56 is that the quantity £0,99982 . 5.99670 ~ 1 0.00313 has to be determined quite accurately. [t would be more convenient to use an NazSO« balance, 0.00018x + 10 = 0,00330y Solving the total balance and the NaSO, balance simultaneously, x = 3210 Ib/hr 32101b| 1gal _ Fi ar TessTe = 385 gal/hr EXAMPLE 2.9 Crystallization The solubility of barium nitrate at 100°C is 34 g/100 gH,0 and 100 g H,0. If you start with 100 g of Ba(NO3), and make a satu water at 100 C, how much water is required? If this solution is cooled ta OFC. how much Ba(NO); is precipitated out of solution? Tie-component Solution: Basis: 100 g Ba(NO3): The maximum solubility of Ba(NO;), in H,O at 100°C is a saturated solution, or 34 g/100 g H,O. Thus the amount of water required at 100°C is 100g H20 | 1008 34g Ba(NOs)2 | If the 100°C solution is cooled to 0°C, the Ba(NO;)2 solution will still be saturated, and the problem now can be pictured as in Fig. The tie element is the water. (NOs)2 — 295g H,0 ’ Originot Final 1 1009 Composition ? Grams Composition 1 1 BatNos), | Knowe i BoINOs), | Keown } eant k it 5 g BolNOs}/100 ¢ H20 ight Known \ | 2959 Weight \ H20 Unknown Crystals | Composition Ba(NO3]2 | Known ? Groms Weight Unknown57 295 g H,0 | 5 g Ba(NO3). . see OOS H;OF = 14.7 8 Ba(NO,)a in final solution original = — final = crystals 100 g Ba(NO;)2 — 14.7 g Ba(NO3), = 85.3 g Ba(NO3), precipitated Algebraic Solution: Let x = gerystals y = final solution H,0 balance ) g H,0 2 final solution Total balance 310 395 — y = 85 g Ba(NO3)2 crystals EXAMPLE 2d0 ‘ Recycle without chemical reaction A distillation coluinn separates 10,000,4b/hr of a 50 percent benztne-50 percent toluene mixture. The product recovered from the condenser at the-top of the column contains 95 percent benzene, and the bottoms from the column contain 96 percent toluene. The vapor stream entering the condenser from the top of the column is 8000 - Ibfhr. A portion of the product is returfed to the colunin as reflux, and the rest is withdrawn for use elsewhere. Assume -the compositions of the streams at the top of the column (V); the product withdrawn-(D), and the reflux (R) are identical. Find the ratio of the amount refluxed to the product withdrawn. Solution: See Fig... . All the compositions are known and two weights are unknown. No tie components are evident in this problem; thus an algebraic solution is mandatory. ea / 8000 lb/hr /” 0 t \, 1 0.95 B24 igyi iv yy 0.08 Tot bavi ! /~Boundary line for 1 balance around Liquid —£_! condenser 10,000 ib/tr | a “ 050 8: \ /—Boundary tine for 0.50 Tol i a overall balonce VL G04 8 104 82 Ligui 096 Tot Lisvid58 By making an overall balance (ignoring the reflux stream for the moment), we can find D. With D known, a balance around the condenser will give us R. Basis; 1 hr Overall material balances: Total material: F=D+W 10,000 = D+ W (a) Component (benzene): FXp = DX + WXw 10,000(0.50) = D(0.95) + W(0.04) (b) Solving (a) and (b) together, 5000 = (0.95)(10,000 — #7) + 0.04W W = 4950 Ib/hr D = 5050 Ib/hr Balance around the condenser: Total material: V=R+D 8000 = R + 5050 R = 2950 lb/hr R _ 2950 _ F = Fag = 0.584 EXAMPLE 2.11 Recycle without chemical reaction Data are presented in Fig. for an evaporator. What is the recycle stream in pounds per hour? 10,000 tb/hr 20% Feed KNO3 Solution Recycle 100°F Soturated Solution Boundary line for balance cround crystallizer Crystals carry off 4% H20 (4% H20 per. Ib’ foto! ‘crystals, + H20) Va Bounddry line for overall balance59 Solution: In this problem the compositions are all known except for the recycle stream, but this can be easily calculated. On the basis of 1 Ib of water, the saturated recycle stream contains (1.0 lb H,O + 0.61b KNO;) = 1.61b total. The recycle stream composition is 0.6 1b KNO; 1lbH,0 _ . TIbH,0 1 16lb.solution ~ 0.375 Ib KNO;/Ib solution With all compositions known, a search for a tie component shows that one exists for an overall balance—the KNO;. Temporarily ignoring the interior streams, we can draw a picture as in Fig. Now we can calculate the amount of crystals (and H,0 evaporated, if wanted). Basis: 1 hr = 10,000 Ib feed 10,000 1b F | 0.20 lb KNO; | _1 Ib crystals Tio F 0.9616 KNO; ~ 2080 Ib/hr crystals 420 Feed 0.20 KNOs aye7™ 080 H20 Crystals 7 * 0.96 KNO3 0.04 H20 To determine the recycle stream R we need, in the absence of an additional tie component, either (a) A balance around the evaporator—or (b) A balance around the crystallizer. The latter is easier since only three rather than four (the recycle added to the feed is the fourth stream) streams are involved. Total balance on crystallizer: M=C+R M =2080+R Component (KNO) balance on crystallizer: Mxy = Cxe + Rep 0.5M = 0.96C + R(0.375) Solving these two equations, 0.5(2080 + R) = 0.375R + 2000 R = 7680 lb/hr60 EXAMPLE 2.J2 Recycle with a chemical reaction Ina proposed process for the preparation of methyl iodide, 2000 Ib/day of hydro- iodic acid are added to an excess of methanol: HI + CH;0H —> CH;I + H,0 If the product contains 81.6 percent CHI along with the unreacted methanol, and the waste contains 82.6 percent hydroiodic acid and 17.4 percent HO, calculate, assuming that the reaction is 40 percent complete in the process vessel: (a) The weight of methanol added per day, (b) The amount of HI recycled. Solution: In this problem all the compositions are known (in weight percent) as well as one weight—the HI input stream. The unknown quantities are the weights of thé CH;0H (M), the product (P) ,the waste (HW), and the recycle stream. An overall balance about the whole process is all that is needed to calculate the M stream, but it will not include the recycle stream—see the dashed line in Fig. CH3] 81.6% Product CH;OH 18.4%. 