HW 26solution

Given Nz compressed adiabatic ally from 1 → 2

then isothermal exaansion from 2- 3

Ep Dsee EFD for more details

.

÷I

1--

-

#Q=o4

µz 4 / / l1 Nz 1

p-

- -

1 kW,→z0,

.lt-

L. Nz

\

i. - - -

-1- -oh-¥98133 !

- - -

-1state , state 2 state 3

P,= 13512Pa Pz= 1100 KR 13=340142

T,

= 280 k Tz =520kTz =520km= 4kgAssumptions D Closed system , 2) DKE =O

,3) BPE=O

4) Nz is an idealgas

at these conditions

5) Process 2 → 3 is internally reversible

Basic Equations .

-

z

Conservation of Mass : M = constantConservation of Energy ( 1st Zaw ) : All = Q - W

Entropy Balance (2nd Law ) : DS = §Eh +6

For an ideal gas ; P ✓ = MRT,

F= Fw,

Mwnizs ,±÷,

SCIP) = 5k ) - Rln(¥a ) ,Ref=IatmSIT ,P ) - s

, CTP ) = silt ) - Sia ) - 12k¥, )

HW 26solution

a) Find 52 - S,

and 53 - SyIs

process1→2 possible if adiabatic ?

stated : T,

= 280 k,

P,

= 13512Pa

state : Tz '

520k, Pz = 1100 KPA

sz - s,

= Sia ) - silt ) - Rmln(¥,)so= ITIdeal Gas Tables

MWNZ

si =

spiffy =189.64M€ 6.769 why - K

28.01 kg/km°1=

Si =

207.712412M€7.4152ko/kg - K

28.01 kg/kmol=

E- s ,=(

7.4152-6.

> 69y¥gk - 823¥, ln(YS0÷mP5a)

sz - s,

=

0.0235µµg . *

"

( KYKJK

52 . s ,=m( sz - s, )= (4.0kg )(

0.0235¥it )

{2-3--0.09411-4

↳ Yes, process

l→2 is possibleif adiabatic because sz - s

,> 0

HW 26solution

StateTz = 520 k,

B = 340 KR

s ; = Si =

7.4152124kg

- K

ss - sztosztsif - III ln(3Y8oY?pa-)

sz - Sz =

0.3485124kg

- K

53 - Sz= m( s }- sz ) = ( 4kg)(

0.3485k¥)

T.z.si/.394k#b) Find Q + W for each process

¥-0,

told I → 2 is adiabatic

1st Law : m(uz- u, ) = Q/→z- W|→z

0

Wpz = m ( U ,- uz)

Ideal Gas Tables : u ,=u(T ,)=u¥If

.

.is:8?otYIdU ,

= 207.07 HAYIn the same

way , Uz= uffwtdn.

= 387.18 KYKG

W|→z = ( 41g ) ( 207.07 - 387.18 ) kYkg

&W,→z=-720.4kYkg-

HW 26solutionProcess 2 → 3 : internally reversible &

isothermal

↳ 62-3=02nd Law : sz - sz=§¥b+0¥→3

Sz - Sz= §¥b = 02=3 0

Tz=T3

Qzsz = I ( 53 - Sz )Qz→ = 520k (1.394k¥ )ftp.724.9/stLaw:U3fUz=Qzs3tWzs3&Wzsz=Qzos3&Wzs3=724.9

HW 27 solution

Give=tEFD_staff

.

Tank A Tank BV , a

= 2.0 m3 YB = SO m3

P, a

= 60012Pa P, ,

= 20012Pa

ValveT ,a= 500k IB = 900k

) opens

gxatef v.

Tank A Tank BV , a

= 2.0 m3 YB = 5.0 m3

Tz Tz

Pz Pz

Find : 12,

K,

0,→z ,

whatcauses 6>0 ?

Assumptions: 1) Closed system ,2) Win :O

3) well insulated Qi→ > =O, 4) Volume of line is negligible

g)Air is an ideal gas ,6) AKE = 0,

7) DPE =O

HW 27 solution

Basic Equations .

