Solution for transport process and separation process

www.elsolucionario.net Supplemental Material for Transport Process and Separation Process Principles Chapter Introduction to Engineering Principles and Units This chapter presents a series of problems that introduce the student to the use of units and methods for expressing different variables such as temperature and composition Likewise, implementation of material balance and energy in fuel cells are illustrated in the following set of problem modules Daniel López Gaxiola Jason M Keith www.elsolucionario.net 1.3-1 Determination of a Solution Density 1.4-1 Gas-Law Constant 1.4-2 Composition of a Gas Mixture 1.5-3 Combustion of Fuel Gas 1.6-1 Pre-heating of Methane and Steam 1.6-2 Heating of an Ethanol Solution 1.6-3 Calculation of Heat Transfer Rate using Steam Tables 1.6-4 Incomplete Combustion of Methane 1.6-5 Standard Enthalpy of Reaction 1.7-1 Cooling of a Fuel Cell 1.7-2 Simultaneous Material and Energy Balances 1.7-3 Oxidation of Woody Biomass Student View www.elsolucionario.net Introduction to Engineering Principles and Units Example 1.3-1: Composition of a Methanol Solution A Direct-Methanol Fuel Cell is being fed with a liquid mixture of 80 mol % methanol and 20 mol % water at a temperature of 20 °C Calculate the mass fractions and the mass of each component of this mixture in g and lbm Strategy We will select a basis of 100 moles of mixture and use the molecular weights of methanol and water to determine the mass Solution We need to use the molecular weight of each component to determine the mass of methanol and water in the mixture, as shown in the following equations: mH O = n H OMH O m CH OH = n CH OH M CH OH 3 Substituting the corresponding quantities into these equations yields: mH O mH O  18 g H O  = ( 20 moles H O )    mol H O  = g H O  32 g CH 3OH  m CH OH = ( _ moles CH 3OH )   mol CH OH    m CH OH = g CH 3OH These values can be converted into lbm by multiplying the results by the conversion factor between lbm and g: mH O mH O   lb m = 360 g H O    _ g  = _ lb m Daniel López Gaxiola Jason M Keith Student View www.elsolucionario.net In the basis of 100 moles of mixture we selected, there will be 80 moles of methanol and 20 moles of water www.elsolucionario.net Supplemental Material for Transport Process and Separation Process Principles  lb m  m CH OH = g CH 3OH    453.59 g  m CH OH = _ lb m To determine the mass fractions of methanol and water, first we need to calculate the mass of mixture by adding the individual weights of its components: m mixture = m H O + m CH OH m mixture = 6.438 lb m Now we can calculate the mass fractions by dividing the mass of each component by the mass of the mixture: xH O = mH O m mixture = lb m 6.438 lb m x H O = 0.123 x CH OH = m CH OH m mixture = _ lb m 6.438 lb m x CH OH = Daniel López Gaxiola Jason M Keith Student View www.elsolucionario.net m mixture = lb m + lb m www.elsolucionario.net Introduction to Engineering Principles and Units Example 1.4-1: Gas-Law Constant Calculate the value of the gas-constant R if the pressure is in mm Hg, the moles are measured in cal g mol units, the volume is in liters and the temperature is in K Convert this value to mol ⋅ K Strategy We can calculate the constant R by solving it from the ideal-gas law at standard conditions Solution R= pV nT At standard conditions, P = 760 mm Hg, V = 22.414 L, n = mol and T = 273.15 K Substituting these values into the ideal gas equation of state, we have: R= ( 760 mm Hg )( 22.414 L ) (1 mol )( 273.15 K ) R = _ To obtain the value of R in R = 62.364 L ⋅ mm Hg mol ⋅ K cal , we can start by converting it to SI units as shown below: mol ⋅ K L ⋅ mm Hg  Pa   m3     mol ⋅ K  760 mm Hg   1000 L  Pa ⋅ m3 R = _ mol ⋅ K A Pascal is defined as: Pa = N m2 If we multiply the Pa by m3, we get: Pa ⋅ m3 = N m = N ⋅ m = Joule(J) m2 