HW4 FunD

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Apr 3, 2024

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Mullis, Ashton Rayne Professor Bowman ChE-201 20240210 Homework 4 1. The attached drawing 01-A-005/1 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 reboiler E303. a) Determine the mass flows missing in the stream table for streams 21, 22, and 23. b) Calculate the mass fractions of each component in streams 21, 22, and 23. c) Suggest an explanation for why the stream compositions are not the same. Solutions (A) System boundary around E-303 and streams 21,22, and 23 in question. General balance equation, In=Out S 21 = S 22 + S 23 S21, ˙ m = ¿ 820,000 lb/hr S 21 ( nC 4 ) = 820,000 − 480,000 − 7,700 − 15,000 = 317,300 lb hr  Round down to 300,000 lb/hr… Balance Equation S22 (C5+) = S21(C5+) – S23(C5+) 15,000 − 3,100 = 11,900 lb hr  Round up to 12,000 lb/hr… S22(C5+) Solve for S22 Total ˙ m (lb/hr) S22( ˙ m ) = ∑ product ❑ S22( ˙ m ) = 200,000+320,000+4,000+12,000 = 536,000 lb/hr

 round up to 540,000 lb/hr… Solve for S23(iC4) Balance Equation S23(iC4) = S21(iC4) – S22(iC4) S23(iC4) = 480,000 – 320,000 S23(iC4) = 160,000 lb/hr Balance Equation S23(C1-C3) S23(C1-C3) S23( ˙ m ) = ∑ streams ❑ S23( ˙ m ) = 276,800 lb/hr  Round up, 280,000 lb/hr (B) Mass Fractions of Components (S21-S23) χ m = m m total S21 (nC4)… χ m = 300,000 820,000 = 0.37 (iC4)… χ m = 480,000 820,000 = 0.59 (C1-C3)… χ m = 7700 820,000 = 0.0094 (C5+)… χ m = 15000 820,000 = 0.018 S22 (nC4)… = 0.37 (iC4)… = 0.59 (C1-C3)… = 0.0074 (C5+)… = 0.022

S23 (nC4)… = 0.39 (iC4)… = 0.57 (C1-C3)… = 0.013 (C5+)… = 0.011 (C) Stream compositions will vary with different boiling points. The boiling points will be higher or lower depending on the components relative mass and density values. Also, since the streams do consist of different chemical varieties, they may be subject to chemical reactions when placed under stress, such as heating, cooling, or mixing. 2. Brown sugar is a mixture of white sugar and molasses. Unrefined brown sugar contains a large portion of molasses. After one round of crystallization of this unrefined sugar, the brown sugar product contains 15.00 wt% molasses and the remainder white sugar. A second round of crystallization reduces the molasses content to 1.10 wt%

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molasses. A plant wishes to produce a brown sugar ideal for baking that contains 6.50 wt% molasses. In order to achieve this concentration of molasses, a plant feeds once-crystallized brown sugar into a recrystallizer. Some of this feed is diverted into a bypass stream that rejoins the sugar that leaves the crystallizer. Pure molasses leaves the crystallizer as a bypass stream. This process is described by the flow chart below. a) If the plant wishes to produce 350.0 g/min of brown sugar today, what is the feed rate, m 1 ? b) What is the flow rate of pure molasses from the crystallizer, m 4 ? c) What is the fraction of the feed that bypasses the crystallizer? Solution (A) ˙ m 1 Sugar, given a feed rate of 350 g/min. m 1 g min = 1.839 total × 350 g min 0.8500 g m 1 = 757.24 g min (B) ˙ m 4 Pure Molasses . 0.1500 g m 1 ( 757 mL min ) = ˙ m 4 + 0.0650 g m 6 ( 350 g min ) 113.55 = ˙ m 4 + 22.75

˙ m 4 = 113.55 − 22.75 = 90.8 g min (C) Molasses Feed Fraction. 0.1500 90.8 = 1.65 E − 3 3. Sea water is to be desalinized by reverse osmosis (RO) using the process shown in Figure 1. The process is designed to purify 1500 lb/hr of sea water which is comprised of 96.9 wt% water with the remainder comprised of various salts. Determine: a) The flow rates of fresh water (m 4 ) and brine waste (m 6 ). b) The percentage of salt rejected by the RO unit. In other words, what percent of the salt fed to the RO unit in m 3 exits the RO unit in m 5 . c) The percentage of brine exiting the RO unit that is recycled. d) How much salt (in lb/hr) would be rejected if the recycle stream was removed (as shown in Figure 2)? Note: 99.8% of the salt is rejected by the RO unit. How does this compare to the process in Figure 1? Can you determine the waste brine and purified water flow rates for the process shown in Figure 2?

