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Energy Engineering - Fundamentals of Chemical Processes

Exercise 1: Solution

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Prof. Gianpiero Groppi – Exercises – Fundamentals of Chemical Processes – A.A. 202 3-202 4 1 EXERCISE 1 Adiabatic flame temperature for propane combustion The combustion of propane with air is carried out in an adiabatic reactor: C3H8+ 5O2 ↔ 3CO 2 + 4H2O Calculate the temperature of the exhaust gas stream at the outlet of the reactor (T OUT ) in the following cases: 1. Stoichiometric feed of the reactants; 2. Air in excess of 30% (molar basis); 3. Considering the case with excess air, assume that due to ill -combustion CO is present in the exhaust, with a selectivity of 10% with respect to the moles o f carbon converted. DATA: C3H8 stream: T = 25°C P = 1 atm AIR stream: T = 25°C P = 1 atm R.H. = 60%, with P SAT H2O (25 °C) = 3.167 kPa ΔH0C,C3H8 (25°C) = -2043 kJ/mol ΔH 0C,CO (25°C) = -283 kJ/ mol Cp equation and parameters from J.M. Smith, H.C. Van Ness, M. M. Abbott, Introduction to Chemical Engineering Thermodynamics, 7th edition, McGraw Hill (2005). Standard formation enthalpies from NIST website. Specie ΔH0F [kJ/mol] ������������,������ ̅̅̅̅̅= ������������������� .(�������+�������.������+ ������������������������) [kJ/mol/K], T in [K] a b c C3H8 -104.7 1.213 2.897E -04 - N2 0 3.280 5.930E -04 -4.000E+03 CO 2 -393.5 5.457 1.045E -03 -1.157E+05 H2O -241.8 3.470 1.450E -03 1.210E+04 O2 0 3.639 5.060E -04 -2.270E+04 CO -110.5 3.376 5.570E -04 -3.100E+03 Prof. Gianpiero Groppi – Exercises – Fundamentals of Chemical Processes – A.A. 202 3-202 4 2 SOLUTION The problem requires the calculation of the temperature of the exhaust gas stream at the outlet of the reactor (T OUT ) in three different cases. CASE 1: Stoichiometric feed of the reactants In this case the propane combustion reaction is the only one occurring in the system, and we can assign to this and extent of reaction ε. The variation in the number of moles of the i -th species with respect to the exte nt of reaction can be defines as ��������̇= �������������� (1) Where νi is the stoichiometric coefficient of the (balanced!) chemical reaction. Once we define as ( IN ) t he inlet of the combustor and (OUT) the outlet, we can obtain from the integration of relation (1) that – once the extent of reaction is known – the molar flowrates exiting the reaction can be defined as: �̇�(��� )= �̇�(��)+ ������������� (2) In the specific case of propane combustion the stochiometric coefficients are -1 per C 3H8, -5 per O 2, +3 per CO 2, +4 per H 2O e 0 per N 2. It is necessary at first to define the (IN) molar streams by using the specifications provided from the text of the e xercise. Since we are not given any extensive property but only intensive one s, first of all we have to define a base of calculus. Th e choice of this base of calculus is completely arbitrary and its choice will not affect the adiabatic temperature nor the composition at the outlet . On the contrary, the value chosen for the base of calculus will affect the extensive properties of the system, such as the molar flowrates exiting the reactor. We can assume as base of calculus and inlet propane mola r flowrate of 1 mol/s. Once propane flowrate is defined, we can calculate all inlet flowrates for all other species. In fact, we know that in this case a stoichiometric amount of oxygen is fed to the system. This means that for each mole of propane fed to the combustor, we have to feed a corresponding amount of oxygen