Lecture28 - ChE 210 Meeting 28 Outline(F&R Sections...

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Unformatted text preview: ChE 210: Meeting 28 March 28, 2011 Outline (F&R Sections 8.0-8.2) Homework Due Friday 04/01/2011 - Questions from Example 7.19, 7.29, 7.42, 7.51 a&b Problem - Hypothetical Process Paths - Sensible Heat and Heat Capacity Example Problem 100 mol/h of a liquid mixture of 25 mol% toluene (T) in acetone (A) at atmospheric temperature and pressure (T = 25 C and Pabs = 1 atm) is fed to an evaporator that operates at 120 C and 1 atm. All of the liquid is vaporized. Heat is provided to the evaporator by condensing saturated steam at 10 bar. Calculate the amount of heat that must be supplied to the evaporator and the mass flow rate of saturated steam. nin = 100 mol/h nout = XAJn = 0.75 mol A/mol xA1out = xTJn = 0.25 mol A/mol xT’out = Out Saturated Steam Water (L) Pabs = 10 bar P = 10 bar Reference T, P and A A hase Hin kJ/mol) Hout (kJ/mol) Given: 1/6 ChE 210: Meeting 28 March 28, 2011 2/6 ChE 210: Meeting 28 March 28, 2011 Hypothetical Process Paths A A U and H are state properties, which means that they only depend on the T, P and phase and not how the species arrived at that state. A A 9 Can calculate U and H relative to any convenient reference state using a hypothetical process path. These calculations combine 5 basic processes: 1. AP at constant T and ¢ (Section 8.2) 2. AT at constant P and (1) (Section 8.3) 3. A4) at constant P and T (Section 8.4) 4. Mixing of two liquids or dissolution of a gas or solid in a liquid at constant T and P (Section 8.5) 5. Chemical reaction at constant T and P (Chapter 9) These steps can be combined in any logical order to move from the reference state to the actual state. In general, AP at constant T and q) is easier to calculate for a liquid or solid than a gas. Liquids and solids at fixed T and q): A A A A A A AV~Oand AU~O,so AH=AU+APVzVAP Ideal gases near and 1 atm: , A Nonideal gases require thermodynamic relationships that are beyond the scope of this course. 3/6 ChE 210: Meeting 28 March 28, 2011 Examples of Hypothetical Process Paths Component: Chloroform ;_ ,1 7 , 5 , , Reference: P = 1 atm, T = 25 C, q) = liquid Actual: P = 10 atm, T = 200 C, 4) = vapor (P =1 atm, T = 25 C, q; = liquid) (P =10 atm, T = 200 C, (l) = vapor) Component: Methanol ‘5 ~ v v <7, ; a , Reference: P = 3 atm, T = 100 C, q) = vapor Actual: P = 3 atm, T = 20 C, (i) = liquid (P = 3 atm, T = 100 C, q) = vapor) (P = 3 atm, T = 20 C, q) = liquid) 4/6 ChE 210: Meeting 28 March 28, 2011 Sensible Heat and Heat Capacity Sensible heat is the amount of heat need to raise or lower the temperature of a substance at constant pressure without a phase change A Closed system: AU = Q and U =f A T,V o o A A A _/\ Open system: AH=Q and HEU+ PV=f T,P,V h w x k W At constantV in a closed system: \ _ m V x i" AU 8U Cv(T) E “m - AU = fCV(T)dT T1 5/6 ChE 210: Meeting 28 March 28, 2011 At constant P in an open system: A A _ AH_ w CP(T)EA!II-T>OE_ 6T P A T2 AH = pr(T)dT T 1 Heat Capacity Formulas (Apgendix 3.2) k I 1 1 —¢rr ‘ L \ ‘7‘? ~ 1, g * ' ~ \ ~ —J 3‘7 )L,,\t/le -\1~~1T,~ l k 1—! 2 / ‘\ - *4 a, ‘1 a- A 5" ‘7 T" x K. :— ) ‘ ..f_'_ ‘ 1! : J a» "" z 5 ,> t--\\ “2 + i ‘1) x “‘1 1 ‘T H \LLL' \VcLJ 1L: (“Shut Ly Tb KP ( . I 4 "1M mm luv-ms LV :3 (‘ .13 I". 4‘1” kkkLL/J. @61st CD I. CV .r \Cd/‘Y‘C‘TJ {755 a 01103 If»): (g; \x/LZI with-UL; 1U" CC”:\V‘HEH a); 135$ f\ (V / “H . . 1 ( \v k") k\C‘f'1\Jl“"LrLCJKAKLZU x j" {Q(_,,La.(j i. i; ) 6/6 5 3 6 S e M a T y t r e P O r P 1 a .w W h P A ENHEHNHSV HZ EEO @003onqu :05 “smméofluoum mo aofimfifiaom H3 UoEmtd» am 2an $53 © doflfim Em fiESuSQfiW ENHSEHD E EQESHSEU NEH» $35:ka 9.38m h:mEHoEHEE .2 .Q Bob ram E Ban“)? Rayth :Hco ENN M H 0 Home 8 5&8 CON; 8% 831 $3 8mm 00 H N Ho? 8 2:28 $35 wNNH mm? M H, M 323 80 0283088 8&8 82$ oNNNn Mine SHE QNN P. H N HQNN 08 @2888 :88 oomHlo $3 NwwN‘ mm? HEN 00 H N H3; Now 22.86 5&8 NRHLRN 20H x HNNHT 39H mHHH M N u HoNH u 8&8 SHEEN 23 x NET 8N V <me M N u 3% 08 €on E68 93% Wow M H 0 OH: N€98 $882 82.28 mmoHlmNN l 22 x SNHI P3 3% M N o SSH 808 2285 8328 ONENN .1 23 x 8.? NHH Name M N 0 S8 88 35:8 558 8N3 onom NNNHu $.mN 3% 9 H w OHS $8 8230: SNHIO gem $.21 WHEN omNN P. H N NSm 2:8 253-: 8N3 $3 Ha.le 2.3 8% on H m Nme 2E0 wfiHfiaH 8N: SE ON.mN\ mmNm 8.: 80 H m NE E ENH 80 H H HHHK £8 0883 wNmumNN quN M H o mHNmH Homfimzv 233m 8288an OONHIO £9? HNis $3 2.3 on H w 8.: £2 maan oowHtmNN 33! 353 Bad MEN H N 82$ 831 Hams SHE 3% .60 A. H w HEN H2 SNHIO ONHWH NET $3 m; 80 H m 3.8 $8 82302 BNHlo EN wwNHl 2.8 8.: 00 H w GET 3: oNNH 0° H H 3% £00048 880$ E M0 NHOH x N on x 9 EH x H moH x a HEB Eome 8% .83 Sign HVSEHHEoo EH:va ‘ ‘ .1. J @809 .V .62 8:688 H8; 8539 $3535 >225 $253. €23:— 329:..— .ENEN 3 88m Ho Massive 9% H33 05 Mom £9.95 32 mohsmmoa Hm ofimonmmm bHoEm 3m woman .00 a. a M 22? «51: x 3.3 + 15.2 x whNHv ‘ 3-2 x 8de + 8E3 n @3088 $3538 THU + ,3 + N 5.68:2 8 50.65ng “N Eom,,..\ KN + E + E + a I HHMHoaSa Ho 50.683218 HH Eom,_% M8 mflsgfl BEL .832 ll usanmo Hfim Na 28% ...
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This note was uploaded on 07/31/2011 for the course CHEM E 210 taught by Professor Shanks during the Spring '02 term at Iowa State.

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Lecture28 - ChE 210 Meeting 28 Outline(F&R Sections...

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