Chapter 3 First Law Thermodynamics

8/9/2019 Chapter 3 First Law Thermodynamics

1/59

CHAPTER

3MEC 451Thermodynamics

First Law ofThermodynamics

Lecture Notes:MOHD HAFIZ MOHD NOH

HAZRAN HUSAIN & MOHD SUHAIRIL

Faculty of Mechanical Engineering

Univeriti !e"nologi MARA# $%$%Shah Ala'# Selangor

For students EM 220 and EM 221 only

1


2/59

Faculty of Mechanical Engineering, UiTM

2

MEC 451 – THEM!"#N$M%C&

FIRS! LA( OF !HERMOD)NAMI*S

ENERGY ANALYSISOF CLOSED SYSTEM


3/59

First La' of Ther(o)yna(ics



3

The First Law is usually referred to as the Law of Conservation

of Energy i!e! energy can neither be created nor destroyed, but

rather transformed from one state to another !

The energy "alance is maintained within the system "eing

studied#defined "oundary!

The various energies associated are then "eing o"served as

they cross the "oundaries of the system!


4/59


4


Energy *alance for Close) &yste(

Heat

+or

z

Close)&yste(

eference -lane, z . /

V

or

E E E in out system− = ∆


5/59


5


$ccording to classical thermodynamics

Q W E net net system− = ∆

The total energy of the system E system is given as

E Internal energy Kinetic energy otential energy

E ! KE E

% & &

% & &

The change in stored energy for the system is

∆ ∆ ∆ ∆ E ! KE E = + +

The first law of thermodynamics for closed systems then can "e

written as

Q W ! KE E net net − = + +∆ ∆ ∆


6/59


'


(f the system does not move with a velocity and has no change in

elevation the conservation of energy e)uation is reduced to

Q W ! net net − = ∆ The first law of thermodynamics can "e in the form of

*+1,,,

*+

2,,,

12

2

1

2

212 "#

$ $ g V V uumW Q net net

−+−+−=−

*#+1,,,

*+

2,,,

12

2

1

2

212 "g "#

$ $ g V V uu%& net net

−+−+−=−

For a constant 'olume (rocess

−+

−+−=−

1,,,

*+

2,,,

12

2

1

2

212


−+−+−=1,,,

*+

2,,,

12

2

1

2

212

$ $ g V V uumQ net


7/59


-


For a constant .ressure .rocess

−+−+−=−

1,,,

*+

2,,,

122

12

212


−

+

−

+−=−− 1,,,

*+

2,,,*+ 12

2

1

2

2

1212

$ $ g V V uumV V Q

net

−+

−+−+−=

1,,,

*+

2,,,*+ 12

2

1

2

21212

$ $ g V V V V uumQ net

−+

−+−=

1,,,

*+

2,,,

12

2

1

2

212

$ $ g V V hhmQ

net


8/59


/MEC 451 – THEM!"#N$M%C&

0igid tan iston cylinder

Example of Closed Systems


9/59



$ closed system of mass 2 g

undergoes an adia"atic .rocess!

The wor done on the system is

3, ! The velocity of the system

changes from 3 m#s to 15 m#s!

uring the .rocess the elevationof the system increases 45 meters!

etermine the change in internal

energy of the system!

E0a(le 231

Solution:

Energy "alance

−+

−+−=−

1,,,

*+

2,,,

12

2

1

2

212


0earrange the e)uation

net Q

( ) ( )

2 2

2 1 2 1

2 1

2 2

2 1 2 1

2 1

2 2

+ *

2,,, 1,,,

+ *

2,,, 1,,,

!/1 4515 33, 2 2 2

2,,, 1,,,

14!451 !!

net

net

V V g $ $ W m u u

V V g $ $ W m u u

u

u "# )ns

− −− = − + + ÷

− −

− = − + + ÷

−− − = ∆ + + ÷ ÷ ÷ ∆ =


10/59


1,MEC 451 – THEM!"#N$M%C&

6team at 11,, a and 2 .ercent

)uality is heated in a rigid container

until the .ressure is 2,,, a! For a

mass of ,!,5 g calculate the amount

of heat su..ly +in * and the total

entro.y change +in #g!7*!

E0a(le 23 Solution:

( )

( )

( )

3

1 1

1 1 1 1

1 1 1 1

1 1 1 1

!

