assignment # 2 Solution

assignment # 2 Solution - A2 Solutions Q1 : W X10093...

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Unformatted text preview: A2 Solutions Q1 : W X10093 -15.0wf% fructose 105.97g total 15190 g fructose(W) lfm g 010% l X10093} -1,?3mi% fructose 13.90 g fmctos W +9003 g fmctos W 180 g fructose 18 g Water Q2: T = 100°C = 212°F = 672°R P = ’?5 pisa (115 113 11311101 min 32 1b] - 3.59 Ibrnolfnun V RT (10.?3ii3psiar’1bmol mks???) , 96.1 39311311101 13 P 3’5 ps1a From ideal gas law, (3.591bmol 96.1 113 min lbmol }- 345 ft 3min At STP: same molar flow rate, molar volume = 359 ft3flbmol (3.59 lbrnol)[359 s3 . -1289 ft33min min bmol Q3: The N2 data are plotted in Figure 1. Alongside the data I plotted the ideal gas equation P = nRTr’V, with n = 1.000 gmol, and V = 22.414 L or 2.24 L. The data lie very close to the line representing the equation; there is a slight deviation only at high P. Nitrogen behaves as an ideal gas. 40 1 l l . 35 I - 30 I - 25 I expedmernz “ g 20 q o. 15 — 10 - - 5 experimm 1 M o W 0 200 400 600 300 1000 1200 Tag Figure 1. Q4: On the block flow diagram, H is hydrogen, M is methane, B is benzene and T is toluene. 933 gas gas Splitter — 938 T ._ y to H se tor fuel para -—)~ Mixer Reactor " (high P) 928 '2 H M T liquid Separator _—’ (“mm”) 98.21 tons)?! 9.6% s 1' (a) 8.21 tonsfh benzene produced (b) liquid stream contains 0.014 mol% hydrogen liquid stream contains 0.062 mol% methane distillation product is 99.6 mol% benzene (c) 35% toluene converted to benzene 35% of gas stream recycled 5 % of gas stream sent to reactor Q5: The flow diagram, with the mixer as system, is Fast Feed 45% G 2% P 1% N 3% T 49% F SuperGro 35% P Broth 15% N 15% G 8% T . 6% P 42% F We" M 6% N Water T F water Formula N 12% P 58% N 30% F From a DOF analysis, we find that there are 19 stream variables, 12 specified stream compositions, 1 specified basis (1 kg broth), and 6 material balances. DOF = 19 - (124-1445) = 0. (a) Since we are given the compositions, but not the quantities, of each of the powders that go into the mix, it is easier to write the stream variables as the (known) mass fraction times the (unknown) total quantity of the powder. We will use “FF”, “SG”, and “FN” to indicate the quantities of the 3 powders. The material balance equations are: Glucose: 0.45FF - 015(1) : 0.15 kg, so FF : 0.333 kg Phosphate: 0.02FF +0.355'G +0.12FN - 0.015(1) - 0.06 kg Nitrate: 0.018” + 0.15.S'G+ 0.58FN :- 0.06(1) 4.06 kg Substituting in the known value for FF and solving simultaneously, we find 56 -0.13 kg, and PW - 0.064 kg Each kg of broth should contain 0.333 kg Fast-Feed, 0.13 kg SuperJGro, and 0.064 kg Formula N, plus 0.4’33 kg water. (b) The cost per kg of broth (excluding any operating costs such as labor or electricity) made from these ingredients is [0.333x$20) +(0.13 x $10)+ (0.064 x $15) =$a92 Since the purchased broth costs $15i'kg. we can save $6,08fkg (at most) by mixing our own. With a 20 kg batch. we save (20 it $6.08) or $121i’batch. If the equipment costs $2000. we need to make up ($2m0i'1131211batch}. or 16 batches total. to pay for the equipment, Q6: Flow diagram: N} H G 3 w w G 1 H B W Mixer Reactor Separator 3 4 H 2 %’ 6 Balanced chemical reaction: C6H1206 +32 "' C6H1406 We’ll solve using each process unit in turn as the system. Components: Glucose (G), water (W), hydrogen (H) and sorbitol ((S). DOFW Variables Constraints Stream 14 S ste_rn I 1 (reaction) I l I Wed flows | I 1 Specified 2 (%glucose in feed compositions solution, ratio of HQ fed to glucose fed} Specified system I 1 (96 conversion in Woe reactor) Material balance I I 11 (2 per unit) DOF: 15-15 =0 Units: all in kgr‘day or wt%. Since we have a reaction of known stoichiometry, we will need to convert mass to moles. Basis: GI 4- WI -100 kgiday Stream composition specifications: GI - 0.3, so G, a- 30 kgiday, W1 - ’30 kgiday G1 4» W1 Hydrogen is fed at stoichiometric ratio to glucose, or 1 mol H, per mole glucose. Our stream variables are in mass flow units, so we need to convert through the molar masses: Hz 1kgmol H2 x 2 kgfkgmol 0.0111 kg H2 , 80 Hz - 0.333 kgfday GI kgmol (36111206 X 180 kgfkgmol kg C6H1206 System performance specification: 600m — - 0.8 Gs amen From reaction stoichiometryr 1 mole H2 is consumed and 1 mole sorbitol is generated per mole glucose consumed. Converting to mass ratios gives: %-L£fl-w Gem iso’Gcm 180 Material balance equations: For mixer: 61 - 63 - 30 kgfday, W1 - W3 - ’3’0 kglday, Hg - W3 s0.333 kgfday For reactor: W3 - W4 - '3‘0 kgiday G3 mm, = G4 = 30 -0.s(3o) = 6 kgfday 2 83 Room H4 0.333 180(24) 0.066 kgs’day 182 5 IS -—24 I242? fda gen 4 180( ) kg 3’ For separator. W4 - W6 - 3’0 kgfday G4 C 66 I 6 kgfday H4 - 35 -0.066 kgfday S4 - S6 :- 242’? kgfday To summarize, 0.333 kglday hydrogen is fed to the process. The total liquid flow rate out is (30+6+24.2?) = 100.3 kgfday, and it contains 69.8 M96 water, 24.2 wt% sorbitol, and 6.0 wt% glucose. ...
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