diff options
Diffstat (limited to '409/CH26')
| -rwxr-xr-x | 409/CH26/EX26.1/Example26_1.sce | 33 | ||||
| -rwxr-xr-x | 409/CH26/EX26.2/Example26_2.sce | 59 | ||||
| -rwxr-xr-x | 409/CH26/EX26.3/Example26_3.sce | 55 | ||||
| -rwxr-xr-x | 409/CH26/EX26.4/Example26_4.sce | 59 | ||||
| -rwxr-xr-x | 409/CH26/EX26.5/Example26_5.sce | 54 |
5 files changed, 260 insertions, 0 deletions
diff --git a/409/CH26/EX26.1/Example26_1.sce b/409/CH26/EX26.1/Example26_1.sce new file mode 100755 index 000000000..b12a393d9 --- /dev/null +++ b/409/CH26/EX26.1/Example26_1.sce @@ -0,0 +1,33 @@ +clear ; +clc; +// Example 26.1 +printf('Example 26.1\n\n'); +//page no. 804 +// Solution + +printf('Table to carry out degree of freedom analysis:\n'); +// Number of variables involved +printf('\nI. Number of variables involved.\n'); +printf(' Species in F1 1 \n'); +printf(' Species in F2 2\n'); +printf(' Specie in F3 5\n'); +printf(' Total stream flows 3\n'); +printf(' Stream temperatures 3\n'); +printf(' Stream pressures 3 \n'); +printf(' Q 1 \n'); +printf(' Extent of reactions 2\n'); +printf('\__________________________________________________________________________________________\n'); +printf(' Total 20\n'); +printf('\n\nII. Number of equations and specifications.\n'); +printf(' Independent species material balances 6\n'); +printf(' Sum of species in each of the two streams 2 \n'); +printf(' Energy balance 1\n'); +printf(' Total stream flows 2\n'); +printf(' Species values(CO) 1\n'); +printf(' Pressures 3 \n'); +printf(' Temperatures 2 \n'); +printf(' O2 to N2 ratio specified in F2 1 \n'); +printf(' Complete reaction, hence the extent of reaction is implied to both reactions 2\n'); +printf('\___________________________________________________________________________________________\n'); +printf(' Total 20\n'); +printf('\n Therefore, by analysing the above table it is clear that degrees of freedom of system is (20 - 20) = 0 \n');
\ No newline at end of file diff --git a/409/CH26/EX26.2/Example26_2.sce b/409/CH26/EX26.2/Example26_2.sce new file mode 100755 index 000000000..fd75626e8 --- /dev/null +++ b/409/CH26/EX26.2/Example26_2.sce @@ -0,0 +1,59 @@ +clear ; +clc; +// Example 26.2 +printf('Example 26.2\n\n'); +//page no. 808 +// Solution + +// Given +// The main reaction is CO(g,1 atm,100 C) + (1/2)O2(g,1 atm,100 C) --> CO2(g,1at,T K) (A) +// Input compounds +m1_CO = 1; // Moles of CO input- [g mol] +m1_O2 = 1 ;// Moles of O2 input - [g mol] +m1_N2 = 3.76 ;// Moles of N2 input - [g mol] +//Output compounds +m2_CO2 = 1 ;// Moles of CO2 output - [g mol] +m2_O2 = .50 ;// Moles of O2 output - [g mol] +m2_N2 = 3.76 ;// Moles of N2 output - [g mol] + +// Additional data is obtained from Appendix D, according to book it is as follows: +// Inputs +H1_fCO = -110520 ;// Heat of formation of CO - [J/g mol] +H1_fO2 = 0 ;// Heat of formation of O2 - [J/g mol] +H1_fN2 = 0 ;// Heat of formation of N2 - [J/g mol] +H1_CO = 2917 - 728 ;// Change in enthalpy during temperature change from 298K to 373 K of CO - [J/g mol] +H1_O2 = 2953 - 732 ;// Change in enthalpy during temperature change from 298K to 373 K of input O2 - [J/g mol] +H1_N2 = 2914 - 728 ;// Change in enthalpy during temperature change from 298K to 373 K of input N2 - [J/g mol] + +H_in = (H1_fCO + H1_CO)*m1_CO + (H1_fO2 + H1_O2)*m1_O2 + (H1_fN2 + H1_N2)*m1_N2;// Enthalpy change for inputs -[J] + +//Outputs - Assume it to be at 2000 K +H2_fCO2 = -393510 ;// Heat of formation of CO2 - [J/g mol] +H2_fO2 = 0 ;// Heat of formation of O2 - [J/g mol] +H2_fN2 = 0 ;// Heat of formation of N2 - [J/g mol] +H2_CO2 = 92466 - 912 ;// Change in enthalpy during temperature change from 298K to 2000 K of CO2 - [J/g mol] +H2_O2 = 59914-732 ;// Change in enthalpy during temperature change from 298K to 2000 K of output O2 - [J/g mol] +H2_N2 = 56902 - 728 ;// Change in enthalpy during temperature change from 298K to 2000 K of output O2 - [J/g mol] + +H1_out = (H2_fCO2 + H2_CO2)*m2_CO2 + (H2_fO2 + H2_O2)*m2_O2 + (H2_fN2 + H2_N2)*m2_N2 ;// Enthalpy change for outputs at 2000 K -[J] + +del_H1 = H1_out - H_in ;// Net enthalpy change of process -[J] + +//Output- Assume it to be at 1750 K +H2_fCO2 = -393510 ;// Heat of formation of CO2 - [J/g mol] +H3_CO2 = 77455 - 912 ;// Change in enthalpy during temperature change from 298K to 1750 K of CO2 - [J/g mol] +H3_O2 = 50555 -732 ;// Change in enthalpy during temperature change from 298K to 1750 K of output O2 - [J/g mol] +H3_N2 = 47940 - 728 ;// Change in enthalpy during temperature change from 298K to 1750 K of output O2 - [J/g mol] + +H2_out = (H2_fCO2 + H3_CO2)*m2_CO2 + (H2_fO2 + H3_O2)*m2_O2 + (H2_fN2 + H3_N2)*m2_N2 ;// Enthalpy change for outputs at 1750 K -[J] + +del_H2 = H2_out - H_in ;// Net enthalpy change of process -[J] + +printf('Heat of above reaction when output is assumed to be at 2000 K is %.0f J.\n',del_H1) ; +printf(' Heat of above reaction when output is assumed to be at 1750 K is %.0f J.\n',del_H2) ; +// Energy balance here reduce to del_H = 0 +printf('\n So we can see that our desired result del_H = 0 is bracketed between 2000 K and 1750 K , hence we will use interpolation to get the theoretical flame temperature.\n') ; +// Use interpolation to get the theoretical flame temperature +del_H = 0 ;// Requred condition +Ft = 1750 + ((del_H - del_H2)/(del_H1 - del_H2))*(2000 - 1750) ;// Interpolation to get Flame temperature(Ft)-[K] +printf(' Theoretical flame temperature by interpolation is %.0f K.\n',Ft) ;
\ No newline at end of file diff --git a/409/CH26/EX26.3/Example26_3.sce b/409/CH26/EX26.3/Example26_3.sce new file mode 100755 index 000000000..5881bd147 --- /dev/null +++ b/409/CH26/EX26.3/Example26_3.sce @@ -0,0 +1,55 @@ +clear ;
+clc;
+// Example 26.3
+printf('Example 26.3\n\n');
+//page no. 811
+// Solution Fig E26.3b
+
+// Given
+v_CH4 = 1000 ;// Volume of CH4 taken - [ cubic feet]
+CH4 = 1 ;// assumed for convenience- [ g mol]
+ex_air = .5 ;// Fraction of excess O2 required
+hp_CaCO3 = 0.130 ;// Heat capacity of CaCO3 -[kJ/g mol]
+hp_CaO = 0.062 ;// Heat capacity of CaO -[kJ/g mol]
+w_CaCO3 = 100.09 ;// Mol. wt. of CaCO3 -[g]
+w_CaO = 56.08 ;// Mol. wt. ofCaO - [g]
+
+// The main reaction are -
+// (a) CaCO3(s,25 C) --> CaO(s,900 C) + CO2(g,500 C)
+// (b) CH4(g,25 C) + 2O2(g,25 C) --> CO2 (g,500 C) + 2H2O (g,500 C)
+
+req_O2 = 2 ;// By eqn. (b), O2 required by CH4 - [g mol]
+ex_O2 = ex_air*req_O2 ;// Excess O2 required - [ g mol]
+O2 = req_O2 + ex_O2 ;// Total O2 entering - [ g mol]
+N2 = O2 *(.79/.21) ;// Total N2 entering - [ g mol]
+
+// By analysis DOF is zero.