1

11,, ,!2

,!,,113 ,!2 ,!1--53 ,!,,1133

,!1'34

-/,!, ,!2 1/,'!3

2441!

2!1-2 ,!2 4!3-44

'!2,4

f fg

m"g

f fg

"*"g

f fg

"# "g K

+tate

at "a ,

' ' , '

u u , u

s s , s

= == +

= + −

=

= +

= +

=

= +

= +

=


11/59


11MEC 451 – THEM!"#N$M%C&

( )

3

2 2

2 2

2

2

2

2,,, ,!1'34

,!15122 245! -!122

,!1'34

,!1-5'/ 311'! -!433-

,!1'34 ,!15122245! 311'! 245!

,!1-5'/ ,!15122

3,3,!42

,!1'34 ,!15122-!122 -!

,!1-5'/ ,!15122

m"g

"# "g

+tate

at "a '

' u s

u s

u

s

= =

− = + − ÷− =

− = + ÷− ( )

!

433- -!122

-!2-, "# "g K

−

=

For a rigid container

'2%'1%,!1'34 m3#g

su.erheated


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( )

( )

2 1

,!,5 3,3,!42 2441!

2!43

Q m u u

"#

= −

= −=

$mount of heat su..lied 8

The change in entro.y 9s

2 1

!

-!2-, '!2,4

1!,-5 "# "g K

s s s∆ = −= −=

F l f M h i l E i i UiTM


13/59



E0a(le 232

$ rigid tan is divided into two e)ual

.arts "y a .artition! (nitially one side

of the tan contains 5 g water at 2,,

a and 25:C and the other side is

evacuated! The .artition is then

removed and the water e;.ands intothe entire tan! The water is allowed to

e;change heat with its surroundings

until the tem.erature in the tan

returns to the initial value of 25:C!

etermine +a* the volume of the tan

+"* the final .ressure +c* the heat

transfer for this .rocess!

Solution:

( )

31

1 < 251

1

3

1

2,, ,!,,1,,3

25

5 ,!,,1,,3

,!,,5

m"g o f -

+tate

"a' '

. -

initial 'olume of half rese'oir

V m'

m

= = ==

=

=

o

5

The initial volume for entire tan

( )3

2 ,!,,5

,!,1

rese'oir V

m==

Com.! li)uid

F lt f M h i l E i i UiTM


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( )net net

net net

Q W m u "e e

Q W

− = ∆ + ∆ + ∆

− m u "e= ∆ + ∆ e+ ∆( )( )2 1net Q m u m u u= ∆ = −

The final .ressure

The heat transfer for this .rocess

3

33

2

2

2

2

25 ,!,,1,,3,!,1

43!34,!,,25

=

> 3!1'

m f "g

mm g "g "g

f g

sat

saturated m

+tate

. - '

''

ch

i,ture

ec" region

' ' 'then "a

= =

== =

< < →= =

o

( )

( )

1 < 25

2 2

2

2

5

5

2

1,4!//

2!3 1,

1,4!// 2!3 1, 23,4!

1,4!3

>

5 1,4!3 1,4!//

,!25

"# "g f -

f fg

f

fg

"# "g

net

u u

u u , u' '

,'

u

.hen

Q

"#

−

−

= =

= +−=

= ×

∴ = + ×

=

= −

=

o

/'e sign indicates heat transfer

into the system

1,4!/3 +23,4!3*

+1,4!//?1,4!/3*

1,4!//



15/59



&ule(entary -ro6le(s 1

1! Two tans are connected "y a valve! Tan $ contains 2 g of car"on

mono;ide gas at --:C and ,!- "ar! Tan @ holds / g of the same gas

at 2-:C and 1!2 "ar! Then the valve is o.ened and the gases are

allowed to mi; while receiving energy via heat transfer from the

surrounding! The final e)uili"rium tem.erature is found to "e 42:C!

etermine +a* the final .ressure +"* the amount of heat transfer! $lso

state your assum.tion! A 210 "a, Q

/342 "# B

2! $ .iston cylinder device contains ,!2 g of water initially at /,, a

and ,!,' m3! ow 2,, of heat is transferred to the water while its

.ressure is held constant! etermine the final tem.erature of the water!$lso show the .rocess on a T?D diagram with res.ect to saturation

lines!

