Analisis Termodinamis Proses Alir

8/17/2019 Analisis Termodinamis Proses Alir

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N LISIS TERMODIN MIS PROSES LI

R

Flow Processes (Proses-proses alir)

Ada 2 konsep fundamental untuk analisis proses-proses alir yaitu:

a. Neraca massa (continuity equation = persamaankontinuitas)


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a. Continuity Equation

Σ. (aliran massa masuk) = Σ (aliran massa

keluar) Σ (mi)masuk = Σ (mi)keluar

(m1 + m2 + …)masuk = (m + m! +

…)keluar m = massa"#aktu

$ila ti%ak a%a percabangan :

m1 = m2


3/47

Σ(m)masuk = Σ (m)keluar

Σ (& ' ρ)masuk = Σ (& ' ρ)keluar

uaspenampang ec. &liran

(pan*ang"#aktu)

apat massa

,ntuk incompressible -ui% : ρ ≅ tetap Σ (&') masuk = Σ (&') keluar

uas tampang saluran sama : & ≅ tetap Σ (ρ') masuk = Σ (ρ') keluar


4/47

ncompressible -ui% %an luas tampang

aliran sama Σ (') masuk = Σ (') keluarb. /ukum 0ermo%inamika (-o# processes)

,1+ 11+(mg)1 + 3m'12+456s = ,2+22+

(mg)2 + 3m'22

atau7/1 + (mg)1 + 3m'1

2 + 456s = /2 + (mg)2 +

3m'22

&ctual'elocityprole

u2 u2

Control'olume

;,;/

u1


5/47


6/47

!e"erapa Alat #teady-Flo$ Proses


7/47

V V 2 1>>

V V 2 1


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B

8ol'ing @or V 2

V h h V 2 1 2 122= − +( )

ample *-+

8team at .D a; FoC; enters an a%iabatic no??le #it9 a lo# 'elocityan% lea'es at .2 a #it9 a quality o@ GH. >in% t9e eAit 'elocity; in m"s.

,ontrol Volume 09e no??le

Property .elation 8team tables

Process &ssume a%iabatic; stea%y5-o#

,onser/ation Principles

,onser/ation of mass

>or one entrance; one eAit; t9e conser'ation o@ mass becomes

m m

m m m

in out ∑ ∑== =1 2


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G

,ontrol Volume 09e turbine.

Property .elation &ssume air is an i%eal gas an% use i%eal gas relations.

Process 8tea%y5state; stea%y5-o#; a%iabatic process



m m

m m m

in out ∑ ∑=

= =1 2

,onser/ation of ener0y

&ccor%ing to t9e sketc9e% control 'olume; mass an% #ork cross t9e controlsur@ace. Neglecting kinetic an% potential energies an% noting t9e processis a%iabatic; #e 9a'e


10/47

1


&ccor%ing to t9e sketc9e% control 'olume; mass crosses t9e controlsur@ace; but no #ork or 9eat trans@er crosses t9e control [email protected] t9e potential energies; #e 9a'e

glecting t9e inlet kinetic energy; t9e eAit 'elocity isV h h2 1 22= −( )

#; #e nee% to n% t9e ent9alpies @rom t9e steam tables.

1 1 2 2

1 2

Superheated Saturated Mix.

300 3067.1 0.2

0.4 0.90

o kJ T C h P MPa hkg

P MPa x

= = = = =

t .2 a hf = D. kI"kg an% hfg = 221.! kI"kg.


11/47

11

2 2= +

= 504.7 + (0.90)(2201.6) = 2486.1

f fg h h x h

kJ

kg

2 2

2 1000 /2(3067.1 2486.1)/

1078.0

kJ m sV kg kJ kg

m

s

= −

=

r

Tur"ines

(pander)

@ #e neglect t9e c9anges in kinetic an% potential energies as -ui% -o#st9roug9 an a%iabatic turbine 9a'ing one entrance an% one eAit; t9econser'ation o@ mass an% t9e stea%y5state; stea%y5-o# rst la# becomes


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12

ample *-*

/ig9 pressure air at 1F -o#s into an aircra@t gas turbine an%un%ergoes a stea%y5state; stea%y5-o#; a%iabatic process to t9e turbineeAit at !! . Calculate t9e #ork %one per unit mass o@ air -o#ingt9roug9 t9e turbine #9en

(a) 0emperature5%epen%ent %ata are use%.(b) C p;a'e at t9e a'erage temperature is use%.