+
+// Carry out elemental balance to get the unknowns
+nG_N2 = N2 ;// N2 balance - [ g mol]
+nG_H2O = 4*CH4/2 ;// H2O balance - [ g mol]
+nG_O2 = ex_O2 ;// [g mol]
+// L = P , from Ca balance ...eqn. (A)
+// 1 + L = nG_CO2 , from C balance ...eqn. (B)
+// 3L + 2*O2 = 2*nG_CO2 + 2*nG_O2 + nG_H2O + P , from O balance ... eqn. (C)
+
+// For energy balance, get required data from software in the CD of book and sensible heat data from Appendix F
+// given data of outputs is taken in array in order CO2(g), O2(g),N2(g),H2O(g) and then CaO(s)
+del_Hi_out = [ -393.250,0,0,-241.835,-635.6] ;// // Heat of formation - [kJ/g mol]
+del_Hf_out = [21.425,15.043,14.241,17.010,54.25] ;//Change in enthalpy during temperature change -[kJ/g mol]
+del_H_out =del_Hi_out + del_Hf_out ;// Change in enthalpy final - [kJ/g mol]
+
+// given data of inputs is taken in array in order CH4(g), CaCO3(s),O2(g) and N2(g)
+del_Hi_in = [ -49.963,-1206.9,0,0] ;// // Heat of formation - [kJ/g mol]
+del_Hf_in = [0,0,0,0] ;//Change in enthalpy during temperature change -[kJ/g mol]
+del_H_in = del_Hi_in + del_Hf_in ;// Change in enthalpy final - [kJ/g mol]
+// Now do energy balance , assume Q = 0 ,
+// del_H_out(1)*nG_CO2 + del_H_out(2)*nG_O2 +del_H_out(3)*nG_N2 + del_H_out(4)*nG_H2O + del_H_out(5)*P = del_H_in(1)*CH4 + del_H_in(2)*L ... eqn. (D)
+// Solve eqn. (A), (B), (C), and (D) to get L ,P , nG_CO2
+a = [1 -1;(del_H_in(2)-del_H_out(5)) -del_H_out(1)] ;// Matrix of coefficients
+b = [-1;(del_H_out(2)*nG_O2 + del_H_out(3)*nG_N2 +del_H_out(4)*nG_H2O-del_H_in(1)*CH4)] ;// Matrix of constants
+x = a\b ;// Matrix of solutions, L = x(1), nG_CO2 = x(2)
+g_CaCO3 = x(1)*w_CaCO3 ;//CaCO3 processed for each g mol of CH4 - [g]
+printf(' CaCO3 processed for each g mol of CH4 is %.0f g.\n',g_CaCO3) ;
+m_CaCO3 = (v_CH4*g_CaCO3)/359.05 ;
+printf(' Therefore, CaCO3 processed per 1000 ft^3 of CH4 is %.0f lb.\n',m_CaCO3) ;
\ No newline at end of file diff --git a/409/CH26/EX26.4/Example26_4.sce b/409/CH26/EX26.4/Example26_4.sce new file mode 100755 index 000000000..0fff42d68 --- /dev/null +++ b/409/CH26/EX26.4/Example26_4.sce @@ -0,0 +1,59 @@ +clear;
+clc;
+// Example 26.4
+printf('Example 26.4\n\n');
+//page no. 815
+// Solution Fig E26.4b
+
+// Given
+SO2_in = 2200 ;// Amount of SO2 entering reactor 2-[lb mol/hr]
+// Basis : 1 lb mol CO entering reactor 1,therefore
+R1_CO_in = 1 ;//CO entering reactor 1-[lb mol]
+air = .80 ;// Fraction of air used in burning
+
+// System- reactor 2
+// Given
+R2_fSO2_in = 0.667 ;// Fraction of SO2 entering reactor 2
+R2_fO2_in = 0.333 ;// Fraction of O2 entering reactor 2
+R2_fSO3_out = 0.586 ;// Fraction of SO3 exiting reactor 2
+R2_fSO2_out = 0.276 ;// Fraction of SO2 exiting reactor 2
+R2_fO2_out = 0.138 ;// Fraction of O2 exiting reactor 2
+// Main Reaction: CO , (1/2)*O2 ---> CO2
+R1_O2_in = (1/2)*air ;// O2 entering reactor 1-[g mol]
+R1_N2_in = R1_O2_in*(79/21) ;// N2 entering reactor 1-[g mol]
+
+//Output of reactor 1
+R1_CO_out = R1_CO_in*(1 - air) ;// [g mol]
+R1_CO2_out = 1*( air) ;// [g mol]
+R1_N2_out = R1_N2_in ;//[g mol]
+
+// By analysis DOF is zero.