A 4211o- B



16/59


1'MEC 451 – THEM!"#N$M%C&


3! $ .iston?cylinder device contains ' g of refrigerant?134a at /,, a

and 5,oC! The refrigerant is now cooled at constant .ressure until ite;ist as a li)uid at 24oC! 6how the .rocess on T?v diagram and

determine the heat loss from the system! 6tate any assum.tion made!

A121025

"# B4! $ ,!5 m3 rigid tan contains refrigerant?134a initially at 2,, a and 4,

.ercent )uality! eat is now transferred to the refrigerant until the .ressure reaches /,, a! etermine +a* the mass of the refrigerant in

the tan and +"* the amount of heat transferred! $lso show the .rocess

on a ?v diagram with res.ect to saturation lines!

A123 "g, 2652

"# B5! $n insulated tan is divided into two .arts "y a .artition! ne .art of

the tan contains ' g of an ideal gas at 5,:C and /,, a while the

other .art is evacuated! The .artition is now removed and the gas

e;.ands to fill the entire tan! etermine the final tem.erature and the

.ressure in the tan!

A07-, 800 "aB

Faculty of Mechanical Engineering UiTM


17/59


1-MEC 451 – THEM!"#N$M%C&

6ome thermodynamic cycle com.oses of .rocesses in which

the woring fluid undergoes a series of state changes such

that the final and initial states are identical!

For such system the change in internal energy of theworing fluid is Gero!

The first law for a closed system o.erating in a

thermodynamic cycle "ecomes

Close) &yste( First La' of a Cycle

Q W !

Q W

net net cycle

net net

− =

=

∆



18/59


1/MEC 451 – THEM!"#N$M%C&

*oun)ary +ors

2

4

5

1

P

V



19/59



No Value of n Process Description Result of !L

1 H isochoric constant volume +D1 % D

2*

2 , isobar ic constant .ressure +1

% 2

*

3 1 isothermal constant tem.erature+T

1 % T

2*

4 1InI J .olytro.ic ?none?

5 J isentropi c constant entro.y +61 % 6

2*

$ccording to a law of constant=nV

2

2

1

1

.

.

=

2

2

1

1

. V

. V =

2211 V V =

1

2

1

1

2

2

1−

=

= n

nn

.

.

V

V



20/59



Darious forms of wor are e;.ressed as follows

Process "oundary #or$

isochoric

isobar ic

isothermal

.olytro.ic

isentropi c

,*+ 1212 =−= V V W

*+ 1212 V V W −=

1

2

1112 ln

V

V V W =

nV V W

−−=

1

112212



21/59



E0a(le 234

6etch a ?D diagram showing the following .rocesses in a cycle

Process 1-2> iso"aric wor out.ut of 1,!5 from an initial volume of ,!,2/

m3 and .ressure 1!4 "ar

Process 2-3> isothermal com.ression and

Process 3-1> isochoric heat transfer to its original volume of ,!,2/ m3

and .ressure 1!4 "ar!

Calculate +a* the ma;imum volume in the cycle in m3 +"* the isothermal wor

in +c* the net wor in and +d* the heat transfer during iso"aric e;.ansion

in !



22/59



Solution:

rocess "y .rocess analysis

( )

( )

( )

12 2 1

2

3

2

1,!5

14, ,!,2/ 1,!5

1

,!1 3

2

,

W V V

V

+ection isobari

V m

c

= − =

=

=

−

−

The isothermal wor

( )

( )

( )

2 2 3 3

3

323 2 2

2

,!1,314, 515

,!,2/

ln

,!,2/14, ,!1,3 ln

,!1,3

1/!-/

2 3

V V

"a

V W V

+ection isothermal

V

"#

=

= = ÷

→ =

= ÷

−

−

=



23/59



The net wor

( )31

12 23 31

,

1,!5 1/!-/

/!2

1

/

3

net

W

W W

+ection isochoric

W W

"#

=∴ = + +

= −

= −

−



24/59



E0a(le 235

$ fluid at 4!15 "ar is e;.anded reversi"ly according to a law D % constant to

a .ressure of 1!15 "ar until it has a s.ecific volume of ,!12 m3#g! (t is then

cooled reversi"ly at a constant .ressure then is cooled at constant volume

until the .ressure is ,!'2 "arK and is then allowed to com.ress reversi"ly

according to a law V n % constant "ac to the initial conditions! The wor

done in the constant .ressure is ,!525 and the mass of fluid .resent is ,!22

g! Calculate the value of n in the fourth .rocess the net wor of the cycle and

setch the cycle on a ?D diagram!



