HW4 FunD

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University of North Dakota *

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201

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Chemistry

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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%
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
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