2000 Ib/doy We first need to convert the product and waste streams to a mole composition in order to make use of the chemical equation. A basis of 100 lb of waste is convenient, although 100 1b of product would also be a satisfactory basis. Basis: 100 lb waste comp. b=% Mw 1b mole HI 82.6 i28 0.646 H,0 17.4 18 0.968 100.0 Basis: 100 Ib product equivalent Ib comp. MW 1b mole mole of CH;QH CHal 81.6 142 0.575 > 0.575 CH,OH 18.4 32 0:575 0.575 1,15061 (a) From the chemical equation we observe that associated with each pound mole of HO in the waste is 1 1b mole of CHI. Consequently the entering HI is (per 100 Ib waste). . Reacted 0.968 Ib mole H.O | 1 Ib molé CHjI| 1b mole HI | 1281b HI _ _1241b HI TOO Ib waste | 1 1b mole H,O | 1 Ib mole CHI [Tb mole Hi ~ T0016 waste Total HI = reacted + unreacted = 124 + 82.6 = 206.6 Ib HI/100 Ib waste On the same basis the CH,OH entering is, 0.968 Ib mole HzO | 1 1b mole CHI 1.150 Ib mole CH;OH| _ 32 1b CH;OH 100 Ib waste T1b mole H,0 | 0.575 1b mole CH3I |11b mole CH;OH _ 61.9 lb CH;OH ="TOOTb waste Basis: 1 day 2000 Ib HI | 100 1b waste | 61.9 1b CHsOH Tday | 206.615 art TOO TB waste ~ 600 1b CHsOH/day (b) To obtain the value of the recycle stream R, a balance must be made about a junction point or about just the reactor itself so as to include the stream R. A balance about junction 1 indicates that (all subsequent bases are 1 day) 2000+ R = reactor feed but this equation is not of much help since we do not know the rate of feed to the reactor. Similarly, for a balance on junction 2, gross product = R+P+W we can compute the value of the product and waste but not that of the gross product. The only additional information available to us that has not yet been used is the fact that the reaction is 40 percent complete on one pass through the reactor. Con- sequently, 60 percent of the HI in the gross feed passes through the reactor unchanged. We can make an HI balance, then, about the reactor: (2000 + R)(0.60) = R + 800 ‘ 400 = 0.40R R = 1000 Ib/day Our material balance around the reactor for the HI is known as a once-through balance and says that 60 percent of the HIthat enters the reactor (as fresh feed plus recycle) leaves the reactor unchanged in the gross product (recycle plus product). Two additional commonly encountered types of process streams are shown in Figs. (a) A bypass stream—one which skips one or more stages of the process and goes directly to another stage. (b) A purge stream—a stream bled off to remove an accumulation of inerts or unwanted material that might otherwise build up in the recycle62 Byposs 8 Process Product Bypass stream. Recycle F Divider} Purge Process Product A recycle stream with purge. stream. The purge rate is adjusted so that the amount of purged material remains below a specified level or so that the { rate of } _ 7 {re of entering material) _ {rate of Peal accumulation! ~~ ~ | and/or production and/or loss Calculations for bypass and purge streams introduce no new principles or tech- niques beyond those presented so far. An example will make this clear. EXAMPLE 2.13 Bypass calculations In the feed-stock preparation section of a plant manufacturing natural gasoline, isopentane is removed from butane-free gasoline. Assume for purposes of simplifica- tion that the process and components are as shown in Fig. What fraction of the butane-free gasoline is passed through the isopentane tower? {rene side stream (S) i-CeHy2 De- 'so- butonizer pentane tower wy n-CsHie (F1 100 tbs [Butane free 12), natural ey feed THT alive plant (7) n=CsHre 80% 90% n-Cete i-CsHy2 20% 10% i-CsHi263 Solution: By examining the flow diagram you can see that part of the butane-free gasoline bypasses the isopentane tower and proceeds to the next stage in the natural gasoline plant. All the compositions are known. What kind of balances can we write for this process? We can write the following (each stream is designated by the letter F, 5, or P): Basis: 100 Ib feed (@) Overall balances (all compositions are known; a tie component exists): Total material balance: in = out 100=S +P Oy Component balance (1-Cs): in = out 100(0.80) = s(0) + P(0.90) Q Using (2), P = 100( S= 100-89 =111b The overall balances will not tell us the fraction of the feed going to the isopentane tower; for this we need another balance. (b) Balance around isopentane tower: Let x = Ib of butane-free gas going to iso- pentane tower. Total material: x = 11 + mCsHy, stream (another unknown, y) @) Component (n-Cs): x(0.80) = y (a tie component) 4) Consequently, combining (3) and (4), x= 11+ 08x x =551b, or the desired fraction is 0.55 Another approach to this problem is to make a balance at points (1) or (2) called mixing points. Although there are no pieces of equipment at those points, you can see that streams enter and leave the junction. (©) Balance around mixing point (2): material into junction = material out * Total material: (100 — x) + y = 89 6) Component (iso-Cs): : . (100 — x)(0.20) + 0 = 89(0.10) © Equation (6) avoids the use of y. Solving, 20 — 0.2x = 8.9. x = 55 Ib, as before64 SUPPLEMENTARY REFERENCES Benson, S. W., Chemical Calculations, 2nd ed., Wiley, New York, 1963. 2. Henley, E. J., and H. Bieber, Chemical Engineering Calculations, McGraw-Hill, New York, 1959. 3, Hougen, 0. A. K. M. Watson, and R, A. Ragatz, Chemical Process Principles, Part I, 2nd ed., Wiley, New York, 1956. _ 4. Whitwell, J. C., and R. K. Toner, Conservation of Mass and Energy, Ginn/Blais- dell, Waltham, Mass., 1969. 5. Williams, E. T., and R. C. Johnson, Stoichiometry for Chemical Engineers, Mc- Graw-Hill, New York, 1958. PROBLEMS 2.4. The steady-state hydrology of the Great Lakes based on average flow values for the period 1900-1960 is shown in Fig. P2.1. Compute the return into lake St. Clair and the evaporation from Lake Ontario based on the data shown. Would the balance for 1 year be a steady-state balance? For 1 month? Prova fuse | { Looe F Passo Ress, (R) Base Ruri ee { Esso (P) Preciotation on Lake ua 1E) Evaporation from Lake ‘SMe Rue Huron | eFS22 Valves - woo cfs 319 — geno Mackinac SFO, I srchigan |Staits 88 Chcoge . Ute re ft St Clan 7 sass ” Prsis ‘ Rem Fig. P21. 