-

Conservation of Mass : M = Mz = M,

= Ma + MBConservation of Energy : Uz - U

,=

Q§z-

WyfszUz = U,

= Uza + Uzis = U, a

+ UYB°

Entropy Balance : sz - S,

= §¥f3+ GL

( szatfy} ) - ( Sia + s, ,D =%

52

Ideal Gas : PV = MRT,

R : Fw

SLT ,P)= 5 CT ) - Rln(¥D ,Ret -

- latm

Solution

÷find mass,

Ma ' FIFE -18.953.595¥ ,= £36 's

same was , mi .'

go.gg#EghYgmoIony=3*k5

HW 27 solution

Next,

find U , t Uz

U ,= Mia U , a t MlbUIB

,a ,=fCT ) - Tables

U ,= §-36kg)(3s9 .SE ) + (3.87kg)( 674.6 ¥y )

U ,= 5616.1 = Uz

Uz = Mzuz = @, A + m, ,)Uz = 5616

.

1 KO

↳ uz= 459.2 KYKG → I

Interpolate in tables to find Tz

Tz - 630k thus'¥=4y!Fs÷yst¥¥ ,↳42=631.8/3

=

MzRgigzI_ =

42.231.310.2874*631.8=7.0m3&Pz=3l7kPa-Next solve entropy balance !

HW 27 solution

s,

= S, a + Sits = M , a sin .

tm ,B sits= m ,a[ Sia - Rey'¥÷d]+m,B[% - Rk

.CI#DSz=MzSz=(miatmiB)Sz=maCsi-RlnfEeDf+m, ,3[ si -

Rhl¥eD]Therefore,

sz - s ,=ma{ se - Sia - Rld¥a)]tm,,3( sirs :B - Rh⇐sD

Table → spa = 5C Sook ) = 2.221 kykg - K

STB = s° ( 900 K ) = 2.85 kykg - K

SE-s°L630K)_ = s°(640K)-s°(63ok)_( 631.8- 630 ) K ( 640 - 630 ) K

Sz° = 2.462 +1,8¥ ( 2.478 - 2.462 ) kykg . K

sz° =

2.4649124kg- K

Sz - S,

= (8.36kg)[

2.4649- 2.221 - 0.287 ln(E¥o)]

t ( 3.87kg ) G.4649- 2.85 - 0.287 ln(3¥o)]sz - s

, =1.568KYK = S#st66=1.568*8↳ Irreversible due to unrestrained expansion +

heat transfer between streams while mixing

HW 28solution

Given Air expands across an adiabatic turbineValves located at turbine inlet + outlet

.

'

inEFII :

. -

ly- T

,

' 800k

|1 P

,= 100012Pa

1-

-

¥-11R '

90012PaI ni 1

11 191in

- wzs }( -

- ¥! l01 } 13 : 14012Pa

: 1

q-

4T-

Py = 100hPaCVZ v

in Ty = 500 k

Find : Wz→3, Qsy , 62-3

Assumptions: 1) steady. state

, 2) uniform flow

3) Adiabatic devices, 4) Air behaves as an ideal gas

5) AKE .-0

, 6) APE=O,

Basic Equations .

-Conservation of Mass : in ;n= inoot = in

, ,z , 3 ,y= in

Conservation of Energy ( 1st Law ) :

steady- state 0 = & - W + afnemislht #¥2) '§↳n(hto¥t¥2)

HW 28solution

Entropy Balance (2nd Law ) : dd¥=E¥b+,§µsms - o§µjmst6

For an ideal gas ; P ✓ = MRT,

R= Fw

SCIP) = 5k ) - Rln(¥a ) ,Ref=latmSZGP ) - s

, CTP ) = silt ) - SICT ) - 12k¥, )

a) Find Was kykgO= Of- Wz . } + inhz - inh

,

to

wz→3 = mi( hz - 43 )wz→ }

= hz - hz

Process 1→2 : Q=W=0,

hz=h ,= h( 800k ) f +

hz = 821.9 why a)tfaeabfsbas

Process 3→4 : Q=W= 0,

h = h = h( 500k )3

h4z= 503.3 KYKG

&Wz;821.9-503.3=318.6kYt-b) Find G- 4

0 = Foth+ s,

-

sy to ,→ . , ( 2nd Law, open ,

SS )Q→y= Sy

- 5,

= sci- si - Ra ... lnC¥ )

& = 2.221 - 2.719 - 0.287 ln(Yo°oo"}-p. )

61×1=0^16312412-0

HW 28solution

c) Find 62-3

In the same way as b),

6z→ ]= 53 - Sz

*= Ss - si - RACE)

0*26201562719- 0.287 In ( Foote, )

d) Majority of entropy is generated by valves

( i.e.,

throttles ) !

G→y - 62-3 > > 62-3- in

entropy generated entropyby valves generated

by turbine