initial commit / add all books

6 files changed, 173 insertions, 0 deletions

diff --git a/409/CH17/EX17.1/Example17_1.sce b/409/CH17/EX17.1/Example17_1.sce
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+clear ;
+clc;
+// Example 17.1
+printf('Example 17.1\n');
+// Page no. 511
+// Solution 
+
+// Basis : F =  1 mol
+F = 1 ;//H2C2O4- [mol]
+ex_O2 =  248 ;//Excess air- [%]
+f_C = 65/100 ;// Fraction of Carbon which convert to CO2
+P = 101.3 ;// Atmospheric pressure-[kPa]
+
+// H2C2O4 + 0.5*O2-->2*CO2  +  H2O
+// H2C2O4 -->2*CO  +  H2O + 0.5*O2
+O2_req = F*0.5 ;// O2 required by the above reaction-[mol]
+O2_in = (1 + ex_O2*F/100)*0.5 ;// Mol. of O2 entering
+
+// Use Elemental balance moles of species in output 
+n_CO2 = f_C*2 ;// [mol]
+n_H2O = (2*F)/2 ;// From 2H balance-[mol]
+n_N2 = ((O2_in*0.79)/(0.21)) ;//  From 2N balance-[mol]
+n_CO = 2-n_CO2 ;// From C balance-[mol]
+n_O2 = ((4 + O2_in*2)-(n_H2O + n_CO + 2*n_CO2))/2 ;// From O2 balance-[mol]
+total_mol = n_CO2 + n_H2O + n_N2 + n_CO + n_O2 ;// Total moles in output stream-[mol]
+y_H2O = n_H2O/total_mol ;// Mole fraction of H2O
+pp_H2O = y_H2O*P ;// Partial pressure of H2O-[kPa]
+
+printf('\nPartial pressure of H2O %.2f kPa.',pp_H2O);
+printf('\nUse partial pressure of H2O to get dew point temperature T from steam table: T  = 316.5 K');
\ No newline at end of file
diff --git a/409/CH17/EX17.2/Example17_2.sce b/409/CH17/EX17.2/Example17_2.sce
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+clear ;
+clc;
+// Example 17.2
+printf('Example 17.2\n');
+//Page no. 517
+// Solution Fig E17.2b
+
+gas =  1 ;// Entering gas-[g mol]
+T = 26 ;// Temperature (for isothermal process)-[degree C]
+// From fig. its clear that at 26 C saturation pressure is at point A
+// Get vapour pressure of benzene from Perry handbook or CD,it is
+vp =  99.7 ;// vapour pressure of benzene at 26 C-[mm of Hg]
+
+// Analysis of entering gas 
+f_C6H6 = 0.018 ;// Mol fraction of benzene
+f_air  =  0.982 ;// Mol fraction of air
+mol_C6H6 = 0.018*gas ;// Moles  of benzene-[g mol]
+mol_air  =  0.982*gas ;// Moles  of air-[g mol]
+
+// Analysis of exit gas
+C6H6_rec =  95/100 ;// Fraction of benzene recovered
+C6H6_out  = 1-C6H6_rec ;//Fraction of benzene in exit stream
+C6H6_out = mol_C6H6*C6H6_out ;//Moles of benzene in exit stream-[g mol]
+air_out = mol_air ;//Moles of air in exit stream-[g mol]
+total_mol = C6H6_out+air_out ;// Total moles in exit stream
+y_C6H6_out = C6H6_out/total_mol ;// Mole fraction of benzene in exit
+P = vp/y_C6H6_out ;// Pressure total of exit
+
+printf('\n Pressure total at exit of compressor %.2e mm of Hg.',P);
\ No newline at end of file
diff --git a/409/CH17/EX17.3/Example17_3.sce b/409/CH17/EX17.3/Example17_3.sce
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index 000000000..f522f54e4
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+clear ;

+clc;

+// Example 17.3

+printf('Example 17.3\n');

+// Page no. 519

+// Solution Fig E17.3b

+

+// Given

+// coal analysis from handbook

+ex_air = .4 ;// Fraction of excess air required

+w_C = 12 ;// Mol. wt. of C-[g]

+mol_C = 71/w_C ;//[kg mol]

+w_H2 = 2.016 ;// Mol. wt. of H2 - [g] 

+mol_H2 = 5.6/w_H2;

+air_O2 = 0.21;// Fraction of O2 in air

+air_N2 = 0.79;// Fraction of N2 in air

+

+// Natural Gas

+// Basis = 1 kg mol C 

+// CH4 + 2O2 --> CO2 + 2H2O .... Eqn. (a)

+CO2_1 = 1 ;//  By Eqn. (a) CO2 produced -[kg mol]

+H2O_1 = 2 ;// By Eqn. (a) H2O produced -[kg mol]

+Req_O2_1 = 2 ;// By Eqn. (a) -[kg mol]

+ex_O2_1 = Req_O2_1*ex_air  ;// Excess O2 required -[kg mol]

+O2_1 = Req_O2_1 + ex_O2_1 ;// Total O2 required - [kg mol]

+N2_1 = O2_1*(air_N2/air_O2) ;//Total N2 required - [kg mol]

+Total_1 = CO2_1 + H2O_1 + N2_1 + ex_O2_1 ;// Total gas produced- [kg mol]

+

+// Coal 

+// C + O2 --> CO2  ..eqn (b)

+// H2 + 1/2(O2) --> H2O.... eqn (c)

+CO2_2 = 1 ;//  By Eqn. (a) CO2 produced -[kg mol]