2.2. By use of an overall steady-state material balance determinine whether or not the petrochemical process indicated in Fig. P2.2 has been properly formu- ated. The block diagrams represent the steam cracking of naphtha into various products, and all flows are on an annual basis, i.e., per year.65 2.3. Figure P2.3 shows the production of aromatics, synthetic fibers, and plastics using naphtha as a feed stock, Check the overall steady-state material balance to see that the annual flows are correctly represented in the diagram. 2.4. Consider a lake as a system, and discuss the material balance on (a) water, (b) salts, and (c) plant life which might be made for such a lake. What quantities can be measured, and what quantities must be calculated? 2.5. A study of the carbon cycle in the biosphere is fundamentally a study of the overall global interactions of living organisms and their environment. The bio- sphere contains a complex mixture of carbon compounds in a continuous state of creation, transformation, and decomposition. This dynamic state is main- tained through the ability of phytoplankton in the sea and plants on land to capture the energy of sunlight and utilize it to transform carbon dioxide (and water) into organic molecules. A fundamental and interesting problem is the effort to grasp the overall balance and flow of material in the worldwide com- munity of plants and animals that has developed in the few billion years since life began. This is ecology in the broadest sense of the word. Figure P2.5 shows the carbon circulation in the biosphere expressed as billions of metric tons per year and also the carbon in various reservoirs. (a) Is the carbon cycle as shown in the steady state? (b) What is the net addition of carbon dioxide to the atmosphere per year? ATMOSPHERE 700 1 Respiration Assimilation of 35 sea Surtoce 5001004 [ev Plants on Land | Phytoplanton { JANI LAND 450 5 | 40 assimitation Exchange 20 i o r a [200 Pionkton,| —}o9 SEA Fish <5 tl s S 20 s/f BlE | decd Orgonic | Dead Organic als Matter Matter “13 700 3000 |= : 34,500 Exchange £13 il ‘Sediments id 8 S50 20,000,000 Fig. P2.5. 2.6. Hydrogen-free carbon in the form of coke is burnt (a) with complete combustion using theoretical air, (b) with complete combustion using 50 percent excess air, and (¢) using 50 percent excess air but with 10 percent of the carbon burning to CO only. In each case calculate the gas analysis which will be found by testing the flue gases-with an Orsat apparatus.2.7. 2.8. 2.9. 2.10. 2.41. 66 In the Deacon process for the manufacture of Cla, a dry mixture of HCl and air is passed over a heated catalyst that promotes the oxidation of HCl to Cl. The air used is 26 percent in excess of that quantitatively required for oxidation according to the reaction 4HCl + 0, + 2Cl, + 2H,0 Compute the composition of the gas that is fed to the reaction chamber. It has been proposed [R. L. Miller and J. D. Winefordner, Environmental Science and Technology, v. 5, p. 445 (May 1971)] that a general relation can be developed for the ratio of O, to N2 in combustion. For example, consider the following cases; the fuel might be cellulose (CsH100s), (1) Stoichiometric case: According to the balanced chemical reaction 10, + 3.76N; + fuels —> 00, + 3.76N, + products the ratio of O to Nz is 0.00. (2) 50 percent excess air case: According to the balanced chemical reaction 1.500, + 5.64N, + fuel —> 0.502 + 5.64Na + products the ratio of O2 to Nz is (0.5/5.64) = 0.0886. (3) 100 percent excess air case: According to the chemical reaction 202 + 7.53Nz + fuel —> 10, + 7.52N2 + products the O2 to N; ratio is 0.133. (4) Infinite excess air case: For this limiting case, the O2 to N2 ratio is simply (1/3.76) = 0.0266. The ratio of O, to Nz, R, and the percentage excess air, P, are related by the equation =~ 3.76P + 376 (a) Is the relation given valid for the ratio of O, to N; for cellulose? For other waste materials? For hexane? (b) Develop a formula in terms of R for the percent excess air for the combustion of 100 Ib of cellulose. If moist hydrogen containing 4 percent water by volume is burnt completely in a furnace with 32 percent excess air, calculate the Orsat analysis of the resulting flue gas. in the analysis of stack gases from incinerators it is desirable to have a con- venient means for measuring the amount of air in excess of the air needed to burn the fuel compleiely, i.e., stoichiometric combustion. The Orsat analysis provides information that can be used to estimate the percentage of excess air. It is proposed that the ratio of O2 to N2 would provide a good means of estimat- ing the percentage-of excess air where the fuel is cellulose (CsH, 0s) and the incinerator temperature is not excessively hot, causing the Nz in the air to oxidize. Determine what the O, to N; ratio is in the stack gas of an incinerator for the stoichiometric burning of cellulose and for 20, 50, and 100 percent excess air. ‘Aviation gasoline is isooctane, CsHy,. If it is burned with 20 percent excess R2.12. 2.13. 2.14. 2.15. 2.16. 