 8.314 ; //[J/molK]
+
+del_g0_rxn = (n_CO2 * del_gf_0_CO2 + n_H2 * del_gf_0_H2 - n_H2O * del_gf_0_H2O - n_CH3OH * del_gf_0_CH3OH) * 10^3 ; // [J/mol]
+K_298 = exp( - del_g0_rxn / (R * T)) ;
+disp(" Example: 9.2 Page no : 444") ;
+printf("\n The equillibrium constant K298 = %.2f ",K_298) ;
diff --git a/527/CH9/EX9.20/9_20exam.sce b/527/CH9/EX9.20/9_20exam.sce new file mode 100755 index 000000000..9d9d89912 --- /dev/null +++ b/527/CH9/EX9.20/9_20exam.sce @@ -0,0 +1,111 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.20
+//Page no :476
+
+clear ; clc
+//Given
+del_g_0_f_CH4 = -2.057 ; //[J/mol]
+del_g_0_f_H2O = -192.713 ; //[J/mol]
+del_g_0_f_CO = -182.494 ; //[J/mol]
+del_g_0_f_CO2 = -203.595 ; //[J/mol]
+del_g_0_f_H2 = 0 ; //[J/mol]
+R = 8.314 ;
+T = 800 ; //[K]
+w = 1 / (R * T) ;
+function Z920 = F920(R)
+ m = R(1) ,
+ n = R(2) ,
+ o = R(3) ,
+ a = R(4), // n_CH4
+ b = R(5), // n_H2O
+ c = R(6), // n_H2
+ d = R(7), // n_CO
+ e = R(8), // n_CO2
+
+Z920(1) = a + d + e - 1 ;
+Z920(2) = 4 * a + 2 * b + 2 * c - 12 ;
+Z920(3) = b + d + 2 * e - 4 ;
+Z920(4) = del_g_0_f_CH4 * w + log(a) - log(a + b + c + d + e) + m + 4 * o ;
+Z920(5) = del_g_0_f_H2O * w + log(b) - log(a + b + c + d + e) + 2 * o + n ;
+Z920(6) = del_g_0_f_H2 * w + log(c)- log(a + b + c + d + e) + 2 * o ;
+Z920(7) = del_g_0_f_CO * w + log(d) - log(a + b + c + d + e) + m + n ;
+Z920(8) = del_g_0_f_CO2 * w + log(e) - log(a + b + c + d + e) + m + 2 * n ;
+
+endfunction ;
+
+
+function [J] = jacob(X)
+
+ m = X(1) ,
+ n = X(2) ,
+ o = X(3) ,
+ a = X(4), // n_CH4
+ b = X(5), // n_H2O
+ c = X(6), // n_H2
+ d = X(7), // n_CO
+ e = X(8), // n_CO
+
+ J(1,1) = 0 ; J(1,2) = 0 ; J(1,3) = 0 ; J(1,4) = 1 ; J(1,5) = 0 ;
+ J(1,6) = 0 ; J(1,7) = 1 ; J(1,8) = 1 ;
+ J(2,1) = 0 ; J(2,2) = 0 ; J(2,3) = 0 ; J(2,4) = 4 ; J(2,5) = 2 ;
+ J(2,6) = 2 ; J(2,7) = 0 ; J(2,8) = 0 ;
+ J(3,1) = 0 ; J(3,2) = 0 ; J(3,3) = 0 ; J(3,4) = 0 ; J(3,5) = 1 ;
+ J(3,6) = 0 ; J(3,7) = 1 ; J(3,8) = 2 ;
