Material Balance and Process Analysis

Added on - 21 Apr 2020

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Material Balance and Process Analysis1PROCESS MATERIAL BALANCE AND ANALYSISA Calculation Paper on Process Analysis ByStudent’s NameName of the ProfessorInstitutional AffiliationCity/StateYear/Month/Day
Material Balance and Process Analysis2INTRODUCTIONThe atomic weight of any element can be gotten by the mass of the atom measured on ascale of the carbon’s isotope whose nucleus has six neutrons and six protons. The weight of themolecule of a compound is the summation of the atom’s atomic weights which is composed ofthe compound’s molecule.Question A1A mixture of air and ethylene (C2H4) is fed to a reactor at a temperatures that is high leading tothe following reactions:The gases exiting the reactor are then cooled and contacted using water contained in the absorberwhere water combines with ethylene form a compound known as ethylene glycol (C2H6O2)The gases escaping the absorber include 6.5% H2O, 4.7% CO2, 9.4% O2, 74.7% N2, and 4.7%C2H2by moles. For every air and ethylene mixture fed to the reactor, 10 moles of water are fed tothe absorber[ CITATION Gor13 \l 2057 ].
Material Balance and Process Analysis3a.The amount of oxygen fed to the reactor can be determined from the using nitrogen as a tie asshown below:By letting the composition of air for Nitrogen to be 79% and for oxygen to be 21%:The amount of gases leaving absorber = 100 molesComposition of gas: N2= 74.7% = 74.7 molesC2H2= 4.7% = 4.7 molesO2= 9.4% = 9.4 molesCO2= 4.7% = 4.7 molesH2O = 6.5% = 6.5 molesN2 in off gas = 74.7 molesThe moles of air introduced into the reactor = 74.7/0.79 = 94.56 molMoles of O2 in this air = 94.56 * 0.21 = 19.86 molb.The component balances around the reactor can be done to determine the quantities of allcomponents in the reactor exit and feed streams as shown below:Moles of O2 in gas at absorber outlet = 9.4 molesO2 consumed in reactor inlet = (x + 4.7) molesAt the absorber outlet, the moles of C2H4 in gas = x + 4.7 – 4.7 = x molesLet y moles of C2H4 takes part in the 1streaction:C2H4+ 3O2C2H4O ..................................... (i)y mol ½ mol y molRemaining (x - y) moles of C2H4 takes part in 2ndReaction[ CITATION Nay15 \l 2057 ]
Material Balance and Process Analysis4C2H4+ 3O22CO2+ 2H2O .............................. (ii)(x-y) 3(x-y) 2(x-y) 2(x-y)Therefore, the total moles of O2consumed = ½ y + 3(x-y) = 10.463x – 2.5y = 10.46 ............................ (iii)The moles of CO2used in Reaction (ii) will be same as the moles of CO2present in gas atabsorber outlet.2(x-y) = 4.7; x-y = 2.35 ........................................................................ (iv)By solving equation 3 and equation 4 simultaneously, x = 9.17, y = 6.82Therefore the moles of C2H4at reactor inlet = x + 4.7 = 9.17 + 4.7 = 13.87Reactor exit stream:Moles of CO2formed = 4.7 molesMoles of H2O formed = 2 (x-y) = 4.7Moles of C2H4O formed = y = 6.82Moles of N2= 74.7 molesMoles of remaining O2= 9.4 molesMoles of remaining C2H4= 4.7 molesc.The total moles of reactor inlet = moles of C2H4 + moles of air13.87 + 94.56 = 108.43 mol
Material Balance and Process Analysis5As for 100 mol of feed to reactor, 10 moles of water fed to absorber, the quantity of water fed toabsorber:Moles of water = 108/100 * 10 = 10.843 molesMass of H2O in gas leaving Absorber = 6.5 molesSo the moles of H2O which vaporize in absorber = 6.5 - 4.7= 1.8 molesd.So remaining H2O moles = 10.843 – 1.8 = 9.043 molesThis H2O will absorb C2H4O completely so ethylene glycol aqueous solution composition willbe:C2H4O = 6.82 moles, H2O = 9.043 molesMol % C2H4O = 6.82/ (6.82 + 9.043) = 0.43H2O = 9.043/ (6.82 + 9.043) = 0.57Question A2The equation of Antoine which gives the saturated vapor pressure as a function of temperature isas shown below:a.The normal boiling point of benzene and the normal boiling point of toluene can bedetermined as shown below:
Material Balance and Process Analysis6Log10(P) = A – B/ (T + C)Normal boiling points at a species is defined at atmospheric pressure:P = 760 mmHgFor Benzene: A = 6.91, B = 1211, C = 221Log10(760) = 6.91 – 1211/ (T + 221)2.88 = 6.91 – 1211/ (T +221)1211/ (T + 201) = 6.91 – 2.881211/ (T + 221) = 403403T = 1211 – 890.634.03T = 320.37T = 79.46oCFor TolueneA = 6.95, B = 1344, C = 219Log 10 (760) = 6.95 – 1344/ (T + 219)2.88 = 6.95 – 1344/ (T + 219)4.07 = 1344/ (T + 219)4.07T + 891.33 = 1344T = 111.22oCb.By letting Benzene be represented by component 1 and toluene be represented bycomponent 2, the Raoult’s law for the components can be written as:YiP = Xi Pisat .......... (1)WhereP1sat is the saturation pressure of component "i"Xi is the mole fraction of component "i" in the liquid streamP is the total pressure of the system[ CITATION FCa12 \l 2057 ]
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