217 67 air and 30 percent of the carbon forms carbon monoxide, what is the flue-gas analysis? In the United States the five most common primary air pollutants (in tons emitted annually) are carbon monoxide, sulfur oxides, hydrocarbons, nitrogen oxides, and particulate matter. If a natural gas that analyzes 80 percent CH,, 10 percent Hz, and 10 percent N; is burned with 40 percent excess air and 10 percent of the carbon forms CO, compute the Orsat (dry basis) analysis of the resulting fiue gas. Will the nitrogen be oxidized? Recommend a method of eliminating the CO formed. A natural gas analyzes CH,, 80.0 percent and N2, 20.0 percent. It is burned under a boiler and most of the CO; is scrubbed out of the flue gas for the production of dry ice. The exit gas from the scrubber analyzes COz, 1.2 percent; O2, 4.9 percent; and N2, 93.9 percent. Calculate the (a) Percentage of the CO, absorbed. (b) Percent excess air used. A natural gas that analyzes 80 percent CH, and 20 percent N2 is burned, and the CO, is scrubbed out of the resulting products for use in the manufacture of dry ice. The exit gases from the scrubber analyze 6 percent O, and 94 percent N32. Calculate the (a) Air to gas ratio. (b) Percent excess air. The U.B.T. Development Company is selling a fuel cell which generates electrical energy by direct conversion of coal into electrical energy. The cell is quite simple —it works on the same principle as a storage battery and produces energy free from the Carnot cycle limitation. To make clear the outstanding advantages of the fuel cell, we can consider the reaction fuel + oxygen = oxidation products qa) Reaction (1) is intended to apply to any fuel and always to 1 mole of the fuel. The fuel cell is remarkable in that it can convert chemical energy directly into work and bypass the wasteful intermediate conversion into heat. A typical fuel consists of Cc 65 H 5 oO 10 Ss 4 Ash 16 ‘Assume that no carbon is left in the ash and that the S is oxidized to SO:. If 20 percent excess air is used for oxidation of the fuel, calculate the composi- tions of all the oxidation products. A steel-annealing furnace burns a fuel oil, the composition of which can be represented as (CH;),, It is planned to burn this fuel with 12 percent excess air. ‘Assuming complete combustion, calculate the Orsat analysis of the flue gas. Repeat this problem on the assumption that 5 percent of the carbon in the fuel is burned to CO only. ‘A producer gas analyzing COz, 4:5 percent; CO, 26 percent; Hz, 13 percent;2.18. 2.19. 2.20. 2.21, 68 CH,, 0.5 percent; and N3, 56 percent, is burned in a furnace with 10 percent excess air. Calculate the Orsat analysis of the flue gas. The Deacon process can be reversed and HCI formed from Cl; and steam by removing the O2 formed with hot coke as follows: 2Cl, + 2H,0 + C—> 4HCl + CO, If chlorine cell gas analyzing 90 percent Cl, and 10 percent air is mixed with 5 percent excess steam and the mixture is passed through a hot coke bed at 900°C, the conversion of Cl, will be 80 percent complete, but all the O2 in the air will react. Calculate the composition of the exit gases from the converter, assuming no CO formation. Solvents emitted from industrial operations can become significant pollutants if not disposed of properly. A chromatographic study of the waste exhaust gas from a synthetic fiber plant has the following analysis in mole percent: cS, 40% sO, 10 H,0 50 It Has been suggested that the gas be disposed of by burning with an excess of air. The gaseous combustion products are then emitted to the air through a smokestack. The local air pollution regulations say that no stack gas is to analyze more than 2 percent SO, by an Orsat analysis averaged over a 24-hr period. Calculate the minimum percent excess air that must be used to stay within this regulation. An industrial gas has the following composition: CO 35% H, 45 Na 10 Or 5 CO, 5 Rather than waste the gas, it is more economical to burn it with sufficient air so that the total amount of oxygen available for combustion is that theoretically required for complete combustion. The exit gases, which are sufficiently hot to maintain all water formed in the vapor state, contain 61.4 percent N2 and 19.4 percent HO. Compute the percent completion of the combustion of the H2. Twelve hundred pounds of Ba(NO;)2 are dissolved in sufficient water to form a saturated solution at 90°C, at which temperature the solubility is 30.6 g/100 g water. The Solution is then cooled to 20°C, at which temperature the solubility is 8.6 g/100 g water. (a) How many pounds of water are required for solution at 90°C, and what weight of crystals is obtained at 20°C? (b) How many pounds of water are required for solution at 90°C, and what weight of crystals is obtained at 20°C, assuming that 10 percent more water is _to be used than necessary for a saturated solution at 90°C? (©) How many pounds of water are required for solution at 90°C, and what2.22, 2.23. 2.24, 2.25. (2.26. 2.27, 69 weight of crystals is obtained at 20°C, assuming that the solution is to be made up 90 percent saturated at 90°C? (d) How many pounds of water are required for solution at 90°C, and what weight of crystals is obtained at 20°C, assuming that 5 percent of the water evaporates on cooling and that the crystals hold satufated solution mechan- ically in the amount of 5 percent of their dry weight? If 100g of Na,SO, is dissolved in 200g of H,O and the solution is cooled until 100 g of Na2SO4-10H20 crystallizes out, find (a) The composition of the remaining solution (mother liquor). (b) The grams of crystals recovered per 100 g of initial solution. The solubility of magnesium sulfate at 20°C is 35.5 g/100 g H,0. How much MgSO,-7H20 must be dissolved in 100 Ib of H20 to form a saturated solution? The solubility of manganous sulfate at 20°C is 62.9 g/100 g H,O. How much MnSO,-5H20 must be dissolved in 100 lb of water to give a saturatedisolution? How much water must be evaporated from 100 gal of Na;CO; solution, con- taining 50 g/l at 30°C, so that 70 percent of the NazCO, will crystallize out when the solution is cooled to 0°C? An evaporator is fed 100 1b/min of 23.1 percent Na,SO, solution and the concen- trated product is passed through a crystallizer where it is cooled to 0°C. How much water must be evaporated per minute so that 90 percent of the sodium sulfate (MW = 142) will crystallize out? Note that at 32.4°C ‘Na,SO, becomes a hydrate: Na2SO, + 10H,0 —> Na,SO,-10H,0 The solubility of Na,SO, and the hydrate are temp., °C solubility, g/100 g HO 100 42.5 32.4 49.5 0 5.0 The feed to a distillation column is separated into net overhead product contain- ing nothing with a boiling point higher than isobutane and bottoms containing nothing with a boiling point below that of propane. See Fig. P2.27. The composi- tion of the feed is mole % Ethylene 2.0 Ethane 3.0 Propylene 5.0 Propane 15.0 Isobutane 25.0 n-Butane 35.0 n-Pentane - 15.0 Total 100.0 The concentration of isobutane in the overhead is 5.0 mole percent, and the concentration of propane in the bottoms is 0.8 mole percent. Calculate the composition of the overhead and bottoms streams per 100 moles of feed.2.28. 