+ J(4,1) = 1 ; J(4,2) = 0 ; J(4,3) = 4 ; J(4,4) = (b+c+d+e)/(a*(a+b+c+d+e)) ; J(4,5) = -1/(a+b+c+d+e) ; J(4,6) = -1/(a+b+c+d+e) ; J(4,7) = -1/(a+b+c+d+e); J(4,8) = -1/(a+b+c+d+e);
+ J(5,1) = 0 ; J(5,2) = 1 ; J(5,3) = 2 ; J(5,4) = -1/(a+b+c+d+e) ;
+ J(5,5) = (a+c+d+e)/(b*(a+b+c+d+e)) ; J(5,6) = -1/(a+b+c+d+e) ;
+ J(5,7) = -1/(a+b+c+d+e) ; J(5,8) = -1/(a+b+c+d+e) ;
+ J(6,1) = 0 ; J(6,2) = 0 ; J(6,3) = 2 ; J(6,4) = -1/(a+b+c+d+e) ;
+ J(6,5) = -1/(a+b+c+d+e) ; J(6,6) = (a+b+d+e)/(c*(a+b+c+d+e)) ;
+ J(6,7) = -1/(a+b+c+d+e) ; J(6,8) = -1/(a+b+c+d+e) ;
+ J(7,1) = 1 ; J(7,2) = 1 ; J(7,3) = 0 ; J(7,4) = -1/(a+b+c+d+e) ;
+ J(7,5) = -1/(a+b+c+d+e) ; J(7,6) = -1/(a+b+c+d+e) ;
+ J(7,7) = (a+b+c+e)/(d*(a+b+c+d+e)) ; J(7,8) = -1/(a+b+c+d+e) ;
+ J(8,1) = 1 ; J(8,2) = 2 ; J(8,3) = 0 ; J(8,4) = -1/(a+b+c+d+e) ;
+ J(8,5) = -1/(a+b+c+d+e) ; J(8,6) = -1/(a+b+c+d+e) ;
+ J(8,7) = -1/(a+b+c+d+e) ; J(8,8) = (a+b+c+d)/(e*(a+b+c+d+e)) ;
+endfunction
+
+// We will use newton Raphson Method to solve the set of equations.
+// Reference : www.infoclearinghouse.com/files/scilab/scilab6a.pdf
+
+function [x] = newtonm(x0,f,J)
+ N = 1000 ;
+ epsilon = 1*10^-10 ;
+ maxval = 1000 ;
+ xx = x0 ;
+
+ while(N>0)
+ JJ = J(xx)
+ // disp(abs(det(JJ)))
+ if abs(det(JJ))<epsilon then
+ error('newtonm-Jacobian is singular- try new x0')
+ abort ;
+ end ;
+ xn = xx -inv(JJ) * f(xx) ;
+ // disp(abs(f(xn)))
+ if abs(f(xn))<epsilon then
+ x = xn ;
+ // disp(100-N) ;
+ // disp((x))
+ return(x) ;
+ end ;
+
+ if abs(f(xn))>maxval then
+ disp(1000-N) ;
+ error('Solution diverges') ;
+ abrot ;
+ end ;
+ N = N -1 ;
+ xx = xn ;
+ end ;
+endfunction ;
+
+x1 = [1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1 ] ; // Initial guess .
+
+[z] = newtonm(x1,F920,jacob) ;
+
+disp("Example 9.20 Page no:476") ;
+printf("\n\n L_c/RT = %f ,\n L_o/RT = %f ,\n L_h/RT = %f ,\n n_CH4 = %f ,\n n_H2O = %f ,\n n_H2 = %f ,\n n_CO = %f ,\n n_CO2 = %f",z(1),z(2),z(3),z(4),z(5),z(6),z(7),z(8)) ;
+//The solutions given in the text book does not satisfy E9.20D, E9.20E,
+// E9.20F and so on .