(c) Cp at F is use%.


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1F

,ontrol Volume 09e turbine.

Property .elation &ssume air is an i%eal gas an% use i%eal gas relations.

Process 8tea%y5state; stea%y5-o#; a%iabatic process



m m

m m m

in out ∑ ∑=

= =1 2


&ccor%ing to t9e sketc9e% control 'olume; mass an% #ork cross t9e controlsur@ace. Neglecting kinetic an% potential energies an% noting t9e processis a%iabatic; #e 9a'e


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1D

0 1 1 2 2

1 2

+ = +

= −

( )

m h W m h

W m h h

out

out

9e #ork %one by t9e air per unit mass -o# is

w W m

h hout out = = −

1 2

Notice t9at t9e #ork %one by a -ui% -o#ing t9roug9 a turbine is equal tot9e ent9alpy %ecrease o@ t9e -ui%.

(a) ,sing t9e air tablesat T 1 = 1F ; h1 = 1FG.G kI"kg

at T 2 = !! ; h2 = !.D kI"kg

w h h

kJ

kg kJ

kg

out = −

= −

=

1 2

1395 97 670 47

7255

( . . )

.


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1

b) ,sing 0able &52(c) at T a'e = GB ; C p, ave = 1.1FB kI"kg⋅

w h h C T T

kJ

kg K

K

kJ

kg

out p ave= − = −

=

⋅

−

=

1 2 1 2

1138 1300 660

7283

, ( )

. ( )

.

c) ,sing 0able &52(a) at T = F ; Cp = 1. kI"kg ⋅

w h h C T T

kJ

kg K K

kJ

kg

out p= − = −

=⋅

−

=

1 2 1 2

1005 1300 660

6432

( )

. ( )

.

mpressors and fans


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1!

Compressors an% @ans are essentially t9e same %e'ices. /o#e'er;compressors operate o'er larger pressure ratios t9an @ans. @ #e neglectt9e c9anges in kinetic an% potential energies as -ui% -o#s t9roug9 ana%iabatic compressor 9a'ing one entrance an% one eAit; t9e stea%y5state;stea%y5-o# rst la# or t9e conser'ation o@ energy equation becomes

ample *-1

Nitrogen gas is compresse% in a stea%y5state; stea%y5-o#; a%iabaticprocess @rom .1 a; 2oC. Juring t9e compression process t9etemperature becomes 12oC. @ t9e mass -o# rate is .2 kg"s; %eterminet9e #ork %one on t9e nitrogen; in k6.

l l 9 ( 9 k 9 % b )


17/47

1

,ontrol Volume 09e compressor (see t9e compressor sketc9e% abo'e)

Property .elation &ssume nitrogen is an i%eal gas an% use i%eal gasrelations

Process &%iabatic; stea%y5-o#



m m

m m m

in out ∑ ∑=

= =1 2


&ccor%ing to t9e sketc9e% control 'olume; mass an% #ork cross t9e controlsur@ace. Neglecting kinetic an% potential energies an% noting t9e processis a%iabatic; #e 9a'e @or one entrance an% one eAit

0 0 0 0 01 1 2 2

2 1

+ + + = − + + +

= −

( ) ( ) ( )

( )

m h W m h

W m h h

in

in

09 k % t9 it i l t % t t9 t9 l i @ t9


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1B

09e #ork %one on t9e nitrogen is relate% to t9e ent9alpy rise o@ t9enitrogen as it -o#s t9roug9 t9e compressor. 09e #ork %one on t9enitrogen per unit mass -o# is

w W

mh hin

in= = −

2 1

&ssuming constant specic 9eats at F @rom 0able &52(a); #e #rite t9e#ork as

w C T T

kJ

kg K K

kJ kg

in p= −

=⋅

−

=

( )

. ( )

.

