University:
The Pennsylvania State University
Course:
CHE 220 | Introduction to Chemical Engineering Thermodynamics
Academic year:

2020
Views:

270

Pages:

16
Author:

study_genius

Topics For Today Empirical Equations for V ( Racket Review of Chapter 2 * * Reading Examples 210,211 review , of CUZ . equation ) The Diﬀerential of VCIP ) In multivariate In Thermo Define Fay : dF= ) 3¥ dV=@÷)pdT+ (0¥) , VCIP ) : dx +2¥, .dy dt : f K Then : calculus f- ( ¥ )p = = ﬀ ) - dY= BVDT * In general p * Special case , : : : wef of thermal wef of isothermal compressibility . , KVIP - ¢K Expansion . are d1nV= p functions of DT T For liquids far from the K Neakfsnctims au f and to be approximately constant . KDP - KP , ( critical point and for solids in general ) of BT , and may be assumed : is heated Calculate the % a) Toluene under const change 12=1 . Volume of from bar 20°C to 30°C . . b) Toluene is heated under const Volume from Calculate the Pressure that develops 20°C . +030°C . . EXAMPEI tp DATA p= : Solution : 1.10×18 K " dlnlk BDT a) const .P .V] pdT=k oust . dpm by Pz dV=O ~D Pz 127.8bar = - bar I KDP - In d1nV=fskT ) ( 8.68×105 K= . part = P, = B- K or th , - ) Y÷=exp( Btzt ) ) , T, ) Pz=P,+B(Tz# < 017126.8 bar ( !) ) = 1.01 or 1% Racket Molar Volume kzcc K= with of liquid sat Th - " > Z⇐P÷ ZTC EXAMPLE C Oz ' Equation @ 2 -1=258 .SK Tc= 304.12 Vc= Zc 94.07 = Tr= CMYMOI 0.274 0.85 (1 VL= - 0.85 ) 2/7 4.43×155 M¥01 94.07*(0-274) NIST : 4.37×155 MY mol Methods Review of * Tables C Steam , Kesler for * hee * Equations of State . Z Ideal gas - - - * other ) Empirical law Vivid Equation Cubic Equations Equations Racket - - Antoine ( SRK ) tables ( steam Most * Ideal + * other ) accurate Nt suitable Gas Eos source for large calwlati scale Very easy to use Valid only in the ideal gas . Truncate Viral * , state ( on ) 7=0.95-1.05 ) Eos Valid outside works only for the ideal vapors gas state ( up to 7=0.85 ) 1.2 When I4 1.0 I 1 I I to 11 ideal use 1 I I gas I I 1.80 I 2.00 1 / 2.50 4.00 1 I . . 1.60 1.40 1.30 0.8 Z (0) 1.20 Tr = 0.85 0.6 1.15 1.10 1.05 0.90 0.4 0.95 1.00 0.2 0.0 0.0 0.5 1.0 Reduced pressure, Pr = P/Pc 1.5 2.0 1.2 When up to to where v use rial I isotherm is 1.0 1.80 I in ear 2.00 2.50 4.00 1.60 1.40 1.30 0.8 Z (0) 1.20 Tr = 0.85 0.6 1.15 1.10 1.05 0.90 0.4 0.95 1.00 0.2 0.0 0.0 0.5 1.0 Reduced pressure, Pr = P/Pc 1.5 2.0 Advanced 7iher/be . Eos Kesler t Benedict . Webb - Rubin SRK BWR¥ ✓ Tabulated ( he version . Kesler PR , C vdw ) ) . Wore * Apnly to gases ank * Require only * for . . . " but all e.g. TSKN " most polar liquids Substances } appropriate for corresponding States non - polar and weakly molecules hydrocarbons h¥ good for : HZO , HZSOY , alcohols . . . Empirical eg Antoine , : Racket can - - * For large similar * For scale Eos estimates Equations are or be very accurate Usually specific to substance computer calculations very useful ( Process design ) SRK or , . quick calculations , use any of the other methods EXAMPLE 3 : A ✓ . For VL : For Vr Racket Try : Tc= 304.12k Tr= 0.85 for VL ldealhasorvinal Pc=73 -74bar Pr or , W=ozzS 0.31 = orskk cannot use be ideal . Kesler gas VL= 4.43×185 MYMOI O.5m3tankcmtains35kgofC0z@258.5k.DetermimthlVo1.fractionoftheliqnjd.PsATe258-Z3.1barSo1ntimiNutk.V m3/mo1 B= 1.747 - ✓ Z=|+BP¥ 0.8121 = py VV= Zvi ytank V= - - Bfa , Vv VL ( 7 . 55×54 he Molar Mass ) ±= - = 0.178 = moYm3 6.28×154 mymol VOI.fr Vl = = XLVLTXVVV 0.0125 1.2 1.0 ; afoot 0.8 Z (0) Vv tuna !Y :# ' easiest Lee Kesler Sood SRK good Tr = 0.85 0.6 0.4 : 0.2 0.0 0.0 0.5 1.0 Reduced pressure, Pr = P/Pc 1.5 2.0 Example 3 ant : 'd : We of remove What is the while COZ maintaining T= 258.5k . pressure ? 15kg Solution Calculate ✓ new 16.85bar 8m=l5kg 81=340 8h mol . MW New molar Volume V= : Egg 3 = It × 'T MHM ) Vv= 7.556×153 m3/mo Vinal Use : RI P= V . . . But is } phase = = B acceptable ? viral 2 - . = It }÷ = 0.863 : within acceptable range Vapor . EXAMPLE 4 : 296,20bar Volume of compressed big . Wz@ . information 8.676×1551<-1 Additional Solution molar the Calculate . k= : Firstcakn1ateVL@PsA5thennuktoca1alateVaCP.Tc-304.12Pc-7374VL.vc.zca-TrMI5GaganYmo1Vc-94.O7Zc-O.274Tr-O.963qdk-kVkP-D1nVp2.nkC.P ;D, ) B) )= V2=ViexPfK¢ - zo bar 0.2918 bar .tn z - 56.89 uiymol Example 5 a) : saturated Is b) c) : steam B of Calculate V of steam @ 212.38°C Vv Ob Vinal Get D= At ask compare steam bar 20 gas @ 12bar , tables ? using the Vinal quatim . P=2 Oba T= @ ideal an Calculate 4 DATA Steam : -2=0.889 212.38C = MYmo| 0.0996 2- It = ,Bz¥ ⇐ ' ) 1¥ : pz= 12bar : Z - = V= From Steam Tables MYMOI Assuming Vinal is valid 20bar @ , 212.38°C : 4.61×157 mymol - = 0.00179 It BIT ZPIT = = 0.9334 0.03139mmol interpolation ) :( 0.1742 MYKG , v= on so my,g > sod agreement EXAMPLE 6 : 0.5 his A vo Solution tank faction ! of 35kg of contains liquid wet Steam @ 200°C What's the . ? From Steam Tables -1=2004 PAI 15.55 V=¥HYg=o XL= Vv te ¥ - bar -01428 = VL= Mats 0.896 VL ¥ ×WL+ a- ×yk = 0.072 0.001157 MYKG V✓= 0.1272 MYKG .