25/59



Solution:

rocess "y .rocess analysis

( )

( )

( )

1 1 2 2

1

3

212 1 1

1

115,!22 ,!12

415

,!,,-32

ln

,!,2'4415 ,!,,-32 l

1

n,!,,-32

3!/ 5

2

+ection isothe

V V

V

m

V W V

V

"

rm l

#

a

=

= ÷

=

=

=

=

−



26/59



( )

( )23 3 2

3

3

,!525

,!525 ,!,2'4115

,!,3

2 3

,-

+ec

W V V

tion iso

"#

V

m

baric

=

=

=

−

− =

+

( )

34

3

,

4+ection iso

W

choric

=

−

( )

( ) ( )

4 1

1 4

1 1 4 441

'2 ,!,,-32

415 ,!,3,-

ln ,!144 ln ,!23'4

1!31/2

1

415 ,!,,-2 '2 ,!,3,-

1 1!31/2

3! 2

1

51 4

4

n

n

V

V

n

n V V

W n

"

+ection olytroic

#

=

÷ = ÷

=

=−=

−−

=−

= −

−

The net wor of the cycle

12 23 34 41

,!,-'

net W W W W W

"#

= + + +=



27/59



&ule(entary -ro6le(s

1! $ mass of ,!15 g of air is initially e;ists at 2 Ma and 35,oC! The air

is first e;.anded isothermally to 5,, a then com.ressed

.olytro.ically with a .olytro.ic e;.onent of 1!2 to the initial state!

etermine the "oundary wor for each .rocess and the net wor of the

cycle!

2! ,!,-/ g of a car"on mono;ide initially e;ists at 13, a and 12,oC! The

gas is then e;.anded .olytro.ically to a state of 1,, a and 1,, oC!

6etch the ?D diagram for this .rocess! $lso determine the value of n

+inde;* and the "oundary wor done during this .rocess! A1289,19 " B



28/59

acu y o ec a ca g ee g, U


3! Two g of air e;.eriences the three?

.rocess cycle shown in Fig! 3?14!

Calculate the net wor!

4! $ system contains ,!15 m3 of air .ressure of 3!/ "ars and 15, C! (t is⁰

e;.anded adia"atically till the .ressure falls to 1!, "ar! The air is then

heated at a constant .ressure till its enthal.y increases "y -, !

6etch the .rocess on a ?D diagram and determine the total wor

done!

se c .%1!,,5 #g!7 and cv%,!-14 #g!7



29/59

y g g,


FIRS! LA( OF !HERMOD)NAMI*S

MASS & ENERGY ANALYSISOF CONTROL VOLUME



30/59

y g g,


Conser7ation of Mass

Conservation of mass is one of the most fundamental

.rinci.les in nature! e are all familiar with this

.rinci.le and it is not difficult to understand it=

For closed system the conservation of mass .rinci.le is

im.licitly used since the mass of the system remain

constant during a .rocess!

owever for control volume mass can cross the

"oundaries! 6o the amount of mass entering and leavingthe control volume must "e considered!



31/59

y g g


Mass an) 8olu(e Flo'ates

Mass flow through a cross?sectional area .er unit time is called themass flow rate! ote the dot over the mass sym"ol indicates a time

rate of change! (t is e;.ressed as

∫ = d)V m ! ρ

(f the fluid density and velocity are constant over the flow cross?

sectional area the mass flow rate is

'oulme s(ecificcalled is

%here

)V )V m

ν

ρ ν

ν

ρ

1=

==



32/59


-rincial of Conser7ation of Mass

The conservation of mass .rinci.le for a control volume can "e

e;.ressed as

in out -V

m m m− =9 9 9

For a steady state steady flow .rocess the conservation of mass

.rinci.le "ecomes

+g#s*in out m m=9 9



33/59


$s the fluid u.stream .ushes mass across the control volume wor

done on that unit of mass is

flo%

flo%

flo%

)W F d* F d* dV ' m

)W

% 'm

δ δ

δ δ

δ

= = = =

= =

Flo' +or 9 The Energy of a Flo'ingFlui)



34/59


The total energy carried "y a unit of mass as it crosses the control

surface is the sum of the internal energy & flow wor & .otential

energy & inetic energy

∑ ++=+++= g$ V

h g$ V

uenergy 22

22

ν

The first law for a control volume can "e written as

∑∑

++−

++=−

in

inin

inin

out

out out

out out net net g$ V

hm g$ V

hmW Q22

2!2!!!