2.29. 70 Feed Recycle Overheod Product (Distillote) Distillation Column Bottoms (residuum) Fig. P2.27. A gas containing 80 percent CH, and 20 percent He is sent through a quartz diffusion tube (see Fig. P.2.28). to recover the helium. Twenty percent by weight of the original gas is recovered, and its composition is 50 percent He. Calculate the composition of the waste gas if 100 lb moles of gas are processed per minute. The initial gas pressure is 17 psia, and the final gas pressure is 2 psig. The baro- meter reads 740 mm Hg. The temperature of the process is 70°F. 80% CHa 20% He [waste gos 50% He recovered gas Fig. P2.28. A and B are immiscible liquids, but they emulsify in one another to give uniform emulsions. See Fig. P2.29. The uniform emulsion is withdrawn from the lower layer and sent to.a settler where the emulsion breaks and is separated. During the addition of 698 kg of A and 1302 kg of B to the extractor, the interface between the layers rises to and stays at a new level of 6 cm higher than the old level. A rise of 1 cm corresponds in volume to 30 kg of A or 40 kg of B. The volumes of A and B are additive in all proportions. The top level in the extrac- 698 kg A 20% P io% Final Level iitiol Level p29 bg b 70% A (302k 8 Winer sate | *O%F W- 1000 kg 22% A 22% Fig. P2.29.2.30. 7] tor remains constant during the operation. What is the composition of the bottom. layer from the settler? To prepare a solution of 50.0 percent sulfuric acid, a dilute waste acid containing 28.0 percent H,SO, js fortified with a purchased acid containing 96.0 percent H,SO.. How many kilograms of the purchased acid must be bought for each 100 kg of dilute acid? 2.31. A fuel composed of ethane (CzH.) and methane (CH,) in unknown proportions 2.32. is burned in a furnace with oxygen-enriched air (50.0 mole percent O2). Your Orsat analysis is: CO, 25 percent; Nz, 60 percent; and Oz, 15 percent. Find (a) The composition of the fuel, i.e., the mole percent methane in the methane- ethane mixture. (b) The moles of oxygen-enriched air used per mole of fuel. In modern U.S. incinerators, refuse burns on moving grates in refractory-lined chambers, and combustible gases and entrained solids burn in secondary com- bustion chambers or zones. Combustion is 85-90 percent complete for the com- bustible materials. A large body of technology has grown up and a variety of mechanical designs of incinerators are available. The scientific principles that underlie the technology often are not well defined, partly because of the hetero- geneous nature of refuse and its variable moisture. Up to three times as much air is introduced into the incinerator as would be needed to supply the oxygen required to oxidize the refuse completely. The temperature in the bed of burning refuse may reach 2500°F or more, and the excess air is required mainly to hold the temperature in the furnace at 1400°-1800°F. Above 1800°F, slag formation in the furnace can become a problem. The table lists the results from one incinerator. (a) Based on the gas analysis, what was the percent excess air used in the incin- erator? (b) Does the solid analysis agree with the gas analysis? \ Gas Analysis Solid Analysis Fraction by Stack Gases Ib/ton of Refuse Volume, Dry Carbon dio; 1,738 6.05% Sulfur dioxide 1 22 ppm Carbon monoxide 10 0.06% Oxygen 2,980 14.32% Nitrogen oxides (NO) 3 93 ppm Nitrogen 14,557 19.51% Total dry gas 19,289 Water vapor 1,400 Total 20,689 Solids, dry basis Grate residue 411 Fly ash 20 Total, Ib/ton of refuse 21,180 source: E, R. Kaiser, “Refuse Reduction Processes,” in Proceedings, The Surgeon Gen- eral's Conference on Solid Waste Management for Metropolitan Washington, U.S. Public Health Service publication No. 1729, Government Printing Office, Washington, D.C., July 1967, p. 93. 172 2.33. Ammonia is a gas for which reliable analytical methods are available to deter- 2.34, 2.35. 2.36. 2.37. 2.38, 2.39. 2.40. mine its concentration in other gases. To measure flow in a natural gas pipeline, pure ammonia gas is injected into the pipeline at a constant rate of 72.3 kg/min for 12 min. Five miles downstream from the injection point, the steady-state ammonia concentration is found to be 0.382 weight percent. The gas upstream from the point of ammonia injection contains no measurable ammonia, How many kilograms of natural gas are flowing through the pipeline per hour? A lacquer plant must deliver 1000 1b of an 8 percent nitrocellulose solution. They have in stock a 5.5 percent solution. How much dry nitrocellulose must be dissolved in the solution to fill the order? Paper pulp is sold on the basis that it contains 12 percent moisture; if the mois- ture exceeds this value, the purchaser can deduct any charges for the excess moisture and also deduct for the freight costs of the excess moisture. A ship- ment of pulp became wet and was received with a moisture content of 22 percent. If the original price for the pulp was $40/ton of air-dry pulp and if the freight is $1.00/100 Ib shipped, what price should be paid per ton of pulp delivered? To meet certain specifications, a dealer mixes bone-dry glue, selling at 25 cents/Ib, with glue containing 22 percent moisture, selling at 14 cents/Ib, so that the mix- ture contains 16 percent moisture. What should be the selling price per pound of the mixed glue? A dairy