\ No newline at end of file diff --git a/527/CH9/EX9.21/9_21exam.sce b/527/CH9/EX9.21/9_21exam.sce new file mode 100755 index 000000000..6f5437380 --- /dev/null +++ b/527/CH9/EX9.21/9_21exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.21
+//Page no :485
+
+clear ; clc ;
+disp(" Example: 9.21 Page no : 485") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression .") ;
\ No newline at end of file diff --git a/527/CH9/EX9.22/9_22exam.sce b/527/CH9/EX9.22/9_22exam.sce new file mode 100755 index 000000000..6dc716e24 --- /dev/null +++ b/527/CH9/EX9.22/9_22exam.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.22
+//Page no :487
+
+clear ; clc ;
+disp(" Example: 9.22 Page no : 487") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression .") ;
\ No newline at end of file diff --git a/527/CH9/EX9.3/9_4exam.sce b/527/CH9/EX9.3/9_4exam.sce new file mode 100755 index 000000000..fc9515672 --- /dev/null +++ b/527/CH9/EX9.3/9_4exam.sce @@ -0,0 +1,37 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.4
+//Page no :447
+
+clear ; clc ;
+//Given
+del_gf_0_CH2O = -110.0 ; //[kJ/mol],From Appendix A.2 & A.3
+del_gf_0_H2 = 0 ; //[kJ/mol],From Appendix A.2 & A.3
+del_gf_0_CH4O = -162.0 ; //[kJ/mol],From Appendix A.2 & A.3
+del_hf_0_CH2O = -116.0 ; //[kJ/mol],From Appendix A.2 & A.3
+del_hf_0_H2 = 0 ; //[kJ/mol],From Appendix A.2 & A.3
+del_hf_0_CH4O = -200.7 ; //[kJ/mol],From Appendix A.2 & A.3n_CH20 = 1 ;
+n_H2 = 1 ;
+n_CH4O = 1 ;
+n_CH2O = 1 ;
+T1 = 298 ;//[K]
+T2 = 873 ; // [K]
+R = 8.314 ; //[J/molK]
+Del_A = 3.302 ;
+Del_B = -4.776 * 10^-3 ;
+Del_C = 1.57 * 10^-6 ;
+Del_D = 0.083 * 10^5 ;
+//Solution (a)
+del_g_rxn_298 = n_CH2O * del_gf_0_CH2O + n_H2 * del_gf_0_H2 - n_CH4O * del_gf_0_CH4O ;
+K_298 = exp( - del_g_rxn_298 * 10^3 / (R * T1)) ;
+disp(" Example: 9.4 Page no : 447") ;
+printf("\n (a) K_298 = %g \n\n As the equilibrium constant is very small very little amount of formaldehyde will be formed .\n",K_298) ;
+
+//Solution(b)
+del_h_rxn_298 = (n_CH2O * del_hf_0_CH2O + n_H2 * del_hf_0_H2 - n_CH4O * del_hf_0_CH4O) * 10^3 ;//[J/mol]
+K_873 = K_298 * exp((-del_h_rxn_298 * (1/T2 - 1/T1)) / R) ;
+printf("\n (b)\n (i) K_873 = %g \n\n",K_873) ;
+
+//Solution(c)
+x = ( -del_h_rxn_298 / R + Del_A * T1 + Del_B / 2 * T1^2 + Del_C /3 * T1^3 - Del_D / T1 ) *(1/T2 - 1/T1) + Del_A * log(T2 / T1) + Del_B / 2 * (T2 -T1) + Del_C / 6 * (T2^2 -T1^2) + Del_D / 2 * (1/(T2^2) -1/(T1^2)) ;
+K_873 = K_298 * exp(x) ;
+printf("\n (ii) K_873 = %g \n\n",K_873) ;
\ No newline at end of file diff --git a/527/CH9/EX9.4/9_4exam.sce b/527/CH9/EX9.4/9_4exam.sce new file mode 100755 index 000000000..fc9515672 --- /dev/null +++ b/527/CH9/EX9.4/9_4exam.sce @@ -0,0 +1,37 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.4