2 1

1 039 125 25

1039


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1G

hrottlin0 de/ices

Consi%er -ui% -o#ing t9roug9 a one5entrance; one5eAit porous plug. 09e-ui% eAperiences a pressure %rop as it -o#s t9roug9 t9e plug. No net #orkis %one by t9e -ui%. &ssume t9e process is a%iabatic an% t9at t9e kinetic

an% potential energies are neglecte%7 t9en t9e conser'ation o@ mass an%energy equations become


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2

09is process is calle% a t9rottling process. 69at 9appens #9en an i%ealgas is t9rottle%K

69en t9rottling an i%eal gas; t9e temperature %oes not c9ange. 6e #illsee later in C9apter 11 t9at t9e t9rottling process is an important processin t9e re@rigeration cycle.

& t9rottling %e'ice may be use% to %etermine t9e ent9alpy o@ saturate%steam. 09e steam is t9rottle% @rom t9e pressure in t9e pipe to ambientpressure in t9e calorimeter. 09e pressure %rop is suLcient to super9eatt9e steam in t9e calorimeter. 09us; t9e temperature an% pressure in t9ecalorimeter #ill speci@y t9e ent9alpy o@ t9e steam in t9e pipe.

l *


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21

ample *-

Mne #ay to %etermine t9e quality o@ saturate% steam is to t9rottle t9esteam to a lo# enoug9 pressure t9at it eAists as a super9eate% 'apor.8aturate% steam at .D a is t9rottle% to .1 a; 1oC. Jetermine t9e

quality o@ t9e steam at .D a.

1 2

09rottling orice

Control8ur@ace

,ontrol Volume 09e t9rottle

Property .elation 09e steam tables

Process 8tea%y5state; stea%y5-o#; no #ork; no 9eat trans@er; neglectkinetic an% potential energies; one entrance; one eAit



m m

m m m

in out ∑ ∑== =1 2


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22


&ccor%ing to t9e sketc9e% control 'olume; mass crosses t9e controlsur@ace. Neglecting kinetic an% potential energies an% noting t9e processis a%iabatic #it9 no #ork; #e 9a'e @or one entrance an% one eAit

0 0 0 0 0 01 1 2 2

1 1 2 2

1 2

+ + + = + + +

=

=

( ) ( )

m h m h

m h m h

h h

22

2

1002675.8

0.1

oT C kJ h

kg P MPa

==

=

09ere@ore;

( )1

1 2

1 @ 0.4

2675.8

f fg P MPa

kJ h h

kg

h x h=

= =

= +


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2F

1

1

2675.8 604.66

2133.4

0.971

f

fg

h h x

h

−=

−=

=

3iin0 cham"ers

09e miAing o@ t#o -ui%s occurs @requently in engineering applications. 09esection #9ere t9e miAing process takes place is calle% a miAing c9amber.

09e or%inary s9o#er is an eAample o@ a miAing c9amber.


24/47

2D

ample *-4

8team at .2 a; FoC; enters a miAing c9amber an% is miAe% #it9 col%#ater at 2oC; .2 a; to pro%uce 2 kg"s o@ saturate% liqui% #ater at .2a. 69at are t9e require% steam an% col% #ater -o# ratesK

8team 1

Col% #ater2

8aturate% #aterF

Controlsur@ace

Mixi!

chambe

r

,ontrol Volume 09e miAing c9amber


Process &ssume stea%y5-o#; a%iabatic miAing; #it9 no #ork



m m

m m m

m m m

in out ∑ ∑=+ =

= −1 2 3

2 3 1


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2


&ccor%ing to t9e sketc9e% control 'olume; mass crosses t9e controlsur@ace. Neglecting kinetic an% potential energies an% noting t9e processis a%iabatic #it9 no #ork; #e 9a'e @or t#o entrances an% one eAit

( )

( ) ( )

m h m h m h

m h m m h m h

m h h m h h

1 1 2 2 3 3

1 1 3 1 2 3 3

1 1 2 3 3 2

+ =

+ − =

− = −

( )

( )m m

h h

h h1 3

3 2

1 2

= −

−

o#; #e use t9e steam tables to n% t9e ent9alpies:

11

1

3003072.1

0.2

oT C kJ h

kg P MPa

==

=


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2!