produces casein which when wet contains 23.7 percent moisture. They sell this for $8.00/100 Ib. They also dry this casein to produce a product contain- ing 10 percent moisture. Their drying costs are $0.80/100 Ib water removed. What should be the selling price of the dried casein to maintain the same margin of profit? In 1969 the Federal Water Pollution Control Administration issued a guidance memorandum indicating that the reduction of phosphate in detergents was desirable. The question, of course, was what replacement substances could be used that would not contribute more pollutional effects to receiving waters. A company manufactures Spic and Spotless detergent, which contains 11.1 percent phosphorus (reported as P) on an as purchased (wet) basis and 14.7 percent on a dry basis. Suppose that all the phosphates (NasP3010, NasP207, Na3PO., etc.) are to be removed from the soap formulation and replaced by NTA (sodium nitrilotriacetate), which is 1.8 times as effective as a water softener as Na3PO., so that the weight percent NTA is 6.0 percent in the new blend of detergent. If NTA costs $0.75/Ib and the phosphates sell for $0.041/lb P, will the new detergent cost more or less than the original detergent? By how much in dollars relative to the original detergent? A paper manufacturer contracts for rosin size containing not more than 20 percent water at 12 cents/Ib f.0.b. the rosin-size plant, a deduction to be made for water above this amount, and the excess freight on the water is to be charged back to the manufacturer at 80 cents/100 Ib. A shipment of 2400 Ib is received which analyzes 26.3 percent water. What should the paper manufacturer pay for the shipment? In anaerobic decomposition, microbial organisms decompose organic compounds in the absence of oxygen to CO2, CH,, HS, etc. Anaerobic digestion is used in2.41, 2.42, 2.43. 2.44, 73 waste-disposal methods but also occurs naturally in many lakes, swamps, etc., where the characteristic smell of H;S is easily noted. In sewage purification where the solids decompose on the bottom of basins, to avoid the smell, the upper layers of the water are aerated so that the H2S is oxidized to less odoriferous compounds. If the Orsat gas analysis above a test basin is SOz, 7.2 percent; 02, 10.1 percent; and N32, 82.7 percent, what percentage of the sulfur in the H2S was oxidized to SO;? A natural gas which is entirely methane, CH,, is burned with an oxygen-enriched. air so that a higher flame temperature may be obtained. The flue gas analyzes COp, 22.2 percent; Oz, 4.4 percent; and Nz, 73.4 percent, Calculate the percent- age of O2 and N2 in the oxygen-enriched air. As superintendent of a lacquer plant, your foreman brings you the following problem: he has to make up 1000 |b of an 8 percent nitrocellulose solution. He has available a tank of a 5.5 percent solution. How much dry nitrocellulose must he add to how much of the 5.5 percent solution in order to fill the order? The Clean Air Act of 1970 requires automobile manufacturers to warrant their control systems as satisfying the emission standards for 50,000 miles. It requires owners to have their engine control systems serviced exactly according to manu- facturers’ specifications and to always use the correct gasoline. In testing an engine exhaust having a known Orsat analysis of 16.2 percent COz, 4.8 percent Oz, and 79 percent Nz at the outlet, you find to your surprise that at the end of the muffler the Orsat analysis is 13.1 percent CO,. Can this discrepancy be caused by an air leak into the muffler? (Assume the analyses are satisfactory.) If so, compute the moles of air leaking in per mole of exhaust gas leaving the engine. Removal of sulfur dioxide from the flue gases of industrial installations is a topic which has attracted worldwide attention during the past decade. Recovery of sulfur dioxide from the flue gases of power plants would reduce air pollution and supplement the recovery of an important industrial chemical. A wide variety of processes is under development throughout the world and selection of the appropriate scheme depends on the size of the installation, the size of the avail- able market for the final commodity produced, transport and handling costs, and the cost of capital plant and operating costs. In one proposal the flue gas from a high-sulfur fuel oil is burned to SO, and then further oxidized to SO; in a series of catalytic converters and absorbers. Each converter is followed by an absorber. The gas entering the first converter is (on a CO2 free basis) SO, 0.80% Or 2.80 N, 96.40 and the gas leaving the first absorber is SO, 0.20% On 2.00 Nz 97.80 Calculate the degree of conversion of SOz in the first converter. What is the analysis of the exit gas after complete conversion and absorption has taken place (on a CO, free basis)? SO, + 40,—»+ SO; SO; + H,O —> H:S0.2.45. 2.46. 2.47, 2.48. 2.49. 2.50. 14 Your boss asks you to calculate the flow through a natural-gas pipeline. Since it is 26 in, in diameter, it is impossible to run the gas through any kind of meter or measuring device. You decide to add 100 1b of CO; per minute to the gas through a small -in. piece of pipe, collect samples of the gas downstream, and analyze them for CO. Several consecutive samples after I hr are time % COr Thr, Omin 2.0 10 min 2.2 20 min 19 30 min 21 40 min 2.0 (a) Calculate the fiow of gas in pounds per minute at the point of injection. (b) Unfortunately for you, the gas at the point of injection of CO, already contained 1.0 percent COz, How much was your original flow estimate in error (in percent)? Note: In part (a) the natural gas is all methane, CHy. A low-grade pyrites containing 32 percent S is mixed with 10 Ib of pure sulfur per 100 Ib of pyrites so the mixture will burn readily, forming a burner gas that analyzes (Orsat) SO, 13.4 percent; O2, 2.7 percent; and Nz, 83.9 percent. No