based on the stoichiometric coefficients. �̇�2 (��)= �̇�3�8 (��) �������2 ������������3�8= 5 ��� /� (3) Since air is used to burn propane , each mol of oxygen entering will bri ng along a corresponding amount of nitrogen (we can assume in this case that air is composed only by N 2, O 2 and H 2O, with a N 2/O 2 ratio of 79/21): Prof. Gianpiero Groppi – Exercises – Fundamentals of Chemical Processes – A.A. 202 3-202 4 3 �̇�2 (��)= �̇�2 (��)0.79 0.21 = 18 .81 ��� /� (4) Concerning H 2O, we are told that the air that we use is humid. If the relative humidity (RH ) is known, we can evaluate the (IN) molar fraction of water once we know the vapor tension of water at the (IN) temperature: ������ = ��2� ��2�0 (25° ������)= � �������2� � �������2�= ������������ ��2�0 (25° �) ������ = 0.0188 (5) One the mol ar fraction is known, based on the definition of molar fraction, we can estimate the molar flowrate of water (remembering that in this case the humidity provided in the text is referred to the air stream only): �������2�= ������̇�2�(��) ∑ ������̇������(��) ������������� = ������̇�2�(��) ������̇�2�(��)+������̇�2(��)+������̇�2(��) → �̇�2�(��)= (�̇�2 (��)+ �̇�2 (��)) �������2� 1−�������2�= 0.455 ��� /� (6) Now that the inlet is fully characterized, we can write the mate rial balances : Specie s IN [mol/s] OUT [mol/s] C3H8 1 1−ε O2 5 5−5ε CO 2 0 0+ 3ε H2O 0.455 0.455+4ε N2 18.81 18.81 Now we have to define the value of the extent of reaction using the specification given regarding propane combustion. In fact , the system is working at complete propane conversion, a nd hence the propane leaving the reactor is null: �̇�3�8 (��� )= 0 ��� /� = 1− ������→ ������= 1 ��� /� (7) Now that the extent o f reaction is known, all (OUT ) mol ar flowrates are known as well: Specie s IN [mol/s] OUT [mol/s] C3H8 1 0 O2 5 0 CO 2 0 3 H2O 0.455 4.45 N2 18.81 18.81 Now we only have to determine the outlet temperature at adiabatic conditions. For this purpose, we can employ an enthalpic balance on the combustor in adiabatic conditions (i.e., no external heat exchange). Prof. Gianpiero Groppi – Exercises – Fundamentals of Chemical Processes – A.A. 202 3-202 4 4 ������̇(��)− ������̇(��� )= 0 (8) Since the system is a mi xture of ideal gas, we can write it as ������̇= ∑ �̇� �� �=1 ������̅�(������,�,������)= ∑ �̇� �� �=1 ������̃�(������) (9) Hence, Eq (8) becomes: ∑ �̇�(��) �� �=1 ������̃�(������(��))− ∑ �̇�(��� ) �� �=1 ������̃�(������(��� ))= 0 (10) Since we are dealing with an ideal gas, the variation of the ent halpic flux with temperature becomes : �������̃�= �̃������,�������� (11) In order to integrate Eq (12) it is necessary to define a reference state, that is intrinsically arbitrary. We can choose as reference state for the calculation for the sp ecific molar enthalpies the elementary species, at a reference temperature and reference pressure. Then we get: ∑ �̇�(��) �� �=1 (∆������̃�,�(��������� )+ ∫ �̃������,�������� �(��) ������������������� )− ∑ �̇�(��� ) �� �=1 (∆������̃�,�(��������� )+ ∫ �̃������,�������� �(��� ) ������������������� )= 0 (12) We can insert the material balances (Eq ( 2)): ∑ �̇�(��) �� �=1 (∆������̃�,�(��������� )+ ∫ �̃������,�������� �(��) ������������������� ) − ∑ (�̇�(��)+ �������������) �� �=1 (∆������̃�,�(��������� )+ ∫ �̃������,�������� �(��� ) ������������������� )= 0 (13) Rearranging Eq. (1 3) we get : ∑ �̇�(��) �� �=1 (∫ �̃������,�������� �(��) ������������������� )− ∑ (�̇�(��)+ �������������) �� �=1 (∫ �̃������,�������� �(��� ) ������������������� ) −������∑ �������∆������̃�,�(��������� ) �� �=1 = 0 (14) Remembering that by definition ∆������̃�(��������� )= ∑ �������∆������̃�,�(��������� ) ���=1 , we obtain : ∑ �̇�(��) �� �=1 (∫ �̃������,�������� �(��) ������������������� )− ∑ (�̇�(��)+ �������������) �� �=1 (∫ �̃������,�������� �(��� ) ������������������� )− ������∆������̃�,�(��������� )= 0 (15) For combustion reactions , ∆������̃�(��������� ) is commonly defined as combustion enthalpy and assumes the notation ∆������̃���� (��������� ). It represents the enthalpy variation associated with the complete combustion of 1 mol of reagent. Prof. Gianpiero Groppi – Exercises – Fundamentals of Chemical Processes – A.A. 202 3-202 4 5 The spec ific he at a t c onstan t pressure is pro vided as polynomial expression as a function of temperature: ∫ �̃������,�������� � ������������������� = ∫ �̃������,�������� � ������������������� = ������(��(������ − ��������� )+ �� 2 (������2− ���������2 )− ��(1 ������ − 1 ��������� )) (16) By substitutin g Eq (16 ) in E q (15) we get : ∑ �̇� (�� ) �� �=1 (��(������(��)− ��������� )+ �� 2 (������(��)2− ���������2 )− ��( 1 ������(��)− 1 ��������� )) − ∑ (�̇� (��)+ �������������) �� �=1 (��(������(��� )− ��������� )+ �� 2 (������(��� )2− ���������2 ) − ��( 1 ������(��� )− 1 ��������� ))− ������∆������̃���� (��������� )= 0 (17) We now have a n on -linear equation in the unknown T OUT , that ca n be numerically solved to ge t TOUT = 2340 K CASE 2: 30% excess of air In this case the solution is very s imilar to that of Case 1. The only difference is t he calculation of the inlet flowrates. In th is case air is fed to the comb ustor with a 30% excess with respect to the stoichiomet ry of the combustion reaction. The excess is defined a s follows: �= �̇�2 (��)− �̇�2,��������� (��) �̇�2,��������� (��) = 0.3 (18) By keeping the same b ase of calculus (1 mol/s of propa ne) we ca n compute the am ount of oxygen fed to the combu stor: �̇�2 (��)= �̇�3�8 (��) �������2 �������3�8 (1+ �)= 6.5 ��� /� (19 ) No w that oxygen is k now n we can evaluate the flowrates o f nitrogen and water at the inlet by applying th e same formul as used for the solut ion of Case 1(Eq s (4) and (5), respectively ). By imposing the same s pecification given in the text of complete propane combustion , the extent of reaction will be also in this case equa l to ������= 1 ��� /�. It is then p ossible t o compute the outlet mol ar flowrates: Prof. Gianpiero Groppi – Exercises – Fundamentals of Chemical Processes – A.A. 202 3-202 4 6 Specie s IN [mol/s] OUT [mol/s] C3H8 1 0 O2 6.5 1.0 CO 2 0 3 H2O 0.592 4.592 N2 24.45 24.45 W e can then ca lculate the outlet temperature by imposing the energy balance at adiabatic condition (E q(8) and following) . The outlet te mperature in this case (TOUT = 1983 K) is lower with respect to what was calculated in C ase 1 , due to the presence of the additional air flow that acts as t hermal dilu ent . CASE 3: 30% excess of air + CO presence in the effluents In the system we have C3H8, O 2, CO 2, H 2O, CO ed N 2. Since one additional mo lecule (CO) is present , and i t is not descri bed by the pro pa ne combustion stoich iometry, we have to saturate an additional degree of freedo m by introducing a n additional reaction. W e can choose any reaction provided that: 1) The two chemical reactions must be line arly indepen dent ; 2) All reacti ng species must appear at least once in a t least o ne of the reaction s A possible choice is that of using the dissociation of CO 2 into CO and O 2: CO 2  CO + 0.5 O 2 The stoichiometric coefficients are i n this case -1 for CO 2, +1 for CO e +0.5 for O2. We