+// Get eqn. to solve by species balance
+//Unknowns - P- exit stream of reactor 2 , F - entry stream of reactor 2 , ex - extent of reaction
+// P*(R2_fSO2_out) - F*0 = 1*ex ... eqn.(a)- By SO3 balance
+// P*(R2_fSO2_out) - F*(R2_fSO2_in) = -1*ex ...eqn.(b) - By SO2 balance
+// By O2 balance we will get eqn. equivalent to eqn. (b), so we need one more eqn.
+
+// Energy balance
+// For energy balance, get required data from software in the CD of book and sensible heat data from Appendix F
+// given data of outputs is taken in array in order CO(g),CO2(g), N2(g),SO2(g),SO3(g) and then O2(g)
+del_Hi_out = [ -109.054,-393.250,0,-296.855,-395.263,0] ; // Heat of formation - [kJ/g mol]
+del_Hf_out = [35.332,35.178,22.540,20.845,34.302,16.313] ;//Change in enthalpy during temperature change -[kJ/g mol]
+del_H_out =del_Hi_out + del_Hf_out ;//[-371.825,15.043,160.781,-449.650,-581.35]// Change in enthalpy final - [kJ/g mol]
+
+// given data of inputs is taken in array in order CO(g),CO2(g), N2(g),SO2(g) and then O2(g)
+del_Hi_in = [ -109.054,-393.250,0,-296.855,0] ;// // Heat of formation - [kJ/g mol]
+del_Hf_in = [17.177,17.753,11.981,0,0] ;//Change in enthalpy during temperature change -[kJ/g mol]
+del_H_in = del_Hi_in+ del_Hf_in ;// Change in enthalpy final - [kJ/g mol]
+// Now do energy balance , assume Q = 0 ,
+// del_H_out(4)*P*R2_fSO2_out + del_H_out(5)*P*R2_fSO3_out - del_H_in(4)*F*R2_fSO2_in + del_Hi_out(6)*P*R2_fO2_out = 0 ... eqn. (c)
+
+// Solve eqn. (a), (b) and (c) to get F ,P , ex
+a = [(R2_fSO3_out) 0 -1;(R2_fSO2_out) -(R2_fSO2_in) 1;(del_H_out(4)*R2_fSO2_out + del_H_out(5)*R2_fSO3_out + del_Hi_out(6)*R2_fO2_out ) -(del_H_in(4)*R2_fSO2_in) 0] ;// Matrix of coefficients
+b = [0;0;(del_H_in(1)*R1_CO_out+del_H_in(2)*R1_CO2_out+del_H_in(3)*R1_N2_out-(del_H_out(1)*R1_CO_out+del_H_out(2)*R1_CO2_out+ del_H_out(3)*R1_N2_out))] ;// Matrix of constants
+x = a\b ;// Matrix of solutions, P = x(1), F = x(2) ,ex = x(3)
+F = x(2) ;//exit stream of reactor 2 - [lb mol]
+R2_SO2_in = R2_fSO2_in*F ;// Moles of SO2 required per lb mol of CO - [lb mol]
+CO = (R1_CO_in*SO2_in)/R2_SO2_in ;//Mole of CO burned in reactor 1 - [lb mol]
+
+printf('Mole of CO burned in reactor 1 is %.0f lb mol.\n',CO) ;
\ No newline at end of file diff --git a/409/CH26/EX26.5/Example26_5.sce b/409/CH26/EX26.5/Example26_5.sce new file mode 100755 index 000000000..e7cb0b861 --- /dev/null +++ b/409/CH26/EX26.5/Example26_5.sce @@ -0,0 +1,54 @@ +clear ;
+clc;
+// Example 26.5
+printf('Example 26.5\n\n');
+//page no. 819
+// Solution
