Solutions Doran

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SOLUTIONS MANUAL Bioprocess Engineering Principles Pauline M. Doran tI .... . iII ' ......... .. oJ 4 i!-402751 RlelilllN "''I' til TEl:1032l88O-7518 FAIl_ E-mail: I co.tnha.ao.ki' -~ --SOLUTIONS MANUAL Bioprocess Engineering Principles Pauline M. Doran University of New South Wales, Sydney, Australia ISBN 0 7334 15474 Pauline M. Doran 1997 Table of Contents Solutions Page Chapter 2 Introduction to Engineering Calculations 1 Chapter 3 Presentation and Analysis of Data 9 Chapter 4 Material Balances 17 ChapterS Energy Balances 41 Chapter 6 Unsteady-State Material and Energy Balances 54 Chapter 7 Fluid Flow and Mixing 76 Chapter 8 Heat Transfer 86 Chapter 9 Mass Transfer ' 98 Chapter 10 Unit Operations 106 Chapter 11 Homogeneous Reactions 122 Chapter 12 Heterogeneous Reactions 139 Chapter 13 Reactor Engineering 151 NOTE All equations, tables, figures, page numbers, etc., mentioned in this manual refer to the textbook, Bioprocess Engineering Principles. Introduction to Engineering Calculations 2.1 Unit conversion (a) From Table A.9 (Appendix A): 1 cP::::: kg m-I $"1 1 m::: lOOcrn Therefore: 1.5 x 10-6 cP "" 1.5 x 10-6 cP ,110-3 k1g ;-1 I = 1.5 x 10-11 kg s-1 cm-1 Answer: 1.5 x 10-11 kg s-1 em-1 (b) From Table A.S (Appendix A): 1 bp (British)::::: 42.41 Btu min-I Therefore: Answer: 5.17 Btu min-1 (e) From Table A.S (Appendix. A): 1 mmHg::::: 1.316 x 10-3 attn From Table A.I (Appendix A): 1 ft::: 0.3048 m From Table A.7 (Appendix A): 11 atm = 9.604 x 10-2 Btu From Table A8 (Appendix A): 1 Btu min-I::::: 2.391 x 10"2 metric horsepower Im=lOOcm 11= lOOOcm3 Ih=60min Therefore: 670mmHgft3 = 670 mmHg ft3 atmI19.604X 1O-2Btul.I.3048m 131 100 em 131 11 I. llatm 1ft 1m lOOOcm3 12.391 x 10-2 metric horsepower 1 1 I h 1 956 10-4 . h h 1 . -60 . = ,x metric orsepower 1 Btu min- mm Answer: 9.56 x 10-4 metric horsepower h (d) From Table A.7 (Appendix A): 1 Btu = 0.2520 keal From Table A.3 (Appendix A): lIb = 453.6 g Therefore: 345 Btu Ib-1 = 345 Btulb-1 .! g I = O.192kcal g-1 Answer: 0.192 kcal g-l 2.2 Unit conversion Case 1 Convert to units of kg, m, s. From Table A.3 (Appendix A), lIb = 0.4536 kg 2 From Table A.2 (Appendix A): 1 ft3 ::: 2.832 x 10-2 m3 From Table A.9 (Appendix A): 1 cP::: 10-3 kg m-l sl 1 rn= lOOcm= lOOOmm Therefore, using Eq. (2.1): Solutions: Chapter 2 (2mm.1 1m 1ft3 Dup l000mmU l00cmU lIb 2.832 x 10 2m3U _ 7 Re::: -p- = I -3 1 -11 - 2.4 x 10 10-0 P 10 kgm s c . 1 cP Answer: 2.4 x 107 ease 2 Convert to units of kg, m, s. From Table Al (Appendix A): 1 in.::: 2.54 x 10-2 m From Table A.9 (Appendix A): 1 Ibm ft-I h-1 ::: 4.134 x 10-4 kg mol sl Ih=3600s Therefore, usingEq. (2.1): Answer: 1.5 x 104 2.3 Dimensionless groups and property data = 1.5 x 104 From the Chemical Engineers' Handbook. the diffusivity of oxygen in water at 2S"C is 2.5 x 10-5 cm2 s-l. Assuming this is the same at 28"C, !lJ= 2.5 x 10-5 cm2 sl, Also, from the Chemical Engineers' Handbook, the density of water at 28"C is PL ::: 0.9962652 g cm-3, and the viscosity of water at 28"C is JlL::: 0.87 cP. The density of oxygen at 28C and 1 atm pressure can be calculated using the ideal gas law. As molar density is the same as n,V. from Eq. (2.32): Temperature in the ideal gas equation is absolute temperature; therefore, from Eq. (2.24): T = (28 + 273.15) K = 301.15 K From Table 2.5, R "" 82.057 cm3 atm K-I gruol-I. Substituting parameter values into the density equation gives: PG "" L "" I atm "" 4.05 x 10-5 gmolcm-3 RT (82.057cm3aImK:""1 gmol 1)(301.15K) From the atomic weights in Table B.l (Appendix B), the molecular weight of oxygen is 32.0. Converting the result for PG to mass tenns: PG "" 4.05xIo-5gmolcm-31 "" L30xlO-3gcm-3 From Table A9 (Appendix A): 1 cP "" 10-2 g cm-1 s-l; from Table Al (Appendix A): 1 ft "" 0.3048 m = 30.48 cm. The parameter values and conversion factors, together with Db "" 2 mm "" 0.2 cm, can now be used to calculate the dimensionless groups in the equation for the Sherwood number. Solutions,' Chapter 2 3 1'0-2 -1 -11 0,87eP. gem s S_J1.L_ leP -349 c - Pc 11 - (0.9962652g em-3)(2.5 x 10 5 em2 s I) -Therefore: From the equation for Sh: k _Sh11_(ll.21(2.5xIO-5cm2s-I)_IA 10-3 -I L---- 02 -.

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