ChE201 HW6

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Apr 3, 2024

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Mullis, Ashton Rayne Professor Bowman ChE-201 Fundementals 20240304 HW6 1) The attached drawing 01-A-004/1 is a process flow diagram (PFD) for one small section of a much larger butane isomerization plant. PFDs are normally printed on 11”x17” sheets of paper. For this problem we will focus on a mass balance around the first isobutane reactor R201, where n-butane is converted into isobutane, along with some C1-C3 side products. a) What is the fractional conversion of n-butane to isobutane? b) What is the fractional conversion of butanes (n-butane and isobutane) to C1-C3? c) What other process units would you suggest be included in the plant to increase the overall conversion? Solutions (A) f i = n i Reacted n i Fed = 520000 − 230000 520000 lb hr = 0.56 n − butane (B) in – out / total out f i = ( 520000 + 12000 ) −( 230000 + 300000 ) ( 230000 + 300000 ) = 0.0038 butanes  c1-c3 (C.) The conversion between A-B is very low, so assuming there is buildup of excess reactant in the system, it would be best to add a separator to remove the reacted product from the excess, and then send the excess stream back via a recycle stream or purge stream, depending on the purity of the recyclable.

2) Solid calcium fluoride (CaF2) reacts with sulfuric acid to form solid calcium sulfate and gaseous hydrogen fluoride (HF). The HF is then dissolved in water to form hydrofluoric acid. A source of calcium fluoride is fluorite ore containing 96.0 wt% CaF2 and 4.0% SiO2. In a typical hydrofluoric acid manufacturing process, fluorite ore is reacted with 93 wt% aqueous sulfuric acid, supplied 15% in excess of the stoichiometric amount. Ninety-five percent of the ore dissolves in the acid. Some of the HF formed reacts with the dissolved silica in the reaction 6HF+SiO2(aq)→H2SiF6(s)+2H2O(l) The hydrogen fluoride exiting from the reactor is subsequently dissolved in enough water to produce 60.0 wt% hydrofluoric acid. a) Draw and label a flow diagram for this process b) Calculate the quantity of fluorite ore needed to produce a metric ton of aqueous hydrofluoric acid. (Some of the given data are not needed to solve the problem) Solutions (A) (B) Basis 0f 100kg Ore Inlet stream 6 mols of HF = 600kg=6E5g required for reaction. Fluorine, 18.998 amu HF, 20.01 g/mol 60% wt HF ratio needed for reaction = 600 kg× 1000 g 1 kg ÷ 20.01 g mol = 3 × 10 4 n HF

96% wt Ca F 2 = 9.6E5 gCa F 2 × ( 20.01 × 2 nF ) 78.07 gmol − 1 Ca F 2 = 4.92E5 g× 0.95 = 4.68E5 g÷ 18.998 Famu = 2.46E4 n F 1 T ore Basis Silica 1.0 – 0.60HF = 40%wt silica Silica, 60.08 g/mol 4E4 gSilica 60.08 gmo l − 1 silica = 6.7E2 nsilica ∈ mixture× 6 nratioS : F = 4.00E3 n F excess 2.46E4nF – 4.00E3nF = 2.06E4n HF 3.00E4 nHF 2.06E4 nHF = 1.46 MT = 1460kg 3) A catalytic reactor is used to produce formaldehyde from methanol in the reaction, CH3OH→HCHO+H2 A single-pass conversion of 60.0% is achieved in the reactor. The methanol in the reactor product is separated from the formaldehyde and hydrogen in a multiple-unit process. The production rate of formaldehyde is 720. kg/h. a) Calculate the required feed rate of methanol to the process (kmol/h) if there is no recycle. b) Suppose the unreacted methanol is recovered and recycled to the reactor and the single- pass conversion remains 60%. Determine the required fresh feed rate of methanol (kmol/h) and the rates (kmol/h) at which methanol enters and leaves the reactor. c) The single-pass conversion in the reactor, Xsp, affects the costs of the reactor (Cr) and the separation process and recycle line (Cs). What effect would you expect an increased Xsp would have on each of these costs for a fixed formaldehyde production rate? (Hint: To get a 100% single-pass conversion you would need an infinitely large reactor, and lowering the single-pass conversion leads to a need to process greater amounts of fluid through both process units and the recycle line.) What would you expect a plot of (Cr+Cs) versus Xsp to look like? What does the design specification Xsp=60% probably represent? Solutions (A)

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˙ n = 750 kg hr × 1 kmol 30.031 gmo l − 1 formaldehyde = 25.0 kmol hr formaldehyde Given :60% methanol conversionrate MethanolConverstion − Feed Rate Required = ˙ n 1 = 25.0 = 60 100 = 25.0 × 100 60 = 41.7 kmol hr methanol (B) Methanol Feed, Input Stream Fresh Feed Rate into Reactor is 1methanol :1 formaldehyde in=out ,25 kmol/hr. Fresh Conversion Rate, 25 25 n 1 + n 3 = 0.60 …n 3 = 16. 6 out (C.) An increase in XSP reactor size would inherently increase the cost since a new larger reactor will be required. However, since the reactor is producing a product of methanol at a higher production rate it would save costs of having to install (or make use of decreases if one exists already) a recycle stream and separator into the system, which would also increase energy costs of the system overall. Assuming the system is at a decreasing conversion and then the overall conversion rate is increased due to the addition of a larger reactor or recycle stream, then the production would begin to increase, translation to y = x 2 , thus it would be a parabola with concavity being positive. 4) One process for the production of methanol is the catalytic reaction of carbon dioxide and hydrogen. The chemical reaction is given by: C O 2 + 3 H 2 →C H 3 OH + H 2 O The diagram for the process is shown below. It is desired to produce 1000 mol/hr of methanol. The single pass conversion of the reactor is 10% while the overall conversion is 80% (both in terms of hydrogen). Note, carbon dioxide in the fresh feed is in 20% excess and all of the product (methanol) and byproduct (water) is removed in the condenser. a) Why is the recycle stream included in the design? b) What is the purpose of the purge stream? In other words, why would this design not work without the purge stream? c) List two reasons (with supporting explanation) why the single pass conversion might be so low.

d) Determine the flow rate and composition of the feed stream (F). e) Determine the flow rate and composition of the purge stream (P). f) What percentage of the flow leaving the condenser enters the purge stream? g) Determine the recycle ratio (R/F). Solutions (A) Since the conversion rate is not to completion on the first pass through, the raw materials are refed back into the process. The purpose of a recycle stream is to reuse reactants, to save costs on fresh feed materials. (B) The purge stream makes this process a steady-state system, since there is accumulation of unprocessed, the purge stream prevents this process from becoming a semi-batch process. (Maintaining reaction equilibrium) (C.) 1. Perhaps the molar ratio of feed reactants and recycle stream is not correct, too much or too little could cause a smaller conversion rate of desired product. 2. Perhaps the systems temperature needs to be adjusted to meet the activation energy requirements of solution.

(D) F = 5250 mol/hr Fa= 3750 mol/hr Fb= 1500 mol/hr (E.) P= 1250 mol/hr Pq= 500 mol/hr Pp= 750 mol/hr (F) P/Cin 1250 mol hr 44000 mol hr × 100 = 2.84% (G) R/F

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