sulfur is left in the cinder. Calculate the percentage of the sulfur fired that burned to SO. (The SO; is not detected by the Orsat analysis.) Pure barytes, BaSO,, is fused with soda ash, NazCO;, and the fusion mass is then leached with water. The solid residue from the leaching analyzes 33.6 percent BaSO, and 66.4 percent BaCO;. The soluble salts in solution analyze 41.9 percent Na,SO, and 58.1 percent Na;CO3. Calculate the composition of the mixture before fusion. j A power company operates one of its boilers on natural gas and another on oil (for peak period operation). The analysis of the fuels are as follows: natural gas oil 96% CH. (CHis)e 4% COr When both boilers are on the line, the flue gas shows (Orsat analysis) 10.0 percent CO2, 4.5 percent O2, and the remainder N2. What percentage of the total carbon burned comes from the oil? Hint: Do not forget the H2O in the stack gas. A power company operates one of its boilers on natural gas and another on oil. The analyses of the fuels show 96 percent CH,, 2 percent C2H2, and 2 percent CO; for the natural gas and C,H) 2, for the oil. The flue gases from both groups enter the same stack, and an Orsat analysis of this combined flue gas shows 10.0 percent CO, 0.63 percent CO, and 4.55 percent Oz. What percentage of the total carbon burned comes from the oil? An automobile engine burning a fuel consisting of a mixture of hydrocarbons is found to give an exhaust gas analyzing 10.0 percent CO; by the Orsat method. It is known that the exhaust gas contains no oxygen or hydrogen. Careful meter- ing of the air entering the engine and the fuel used shows that 12.4 Ib of dry air enter the engine for every pound of fuel used. (a) Calculate the complete Orsat gas analysis.2.51. 2.52. 2.53. 75 (b) What is the weight ratio of hydrogen to carbon in the fuel? A fuel oil and a sludge are burned together in a furnace with dry air. Assume the fuel oil contains only C and H. fuel oil sludge fue gas wet C=2% — water = 50% SO, = 1.52% solids = 50 COz = 10.14 H=2% 2.02 On = 4.65 No = 81.67 (a) Determine the weight percent composition of the fuel oil. (b) Determine the ratio of pounds of sludge to pounds of fuel oil. A solvent dewaxing unit in an oil refinery is separating 3000 bbl/day of a lubricat- ing distillate into 23 vol percent of slack wax and 77 vol percent of dewaxed oil. The charge is mixed with solvent, chilled, and filtered into wax and oil solution streams. The solvent is then removed from the two streams by two banks of stripping columns, the bottoms from each column in a bank being charged to the next column in the bank. The oil bank consists of four columns, and the wax bank of three. A test on the charge and bottoms from each column gave the following results: percent solvent by volume to Ist no. 1 no. 2 no. 3 no. 4 column bottoms bottoms — bottoms —_— bottoms Pressed oil 83 70 27 4.0 0.8 Wax 83 7 23 0.5 = Calculate the following: (a) Total solution per day charged to the whole unit. (b) Percentage of total solvent in oil solution removed by each column in oil bank. (© Percentage of total solvent in wax solution removed by each column in wax bank. (b) Barrels of solvent lost per day (in bottoms from last column of each bank). The price of crude oil is based on its API-gravity; the highest gravities com- mand the best prices. It is often possible to blend two crudes advantageously so that the price obtained for the mixture is greater than the price of the separate crudes. This is possible because the price vs. API function is discontinuous (like income tax rates). A crude oil with a 34.4°API gravity is to be mixed with 50,000 bb! of a 30.0°API gravity crude to give a mixture which has a gravity of 31.0°API. What is the increased selling price of the mixed crude over that of the separate components? Data: (1) Crude prices: API S/bbl 29.0-30.9 $2.60 31.0-32.9 2.65 33.0-34.9 2.7076 (2) API gravities are not additive on any basis. (3) Sp gr at 60°F = 141.5/(131.5 + API) 2.54.* It is desired to mix three L.P.G. (liquified petroleum gas) products in certain 2.55. 2.56. 2.57. proportions in order that the final mixture will meet certain vapor pressure specifications. These specifications will be met by a stream of composition D below. Calculate the proportions in which streams 4, B, and C must be mixed to give a product with a composition of D. The values are liquid volume percent. component stream A B Cc D ey 5.0 14 ce 90.0 10.0 31.2 is0-Cy 5.0 85.0 8.0 53.4 ne 5.0 80.0 126 iso-Cy 12.0 14 100.0 100.0 100.0 100.0 A coal analyzes 74 percent C and 12 percent ash (inert). The flue gas from the combustion of the coal analyzes COz, 12.4 percent; CO, 1.2 percent; O2, 5.7 percent; and Nz, 80.7 percent. Calculate the following: (a) The pounds of coal fired per 100 moles of flue gas. (b) The percent excess air. (c) The pounds of air used per pound of coal Assume that there is no nitrogen in the coal. Sea water is to be desalinized by reverse osmosis using the scheme indicated in Fig. P2.56. Use the data given in the figure to determine Brine Recycle Reverse Osmosis, Cell Brine Waste (8) 5.25% Salt o Desolinised Woter 500 ppm Salt Fig. P2.56. (a) The rate of waste brine removal (B). (b) The rate of desalinized water (called potable water) production (D). (c) The fraction of the brine leaving the osmosis cell (which acts in essence as a separator) that is recycled. Note: ppm designates parts per million. In the production of NH3, the mole ratio of the N; to the H; in the feed to the whole process is 1 N; to 3 Hz. Of the feed to the reactor, 25 percent is con- verted to NH. The NH; formed is condensed to a liquid and completely removed from the reactor, while the unreacted Nz and H; are recycled back to mix with the feed to the process. What is the ratio of recycle to feed in vound recvcle71 per pound feed? The feed is at 100°F and 10 atm, while the product is at 40°F and 8 atm. 2.58.