Daniel López Gaxiola Jason M Keith Student View www.elsolucionario.net The ideal-gas equation of state can be solved for R to yield: www.elsolucionario.net Supplemental Material for Transport Process and Separation Process Principles Now we can use a conversion factor between Joules and calories to determine the value of R in the desired units: R = cal mol ⋅ K www.elsolucionario.net R = 1.987  J  cal   mol ⋅ K  J  Daniel López Gaxiola Jason M Keith Student View www.elsolucionario.net Introduction to Engineering Principles and Units Example 1.4-2: Composition of a Gas Mixture Hydrogen can be produced by an ethanol reforming process The gas produced by the reformer is exiting at a pressure of 2144 kPa and has the following composition Component Molar Fraction H2 0.392 H2 O 0.438 CO 0.081 CO2 0.080 CH4 0.009 Strategy The definition of partial pressure will allow us to solve this problem Solution To calculate the partial pressure of each gas in the stream exiting the reformer, we can use the following equation: Pi = y i P where: yi = molar fraction of the species i present in the gas mixture Pi = Partial pressure of the species i present in the gas mixture P = absolute pressure of the system Substituting the corresponding molar fraction and the absolute pressure of the system into the definition of partial pressure, we have: PH = y H P = 0.392 ( _ kPa ) 2 PH = 840.45 kPa PH O = y H O P = 0.438 ( _ kPa ) PH O = kPa Daniel López Gaxiola Jason M Keith Student View www.elsolucionario.net Determine the partial pressure of each component www.elsolucionario.net Supplemental Material for Transport Process and Separation Process Principles PCO = yCO P = _ ( _ kPa ) PCO = 173.66 kPa PCO = y CO P = _ ( _ kPa ) 2 PCO = kPa PCH = y CH P = ( _ kPa ) 4 www.elsolucionario.net PCH = kPa Daniel López Gaxiola Jason M Keith Student View www.elsolucionario.net Introduction to Engineering Principles and Units Example 1.5-3: Combustion of Fuel Gas Component Mol % H2 41.9 H2 O 5.1 CO 1.7 CO2 41.9 CH4 9.4 The combustion reactions occurring inside the firebox are shown below The combustion of methane is only 87 % complete 1) CH4 + 2O2 2) CO + 3) H2 + O2 O2 CO2 + 2H2O CO2 H2 O Determine the composition of the gas produced by the combustion reaction assuming an air composition of 21 mol % oxygen and 79 mol % nitrogen Strategy We can perform molecular material balances around the combustion chamber to determine the amounts of each species in the exhaust gases A basis of 100 moles of fuel will be selected for simplicity Solution We can start by performing a methane balance around the combustion chamber: CH4 balance Input = Output + Consumption Input = 9.4 moles CH Daniel López Gaxiola Jason M Keith Student View www.elsolucionario.net In the steam-methane reforming process for producing hydrogen, part of the reformer exit gas is being burned with 25 % excess oxygen from air in order to supply heat for the reforming reaction to occur The fuel being burned has the following composition: www.elsolucionario.net Supplemental Material for Transport Process and Separation Process Principles The amount of methane exiting in the flue gas can be determined using the definition of fractional conversion: n CH ,in n CH Solving for the n CH n CH n CH n CH − n CH 4 ,out ,in and substituting the corresponding quantities yields: ,out ,out = n CH ,out = 9.4 moles − ( 9.4 moles ) ,out = moles 4 ,in − x CH n CH 4 ,in Since no information is given about the fractional conversion for reactions 2) and 3), complete combustion will be assumed Thus, n CO,out = n H ,out =0 We can proceed to perform a material balance on carbon dioxide as follows: CO2 balance Input + Generation = Output Input = moles By looking at the chemical reactions, it can be seen that both reactions 1) and 2) are generating carbon dioxide From the stoichiometric coefficients of these