ChE 220Lecture 17

Topics for Today

* Examples with ideal gases* Irreversible processes

* Review of Ch3

Reading : 38,39

EXAMPLE l :

5mo1ofN2@5bar.300K.ismixedwithl0molesofNzac5bar.5o0K.a4abatica11yaeconst.P.

Determine final and enevsy balances.

Assumeidealgasmthq >= 29µmol . K .

Solution.

. Adiabatic : Q=0I const . P : Q=oHtt

€t ,

FTZ

h ,N nz initial A : n ,H(T , ,P)th2H(Tz,P )

final It ( nitric ) HCTE,

P )

OH = ( nitm )Hf - h ,H , -

mHz=O nets

mi ( HF - H ,)+m(He - Hz )=0

in x '

DHFCPCTF - T , ) oHz=cp( TF - Tz )

~D Mcptf - T, ) tnzcp ( Tf - Tz ) - ON >

Tf= mi¥T2=

433.3kNitmz

Example I ( cent 'd ) Method 2 : calculate all H from a ref . state :

choose : HCT , ,P)=0

initial :

Hiftn' HKTBHH 't " " )

} � Hinitiakhvptz 'T )

H÷lz = Hot ( p ( Tz - T , )

final : HE( nitnz ) HCTF,

P )

HLT . ,P)= cpt ,. T , )

) � Htm' - MHM ) soft . t , )

OH =0⇒ mcptz - T, ) + ( nitnz ) ftp.T/=0XTf=niT+mTzmitnz

Exercise : Change the ret . State to : HCTF,PHO

and repeat calculations.

Example 1 ( an 'd ) Methot 3

Them, moles of compartment I change from

th,P ) -w( Tf ,P)

OH 'TEn , got ,= - T )

Then, moles of compartment I change from th,P)to (TIP)

DH ztokmzg > ( TF - T, )

Total Change; t¥oH%oHEr=m , cptf , ) tmcptt - h )=0� Tf =

MT tnztz

mitmz

Example 1 cont 'd Energy balances

Dlttk N , ( ✓ ( Tf - T , ) tnzcv ( TF - Tz ) = 0

Qtr = 0

Wtrkoutot - Q = o

Horris this possible ?

* Compartment l expands and produces work.

But

compartment Z contracts and ansumes work .

The two amount 1 cancel each other.

Before

mixingV, = RpI= 4.988×15331 m

: NNHNZVZ = 0.108 m3Vz= 12¥.

8.314.16 Mymo , Mno volume change

trAfter mixing : Vf = My = 7.205.503%1

,-7 Cnithz )VF= 0.108ms

EXAMPLE Z : Repeat Example 1- with Cp= function of T.

92/12=90+9 ,T+9#+a5P+qyT4 at 3.53g9 , = - 0.000261

Solution 92=7.0×10893=1.3053×159

Working equation is same at before : 94=9.9×10-13TF

oHt±o=n,§FpHdT+nzJcptHT=oTz

MR :{ not . - T . )+ai( II . TI )+ . . . }+ nz{ao( TF - Tz ) tai ( T¥ . Tzl )+ . . . }=0

Sdvefov TF ~D TF = 434k

Resultnomucw different,

but that's only bkthegssot Nzdoes mol change much with 't

.

EXAMPLE 3 : Compartment 1 : Nz @ T , = 500k , P,= 5bar

,VE 0.25 m3

Compartment 2 : Vaccum V2= 0 . I m3

Remove partition and allow to equilibrate .

Determine final state & perform balances.

Assume ideal gas with g , = 29J And . K.

Solution

Hit?! vac. Before

on 't ¢¥wFri=o� Tz =Tl

Iii.FI?if . AterFinal Pressure :

R¥, = Pz¥k) � Pz =P ,

I= 3.57 bar

/ # Vitvz

* Process is irreversible

* Gas expands w/o producing work .

Example 3 ( cont 'd ) Perform the expansion reversibly andcalculate the energy balances

.

It.si?eiiiExpand against opposing pressure .

Reversible Adiabatic Process :

Final Pressure

P, .VE Pzckth )t� Pz= Pi (¥-3.12 barV it Vz

f- Cpkv = li 4

Final Temperature

Tz=Ti ( P⇒PkP= 436.8k

Work:W= on . ¢°= cvtz- T, ) = - 1308 Jlmol } � Wttk

39,33g JN= RVYPT

,= 30 .

1 mol