Solution (A) Balance on m 4 ∧ m 6 Equation IN=Out m 1 = m 4 + m 6 -- 1500 lb hr ( 0.031 ) = m 4 ( 500 ppm ) + m 6 ( 0.0525 ) — 46.5 lb hr = m 4 ( 0.0005 ) + m 6 ( 0.0525 ) System of two Equations (1)(0.0525)-(2) 78.75 = 0.0525 m 4 + 0.0525 m 6 -(46.5 = m 4 0.0005 + m 6 0.0525 ¿ 32.25 = 0.052 m 4 m 4 = 620.1923077 620.2 lb hr 1500 lb hr = 620.1923077 + m 6 m 6 = 879.8076923 879.8 lb hr (B) Balance In=out m 1 + m 2 = m 3

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∵ m 1 = 1500 lb hr ∴ 1500 = m 3 − m 2 —1 Salt Balance m 1 ( 0.031 ) + m 2 ( 0.0525 ) = m 3 ( 0.04 ) 46.5 + m 2 ( 0.0525 ) = m 3 ( 0.04 ) 46.5 = m 3 ( 0.04 ) − m 2 ( 0.0525 ) –-2 Systems of two equations (1)(0.0525)-(2) 78.75 = 0.0525 m 3 − 0.0525 m 2 − ¿ ) 32.25 = 0.0525 − 0.04 m 3 m 3 = 2580 lb hr 1500 = 2580 − m 2 m 2 = 1080 lb hr Salt Rejected 4%( m 3 ¿ 2580 × 0.04 = 103.2 lbs Salt Rejected 5.25% ( m 5 ) m 5 = m 6 + m 2 m 5 = 879.8 + 1080 ( lb hr ) m 5 = 1959.8 lb hr ( 0.0525 ) = 102.89

Total Salt Rejected % 102.89 103.2 × 100 = 99.7% (C.) m 2 m 5 × 100% 1080 1959.8 × 100 = 55.1% (D) Salt Rejected 1500 lb hr × 0.31 wt % salt = 46.5 lb hr RO Recycle Unit Output = 500ppm*100% = 0.05 m 4 = 46.5 0.05 = 930 lb hr m 6 = m 1 − m 4 m 6 = 1500 − 930 m 6 = 570 lb hr 4. Carbon dioxide (along with other gases) has been implicated in global warming. A process is being developed which will separate CO 2 from flu gas in a coal fired power plant and subsequently sequester the CO 2 underground. A schematic of the proposed process is shown below.

A cleaned and dry flu stream (m 1) comprised of 26 wt% CO 2 and 74 wt% N 2 , enters into an absorber at a flow rate of 5.0 × 10 4 kg/hr where it contacts liquid-phase amine solvent (MEA) in an absorber. The amine solvent selectively absorbs CO 2 . The flu gas stream exiting the absorber (m 2 ) is composed of 98 wt% N 2 and 2.0 wt% CO 2 . The liquid stream exiting the absorber (m 3 ) contains only amine and CO 2 . The liquid stream is subsequently fed to a stripper where it is heated and 95% of the available CO 2 is desorbed from the amine. The CO 2 exiting the stripper (m 4 ) is subsequently pressurized and injected into subterranean rock formations. The MEA exiting the stripper (m 5 ), comprised of 99 wt% amine and 1.0 wt% CO 2 , is recycled back into the absorber. a) What percent of the CO 2 in stream m 1 is removed? b) What is the concentration (in wt%) of CO 2 in stream m 3 ? c) What is the flow rate (in kg/hr) of the recycle stream (m 5 )? d) Tests conducted on experimental apparatus reveal that there will be a trace amount of amine in the pure CO 2 stream. If the concentration of amine in stream m 4 is 25 ppm (not 0 as the figure shows) determine the required flow rate of make-up MEA (in kg/hr) required in the process. For part d) please assume that the total flow rate of m 4 to be 15,000 kg/hr. Solution m 1 CO 2 = 0.26 ∗ 5.0E4 kg hr = 13,000 kg m 1 N 2 = 0.74 ∗ 5.0E4 kg hr = 37,000 kg

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m 2 = m 1 N 2 0.98 m 2 = 37,000 0.98 = 37,755.10204 kg 37,755.1 kg m 2 C O 2 = m 2 ( 0.02 ) m 2 C O 2 = 37,755.1 × 0.02 = 755.102 kg C O 2 Removed = 13,000 − 755.102 = 12244.979 kg 12,245.0 kg C O 2 % Removed = 12,245.0 5E4 × 100 = 24.5% (B.) C % M 3 m 5 = 0.99% amine, 0.01% C O 2 mC O 2 , 0.01 m 5 = 0.05 ( 0.01 m 5 + 12244.979 ) kg / hr 0.01 m 5 = 0.0005 m 5 + 612.24895 0.0095 m 5 = 612.24896 ( C ) −− →m 5 = 64447.3 kg /hr m 3 C O 2 = 0.01 ( 64447.3 ) + 12244.979 m 3 C O 2 = 12889.5 kg MEA 0.99 ∗ 64447.3 = 63802.8 kg (B)-  C O 2 WT % = 12889.5 12889.5 + 63802.8 × 100 = 16.8% (D)

0.0025 MEA ppm, ˙ m 4 = 15,000 kg hr 0.0025 MEA× 15000 kg hr = 37.5 kg hr MEA m 3 = m 4 ∗ 25 ppm + m 5 ( 0.99 ) m 3 = 15,000 ∗ 0.0025 + 64447.3 ( 0.99 ) m 3 = 63,840.3 k g /hr