reactions, we can see that one mole of fuel is producing one mole of CO2 Thus, n CO ,generated = n CH ,reacted + n CO,reacted Substituting numeric values into this equation, we get: n CO n CO ,generated = ( 9.4 moles ) + moles ,generated = 9.88 moles CO We can substitute this result into the material balance equation for CO2 to yield: n CO ,out = moles + 9.88 moles Daniel López Gaxiola Jason M Keith Student View www.elsolucionario.net x CH = www.elsolucionario.net Drying of Process Materials Example 9.3-2: Use of Humidity Chart kg H O Use the kg Air humidity chart to determine the percentage humidity HP, humid volume υH , humid heat CS, and dew point of this air/steam mixture The air in a fuel cell has a dry – bulb temperature of 80°C and a humidity of 0.035 Strategy We can locate the given information in the humidity chart to determine the values required to solve this problem kg H O at a temperature of kg Air 80°C From this point we can move horizontally to the left until reaching the saturation line (HP = 100%) The temperature in this point corresponds to the dew point, found to be: First we locate the point that corresponds to a humidity of H = 0.035 Tsat = _ °C At the point we located initially for the dry bulb temperature and absolute humidity, we can read directly the percentage humidity For the air in the fuel cell, we find that: H P = _ % In section 9.3B of Geankoplis, we are given the following equations for the humid heat and volume as a function of the absolute humidity: υH = ( 2.83 × 10−3 ) + ( 4.56 × 10 −3 H )  T CS = m3 kg H O kJ In these equations, υH is in , H is in , T is in K, and CS is in kg dry air kg Air kg dry air ⋅ K Entering the humidity and the temperature into these equations, we get:    kg H O    υH = ( 2.83 × 10 −3 ) +  4.56 × 10−3  0.035    ( K ) kg Air       υH = Daniel López Gaxiola Jason M Keith m3 kg dry air Student View www.elsolucionario.net Solution www.elsolucionario.net Supplemental Material for Transport Process and Separation Process Principles  kg H O  CS = 1.005 + 1.88  0.035  kg Air   CS = _ kJ kg dry air ⋅ K The following figure illustrates how to determine the saturation temperature and percentage humidity from the chart: Adiabatic Saturation Curve 100 % T = °C Initial point corresponding to kg H O T = 80°C and H = 0.035 kg Air This point is approximately at humidity of 6.25% Daniel López Gaxiola Jason M Keith also located a percentage Student View www.elsolucionario.net Percentage Humidity Lines www.elsolucionario.net Drying of Process Materials Example 9.3-3: Adiabatic Saturation of Feed Air for Proton – Exchange Membrane Fuel Cells kg H O enters an adiabatic kg Air saturator before being fed to the cathode side of the fuel cell The air enters the saturator at a dry bulb kg H O temperature of 60 °C and must enter the fuel cell with a humidity of 0.063 Determine the kg Air final temperature and percent humidity of the air Air to be used as reactant in a PEMFC with an initial humidity of 0.057 Strategy Both HP and the temperature of the air entering the fuel cell can be obtained using the humidity chart First we need to locate the point that corresponds to the temperature of 60°C and humidity of kg H O 0.057 Once we located this point in the humidity chart, we move parallel to the adiabatic kg Air saturation curves, until reaching a point where the absolute humidity is 0.063 kg H O kg Air Now we are at the point where the air has the conditions required for entering the fuel cell and therefore we can read the temperature and percentage humidity to be: T ≈ °C H P ≈ % In the following page we can see a chart indicating the method to determine these values Daniel López Gaxiola Jason M Keith Student View www.elsolucionario.net Solution www.elsolucionario.net Supplemental Material for Transport Process and Separation Process Principles The next chart