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Material Balance and Process Analysis
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Added on 2020-04-21
Material Balance and Process Analysis
Added on 2020-04-21
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Material Balance and Process Analysis 1PROCESS MATERIAL BALANCE AND ANALYSISA Calculation Paper on Process Analysis ByStudent’s NameName of the ProfessorInstitutional AffiliationCity/StateYear/Month/Day
Material Balance and Process Analysis 2INTRODUCTIONThe atomic weight of any element can be gotten by the mass of the atom measured on a scale of the carbon’s isotope whose nucleus has six neutrons and six protons. The weight of the molecule of a compound is the summation of the atom’s atomic weights which is composed of the compound’s molecule. Question A1A mixture of air and ethylene (C2H4) is fed to a reactor at a temperatures that is high leading to the 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% C2H2 by 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 Analysis 3a.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 all components 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 1st reaction:C2H4 + 3O2C2H4O ..................................... (i)y mol ½ mol y molRemaining (x - y) moles of C2H4 takes part in 2nd Reaction[ CITATION Nay15 \l 2057 ]
Material Balance and Process Analysis 4C2H4 + 3O2 2CO2 + 2H2O .............................. (ii)(x-y) 3(x-y) 2(x-y) 2(x-y)Therefore, the total moles of O2 consumed = ½ y + 3(x-y) = 10.463x – 2.5y = 10.46 ............................ (iii)The moles of CO2 used in Reaction (ii) will be same as the moles of CO2 present in gas at absorber 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 C2H4 at reactor inlet = x + 4.7 = 9.17 + 4.7 = 13.87Reactor exit stream:Moles of CO2 formed = 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 air 13.87 + 94.56 = 108.43 mol
Material Balance and Process Analysis 5As for 100 mol of feed to reactor, 10 moles of water fed to absorber, the quantity of water fed to absorber: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 will be:C2H4O = 6.82 moles, H2O = 9.043 molesMol % C2H4O = 6.82/ (6.82 + 9.043) = 0.43 H2O = 9.043/ (6.82 + 9.043) = 0.57Question A2The equation of Antoine which gives the saturated vapor pressure as a function of temperature is as shown below: a.The normal boiling point of benzene and the normal boiling point of toluene can be determined as shown below:
Material Balance and Process Analysis 6Log 10 (P) = A – B/ (T + C)Normal boiling points at a species is defined at atmospheric pressure:P = 760 mmHg For Benzene: A = 6.91, B = 1211, C = 221Log 10 (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 by component 2, the Raoult’s law for the components can be written as:YiP = Xi Pisat .......... (1)Where P1sat is the saturation pressure of component "i" Xi is the mole fraction of component "i" in the liquid stream P is the total pressure of the system[ CITATION FCa12 \l 2057 ]
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