diff options
Diffstat (limited to '409/CH13')
| -rwxr-xr-x | 409/CH13/EX13.1/Example13_1.sce | 12 | ||||
| -rwxr-xr-x | 409/CH13/EX13.2/Example13_2.sce | 12 | ||||
| -rwxr-xr-x | 409/CH13/EX13.3/Example13_3.sce | 24 | ||||
| -rwxr-xr-x | 409/CH13/EX13.4/Example13_4.sce | 20 | ||||
| -rwxr-xr-x | 409/CH13/EX13.5/Example13_5.sce | 23 | ||||
| -rwxr-xr-x | 409/CH13/EX13.6/Example13_6.sce | 21 | ||||
| -rwxr-xr-x | 409/CH13/EX13.7/Example13_7.sce | 68 | ||||
| -rwxr-xr-x | 409/CH13/EX13.8/Example13_8.sce | 34 |
8 files changed, 214 insertions, 0 deletions
diff --git a/409/CH13/EX13.1/Example13_1.sce b/409/CH13/EX13.1/Example13_1.sce new file mode 100755 index 000000000..4db43d9df --- /dev/null +++ b/409/CH13/EX13.1/Example13_1.sce @@ -0,0 +1,12 @@ +clear; +clc; +// Example 13.1 +printf('Example 13.1\n\n'); +//Page No. 404 +// Solution + +m_CO2 = 40 ;// Mass of CO2-[kg] +mol_wt_CO2 = 44 ;// Molecular mass of 1kmol CO2 -[kg] +mol_V = 22.42 ;// Molar of ideal gas at standard condition-[cubic metre/kg mol] +V_CO2 = (m_CO2 * mol_V)/(mol_wt_CO2);// volume of CO2-[cubic metre] + printf('Volume occupied by 40 kg CO2 at standard condition is %.1f cubic metre.',V_CO2);
\ No newline at end of file diff --git a/409/CH13/EX13.2/Example13_2.sce b/409/CH13/EX13.2/Example13_2.sce new file mode 100755 index 000000000..a73df1f29 --- /dev/null +++ b/409/CH13/EX13.2/Example13_2.sce @@ -0,0 +1,12 @@ +clear ; +clc; +// Example 13.2 +printf('Example 13.2\n\n'); +//Page No. 405 +// Solution + +p =1 ;// Pressure -[atm] +V = 22415 ;// Molar valume -[cubic centimetre/g mol] +T = 273.15 ;// Temperature-[K] +R = (p*V/T);// Universal gas constant-[(cubic centimetre.atm)/(K.g mol)] + printf('Universal gas constant is %.2f (cubic centimetre*atm)/(K*g mol). ',R);
\ No newline at end of file diff --git a/409/CH13/EX13.3/Example13_3.sce b/409/CH13/EX13.3/Example13_3.sce new file mode 100755 index 000000000..8c444afe7 --- /dev/null +++ b/409/CH13/EX13.3/Example13_3.sce @@ -0,0 +1,24 @@ +clear ; +clc; + // Example 13.3 + printf('Example 13.3\n\n'); + // Page No.406 + // Solution + +m_CO2 = 88 ;// Mass of CO2-[lb] +mol_wt_CO2 = 44 ;// Molecular mass of 1 lb mol CO2 -[lb] +mol_V = 359 ; // Molar volume-[cubic feet] + + // State 1-standard condition +P1 = 33.91 ; // Pressure -[ft of water] +T1 = 273 ;// Temperature-[K] + + // State 2 +P2 = 32.2 ;// Pressure -[ft of water] +Tc = 15 ;// Temperature-[degree C] +T2 = Tc+273 ;// Temperature-[K] + + // Use eqn. 13.2 to get final volume +V1 = (m_CO2 * mol_V) / (mol_wt_CO2); +V2 = (V1 * T2 * P1) / (T1 * P2); + printf('The volume occupied 88 lb of CO2 at given condition is %.0f cubic feet.',V2);
\ No newline at end of file diff --git a/409/CH13/EX13.4/Example13_4.sce b/409/CH13/EX13.4/Example13_4.sce new file mode 100755 index 000000000..209c3ac9c --- /dev/null +++ b/409/CH13/EX13.4/Example13_4.sce @@ -0,0 +1,20 @@ +clear ; +clc; +// Example 13.4 +printf('Example 13.4\n\n'); +//Page No. 408 +// Solution + +mol_wt_N2 = 28 ;// Molecular mass of 1 kg mol N2 -[kg] +mol_V = 22.42 ;// Molar of ideal gas at standard condition-[cubic metre/kg mol] +Tc = 27 ;// Temperature-[degree C] +T = Tc + 273 ;//Temperature-[K] +P = 100 ;//Pressure-[kPa] + +//Standard condition +Ps = 101.3 ;// Pressure -[kPa] +Ts = 273 ;//Temperature-[K] + +V = (T * Ps * mol_V)/(Ts * P) ;// Volume occupied by N2-[cubic metre] +D_N2 = mol_wt_N2/V ;// Density of N2 at given condition-[kg/cubic metre] + printf(' Density of N2 at given condition is %.3f kg/cubic metre.',D_N2);