2

2 @ 202

2083.91

0.2 o

o

f C

T C kJ h h

kg P MPa

=≈ =

=

3 21 3

1 2

( )( )

(504.7 83.91) /20

(3072.1 83.91) /

2.82

h hm mh h

kg kJ kg

s kJ kg

kg

s

−=−

−=

−

=

N N

( . )

.

m m m

kg

s

kg s

2 3 1

20 2 82

1718

= −

= −

=


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2

Heat echan0ers

/eat eAc9angers are normally #ell5insulate% %e'ices t9at allo# energyeAc9ange bet#een 9ot an% col% -ui%s #it9out miAing t9e -ui%s. 09epumps; @ans; an% blo#ers causing t9e -ui%s to -o# across t9e control

sur@ace are normally locate% outsi%e t9e control sur@ace.

ample *-5


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2B

ample *-5

&ir is 9eate% in a 9eat eAc9anger by 9ot #ater. 09e #ater enters t9e 9eateAc9anger at DoC an% eAperiences a 2oC %rop in temperature. &s t9eair passes t9roug9 t9e 9eat eAc9anger; its temperature is increase% by

2oC. Jetermine t9e ratio o@ mass -o# rate o@ t9e air to mass -o# rate o@t9e #ater.1&irinlet

26atereAit2

&ir eAit

16aterinlet

Controlsur@ace

ntrol Volume 09e 9eat eAc9anger

operty .elation &ir: i%eal gas relations6ater: steam tables or incompressible liqui% results

ocess &ssume a%iabatic; stea%y5-o#


29/47

2G

,onser/ation Principles ,onser/ation of mass

( / )m m m kg sin out system− = ∆(stea%y)

r t#o entrances; t#o eAits; t9e conser'ation o@ mass becomes

, , , ,

m m

m m m m

in out

air w air w

=

+ = +1 1 2 2

>or t#o -ui% streams t9at eAc9ange energy but %o not miA; it is better toconser'e t9e mass @or t9e -ui% streams separately.

, ,

, ,

m m m

m m m

air air air

w w w

1 2

1 2

= =

= =


&ccor%ing to t9e sketc9e% control 'olume; mass crosses t9e controlsur@ace; but no #ork or 9eat trans@er crosses t9e control [email protected] t9e kinetic an% potential energies; #e 9a'e @or stea%y5-o


30/47

F

E E E kW in out system− =

"ate #$ et eer!% tra&$er '% heat, #r, ad *a&&

"ate ha!e i itera, ieti, p#tetia, et., eer!ie&

( )

∆(stea%y)

( ) ( )

, , , , , , , ,

, , , ,

E E m h m h m h m h

m h h m h h

in out

air air w w air air w w

air air air w w w

=+ = +

− = −

1 1 1 1 2 2 2 2

1 2 2 1

( )

( )

, ,

, ,

m

m

h h

h h

air

w

w w

air air

= −

−

2 1

1 2

6e assume t9at t9e air 9as constant specic 9eats at F ; 0able &52(a)(#e %onOt kno# t9e actual temperatures; *ust t9e temperature %iPerence).$ecause #e kno# t9e initial an% nal temperatures @or t9e #ater; #e canuse eit9er t9e incompressible -ui% result or t9e steam tables @or itsproperties.

,sing t9e incompressible -ui% approac9 @or t9e #ater; 0able &5F;C p, w = D.1B kI"kg⋅.


31/47

F1

( )

( )

, , 2 ,1

, ,1 , 2

( )

( )

4.18 20

1.005 25

/3.33

/

p w w wair

w p air air air

w

air

air

w

C T T m

m C T T

kJ K

kg K kJ

K kg K

kg s

kg s

−=

−

−

⋅=−

⋅

=

N

N

& secon% solution to t9is problem is obtaine% by %etermining t9e 9eattrans@er rate @rom t9e 9ot #ater an% noting t9at t9is is t9e 9eat trans@errate to t9e air. Consi%ering eac9 -ui% separately @or stea%y5-o#; oneentrance; an% one eAit; an% neglecting t9e kinetic an% potential energies;t9e rst la#; or conser'ation o@ energy; equations become

