Exam Solution Chemical Processes

1The UNIVERSITY OF BRITISH COLUMBIA Department of CHEMICAL and BIOLOGICAL ENGINEERING CHBE 241: MATERIAL and ENERGY BALANCES MIDTERM – EXAMINATION October 23, 2013: 3 – 5 PM Instructions: - Hand calculators (including programmable calculators) are allowed The only allowed material is the one page of formula sheet provided to you with this exam. You can detach the formula sheet for easier handling. Solve individually all 2 problems. If any of these rules are not respected, it will be dealt with according to University Policy on student ethics during examination. *** use a maximum of two significant figures after the decimal point **** 2 Formula Sheet 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝐶𝑜𝑛𝑣𝑒𝑟𝑠𝑖𝑜𝑛 𝑓 (𝐴) 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑨 𝐶𝑂𝑁𝑆𝑈𝑀𝐸𝐷 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 = 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑨 𝐹𝐸𝐷(𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑) 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 𝑆𝑒𝑙𝑒𝑐𝑡𝑖𝑣𝑖𝑡𝑦 () = 𝑌𝑖𝑒𝑙𝑑(𝑌) = 𝑚𝑜𝑙𝑒𝑠𝑜𝑓 𝐷𝐸𝑆𝐼𝑅𝐸𝐷 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑨 𝐶𝑂𝑁𝑆𝑈𝑀𝐸𝐷 𝑖𝑛 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 𝑚𝑜𝑙𝑒𝑠𝑜𝑓 𝐷𝐸𝑆𝐼𝑅𝐸𝐷 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑨 𝐹𝐸𝐷 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑎𝑙𝑅𝑒𝑐𝑜𝑣𝑒𝑟𝑦 𝑜𝑓 𝑨 = 𝑎𝑚𝑜𝑢𝑛𝑡𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑨 𝑖𝑛 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑜𝑟 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑠𝑡𝑟𝑒𝑎𝑚 𝒋 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑨 𝑓𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑜𝑟 (Accumulation in the system) = (IN) – (OUT) + (Generated in the system) – (Consumed in the system) 1 atm = 1.013x105 Pa = 760 mmHg R = 8.314 J / (mol K) 3 Problem 1 (60% of the mark): Canexus corporation at its plant in North Vancouver is producing NaOH, Cl2,(g) and H2,(g) by electrolysis of NaCl in water (i.e. a brine solution). As a recently hired process engineer, after the first week on the job, your manager asks you to submit a report that contains the following material balance analysis of their process. In the process, salt (solid) is mixed with water to produce a 30 %wt. salt solution. This salt solution is fed to the electrolysis unit, where NaOH solution, Cl2,(g) and H2,(g) are produced, each in separate streams. The aqueous solution containing NaOH and un-reacted NaCl is introduced to an evaporator where some of the water is removed as steam. The concentrated product solution coming out from the evaporator has the following composition (in weight %): 50 % NaOH, 7 % NaCl and 43 % H2O. The plant has a production rate of 5000 kg/h product solution. The atomic masses are (in [kg/kmol]): Na = 23, Cl = 35.5, H = 1, O = 16. Perform the following assuming steady-state: A) [10%] Draw the block flow diagram, label streams and write down the balanced reaction occurring in the electrolysis unit between salt and water. B) [10%] Calculate the extent of reaction in [kmol/h]. C) [20%] Calculate the fractional conversion of NaCl and the Cl2,(g) volumetric production rate in [m3/h] at 40 oC and 1.5 atm absolute pressure. Note: 1 atm = 1.013x105 Pa. R = 8.314 J/(mol K) D) [20%] How much water is removed in the evaporator? 4 Product soln. A: [10%]) NaCl 4 1 H2O Cl2,(g) H2,(g) Solution Problem 1: 30 %wt NaCl Mixer NaOH, NaCl, 6 H2O Electrolysis 3 2 7 5 5000 kg/h Evaporator 8 NaOH 50% NaCl 7% H2O H2O 43% 2NaCl + 2H2O  2NaOH + H2 + Cl2 B: [10%]) = 𝑚̇ NaOH mass balance for the overall process: 𝑚̇ 𝑀 where, 𝑚̇ = 𝑚̇ = Thus, = = 𝑚̇ = = 𝑀 = 𝑔 𝑚𝑜𝑙 . C: [20%]) NaCl fractional conversion: 𝑓 = 𝑛̇ Thus, 𝑛̇ 𝑓 𝑛̇ ( ̇ = ) Cl2 mole balance: 𝑛̇ = Use ideal gas law: ̇ = = 𝑛̇ = 𝑚̇ 𝑀 = = 𝑚𝑜𝑙ℎ ( = -we need the NaCl molar feed rate. 