can assi gn a n extent of reaction ������2 to this reaction , while we will define the extent of reaction of propane combustion as ������1in this case . The inlet condit ions to t he combustor are th ose calculated when solving Case2. For the calculation of the outlet con ditions we can gen eraliz e the material ba lance as follo ws in the case that NR reaction s are occurring: �̇� (��� )= �̇� (�� )+ ∑ �������,�������� ������� � (20) Si ottiene, quindi, la seguente tabella che rappresenta i bilanci materiali sulle specie molecolari: Specie s IN [mol/s] OUT [mol/s] C3H8 1 1− ������1 O2 6.5 6.5− 5������1+ 0.5 ������2 CO 2 0 3������1− ������2 H2O 0.592 0.592 + 4������1 CO 0 ������2 Prof. Gianpiero Groppi – Exercises – Fundamentals of Chemical Processes – A.A. 202 3-202 4 7 N2 24.45 24 .45 The outlet composit ion calculation requir es the definition of both the extents of reactions . In order to do this we have to use two constraints : the first one will impose that propane conversion is complete (21) , the second one will impose that 10% of the convert ed carbon will lead to CO (22) . �̇�3�8 (2) = 0 ��� /� = 1− ������1→ ������1= 1 ��� /� (21) ��������� = ��� ������� ��� ��������� ����� ������ = �̇��(��� ) 3(�̇�3�8 (��)− �̇�3�8 (��� )) = ������2 3∙������1 (22) Note that s ince th e speci fication on CO selec tivity is gi ven on a carbon basis , the mol of propane converted in the system ha ve to be multiplied by 3, as propane contains 3 carbon atoms. From the two con straints we g et that ������1= 1 ��� /� and ������2= 0.3 ��� /� and hence the outlet flowrates are known. Now we can proceed with the evaluation of the outlet temper ature, by impo sing the enthalpic balance with th e s ame notation used for the two previous cases, generalized in the case of NR reactions. ∑ �̇� (1) �� �=1 (��(������(��)− ��������� )+ �� 2 (������(��)2− ���������2 )− ��( 1 ������(��)− 1 ��������� )) − ∑ (�̇� (��)+ �������������) �� �=1 (��(������(��� )− ��������� )+ �� 2 (������(��� )2− ���������2 ) − ��( 1 ������(��� )− 1 ��������� ))− ∑ �������∆������̃�,�(��������� ) ������� �=1 = 0 (25) Da cui: ∑ �̇� (�� ) �� �=1 (��(������(��)− ��������� )+ �� 2 (������(��)2− ���������2 )− ��( 1 ������(��)− 1 ��������� )) − ∑ (�̇� (��)+ �������������) �� �=1 (��(������(��� )− ��������� )+ �� 2 (������(��� )2− ���������2 ) − ��( 1 ������(��� )− 1 ��������� ))− ������1∆������̃�,1(��������� )− ������2∆������̃�,2(��������� )= 0 (26) The text provides the ∆������̃���� (��������� ) for propane and for CO. Conseq uently, the values o f ∆������̃�,1 e ∆������̃�,2 will be equal to ∆������̃�����3�8 (��������� ) e −∆������̃������ (��������� ), re specti vely. Note that the ∆������̃� of the second reaction is the combustion enthaply with opposite sign. This is due to the fact that ∆������̃������ (��������� ) is gi ve n with respect to the CO combustion reacti on, whi ch is the opposite reaction wi th res pect to what we have consi dered. In fact, CO combustion is: Prof. Gianpiero Groppi – Exercises – Fundamentals of Chemical Processes – A.A. 202 3-202 4 8 CO + 0.5 O 2  CO 2 → ∆������̃������ (��������� ) While i f we take the opposite reaction: CO 2  CO + 0.5 O 2 → −∆������̃������ (��������� ) By imposing the entha lpic balance we finally get that ������(��� )= 1923 ������. Note that, for all three ca ses, as the f ormat ion entha lpies at reference conditions are also provided in the te xt, by calculating the entha lpic balances using Eq. (12 ) it is possible to avoid taking into account the ∆������̃���� (��������� ) data while getting to the sa me results .