,1 ,1 , , 2 , 2

,1 ,1 , , 2 , 2

, ,

-

-

in out

air air in air air air

w w out w w w

in air out w

E E

air m h Q m h

water m h Q m h

Q Q

=+ =

= +

=

N N

NN N

NN N

N N

pe an uc o$


32/47

F2

p

09e -o# o@ -ui%s t9roug9 pipes an% %ucts is o@ten a stea%y5state; stea%y5-o# process. 6e normally neglect t9e kinetic an% potential energies79o#e'er; %epen%ing on t9e -o# situation; t9e #ork an% 9eat trans@er mayor may not be ?ero.

ample *-78

n a simple steam po#er plant; steam lea'es a boiler at F a; !oC; an%enters a turbine at 2 a; oC. Jetermine t9e in5line 9eat trans@er @romt9e steam per kilogram mass -o#ing in t9e pipe bet#een t9e boiler an%t9e turbine.

Controlsur@ace

18team@romboiler

8teamtoturbine2

,ontrol Volume ipe section in #9ic9 t9e 9eat loss occurs.


Process 8tea%y5-o#


Qout

, i f


33/47

FF


( / )in out systemm m m kg s− = ∆(stea%y)

r one entrance; one eAit; t9e conser'ation o@ mass becomes

m mm m m

in out == =1 2


&ccor%ing to t9e sketc9e% control 'olume; 9eat trans@er an% mass cross

t9e control sur@ace; but no #ork crosses t9e control sur@ace. Neglectingt9e kinetic an% potential energies; #e 9a'e @or stea%y5-o#

"ate #$ et eer!% tra&$er "ate ha!e i itera, ieti, '% heat, #r, ad *a&& p#tetia, et., eer!ie&

( )in out system E E E kW − = ∆ 1 D 2 DF 1D2 DF

(stea%y)

%etermine t9e 9eat trans@er rate per unit mass o@ -o#ing steam as

( )

m h m h Q

Q m h h

q

Q

m h h

out

out

out

out

1 1 2 2

1 2

1 2

= +

= −

= = −

t9 t t bl t % t i t9 t9 l i t t9 t t t


34/47

FD

use t9e steam tables to %etermine t9e ent9alpies at t9e t#o states as

11

1

6003682.8

3

oT C kJ h

kg P MPa

==

=

2

22

5003468.3

2

oT C kJ h

kg P MPa

==

=

1 2

(3682.8 3468.3)

214.5

out q h h

kJ

kg

kJ kg

= −

= −

=

ample *-77

&ir at 1oC; .1 a; D m"s; -o#s t9roug9 a con'erging %uct #it9 amass -o# rate o@ .2 kg"s. 09e air lea'es t9e %uct at .1 a; 11F.! m"s. 09e eAit5to5inlet %uct area ratio is .. >in% t9e require% rate o@ 9eattrans@er to t9e air #9en no #ork is %one by t9e air.


35/47

F

Controlsur@ace

1&ir inlet

&ir eAit2

Qin

,ontrol Volume 09e con'erging %uct

Property .elation &ssume air is an i%eal gas an% use i%eal gas relations

Process 8tea%y5-o#



( / )m m m kg sin out system− = ∆(stea%y)

r one entrance; one eAit; t9e conser'ation o@ mass becomes

m m

m m m

in out =

= =1 2



36/47

F!

n t9e rst la# equation; t9e @ollo#ing are kno#n: P1; T

1 (an%

h1); ; ; ; an% A2" A1. 09e unkno#ns are ; an% h2 (or T 2). 6e use

t9e rst la# an% t9e conser'ation o@ mass equation to sol'e @or t9e t#ounkno#ns.