+//Page no :447
+
+clear ; clc ;
+//Given
+del_gf_0_CH2O = -110.0 ; //[kJ/mol],From Appendix A.2 & A.3
+del_gf_0_H2 = 0 ; //[kJ/mol],From Appendix A.2 & A.3
+del_gf_0_CH4O = -162.0 ; //[kJ/mol],From Appendix A.2 & A.3
+del_hf_0_CH2O = -116.0 ; //[kJ/mol],From Appendix A.2 & A.3
+del_hf_0_H2 = 0 ; //[kJ/mol],From Appendix A.2 & A.3
+del_hf_0_CH4O = -200.7 ; //[kJ/mol],From Appendix A.2 & A.3n_CH20 = 1 ;
+n_H2 = 1 ;
+n_CH4O = 1 ;
+n_CH2O = 1 ;
+T1 = 298 ;//[K]
+T2 = 873 ; // [K]
+R = 8.314 ; //[J/molK]
+Del_A = 3.302 ;
+Del_B = -4.776 * 10^-3 ;
+Del_C = 1.57 * 10^-6 ;
+Del_D = 0.083 * 10^5 ;
+//Solution (a)
+del_g_rxn_298 = n_CH2O * del_gf_0_CH2O + n_H2 * del_gf_0_H2 - n_CH4O * del_gf_0_CH4O ;
+K_298 = exp( - del_g_rxn_298 * 10^3 / (R * T1)) ;
+disp(" Example: 9.4 Page no : 447") ;
+printf("\n (a) K_298 = %g \n\n As the equilibrium constant is very small very little amount of formaldehyde will be formed .\n",K_298) ;
+
+//Solution(b)
+del_h_rxn_298 = (n_CH2O * del_hf_0_CH2O + n_H2 * del_hf_0_H2 - n_CH4O * del_hf_0_CH4O) * 10^3 ;//[J/mol]
+K_873 = K_298 * exp((-del_h_rxn_298 * (1/T2 - 1/T1)) / R) ;
+printf("\n (b)\n (i) K_873 = %g \n\n",K_873) ;
+
+//Solution(c)
+x = ( -del_h_rxn_298 / R + Del_A * T1 + Del_B / 2 * T1^2 + Del_C /3 * T1^3 - Del_D / T1 ) *(1/T2 - 1/T1) + Del_A * log(T2 / T1) + Del_B / 2 * (T2 -T1) + Del_C / 6 * (T2^2 -T1^2) + Del_D / 2 * (1/(T2^2) -1/(T1^2)) ;
+K_873 = K_298 * exp(x) ;
+printf("\n (ii) K_873 = %g \n\n",K_873) ;
\ No newline at end of file diff --git a/527/CH9/EX9.5/9_5eaxm.sce b/527/CH9/EX9.5/9_5eaxm.sce new file mode 100755 index 000000000..a9f54493a --- /dev/null +++ b/527/CH9/EX9.5/9_5eaxm.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.5:
+//Page no :450
+
+clear ; clc ;
+disp(" Example: 9.5 Page no : 450") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression .") ;
\ No newline at end of file diff --git a/527/CH9/EX9.6/9_6eaxm.sce b/527/CH9/EX9.6/9_6eaxm.sce new file mode 100755 index 000000000..fe9738af2 --- /dev/null +++ b/527/CH9/EX9.6/9_6eaxm.sce @@ -0,0 +1,25 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.6
+//Page no :451
+
+clear ; clc ;
+del_g0_f_C6H6 = -32.84 ; //[kJ/mol] , From Table E9.6
+del_g0_f_C2H4 = 68.15 ; //[kJ/mol] , From Table E9.6
+del_g0_f_H2 = 0 ; //[kJ/mol] , From Table E9.6
+del_h0_f_C6H6 = -84.68 ; //[kJ/mol] , From Table E9.6
+del_h0_f_C2H4 = 52.26 ; //[kJ/mol] , From Table E9.6
+del_h0_f_H2 = 0 ; //[kJ/mol] , From Table E9.6
+T1 = 298.2 ;//[K]
+P = 1 ;//[bar]
+R = 8.31 ;
+T2 = 1273 ; // [K]
+del_g0_f_rxn = del_g0_f_C2H4 + del_g0_f_H2 - del_g0_f_C6H6 ;
+K_298 = exp ( - (del_g0_f_rxn * 10^3) / (R * T1)) ;
+
+del_h0_f_rxn = (del_h0_f_C2H4 + del_h0_f_H2 - del_h0_f_C6H6) * 10^3 ;