indicates the procedure followed to determine the temperature and percentage humidity: Percentage Humidity Lines Adiabatic Saturation Curve H = 0.063 kg H O kg Air T = °C Initial point corresponding to kg H O T = 60°C and H = 0.057 kg Air Point corresponding to kg H O H = kg Air Also we can read the percent humidity to be approximately _ % Daniel López Gaxiola Jason M Keith Student View www.elsolucionario.net 80% 70% www.elsolucionario.net Drying of Process Materials Example 9.3-4: Wet Bulb Temperature and Humidity Estimate the humidity of the reactant air in a proton – exchange membrane fuel cell if it has a dry and wet bulb temperature of 80 °C and 42.5 °C, respectively Strategy The humidity of the air can be determined from the humidity chart using the temperature data given in the problem statement The wet bulb temperature corresponds to the temperature of the air when it has 100 % humidity Hence, we need to move vertically from the temperature axis in the chart until reaching the curve corresponding to 100 % humidity From this point we move downwards to the right, parallel to the adiabatic saturation curves until we reach the vertical line for T = 80°C Now we can read the humidity of the air in the fuel cell to be: H = kg H O kg Air The method to determine the previous humidity value was obtained as follows: Percentage Humidity Lines Adiabatic Saturation Curve 100% H = Twet = 42.5 °C Initial point Daniel López Gaxiola Jason M Keith kg H O kg Air Tdry = 80 °C Student View www.elsolucionario.net Solution www.elsolucionario.net Supplemental Material for Transport Process and Separation Process Principles Chapter 11 Vapor – Liquid Separation Processes Separation processes in Chemical Engineering are used to transform a mixture of substances into two or more different products In this chapter, the following problem modules illustrate the principles of chemical processes in which the separation involves the vapor and liquid phases www.elsolucionario.net 11.1-1 Use of Raoult’s Law for Methanol – Water Equilibrium Data 11.3-1 Relative Volatility of Methanol – Water Mixture Daniel López Gaxiola Jason M Keith Student View www.elsolucionario.net Vapor – Liquid Separation Processes Example 11.1-1: Use of Raoult’s Law for Methanol – Water Equilibrium Data T (°C) PHsat2O (kPa) sat PCH (kPa) 3OH 64.5 65 66 67.6 69.3 71.2 73.1 75.3 78 81.7 84.4 87.7 89.3 91.2 93.5 96.4 100 24.521 25.03 26.57 28.233 30.328 32.964 35.772 39.109 43.866 50.94 56.577 64.477 68.417 73.598 80.227 89.254 101.325 100.74 102.66 106.73 113.68 121.06 129.41 139.55 151.28 167.02 190.44 209.96 234.53 248.34 264.85 285.04 313.83 351.38 Strategy We need to use Raoult’s Law to obtain the composition of both phases at the given temperature and pressure Solution The following equation describes Raoult’s Law: yA = _ P Wankat, P.C., Separation Process Engineering, Second Edition, Prentice Hall, 2007 Methanex Corporation "Technical Information & Safe Handling Guide for Methanol” September 2006 Accessed: February 2011 http://www.methanex.com/products/documents/TISH_english.pdf Daniel López Gaxiola Jason M Keith Student View www.elsolucionario.net A methanol – water mixture is assumed to be in equilibrium with the vapor phase in the reservoir for a Direct – Methanol Fuel Cell Using the following equilibrium data [1,2] and Raoult’s Law determine the composition of the vapor and liquid phases at 72°C and 101.325 kPa The vapor pressure data for water was obtained from Appendix A.2 of Geankoplis www.elsolucionario.net Supplemental Material for Transport Process and Separation Process Principles In this problem, A = Methanol and B = Water As we can see, we cannot yet determine the composition of the vapor phase because we need to calculate the composition of the liquid phase first This can be done using Equation 11.1-3 of Geankoplis, given by: PA x A + PB (1 − x A ) = We can use the pressures PA and PB, using linear interpolation of the given data, at the temperature of 72°C Thus, PA @ T = 72° C − PA@ T =71.2°C PA@ T =73.1°C − PA @ T = 71.2°C = _ °C − _ °C 73.1°C − _ °C PA@ T =72°C = _ kPa Using the same procedure we can determine the pressure of water at 72°C to be: PB@ T = 72°C = kPa Now we can substitute the pressure values into the equation for the total pressure, and solve for xA to yield: _ x A + _ (1 − x A ) = 101.325 99.53x A = _ x A = _ Hence the molar fraction of water is given by: x B = − x A = − _ x B = _ Now we can enter the value of xA into Raoult’s Law to determine yA and yB Thus, yA = _ kPa ( _ ) 101.325 kPa y A = _ y B = − y A = − _ y B = 0.109 Daniel López Gaxiola Jason M Keith Student View www.elsolucionario.net Solving for the pressure PA@ T =72°C we determine the pressure to be: www.elsolucionario.net Vapor – Liquid Separation Processes Example 11.3-1: Relative Volatility of Water – Methanol Mixture Determine the relative volatility of the methanol – water mixture described in problem 11.1-1 at the temperature of 72°C Strategy Section 11.3B of Geankoplis gives an Equation for determining the relative volatility Solution The relative volatility of a mixture is defined by equation 11.3-3 of Geankoplis, shown below: PA PB We can substitute the saturation pressures at the temperature of 72°C, which are given in Example 11.1-1 Thus, the relative volatility is found to be: α AB = _ kPa _ kPa α AB = _ Daniel López Gaxiola Jason M Keith Student View www.elsolucionario.net α AB = www.elsolucionario.net Supplemental Material for Transport Process and Separation Process Principles Chapter 12 Liquid – Liquid and Fluid – Solid Separation Processes This chapter includes examples of adsorption processes where one or more components of a gas or liquid stream are adsorbed on the surface of an adsorbent material The following problem module illustrates pressure swing adsorption process for purifying hydrogen for proton – exchange membrane fuel cells www.elsolucionario.net 12.3-1 Hydrogen Purification in Pressure Swing Adsorption Process Daniel López Gaxiola Jason M Keith Student View www.elsolucionario.net Liquid – Liquid and Fluid – Solid Separation Processes Example 12.3-1 Hydrogen Purification in Ethanol – Reforming Process Use the data in the following table to determine the time required to reach the break – point  C  = 0.01 , the time equivalent to the total capacity of the bed, and the time concentration   C0  equivalent to the usable capacity of the bed up to the break – point time Also calculate the length and capacity of unused bed after the break-point time What is the saturation capacity of the bed? Breakthrough Concentration of CO2 in the packed bed t(s) 27 187 480 600 613 640 667 747 853 973 1067 1200 1360 C/C0 0.0063 0.0125 0.0358 0.1391 0.3487 0.4249 0.4884 0.5600 0.6661 0.7168 0.7611 0.8055 0.8561 t(s) 1493 1640 1827 1933 2107 2240 2373 2560 2707 2853 3120 3280 3440 3600 C/C0 0.8814 0.9066 0.9318 0.9444 0.9569 0.9695 0.9757 0.9819 0.9817 0.9879 0.9940 0.9938 1.0000 1.0000 Strategy The break-through and capacity of the packed bed can be determined using the design equations given in Section 12.3D of Geankoplis Solution First, the break – point time can be obtained from the tabulated data, at the point where C = 0.01 to C0 be equal to tb = s Lopes, F.V.S., Grande, C.A., Rodrigues, A.E., Chemical Engineering Science, 66, 303 – 317 (2011) Daniel López Gaxiola Jason M Keith Student View www.elsolucionario.net The dry reformate stream coming from an ethanol reforming process contains 70 mole % H2 and 30 mole % CO2 Lopes et al [1] study the adsorption of CO2 on a packed bed of activated carbon with a length of 0.267 m The bed contains 245.6 kg of adsorbent material consisting of activated carbon, −5 m with a particle diameter of 