+
+// Given
+CA = 10000 ;// Produced citric acid - [kg]
+f_glucose = .30 ;// Fraction of glucose in solution
+con_glucose = .60 ;// Fraction of glucose consumed
+w_glucose = 180.16 ;// Mol. wt. of d,alpha glucose -[g]
+H_glucose = -1266 ;// Specific enthalpy change of glucose - [kJ/g mol]
+w_CA = 192.12; // Mol. wt. of citric acid -[g]
+H_CA = -1544.8 ;// Specific enthalpy change of citric acid - [kJ/g mol]
+w_BM = 28.6 ;// Mol. wt. of biomass -[g]
+H_BM = -91.4 ;// Specific enthalpy change of biomass - [kJ/g mol]
+H_CO2 = -393.51 ;// Specific enthalpy change of CO2 - [kJ/g mol]
+
+// Main reaction is :
+// 3 * glucose + 7.8*O2 ---> 5.35*BM + 2.22*CA + 4.50*CO2 ..reaction (a)
+
+// Material Balance
+mol_CA = CA/w_CA ;// Mole of citric acid produced - [kg mol]
+g_soln = (mol_CA*(3/2.22)*w_glucose*1)/(con_glucose*f_glucose) ;//Mass of 30 % glucose solution introduced -[kg]
+i_glucose = g_soln* f_glucose / w_glucose ;// Initial moles of glucose - [kg mol]
+f_glucose = (1 - con_glucose)*i_glucose ;// Final moles of glucose - [kg mol]
+f_CA = mol_CA ;// Final moles of citric acid - [kg mol]
+f_BM = f_CA*(5.35/2.22) ;// Using the reaction (a)- Final moles of biomass - [kg mol]
+i_O2 = i_glucose*(7.8/3) ;// Using the reaction (a)- Initial moles of O2 - [kg mol]
+f_CO2 = i_glucose*(4.5/3)*con_glucose ;// Using the reaction (a) - Final moles of CO2 - [kg mol]
+
+// Energy balance
+// For closed system - del_U = Q + W
+power = 100 ;// Power of aerator -[hp]
+time = 220 ;// Time taken for reaction - [ hr ]
+W = (power*745.7*time*3600)/1000 ;// Work done by aerator - [kJ]
+
+// Assume del_U = del_H , pv work is equal to zero ,hence
+// Q = del_H - W
+
+Hi_glucose = i_glucose*H_glucose*1000 ;// Enthalpy change of glucose input - [kJ]
+Hi_O2 = i_O2*0*1000 ;// Enthalpy change of O2 input - [kJ]
+H_in = Hi_glucose + Hi_O2 ;// Enthalpy change of input - [kJ]
+
+Hf_glucose = f_glucose*H_glucose*1000 ;// Enthalpy change of glucose output - [kJ]
+Hf_BM = f_BM * H_BM*1000 ;//Enthalpy change of biomass output - [kJ]
+Hf_CA = f_CA *H_CA*1000 ;//Enthalpy change of citric acid output - [kJ]
+Hf_CO2 = f_CO2 *H_CO2*1000 ;//Enthalpy change of CO2 output - [kJ]
+H_out = Hf_glucose + Hf_BM +Hf_CA + Hf_CO2 ;// Enthalpy change of output - [kJ]
+del_H = H_out - H_in ;// Total enthalpy change in process - [kJ]
+Q = del_H - W ;// Heat removed - [kJ]
+
+printf('Heat exchange from the fermentor during production of 10,000 kg citric acid is %.2e kJ(minus sign indicates heat is removed).\n',Q) ;
\ No newline at end of file |