\ No newline at end of file diff --git a/409/CH13/EX13.5/Example13_5.sce b/409/CH13/EX13.5/Example13_5.sce new file mode 100755 index 000000000..6f5faa443 --- /dev/null +++ b/409/CH13/EX13.5/Example13_5.sce @@ -0,0 +1,23 @@ +clear ; +clc; +// Example 13.5 +printf('Example 13.5\n\n'); +//Page No. 409 +// Solution + +mol_wt_N2 = 28 ;// Molecular mass of 1 lb mol N2 -[lb] +mol_wt_air = 29 ;// Molecular mass of 1 lb mol air -[lb] +mol_V = 359 ;// Molar volume of ideal gas-[cubic feet] +//Given condition +Tf = 80 ;// Temperature-[degree F] +T = Tf + 460 ;//Temperature-[degree Rankine] +P = 745 ;//Pressure-[mm of Hg] + +//Standard condition +Ps = 760 ;// Pressure -[mm of Hg] +Ts = 492 ;//Temperature-[degree Rankine] + +D_air = (Ts * P * mol_wt_air)/(T * Ps * mol_V) ;// Density of air at given condition-[lb/cubic feet] +D_N2 = (Ts * P * mol_wt_N2)/(T * Ps * mol_V) ;// Density of N2 at given condition-[lb/cubic feet] +sg_N2 = D_N2/D_air ;// Specific gravity of N2 compared to air at given condition + printf(' Specific gravity of N2 compared to air at given condition is %.3f .',sg_N2);
\ No newline at end of file diff --git a/409/CH13/EX13.6/Example13_6.sce b/409/CH13/EX13.6/Example13_6.sce new file mode 100755 index 000000000..baa2d488f --- /dev/null +++ b/409/CH13/EX13.6/Example13_6.sce @@ -0,0 +1,21 @@ +clear ; +clc; +// Example 13.6 +printf('Example 13.6\n\n'); +//Page No. 414 +// Solution + +F_gas = 1 ;// Flue gas [kg mol] +mf_CO2 = 14/100 ;// [mol fraction] +mf_O2 = 6/100 ;// [mol fraction] +mf_N2 = 80/100 ;// [mol fraction] +P = 765 ;//Pressure-[mm of Hg] +T = 400 ;// Temperature-[degree F] +p_CO2 = P * mf_CO2 ;// Partial pressure of CO2-[mm of Hg] +p_O2 = P * mf_O2 ;// Partial pressure of O2-[mm of Hg] +p_N2 = P * mf_N2 ;// Partial pressure of N2-[mm of Hg] + + printf(' Component pi(Partial pressure-[mm of Hg]) \n'); + printf(' CO2 %.1f mm of Hg\n ',p_CO2); + printf(' O2 %.1f mm of Hg\n ',p_O2); + printf(' N2 %.1f mm of Hg\n ',p_N2);
\ No newline at end of file diff --git a/409/CH13/EX13.7/Example13_7.sce b/409/CH13/EX13.7/Example13_7.sce new file mode 100755 index 000000000..27b9c3e48 --- /dev/null +++ b/409/CH13/EX13.7/Example13_7.sce @@ -0,0 +1,68 @@ +clear ; +clc; +// Example 13.7 +printf('Example 13.7\n\n'); +//Page no. 416 +// Solution fig E13.7 + +G = 100 ;// Basis: Pyrolysis Gas-[lb mol] +ub_CO = 10/100 ;// fraction of CO left unburnt +ex_air = 40/100 ;;// fraction of excess air +m_vol = 359 ;// molar volume of gas at std. cond.-[cubic feet] +Ts = 492 ;// Standard temperature -[degree Rankine] +Ps = 29.92 ;//Standard pressure -[in. Hg] + +// Analysis of entering gas of entering gas +Tf1 = 90 ;// Temperature of gas-[degree F] +T_gas = Tf1 + 460 ;//Temperature of gas-[degree Rankine] +P_gas = 35 ;//Pressure-[in. Hg] +CO2 = 6.4/100 ;// mol fraction of CO2 +O2 = 0.1/100 ;// mol fraction of O2 +CO = 39/100 ;// mol fraction of CO +H2 = 51.8/100 ;// mol fraction of H2 +CH4 = 0.6/100 ;// mol fraction of CH4 +N2 = 2.1/100 ;// mol fraction of N2 + +// Analysis of entering air +Tf2 = 70 ;// Temperature of air -[degree F] +T_air = Tf2 + 460 ;//Temperature of air-[degree Rankine] +P_air = 29.4 ;//Pressure of air [in. Hg] +f_N2 = 79/100 ;// mol fraction of N2 +f_O2 = 