&ccor%ing to t9e sketc9e% control 'olume; 9eat trans@er an% mass crosst9e control sur@ace; but no #ork crosses t9e control sur@ace. /ere keep t9ekinetic energy an% still neglect t9e potential energies; #e 9a'e @or stea%y5

state; stea%y5-o# process

"ate #$ et eer!% tra&$er "ate ha!e i itera, ieti, '% heat, #r, ad *a&& p#tetia, et., eer!ie&

( )in out system E E E kW − = ∆ 1 D2 DF 1D2 DF

(stea%y)

1V V 2 mN

Qin

( / )m m kg s=N N


37/47

F

1 2

1 1 2 2

1 2

1 2

1 1 2 21 2

( / )

1 1

m m kg s

V A V Av v

P P

V A V A T T

=

=

=

N N

r r

r r

8ol'ing @or T 2

ssuming C p = constant; h2 5 h1 = C p(T 2 5 T 1)

ooks like #e ma%e t9e #rong assumption @or t9e %irection o@ t9e 9eat


38/47

FB

g ptrans@er. 09e 9eat is really lea'ing t9e -o# %uct. (69at type o@ %e'ice ist9is any#ayK)

.Q Q kW out in= − = 2 87

9i:uid pumps

09e #ork require% #9en pumping an incompressible liqui% in an a%iabaticstea%y5state; stea%y5-o# process is gi'en by

9e ent9alpy %iPerence can be #ritten as

h h u u Pv Pv2 1 2 1 2 1− = − + −( ) ( ) ( )

>or incompressible liqui%s #e assume t9at t9e %ensity an% specic 'olume


39/47

FG

>or incompressible liqui%s #e assume t9at t9e %ensity an% specic 'olumeare constant. 09e pumping process @or an incompressible liqui% isessentially isot9ermal; an% t9e internal energy c9ange is approAimately?ero (#e #ill see t9is more clearly a@ter intro%ucing t9e secon% la#). 09us;t9e ent9alpy %iPerence re%uces to t9e %iPerence in t9e pressure5specic

'olume pro%ucts. 8ince v 2 = v 1 = v t9e #ork input to t9e pump becomes

W

,W W in pump= − is t9e net #ork %one by t9e control 'olume; an% it is note% t9at #ork isinput to t9e pump7 so; .

e neglect t9e c9anges in kinetic an% potential energies; t9e pump #ork becom

− − = −

= −

( ) ( ) ( )

( )

,

,

W m v P P kW

W m v P P

in pump

in pump

2 1

2 1

6e use t9is result to calculate t9e #ork supplie% to boiler @ee%#aterpumps in steam po#er plants.

@ #e apply t9e abo'e energy balance to a pipe section t9at 9as no pump (); #e obtain.

0W =N

2 2V V


40/47

D

2 12 1 2 1

2 2

2 12 1 2 1

2 2

2 2 1 12 1

( ) ( ) ( )2

0 ( ) ( )2

1

2 2

V V W m v P P g ! ! kW

V V m v P P g ! !

v

P V P V ! !

g g

ρ

ρ ρ

−− = − + + −

−= − + + −

=

+ + = + +

N N

r r

N

r r

09is last equation is t9e @amous $ernoulliQs equation @or @rictionless;

incompressible -ui% -o# t9roug9 a pipe.

;niform-#tate< ;niform-Flo$ Pro"lems

Juring unstea%y energy trans@er to or @rom open systems or control'olumes; t9e system may 9a'e a c9ange in t9e store% energy an% mass.

8e'eral unstea%y t9ermo%ynamic problems may be treate% as uni@orm5state; uni@orm5-o# problems. 09e assumptions @or uni@orm5state; uni@orm5-o# are

•09e process takes place o'er a specie% time perio%


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D1

• 09e process takes place o'er a specie% time perio%.• 09e state o@ t9e mass #it9in t9e control 'olume is uni@orm at any instanto@ time but may 'ary #it9 time.• 09e state o@ mass crossing t9e control sur@ace is uni@orm an% stea%y. 09emass -o# may be %iPerent at %iPerent control sur@ace locations.