Total Energy of a Flo'ing Flui)



35/59


Total Energy of a Flo'ing Flui)

The steady state steady flow conservation of mass and first law of

thermodynamics can "e e;.ressed in the following forms

*+1,,,

*+

2,,,

12

2

1

2

212

!!!

"W $ $ g V V

hhmW Q net net

−+

−+−=−

*+1,,,

*+

2,,,

12

2

1

2

212 "#

$ $ g V V hhmW Q net net

−+−+−=−

*#+

1,,,

*+

2,,,

12

2

1

2

212 "g "#

$ $ g V V hh%& net net

−+−+−=−



36/59


&tea)y;o' Engineering "e7ices



37/59


No


38/59


Energy "alance +noGGle O diffuser*>

∑∑

++++=

++++

out

out out

out out out out

in

inin

inininin g$V

hmW Q g$V

hmW Q22

2!!!

2!!!

+=

+

22

2!2!out

out out in

inin

V hm

V hm

+=

+22

2

22

2

11

V hV h



39/59


E0a(le 23>

6team at ,!4 Ma 3,,PC

enters an adia"atic noGGle witha low velocity and leaves at ,!2

Ma with a )uality of ,Q!

Find the e;it velocity!

Solution:

6im.lified energy "alance>

2

1

1

V

h +

( )

2

22

11

1

2 22

22

2 2

1

3,'-!1,!4

3,, su.2

,!2

24/'!1,!

"# "g

o

f fg

"# "g

V

h

+tate

h Ma

. - erheated +tate

h h , h Ma

h ,

÷ = + ÷ ÷

==

=

= += == 1 2

1 2

1

,!4 ,!2

3,, ,!

1 2

,

o

Ma Ma

. - ,

+tate + te

V

ta

= =

= =5

E;it velocity>

( )2 2,,, 3,'-!1 24/'!1

1,-/ #

V

m s

= −

=



40/59


E0a(le 23?

$ir at 1,:C and /, a enters the

diffuser of a Net engine steadilywith a velocity of 2,, m#s! The

inlet area of the diffuser is ,!4 m2!

The air leaves the diffuser with a

velocity that is very small

com.ared with the inlet velocity!

etermine +a* the mass flow rate

of the air and +"* the tem.erature

of the air leaving the diffuser!

1 2

1

1

2

1

/, ,

1

1

,

2,, #

,!

2

4

o

"a V

. -

V m s

+t

)

ate +tate

m

=

==

=

5

Solution:

2221

1 22

V V h h

+ = + ÷

,

2

÷ ÷

6im.lified energy "alance>

From (deal Ras Law>

311

1

1!,15 m"g

:. '

= =



41/59


Mass flow rate

( ) ( )

1 1

1

1

12,, ,!4

1!,15

-/!/"g

s

m V )'=

= ÷

=

9

Enthal.y at state 1

( )1 1 1!,,5 2/3

2/4!42

(

"# "g

h - . = =

=

From energy "alance>

2

1

2 1

2

22

2,,,

2,,2/4!42

2,,,

3,4!42

3,4!42

1!,,5

3,2!

"# "g

(

V

h h

h. -

K

= +

= +

=

=

=

=



42/59


Tur6ine 9 Co(ressor

Tur"ine S a wor .roducing device through the e;.ansion of a

fluid!

Com.ressor +as well as .um. and fan* ? device used to increase

.ressure of a fluid and involves wor in.ut!

8 % , +well insulated* 9E % , 97E % , +very small com.are

to 9enthal.y*!



43/59


Energy "alance> for tur"ine

∑∑

++++=

++++

out

out out

out out out out

in

inin

inininin g$

V hmW Q g$

V hmW Q

22

2!!!

2!!!

( ) ( )out out out inin hmW hm!!!

+=

( )21!!

hhmW out −=



44/59


Energy "alance> for com.ressor .um. and fan

∑∑

++++=

++++

out

out out

out out out out

in

inin

inininin g$

V hmW Q g$

V hmW Q

22

2!!!