21/100 ;// mol fraction of O2 + +// Get O2 required for combustion of CO,H2 & CH4 according to the following equation +// CO + 1/2O2-->CO2 +//H2 + 1/2O2-->H20 +//CH4 + 2O2--> CO2 + 2H2O +O2r_O2 = O2 * G ;// O2 required by O2-[lb mol] +O2r_CO = CO * G/2 ;// O2 required by CO-[lb mol] +O2r_H2 = H2 * G/2 ;// O2 required by H2-[lb mol] +O2r_CH4 = G * CH4 * 2 ;// O2 required by CH4-[lb mol] +O2r_total = O2r_O2 + O2r_CO + O2r_H2 + O2r_CH4 ;// Total O2 required-[lb mol] +ex_O2 = ex_air * O2r_total ;// Excess O2-[lb mol] +total_O2 = ex_O2 + O2r_total ;// Total amt of O2 in air-[lb mol] +total_N2 = total_O2 * (f_N2/f_O2);// Total amt of in air-[lb mol] +air = total_O2 + total_N2 ;// Total air entering -[lb mol] + +// Product analysis +P_CO = ub_CO * CO * G ;//Unburnt CO in P-[lb mol] +//Element balance of 2N +P_N2 = N2 * G + total_N2 ;// N2 in P-[lb mol] +//Element balance of C +P_CO2 = (CO2 + CO + CH4) * G - 1 * P_CO;//CO2 in P-[lb mol] +// Element balance of 2H +P_H2O = (H2 + 2 * CH4) * G ;// H2 in P-[lb mol] +// Element balance of 2O +P_O2 = (CO2 + O2 + 0.5 * CO) * G + total_O2 -P_CO2-0.5 * (P_H2O + P_CO);// O2 in P-[lb mol] +P = P_CO + P_N2 + P_CO2 + P_H2O + P_O2 ;// Product-[lb mol] +Tf3 = 400 ;// Temperature of product-[degree F] +T_prod = Tf3 + 460 ;//Temperature of product-[degree Rankine] +P_prod = 35 ;// Pressure of product -[in.Hg] +V_gas = (G * m_vol * T_gas * Ps)/(Ts * P_gas); +V_air = (air * m_vol * T_air * Ps)/(Ts * P_air); +V_prod = (P * m_vol * T_prod * Ps)/(Ts * P_prod); +air_ft3 = V_air/V_gas ;//Air supplied per ft^3 of gas entered-[cubic feet] +P_ft3 = V_prod/V_gas ;//Product gas produced per ft^3 of gas entered-[cubic feet] + +printf(' Air supplied per ft^3 of gas entered %.2f cubic feet.\n ',air_ft3); +printf(' Product gas produced per ft^3 of gas entered %.2f cubic feet.\n',P_ft3); diff --git a/409/CH13/EX13.8/Example13_8.sce b/409/CH13/EX13.8/Example13_8.sce new file mode 100755 index 000000000..ef84c3a4c --- /dev/null +++ b/409/CH13/EX13.8/Example13_8.sce @@ -0,0 +1,34 @@ +clear ; +clc; +//Page No. 419 +// Example 13.8 +printf('Example 13.8\n\n'); +// Solution fig E13.8 + +T1c = 15 ;// Temperature of F & P -[degree C] +T1 = 273 + T1c ;// Temperature of F & P -[K] +P1 = 105 ;// Pressure of F & P -[kPa] +// F analysis +F_CO2 = 1.2/100 ;// Volume fraction +F_odr = 98.8/100 ;// Volume fraction + +// P analysis +P_CO2 = 3.4/100 ;// Volume fraction +P_odr = 96.6/100 ;// Volume fraction + +Tc_CO2 = 7 ;//Temperature CO2 -[degree C] +T_CO2 = 273 + Tc_CO2 ;// Temperature CO2 -[K] +P_CO2 = 131 ;// Pressure of CO2 -[kPa] +CO2 = 0.0917 ;// Volume flow rate of CO2-[cubic metre/min] +// Convert given volume flow rate of CO2 at temperature of F & P +nw_CO2 = (CO2 * T1 * P_CO2)/(T_CO2 * P1) ;// volume flow rate of CO2 at temperature of F & P-[cubic metre] +// Solve P & F by following eqns. obtained by component balance of CO2 and total balance +// F(F_odr) = P(P_odr) - others balance +// F + nw_CO2 = P - Total balance +// Solving by matrix method +a = [F_odr -P_odr;1 -1];// Matrix formed by coefficients of unknown +b = [0;-nw_CO2] ;// Matrix formed by constants +x = a\b ;// matrix of solution, x(1) = F;x(2) = P +F = x(1) ;//Volume flow rate of entering gas-[cubic metre/min] +P = x(2) ;//Volume flow rate of product [cubic metre/min] +printf('Volume flow rate of entering gas is %.2f cubic metre/min',F); |