+K_1273 = K_298 * exp( - del_h0_f_rxn / R * (1/T2 - 1/T1)) ;
+
+x = sqrt( K_1273 / ( K_1273 + P)) ;
+
+disp(" Example: 9.6 Page no : 451") ;
+printf("\n n_C2H6 = %.2f mol\n\n n_C2H4 = %.2f mol\n\n n_H2 = %.2f mol",1-x ,x,x) ;
\ No newline at end of file diff --git a/527/CH9/EX9.7/9_7exam.sce b/527/CH9/EX9.7/9_7exam.sce new file mode 100755 index 000000000..db344931e --- /dev/null +++ b/527/CH9/EX9.7/9_7exam.sce @@ -0,0 +1,87 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.7
+//Page no :453
+
+clear ; clc ;
+//Given
+del_h0_f_NH3 = -46.11 ; // [kJ/mol],From table E9.7
+del_h0_f_N2 = 0 ; // [kJ/mol],From table E9.7
+del_h0_f_H2 = 0 ; // [kJ/mol],From table E9.7
+del_g0_f_NH3 = -16.45 ; // [kJ/mol],From table E9.7
+del_g0_f_N2 = 0 ; // [kJ/mol],From table E9.7
+del_g0_f_H2 = 0 ; // [kJ/mol],From table E9.7
+n_NH3 = 2 ;
+n_N2 = -1 ;
+n_H2 = -3 ;
+A_NH3 = 3.578 ; B_NH3 = 3.02 * 10^-3 ; D_NH3 = -0.186 * 10^5 ;
+A_N2 = 3.280 ; B_N2 = 0.593 * 10^-3 ; D_N2 = 0.040 * 10^5 ;
+A_H2 = 3.249 ; B_H2 = 0.422 * 10^-3 ; D_H2 = 0.083 * 10^5 ;
+R = 8.314 ;
+T = 298 ;
+T2 = 773 ;
+P = 1 ; //[bat]
+
+Del_h0_rxn = (n_NH3 * del_h0_f_NH3 + n_N2 * del_h0_f_N2 + n_H2 * del_h0_f_H2) * 10^3 ;
+Del_g0_rxn = (n_NH3 * del_g0_f_NH3 + n_N2 * del_g0_f_N2 + n_H2 * del_g0_f_H2) * 10^3 ;
+del_A = n_NH3 * A_NH3 + n_N2 * A_N2 + n_H2 * A_H2 ;
+del_B = n_NH3 * B_NH3 + n_N2 * B_N2 + n_H2 * B_H2 ;
+del_D = n_NH3 * D_NH3 + n_N2 * D_N2 + n_H2 * D_H2 ;
+
+K_298 = exp( - Del_g0_rxn / ( R * T)) ;
+K_T = K_298 * exp( - Del_h0_rxn / R * (1 / T2 - 1 / T)) ;
+A = K_T * P^2 *27 -16 ;
+B = 64 - K_T * P^2 * 108 ;
+C = -64 + K_T * P^2 * 162 ;
+D = -108 * K_T * P^2 ;
+E = 27 * K_T * P^2 ;
+
+//(a)
+mycoeff =[E , D ,C , B ,A];
+p = poly(mycoeff , "x","coeff") ;
+M = roots(p);
+
+for i = 1:3
+ isreal(M(i,1)) ;
+ if ans == %f then
+ y = M(i,1) / M(i+1,1) - 1 ;
+ sign(y) ;
+ if ans == %t then
+ x = M(i,1) ;
+ else
+ x = M(i+1,1) ;
+
+ end
+ end
+end
+disp(" Example: 9.7 Page no : 453") ;
+printf("\n (a)\n Extent of reaction = %.3f \n",x);
+
+//(b)
+X = (-Del_h0_rxn / R + del_A * T + del_B / 2 * T^2 - del_D / T) * (1/T2 - 1/T) + del_A * log(T2 / T) + del_B / 2 * (T2 - T) + del_D / 2 * (1/(T2^2) - 1/(T^2) );
+K_T = K_298 * exp(X) ;
+
+A = K_T * P^2 *27 -16 ;
+B = 64 - K_T * P^2 * 108 ;
+C = -64 + K_T * P^2 * 162 ;
+D = -108 * K_T * P^2 ;
+E = 27 * K_T * P^2 ;
+
+mycoeff =[E , D ,C , B ,A];
+p1 = poly(mycoeff , "x","coeff") ;
+M1 = roots(p1);
+
+for i = 1:3
+ isreal(M1(i,1)) ;
+ if ans == %f then
+ y = M1(i,1) / M1(i+1,1) - 1 ;
+ sign(y) ;
+ if ans == %t then
+ x1 = M1(i,1) ;
+ else
+ x1 = M1(i+1,1) ;
+
+ end
+ end
+end
+printf(" (b)\n Extent of reaction = %.3f\n",x1);
+disp(" Under these conditions we do not expect to produce an appreciable amount of ammonia .")