2.9 mm The average flow rate of the dry reformate is ×10 and has s kg a density of 0.587 m www.elsolucionario.net Supplemental Material for Transport Process and Separation Process Principles The time tt for the total capacity of the bed can be calculated using the integral of the C curve as a C0 function of time: ∞ C t t = ∫  − dt = A1 + A  C0  where A1 and A2 are the shaded areas shown in the following figure: 1.0 0.9 0.8 0.7 C/C0 0.6 0.5 0.4 0.3 0.2 0.1 0.0 500 1000 1500 2000 2500 3000 3500 4000 t(s) A1 Cb As it can be seen in this figure, we can determine the break – through time by calculating the areas A1 and A2 Thus, A1 = ( _ s − 0s )(1 − _ ) A1 = _ s Daniel López Gaxiola Jason M Keith Student View www.elsolucionario.net A2 www.elsolucionario.net Liquid – Liquid and Fluid – Solid Separation Processes 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 A2 A3 500 1000 1500 2000 t(s) 2500 3000 3500 4000 A1 The area of the rectangle to the right of the break point can be calculated as follows: A = ( s − _ s )(1 − _ ) A = 3413 s The results for the numerical integration using trapezoidal method will yield: A3 = _ s Now we can obtain the area A2 to be: A2 = A − A3 = 3413 s − _ s A2 = _ s We can substitute the areas A1 and A2 into the equation for the break-through time tt, to obtain: t t = s + _ s t t = _ s Daniel López Gaxiola Jason M Keith Student View www.elsolucionario.net C/C0 To determine the area A2, we can use the trapezoidal method to calculate the area under the curve and subtract it from the total area A of the rectangle, as shown below: www.elsolucionario.net Supplemental Material for Transport Process and Separation Process Principles The time equivalent to the available capacity of the bed before the break – point time can be obtained as shown in the following steps: tb  C t u = ∫ 1 − dt =  C0  t u = _ s The length of the unused bed can be calculated using Equation 12.3-4 of Geankoplis:   HT  where: HT = Total length of the packed bed Substituting the times we calculated and the length of the packed bed into this equation, we get:  _ s  H UNB = 1 −  ( m )  _ s  H UNB = 0.212 m Finally, to determine the saturation capacity of the activated carbon in the bed, we need to obtain the moles of carbon dioxide adsorbed on the bed The carbon dioxide can be obtained by multiplying the initial concentration of CO2 in the gas by the mass of gas in the time tt of _ s Thus, 0.30 Total CO adsorbed = kmol CO  kg gas   −5 m  × 10   ( s )  0.587  kmol gas  s  m3   kg CO     kg gas kmol CO   _ kmol gas Total CO adsorbed = kg CO The molecular weight of the gas mixture was obtained by multiplying the molar fraction of each component by its corresponding molecular weight and adding the results, as shown below: Daniel López Gaxiola Jason M Keith Student View www.elsolucionario.net  t H UNB = 1 − u  tt www.elsolucionario.net Liquid – Liquid and Fluid – Solid Separation Processes   kg H  kg CO  M gas = y H2 M H2 + yCO2 M CO2 =   + 0.3  _  kmol H  kmol CO    M gas = kg kmol Now we can divide the total CO2 adsorbed by the mass of activated carbon on the bed Hence, Saturation Capacity = kg CO 245.6 kg adsorbent kg CO kg adsorbent www.elsolucionario.net Saturation Capacity = _ Daniel López Gaxiola Jason M Keith Student View ... Material for Transport Process and Separation Process Principles Example 1.6-1: Pre-heating of Methane and Steam a) The steam used for producing hydrogen by steam-methane reforming process is... plates and the initial and final temperatures of the fuel cell into the equation for ∆H fuel cell we get: www.elsolucionario.net Supplemental Material for Transport Process and Separation Process. .. for Transport Process and Separation Process Principles Example 1.6-3: Calculation of Heat Transfer Rate using Steam Tables Liquid water at 30 °C is fed into a steam-methane reforming plant for