𝑛̇ = 𝑛̇ = = ) = 𝑛̇ NaCl overall process mole balance: 𝑛̇ ( = ) 𝑚𝑜𝑙ℎ 𝑅𝑇; ̇ = = 𝑚ℎ 𝑚𝑜𝑙ℎ 5 D: [20%]) H2O overall process mass balance: = 𝑚̇ 𝑚̇ 𝑚̇ 𝑀 We want 𝑚̇ but we also do not know 𝑚̇ The latter can be easily calculated from a water mass balance on the mixer: Hence: 𝑚̇ = 𝑀 𝑛̇ = = ̇ ̇ = 𝑔ℎ Thus, the water removed in the evaporator: 𝑚̇ = answer is acceptable. = 𝑔ℎ or 337.36 kmol/h. Either ----------------------------------------------------------------------------------------Verify the calculations by performing a total mass balance on the process: 𝑚̇ 𝑚̇ = 𝑚̇ Note that: 𝑚̇ 𝑚̇ = = 𝑚̇ = 𝑚̇ 𝑚̇ 𝑔ℎ and 𝑚̇ = = = 𝑔ℎ = Thus: = (𝑣𝑒𝑟𝑦 𝑔𝑜𝑜𝑑 𝑚𝑎𝑠𝑠 𝑎𝑙𝑎𝑛𝑐𝑒 𝑐𝑙𝑜𝑠𝑢𝑟𝑒) ------------------------------------------------------------------------------------------ 𝑔ℎ 6 Problem 2 (40% of the mark): Ethylene oxide (C2H4O) is a high volume chemical intermediate that is used to produce glycol and polyethylene glycol. Ethylene oxide is produced in a catalytic reactor by the partial oxidation of ethylene (C2H4) using air: C2H4 + 1/2O2  C2H4O [1] In addition to reaction [1], a secondary reaction takes also place in the same reactor: C2H4 + 3O2  2CO2 + 2H2O [2] The product gas leaving the reactor has the following composition [in % kmol]: 40% C2H4O, 3% CO2, 3% H2O (as steam), 10% C2H4 and the rest (to 100%) is air. The plant has a production capacity of 105 kmol/year C2H4O. The air molar composition is: 21% O2 and 79% N2. Perform the following assuming steady-state: A) [10%] Perform a DOF (degree of freedom) analysis on the reactor and comment briefly on your result. B) [10%] Calculate the ethylene (C2H4) feed to the reactor expressed in [lb/h] necessary to assure the plant production of ethylene oxide. The molar mass [kg/kmol]: C2H4O = 44, C2H4 = 28. C) [10%] Prove by calculation whether C2H4 or O2 is the limiting reactant in the reactor D) [10%] Determine the selectivity and yield of C2H4O with respect to the limiting reactant. 7 Solution Problem 2: Product gas Reactor C2H4 Air (21 % O2 + 79% N2) C2H4 + 1/2O2  C2H4O C2H4 + 3O2  2CO2 + 2H2O C2H4O 40%; 105 kmol/year CO2 3% H2O 3% C2H4 10% Air (21% O2 + 79% N2) 44% A: 10%) DOF = Nvariables - Nequations Nvariables: Stream components = 3 (inlet) + 6 (outlet) = 9 System variables = 2 (reactions) Nvariables = 11 Nequations: Basis flow = 1 (C2H4O production rate) Stream compositions = 4 (independent out compositions) + 1 (indep. composition air in) + 1 (indep. composition air out) = 6 System specifications = 0 MBE = 6 (C2H4, oxygen, nitrogen, CO2, H2O, C2H4O) Nequations = 13 DOF = 11 – 13 < 0 (overspecified: problem can be solved). [A good question would be how can the outlet air composition remain the same when O 2 is consumed whereas N2 is not consumed in the reactor? Because some of the N2 can be separated and purged from the system, such that to maintain a constant air composition (i.e. O2 and N2 ratio). However, this long story is not essential here because N2 is an inert species so it does not matter. You need to utilize the specified air composition. That’s all. Originally, this was a reaction, separation and purge problem but I simplified it for you guys to a reactor problem only, not to give you at this point a purge problem and to shorten the calculations in the Midterm.] 8 B: 10%) = 𝑛̇ C2H4 mole balance: 𝑛̇ ; Need the two extents of reactions: use 2 key products for each reaction = C2H4O mole balance: CO2 mole balance: = 𝑛̇ ; hence 𝑛̇ = 𝑚𝑜𝑙𝑦𝑒𝑎𝑟 ; (alternatively, it would be the same with H2O mole balance) where from the OUT stream composition observe that: ̇ = ̇ 𝑛̇ Further: 𝑛̇ = = Thus: = 𝑛̇ = = 𝑚𝑜𝑙ℎ; hence 𝑛̇ = 𝑚𝑜𝑙 = 𝑦𝑒𝑎𝑟 = = 𝑚𝑜𝑙𝑦𝑒𝑎𝑟 = 𝑚𝑜𝑙ℎ 𝑚𝑜𝑙𝑦𝑒𝑎𝑟, 𝑛̇ 𝑛̇ = = 𝑔 = ℎ 𝑙 ℎ C: 10%) Limiting reactant: need the ratio ̇ ̇ , compared to stoichiometric ratio in the reactor: ̇ = ̇ = 𝑛̇ O2 mole balance: 𝑛̇ = Thus: ̇ ̇ 𝑛̇ 𝑛̇ , where = = = 𝑛̇ = 𝑚𝑜𝑙𝑦𝑒𝑎𝑟 ; 𝑛̇ = = = 𝑚𝑜𝑙𝑦𝑒𝑎𝑟 Thus, O2 is the limiting reactant in the reactor. D: 10%) Selectivity and Yield with respect to O2 which is the limiting reactant. Desired product: C2H4O 9 = 𝑌= = = = 𝑚𝑜𝑙 𝐶 𝑂 𝑚𝑜𝑙𝐶 𝑂 𝑚𝑜𝑙𝑂 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑚𝑜𝑙𝑂 𝑓𝑒𝑒𝑑 Note: the theoretical best selectivity with respect to O2 is, based on stoichiometry of reaction [1], = 1 / 0.5 = 2. In other words, without the secondary un-desired reaction in the system. Clearly, the practical selectivity is lower than the theoretical. --------------------------------------------------------------------------------------------------------------------Verify calculations: Total Mass In = Total Mass Out; Note: N2 can be neglected from the reactor because it is not reacting. 𝑛̇ 𝑛̇ = 𝑛̇ 𝑛 𝑛̇ =( 6,304,200 = 6,304,200  100% mass balance closure! 𝑛̇ 𝑛̇ )