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Lecture05 - ChE 210 Meeting 5 Outline(F&R Sections...

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Unformatted text preview: ChE 210: Meeting 5 January 21, 2011 Outline (F&R Sections 3.4-42) Homework Due Friday 01/28/2011 - Measurement of Fluid 3.15, 3.16, 3.19, 3.32, 3.40, 3.41, Pressure 3.48, 4.4, 4.7a-c, 4.23 - Temperature - Batch, Semibatch and Continuous Processes - Balance Equation Measurement of Fluid Pressure P2 Manometers have one end of a U-shaped tube exposed ll d2 to the fluid of interest, and the other end either sealed, exposed to the atmosphere or exposed to the same fluid of interest. 1/7 ChE 210: Meeting 4 January 19, 2011 exposed to the atmosphere or exposed to the same fluid of interest. X 1‘ “ltd PTLfilUC Nib to L32 “HM? SAME cm H '1 C‘ sauna hi5” {QM} UL “HY; Ostmnfl (dimtli‘m mart) r “19‘3“! 0* d“ (:39 ”+2 (1-5. fl"){,14"‘><i~?t";‘m‘da” r) “ 4/5 ChE 210: Meeting 5 January 21, 2011 Temperature Temperature is a measure of the average kinetic energy of molecules, which cannot be measured directly. Temperature is determined indirectly by measuring a physical property of a substance that depends on temperature. The most common temperature scales are defined by the freezing and boiling point of water. 2/7 ChE 210: Meeting 4 January 19, 2011 Temperature Temperature is a measure of the average kinetic energy of molecules, which cannot be measured directly. Temperature is determined indirectly by measuring a physical property of a substance that depends on temperature. The most common temperature scales are defined by the freezing and boiling point of water. i" J l b l Off L It A "l: 2;} ' {Cf fin " 9kg F: fill“ F — ‘2 )1)" .M,‘ ‘ * :x "r" ‘ M I u “ gig ;"'"’"""TT‘ 5 it”: ”a la. 8 l T l C lL/ 3C H C C NBC/CC X Jrzf flhflfrlffiwf di {QBFC/‘QQS* Rbiscixue times we clglrkmd (so 2am ClbWSr-“mc” la; “b mMEL-cciar mm or“ 0K 1- 0C f aqah‘fi ¢R = OF+ Ltficlun ,- ,_ ‘ ‘ m r” \43 ComEruam art on gun oi T62“ 5/5 ChE 210: Meeting 5 January 21, 2011 Batch Semibatch and Continuous Processes t If batch O reactants products :. teed rag; rna'btfi-flaiu : l : : I : 2i run (fwcess (mm, I‘”H’c‘?d.<’tC) (.5 :’ airbus/é" fang/luff reactants ,‘ K: on 77.093313 9.,— /. nus; ,?‘)Ll“7‘€/I(,US flux,- «1‘ (420121! aanfuw 1-3 +"f‘ucz add I .. products reactants Show [Egg/N) / » raw ,nm‘ififxzd HQOS n" 2 “/Le‘ aéém‘ Ni: ; fimwf’d OR K - n , _|/ )‘f< — roducts , , , p i, /’ 0L1; 41_)_I{e/)¢L( 1‘3 mafia-fr” 2 (dB/1L1!!!“ I hawk? 0011’ 3/7 ChE 210: Meeting 5 , January 21, 2011 Steady State = NO change in process variables (T, P, V, m, etc) with time XCBCU') 0:)IL/ OCch‘r' 1/1 H clam/max frat” (13:; Unsteady State (or Transient) = Process ,variables change with time '3!— Udf) ULUUJ’ ,1? bQ-fl 1"), (QC/77.45111 ffi’fl'} C‘s" CU/Wh/VL-(CLLS Flu 8’11; (x (HSpCDC/U/jy (may)? yarJ’up m1 ,mbmxacrfl General Balance Eguation appncir-> #L mas-5, energy £217.41 x‘7’lniif79cj/77Ldém Input + Generation — Output — Consumption = Accumulation General Rules for Simplifying the General Balance Equation: 1. If balancing total mass, generation and consumption = O , _ ‘ , {067333 (12101707; be Grate-RT] (gr arkas'fl’DC/é'd (exec/Ff h“) Hue/6dr rmc +10 J) 2. If balancing a non-reacting species, generation and consumption = O 3. If a process is at SS, accumulation = O for total mass and species balances 4/7 ChE 210: Meeting 5 ' January 21, 2011 Total Mass: Continuous Semibatch Reactive Species: N 4 Continuous in + Jim .03; V Lox» : L) Semibatch 5/7 ChE 210: Meeting 5 January 21, 2011 Example 4.2-2 (Problem-Solving Process) One thousand kilograms per hour of a mixture of benzene (B) and toluene (T) containing 50% benzene by mass is separated by distillation into two fractions. The mass flow rate of benzene in the top stream is 450 kg B/h and that of toluene in the bottom stream is 475 kg T/h. The operation is at steady-state. Write balances on benzene and toluene to calculate the , unknown component flow rates in the output streams. 1. Read the entire problem statement. 2. Draw a process flow diagram. 3. Label the diagram with all of the known information. mB’out-I- = 450 kg/h mT,outT = Out, Top min = 1000 kg/h In XB’in = 0.5 XT,in = 0.5 . Out, Bottom mB,outB = m T,outB = 475 kg/h timwcnhm as to \lel \lOflQb‘Q- ' " J ’ LIE-vmpl‘ntm[SWE'CA-J’n X (308er comet be labeled (mum cmmsimm of QM (odesfl W Wall/Uflfi C'Ol’WQ/liefl‘l 6/7 ...
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