2!!!

( ) ( )out out ininin hmhmW !!!

=+

( )12!!

hhmW in −=



45/59


E0a(le 23@

The .ower out.ut of an adia"atic steam tur"ine is 5 M! Com.arethe magnitudes of 9h 9e and 9.e! Then determine the wor done

.er unit mass of the steam flowing through the tur"ine and calculate

the mass flow rate of the steam!

ata > (nlet + % 2 Ma T % 4,,PCv % 5, m#s G % 1, m*

E;it + % 15 a ; % ,Q v % 1/, m#s G % 'm*



46/59


Solution:

( )

1

11

2

2

2 2 2 2

1

su.2

324-!'4,,

2

15

!,!

225!4 ,! 23-3!1

23'1!-3

o "# "g

f fg

"# "g

+tate

erheated ( Ma

h. -

+tate

"a

sat mi,ture ,

h h , h

= ==

= =

= +

= +

=

!

inQ

!

inW +

2!

2! ! !

2

2

inin in in

in

out out out out out out

out

V m h g$

V Q W m h g$

+ + + = ÷

+ + + + ÷

∑

∑

From energy "alance>

( )

2 1

2 2

2 1

2 1

//5!/-

14!5

2,,,

,!,41,,,

"# "g

"# "g

"# "g

h h h

V V KE

g $ $ E

∆ = − = −

−∆ = =

−∆ = = −

6olve the e)uation>

324/!4

+23-2!3*

23'1!,1

?//-!3



47/59


the wor done .er unit mass

( ) ( )2 2

1 21 21 2

2,,, 1,,,

//5!/- 14!5 ,!,4

/-,!'

out

"# "g

g $ $ V V W h h − −= − + + ÷ ÷

= − +=

The mass flow rate

5,,,5!-4

/-,!'

"g out

s

out

W m

W

= = =9

9

//-!3

/-2!4/

/-2!4/

5!-3



48/59


E0a(le 23A

$ir at 1,, a and 2/, 7 is

com.ressed steadily to ',,a and 4,, 7! The mass

flow rate of the air is ,!,2

g#s and a heat loss of 1'

#g occurs during the .rocess! $ssuming the

changes in inetic and

.otential energies are

negligi"le determine thenecessary .ower in.ut to the

com.ressor!

Solution:

sim.lified energy "alance>

( )( )

2 1

2 1

in out

out

W m h h Q

m h h m&= − += − +

99 9

9 9

1

11

2

22

1

1,,

2/,!132/,

2

',,

4,,!/4,,

"# "g

"# "g

+tate

air "a

h. K

+tate

air "a

h. K

= ==

= ==



49/59


Thus

( ),!,2 4,,!/ 2/,!13 1'2!-4

inW

"W = − +

=9



50/59


Throttling 8al7e

Flow?restricting devices that

cause a significant .ressure dro.

in the fluid!

6ome familiar e;am.les are

ordinary adNusta"le valves and

ca.illary tu"es!



51/59


6team enters a throttling valve at

/,,, a and 3,,:C and leavesat a .ressure of 1',, a!

etermine the final tem.erature

and s.ecific volume of the

steam!

E0a(le 231/

( ) ( )

1

11

2

2 1

2 2 2 2 2

1

su./,,,

2-/'!53,,

2

1',,int

15,, 1/!2 ,!,,1154 ,!131-1, /44!55 2-1

1',,

1-5, 2,5!-2 ,!,,11'' ,!11344, /-/!1' 2-5!2

o "# "g

o f g f g

f g f g

+tate

erheated "a

h. -

+tate

"ama"e er(olation

h h

' ' h h "a . -

. ' ' h h

= ==

= =



52/59


2 2,1!3o

sat . . - = =

$t state 2 the region is sat!

mi;ture

Retting the )uality at state 2

2 2

2

2 2

2-/'!5 /5-!4

2-2!'/ /5-!4

,!-

f

g f

h h ,h h

−=−

−=

−=

6.ecific volume at state 2

( )3

2 2 2 2

,!,,115//

,!- ,!1244,2 ,!,,115//

,!124,

f fg

m

"g

' ' , '= += +

−

=



53/59

53


The section where the mi;ing .rocess

taes .lace!