diff --git a/527/CH9/EX9.8/9_8eaxm.sce b/527/CH9/EX9.8/9_8eaxm.sce new file mode 100755 index 000000000..391e9cc53 --- /dev/null +++ b/527/CH9/EX9.8/9_8eaxm.sce @@ -0,0 +1,7 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.8
+//Page no :454
+
+clear ; clc ;
+disp(" Example: 9.8 Page no : 454") ;
+disp(" The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression .") ;
\ No newline at end of file diff --git a/527/CH9/EX9.9/9_9exam.sce b/527/CH9/EX9.9/9_9exam.sce new file mode 100755 index 000000000..0cd7a0099 --- /dev/null +++ b/527/CH9/EX9.9/9_9exam.sce @@ -0,0 +1,48 @@ +//Engineering and Chemical Thermodynamics
+//Example 9.9
+//Page no :454
+
+clear ; clc;
+//Given
+K_T = 1.51 * 10^-5 ;
+P = 300 ; //[bar]
+T = 500 + 273.2 ; //[K]
+R = 8.314 ;
+
+function y = f991(k),
+ y = ((2 * k)^2 * (4 - 2 * k)^2 / ((1 - k) * (3 - 3*k)^3)) * P^-2 - K_T
+endfunction
+
+z1 = fsolve([0.3],f991) ;
+
+disp(" Example: 9.9 Page no : 454") ;
+printf("\n (a)\n Extent of reaction = %.2f \n",z1);
+
+//(b)
+P_c = [111.3 * 101325 , 33.5 * 101325 , 12.8 * 101325] ;
+T_c = [405.5 , 126.2 , 33.3] ;
+
+for i = 1:3
+ a(1,i) = 27 / 64 * (R * T_c(1,i))^2 / P_c(1,i) ;
+ b(1,i) = (R * T_c(1,i)) / (8 * P_c(1,i)) ;
+
+ function y = f992(v) ,
+ y = (R * T) / (v - b(1,i)) - a(1,i) / (v^2) - P * 100000 ;
+ endfunction
+
+ V(1,i) = fsolve([0.0002],f992) ;
+
+ sai(1,i) = exp( - log((V(1,i) - b(1,i)) * P * 10^5/ ( R * T)) + b(1,i) / (V(1,i) - b(1,i)) - 2 * a(1,i) / (R * T * V(1,i))) ;
+
+end
+
+function y = f993(k),
+ y = ((2 * k)^2 * sai(1,1)^2 * (4 - 2 * k)^2 * 3 / ((1 - k) * sai(1,2)* (3 - 3*k)^3 * sai(1,3)^3 ))* P^-2 - K_T
+endfunction
+
+z2 = fsolve([0.3],f993) ;
+
+x = (z1 - z2) / z1 * 100 ;
+
+printf(" (b)\n Extent of reaction = %.2f \n",z2);
+printf("\n A correction of about %d%% is observed from accounting for nonideal behaviour . ",x)
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