+H2O_2 = mol_H2/mol_C ;// By Eqn. (a) H2O produced -[kg mol]

+Req_O2_2 = 1 + (mol_H2/mol_C)*(1/2) ;// By Eqn. (b) and (c) -[kg mol]

+ex_O2_2 = Req_O2_2*ex_air  ;// Excess O2 required -[kg mol]

+O2_2 = Req_O2_2 + ex_O2_2; // Total O2 required - [kg mol]

+N2_2 = O2_2*(air_N2/air_O2); //Total N2 required - [kg mol]

+Total_2 = CO2_2 + H2O_2 + N2_2 + ex_O2_2 ;// Total gas produced- [kg mol]

+

+// Let P (total pressure) = 100 kPa

+P = 100 ;// Total pressure -[kPa]

+p1 = P*(H2O_1/Total_1) ;// Partial pressure of water vapour in natural gas - [kPa]

+Eq_T1 = 52.5  ;// Equivalent temperature -[degree C]

+p2 = P*(H2O_2/Total_2) ;// Partial pressure of water vapour in coal - [kPa]

+Eq_T2 = 35  ;// Equivalent temperature -[degree C]

+printf('                                Natural gas                         Coal\n')

+printf('                            ----------------------             --------------------\n')

+printf('Partial pressure:                %.1f kPa                            %.1f kPa\n',p1,p2 ) ;

+printf('Equivalent temperature:          %.1f C                              %.1f C\n',Eq_T1,Eq_T2 );
\ No newline at end of file
diff --git a/409/CH17/EX17.4/Example17_4.sce b/409/CH17/EX17.4/Example17_4.sce
new file mode 100755
index 000000000..eccc583d1
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+++ b/409/CH17/EX17.4/Example17_4.sce
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+clear ;
+clc;
+// Example 17.4
+printf('Example 17.4\n\n');
+//Page no. 522
+// Solution Fig E17.4
+
+F = 30 ;// Volume of initial gas-[m^3]
+P_F =  98.6 ;// Pressure of gas-[kPa]
+T_F =  273+100 ;// Temperature of gas-[K]
+P_p = 109 ;//[kPa]
+T_p = 14+273 ;// Temperature of gas-[K]
+R = 8.314 ;// [(kPa*m^3)/(k mol*K)] 
+// Additional condition
+vpW_30 = 4.24 ;//Vapour pressure-[kPa]
+vpW_14 = 1.60 ;//Vapour pressure-[kPa]
+n_F = (P_F*F)/(R*T_F) ;// Number of moles in F
+
+// Material balance to calculate  P & W
+P = (n_F*((P_F-vpW_30)/P_F))/((P_p-vpW_14)/P_p) ;// P from mat. bal. of air -[kg mol]
+W = (n_F*(vpW_30/P_F))- P*(vpW_14/P_p); // W from mat. bal. of water -[kg mol]
+iW = n_F*(vpW_30/P_F) ;// Initial amount of water -[kg mol]
+fr_con = W/iW ;//Fraction of water condenseed 
+
+printf('\n Fraction of water condenseed %.3f.',fr_con);
\ No newline at end of file
diff --git a/409/CH17/EX17.5/Example17_5.sce b/409/CH17/EX17.5/Example17_5.sce
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index 000000000..e5b16face
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+clear;
+clc;
+// Example 17.5
+printf('Example 17.5\n');
+//Page no. 527
+// Solution Fig E17.5
+
+P =  100 ;// Pressure of air-[kPa]
+T =  20 + 273 ;// Temperature of air-[K]
+R = 8.314 ;// [(kPa*m^3)/(k mol*K)] 
+EOH =  6 ;// Amount of ethyl alcohol to evaporate-[kg]
+mw_EOH = 46.07 ;// Mol.wt. of 1 k mol ethyl alcohol-[kg]
+// Additional data needed
+vp_EOH = 5.93 ;// Partial pressure of alcohol at 20 C-[kPa]
+vp_air = P-vp_EOH ;// Partial pressure of air at 20 C-[kPa]
+n_EOH  = EOH/mw_EOH ;//Moles of ethyl alcohol -[kg mol]
+n_air = (n_EOH*vp_air)/vp_EOH ;// Moles of air -[kg mol]
+V_air = n_air*R*T/P ;// Volume of air required
+
+printf('\n Volume of air required to evaporate 6 kg of ethyl alcohol is %.1f cubic metre . \n',V_air);
\ No newline at end of file
diff --git a/409/CH17/EX17.6/Example17_6.sce b/409/CH17/EX17.6/Example17_6.sce
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index 000000000..ccf7fc101
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+++ b/409/CH17/EX17.6/Example17_6.sce
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+clear ;
+clc;
+// Example 17.6
+printf('Example 17.6\n\n');
+//Page no. 529
+// Solution 
+
+P =  760 ;// Pressure -[ mm of Hg]
+// Get vapour pressure of n-heptane from Perry, 40 mm of Hg
+vp = 40 ;// vapour pressure of n-heptane-[mm of Hg]
+
+// Use the 2nd relation given in problem to find K
+K = 10^((log10(vp/P)-0.16)/1.25) ;
+
+// Get t using the 1st relation in the question
+// For t_half
+x = 0.5 ;// mole fraction after  t_half
+x0 = 1 ;// initial mole fraction 
+t_half = (log(x/x0))/(-K);// Time required to reduce the concentration to one-half-[min]
+printf('Time required to reduce the concentration to one-half is %.1f min. \n',t_half);
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