$n ordinary T?el"ow or a ?el"ow in

a shower for e;am.le serves as the

mi;ing cham"er for the cold? and

hot?water streams!

Mi0ing Cha(6er



54/59

54


Mi0ing Cha(6er

Energy @alance>

( )

( ) ( )

1 1 2 2 3 3

1 1 3 1 2 3 3

1 1 2 3 3 2

3 21 3

1 2

m h m h m h

m h m m h m h

m h h m h h

h hm m

h h

+ =

+ − =

− = − −= ÷−

9 9 9

9 9 9 9

9 9

9 9



55/59

55


evices where two moving fluid

streams e;change heat without

mi;ing!

eat e;changers ty.ically involve

no wor interactions +w % ,* and

negligi"le inetic and .otentialenergy changes for each fluid

stream!

Heat E0changer



56/59

5'


Li)uid sodium flowing at 1,,

g#s enters a heat e;changer at

45,:C and e;its at 35,:C! The

s.ecific heat of sodium is 1!25

#g!oC! ater enters at 5,,,

a and 2,oC! etermine the

minimum mass flu; of the waterso that the water does not

com.letely va.oriGe! eglect the

.ressure dro. through the

e;changer! $lso calculate the

rate of heat transfer!

E0a(le 2311 Solution:

sim.lified energy "alance>

( ) ( )

( ) ( )

1 1 2 2

1 2 2 1

1 2 2 1

s s % % s s % %

s s s % % %

s ( s s s % % %

m h m h m h m h

m h h m h h

m - . . m h h

+ = +− = −

− = −

9 9 9 9

9 9

9 9

1

11

2

2

1>

!5,,,

//!'12,

2 >

5,,,

2-4!2

o "# % "g

"# % "g

+tate %ater com( li&uid "a

h. -

+tate %ater

"a

h

= ==

==

Assume a sat.vapor state toobtain the max.allowable exitingenthalpy.



57/59

5-


the minimum mass flu; of the water

so that the water does not

com.letely va.oriGe

( )

( )

( ) ( )

1 2

2 1

1,, 1!25 45, 35,

2-4!2 //!'1

4!'2

s ( s s s

%

% %

"g

s

m - . . m

h h

−=

−

−= −=

9

the rate of heat transfer

( )( )

2 1

4!'2 2-4!2 //!'1

12!5

% % % %Q m h h

MW

= −= −

=

9 9



58/59

5/



1! $ir flows through the su.ersonic noGGle ! The inlet conditions are - a

and 42,:C! The noGGle e;it diameter is adNusted such that the e;itingvelocity is -,, m#s! Calculate + a * the e;it tem.erature + " *the mass flu;

and + c * the e;it diameter! $ssume an adia"atic )uasie)uili"rium flow!

2! 6team at 5 Ma and 4,,:C enters a noGGle steadily velocity of /, m#s

and it leaves at 2 Ma and 3,,:C! The inlet area of the noGGle is 5, cm 2and heat is "eing lost at a rate of 12, #s! etermine +a* the mass flow

rate of the steam +"* the e;it velocity of the steam and +c* the e;it area

noGGle!

3! 6team enters a tur"ine at 4,,, a and 5,,o

C and leaves as shown in Fig$ "elow! For an inlet velocity of 2,, m#s calculate the tur"ine .ower

out.ut! + a *eglect any heat transfer and inetic energy change + " *6how

that the inetic energy change is negligi"le!



59/59

5

Figure $

4! Consider an ordinary shower where hot water at ',:C is mi;ed with cold

water at 1,:C! (f it is desired that a steady stream of warm water at 45:C "e su..lied determine the ratio of the mass flow rates of the hot to cold

water! $ssume the heat losses from the mi;ing cham"er to "e negligi"le

and the mi;ing to tae .lace at a .ressure of 15, a!

5! 0efrigerant?134a is to "e cooled "y water in a condenser! The refrigerant

enters the condenser with a mass flow rate of ' g#min at 1 Ma and -,PC

and leaves at 35:C! The cooling water enters at 3,, a and 15:C and

leaves at 25PC! eglecting any .ressure dro.s determine +a* the mass

flow rate of the cooling water re)uired and +"* the heat transfer rate from

the refrigerant to water!

Documents

Chapter 3 First Law Thermodynamics