0o n% t9e amount o@ mass crossing t9e control sur@ace at a gi'en location;#e integrate t9e mass -o# rate o'er t9e time perio%.

e c9ange in mass o@ t9e control 'olume in t9e time perio% is

e uni@orm5state; uni@orm5-o# conser'ation o@ mass becomes

m m m mi e CV − = −∑∑ ( )2 1 c9ange in internal energy @or t9e control 'olume %uring t9e time perio% is

energy crossing t9e control sur@ace #it9 t9e mass in t9e time perio% is


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D2

energy crossing t9e control sur@ace #it9 t9e mass in t9e time perio% is

9ere

* =i; @or inletse; @or eAits

9e rst la# @or uni@orm5state; uni@orm5-o# becomes

69en t9e kinetic an% potential energy c9anges associate% #it9 t9e control'olume an% t9e -ui% streams are negligible; it simplies to

ample *-72


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DF

ample *-72

Consi%er an e'acuate%; insulate%; rigi% tank connecte% t9roug9 a close%'al'e to a 9ig95pressure line. 09e 'al'e is opene% an% t9e tank is lle%#it9 t9e -ui% in t9e line. @ t9e -ui% is an i%eal gas; %etermine t9e nal

temperature in t9e tank #9en t9e tank pressure equals t9at o@ t9e line.

,ontrol Volume 09e tank

Property .elation %eal gas relations

Process &ssume uni@orm5state; uni@orm5-o#



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DD

m m m mi e CV − = −∑∑ ( )2 1

; @or one entrance; no eAit; an% initial mass o@ ?ero; t9is becomesm mi CV = ( )2


>or an insulate% tank Q is ?ero an% @or a rigi% tank #it9 no s9a@t #ork W is?ero. >or a one5inlet mass stream an% no5eAit mass stream an% neglectingc9anges in kinetic an% potential energies; t9e uni@orm5state; uni@orm5-o#conser'ation o@ energy re%uces to

or

m h m u

h u

u Pv u

u u Pv

C T T Pv

i i CV

i

i i i

i i i

v i i i

=

=

+ =

− =

− =

( )

( )

2 2

2

2

2

2



C T T T( )


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D

C T T T

T C

C T

C

C T

kT

v i i

v

v

i

p

v

i

i

( )2

2

− =

= +

=

=

@ t9e -ui% is air; k = 1.D an% t9e absolute temperature in t9e tank at t9enal state is D percent 9ig9er t9an t9e -ui% absolute temperature in t9esupply line. 09e internal energy in t9e @ull tank %iPers @rom t9e internalenergy o@ t9e supply line by t9e amount o@ -o# #ork %one to pus9 t9e-ui% @rom t9e line into t9e tank.tra Assi0nment

e#ork t9e abo'e problem @or a 1 mF tank initially open to t9eatmosp9ere at 2oC an% being lle% @rom an air supply line at G psig;2oC; until t9e pressure insi%e t9e tank is psig.


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Pum,'

• Li/ui' *re u'u*ll0 move -0 ,um,'1 The '*me e/u*$ion' *,,l0 $o *i*-*$i# ,um,' *' $o *i*-*$i# #om,re''or'1

• 2or *n i'en$ro,i# ,ro#e''.

• &i$h

• 2or li/ui3

– –

–

( ) ∫ =∆= 2

1

)( P P

" s V#P $ isentropi%W

( ) )()( 12 P P V $ isentropi%W " s −=∆=

#P T V #T C #$ P )1( β −+= V#P T

#T C #" P β −=

P T V T C $ P ∆−+∆=∆ )1( β

P V

T

T C " P ∆−=∆ β

1

2

ater at 45 ad 10 a eter& a adia'ati pu*p ad i& di&har!ed


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ater at 45 ad 10 a eter& a adia'ati pu*p ad i& di&har!ed

at a pre&&ure #$ 8600 a. &&u*e the pu*p e$$iie% t# 'e 0.75.

auate the #r #$ the pu*p, the te*perature ha!e #$ the ater,

ad the etr#p% ha!e #$ ater.

kg

%mV

3

1010=he &aturated iuid ater at 45- K

110425 6−×=β

K kg

kJ C P

⋅= 178.4

( ) )()( 12 P P V $ isentropi%W " s −=∆=

kg

kJ

kg

%mkPaisentropi%W s 676.810676.8)108600(1010)(

36 =×=−×=

kg

kJ $

isentropi%W W s s 57.11

)(=∆==

η P T V T C $ P ∆−+∆=∆ )1( β

K T 97.0=∆

P V T

T C " P ∆−=∆ β

1

2

kJ

Analisis Termodinamis Proses Alir

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