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

2020
Views:

494

Pages:

8
Author:

study_genius

Topics for Reading : * Examples * Irreversible * Review of 38,39 with Today ideal processes Ch3 gases EXAMPLE l : 5mo1ofN2@5bar.300K.ismixedwithl0molesofNzac5bar.5o0K.a4abatica11yaeconst.P . final Determine and idealgasmthq Solution 29µmol >= t h F , N , const TZ initial nz final OH A n : It ( ( = ,H(T )Hf P . Tf= - T, ) tnzcp mi¥T2 Nitmz ( Tf = Q=oHtt HCTE P ) , - - h ,H , - - Tz ) - 433.3k nets He Hz )=0 H, - DHFCPCTF Mcptf Q=0 : : in ~D . )+m( mHz=O ( HF Assume . ,P)th2H(Tz,P ) , ) nitric nitm mi K . Adiabatic . . I € balances enevsy ON T, - > ) x ' oHz=cp( TF - Tz ) Example I choose initial : : ( HCT Method 2 ) cent 'd , ( nitnz mcptz =0 Exercise : Change and 't HKTBHH HLT . ,P)= OH from H a ref Hiftn HE ' = : all . state : ,P)=0 H÷ lz Hot final calculate : - ( p ( Tz the repeat ) + , ( ret . T, nitnz . ) } Hinitiakhvptz 'T ) ) HCTF , P ) cpt T, T, - " " ) ) State calculations ) Htm ' - MHM ) soft . ftp.T/=0XTf=niT+mTz mitnz to : . HCTF , PHO t, ) ) Example ( 1 Them , an 'd ) , of moles OH Then Methot , of moles Change th I compartment 'TEn got ,= DH Total 3 - T > ( TF change from I - T, MT = ) -w( Tf ,P) to (TIP) ,P) th ) ; t¥oH%oHEr=m Tf ,P ) compartment ztokmzg change from tnztz mitmz , cptf ) tmcptt , - h )=0 Example 1 Dlttk Qtr N = 'd cont ( Tf ( ✓ , balances ) ( TF T, - - Tz ) = 0 - Q o = possible ? this * Compartment The Before expands l compartment Z contracts 1 amount two : ansumes each other : Vf My = 7.205.503%1, . work . . m NNHNZVZ 12¥ 8.314.16 Mymo , = But work and RpI= 4.988×15331 = . mixing cancel and produces mixing V, Vz= After tnzcv 0 Wtrkoutot Horris Energy no -7 Cnithz 0.108 = M volume tr m3 change )VF= 0.108ms EXAMPLE Z Repeat : Example 92/12=90+9 with 1- function of Cp= ,T+9#+a5P+qyT4 at 9 3.53g = , - 0.000261 92=7.0×108 Solution is Working equation before at same 93=1.3053×159 94=9.9×10-13 : TF oHt±o=n,§FpHdT+nzJcptHT=o Tz MR + Sdvefov TF :{ mol not nz{ao( ~D TF Resultnomucw does T change . TF = - - Tz T II TI )+ )+ai( . ) tai . ( T¥ Tzl )+ . . } . . . . . }=0 434k diﬀerent much but , with that's 't . only bkthegssot Nz . EXAMPLE 3 Compartment : Compartment Remove 1 2 : Nz @ T , partition Solution !Hit ? I 't perform g, .FI?if Iii Ater , = 5bar , Final / * Process is * Gas expands w/o producing = V2= 0 balances 29J And K . . I m 3 . . . : Pz¥k) # irreversible work 0.25 m3 =Tl Pressure R¥, = equilibrate VE ¢¥wFri=o Tz . P to allow and on Before . , Vaccum : Determine state & final with ideal Assume gas vac 500k = . Pz =P = , Vitvz 3.57 bar Example 3 ( cont 'd ) Perform the the calculate Expand .si?eiii It Final , fFinal balances . against opposing pressure Reversible Adiabatic ¥-3.12 Pzckth )t Cpkv Process . : Pi Pz= ( V it bar Vz li 4 = Temperature Tz=Ti Work energy and reversibly Pressure .VE P expansion :W= N= ( P PkP= on . 436.8k ¢°= RVYPT = , cvtzT ) - 30 1 . mol , = - 1308 Jlmol } Wttk 39,33g J