* An isomerizer is a catalytic reactor which simply tends to rearrange isomers. The number of moles entering an isomerizer is equal to the number of moles leaving. A process, as shown Fig. P2.58, has been designed to produce a p- Product A- 40 A- 70 t=S00°F ju es . 8-298 B- 27.0 50 psio Crystatizes, Oo Sao aaa Rs fea 12.6 = 21.6 al Feed 54 Z A= 15.0 - Distilletion g 7200 | Product serie tower D- 15.0 Fig. P2.58. xylene-rich product from an aromatic feed charge. All compositions on the flow sheet are in mole percent. The components are indicated as follows: A ethyl benzene B o-xylene Co mexylene D p-xylene Eighty percent of the ethyl benzene entering the distillation tower is removed in the top stream from the tower. The ratio of the moles fresh feed to the process as a whole to the moles of product from the crystallizer is 1.63. Find the (a) Reflux ratio (ratio of moles of stream from the bottom of the distillation tower per mole of feed to the tower). (b) Composition (in mole percent) of the product from the crystallizer. (©) Moles ieaving the isomerizer per mole of feed. 2.59, In an attempt to provide a means of generating NO cheaply, gaseous NH; is burned with 20 per cent excess O2: 4NH; + 50; —> 4NO + 6H20 The reaction is 70 percent complete. The NO is separated from the unreacted NH, and the latter recycled as shown Fig..P2.59.Compute the (a) Moles of NO formed per 100 moles of NH; fed. (b) Moles of NH recycled per mole of NO formed. Reactor Recycle NHs Oe NQ Oe NEG} H,0 Fig. P2.59.2.60. 2.61. 2.62. 78 Acetic acid is to be generated by the addition of 10 percent excess sulfuric acid to calcium acetate. The reaction Ca(Ac), + H:SO.——> CaSO, + 2HAc goes to 90 percent completion. The unused Ca(Ac), and the H2SO, are separated from the products of the reaction, and the excess Ca(Ac), is recycled. The acetic acid is separated from the products. Find the amount of recycle per hour based on 1000Ib of feed per hour, and also the pounds of acetic acid manufactured per hour. See Fig. P2.60. HpS04 Hac — Reactor UL H2504 + CoSO, Col Ache Fig. P2.60. Iodine can be obtained commercially from kelp by treating the seaweed with MnO, and 20 percent excess H2SO, according to the reaction 2Nal + MnO, + 2H,SO,—> Na,SO. + MnSO, + 2H,0 + In All the H,0, Na,SO,, MnSO,, Ip, and inerts are removed in the separator. The kelp contains 5 percent Nal, 30 percent H20, and the rest can be considered to be inerts. The product contains 54 percent I, and 46 percent H2O. See Fig. P2.61. Assuming the reaction to be 80 percent complete, calculate the following: (a) The pounds of I; produced per ton of kelp used. (b) The percentage composition of the waste. (c) The percentage composition of the recycle. [_54% Ig HySOq Mn0,—>| Reactor Kelp —| Separator | 46% HO Woste #0 : Recy MnSOa NopSOq Inerts Fig. P2.61. ‘A solution containing 10 percent NaCl, 3 percent KCl, and 87 percent water is fed to the process shown in Fig. P2.62 at the rate of 18,400 kg/hr. The composi- tions of the streams are as follows: NaCl KCI H,0 Evap. product, P 16.8 21.6 61.6 Recycle, R 18.9 12.3 68.8 Calculate the kilograms per hour of solution P passing from the evaporator to the crystallizer and the kilograms per hour of the recycle R.2.63. 79 [———~ Water Vapor Feed e a Evoporotor ceystotzer IoC KCI Fig. P2.62. A natural gasoline plant at Short Junction, Oklahoma produces gasoline by removing condensable vapors from the gas flowing out of gas wells. The gas from the well has the following composition: component mole % CH, 713 C2Hs 14.9 C3Hs 3.6 iso- and n-CyHio 1.6 Cs and heavier 0.5 Na 100.0 This gas passes through an absorption column called a scrubber (see Fig. P2.63), where it is scrubbed with a heavy, nonvolatile oil. The gas leaving the scrubber has the following analysis: component mole % ‘ CH 92.0 C2He 5.5 Na 2.5 The scrubbing oil absorbs none of the CH. or Nz, much of the ethane, and all of the propane and higher hydrocarbons in the gas stream. The oil stream is then sent to a stripping column, which separates the oil from the absorbed hydrocarbons. The overhead from the stripper is termed natural gasoline, while the bottoms from the stripper is called Jean oil. The lean-oil stream is cooled and returned to the absorption column. Assume that there are no leaks in the system. Gas ftom the wells is fed to the absorption column at the rate of 52,000 Ib moles/day and the fiow rate of the rich oil going to the stripper is 1230 Ib/min (MW = 140). Calculate the following: (a) The pounds of CH, passing through the absorber per day. (b) The pounds of C,H. absorbed from the gas stream per day. (©) The weight percentage of propane in the rich oil stream leaving the scrubber.Product 80 Poraffin Oil (P) in Chiorex ( Undesirable (U} L) in Chlorex Plus 1 some P Chlorex (C) Feed (A) 2.68. 70% P 30%U Fig. P2.67. (b) What weight of undesirable material, including the paraffin oil retained with the U, is recovered per hour? (©) If 700 Ib of paraffin oil containing 0.2 1b of U per pound of paraffin oil flow per hour from settler I to settler II and 670 Ib of paraffin oil containing 0.05 Ib of U per pound of paraffin oil flow per hour from settler II to settler II, what weights of undesirable material U are being transferred per hour from settlers III to II and from I to 1? Two thousand kilograms per hour of solid industrial waste containing a toxic organic compound are extracted with 20,000 kg/hr of pure solvent in a two- stage countercurrent extraction system. See Fig. P2.68. The solid waste contains 10 percent by weight of extractable compound. After exposure to the solvent, the waste retains 1.5 kg of solvent per kilogram of waste (on a toxic compound- free basis). What fraction of the toxic compound is recovered per hour? 2000 kg Waste Plus Toxic Compound —> Solvent Plus Toxica——4 Solvent Compound 20,000 kg /hr LY x} Waste Plus Toxic Compound