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8/9/2019 Chapter 3 First Law Thermodynamics
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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
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MEC 451 – THEM!"#N$M%C&
FIRS! LA( OF !HERMOD)NAMI*S
ENERGY ANALYSISOF CLOSED SYSTEM
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First La' of Ther(o)yna(ics
MEC 451 – THEM!"#N$M%C&
Faculty of Mechanical Engineering, UiTM
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!
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MEC 451 – THEM!"#N$M%C&
Energy *alance for Close) &yste(
Heat
+or
z
Close)&yste(
eference -lane, z . /
V
or
E E E in out system− = ∆
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MEC 451 – THEM!"#N$M%C&
$ccording to classical thermodynamics
Q W E net net system− = ∆
The total energy of the system E system is given as
E Internal energy Kinetic energy otential energy
E ! KE E
% & &
% & &
The change in stored energy for the system is
∆ ∆ ∆ ∆ E ! KE E = + +
The first law of thermodynamics for closed systems then can "e
written as
Q W ! KE E net net − = + +∆ ∆ ∆
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MEC 451 – THEM!"#N$M%C&
(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
$ $ g V V uumW Q net net
−+−+−=1,,,
*+
2,,,
12
2
1
2
212
$ $ g V V uumQ net
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Faculty of Mechanical Engineering, UiTM
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MEC 451 – THEM!"#N$M%C&
For a constant .ressure .rocess
−+−+−=−
1,,,
*+
2,,,
122
12
212
$ $ g V V uumW Q net net
−
+
−
+−=−− 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
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/MEC 451 – THEM!"#N$M%C&
0igid tan iston cylinder
Example of Closed Systems
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Faculty of Mechanical Engineering, UiTM
MEC 451 – THEM!"#N$M%C&
$ 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
$ $ g V V uumW Q net net
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
− −− = − + + ÷
− −
− = − + + ÷
−− − = ∆ + + ÷ ÷ ÷ ∆ =
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1,MEC 451 – THEM!"#N$M%C&
6team at 11,, a and 2 .ercent
)uality is heated in a rigid container
until the .ressure is 2,,, a! For a
mass of ,!,5 g calculate the amount
of heat su..ly +in * and the total
entro.y change +in #g!7*!
E0a(le 23 Solution:
( )
( )
( )
3
1 1
1 1 1 1
1 1 1 1
1 1 1 1
!
1
11,, ,!2
,!,,113 ,!2 ,!1--53 ,!,,1133
,!1'34
-/,!, ,!2 1/,'!3
2441!
2!1-2 ,!2 4!3-44
'!2,4
f fg
m"g
f fg
"*"g
f fg
"# "g K
+tate
at "a ,
' ' , '
u u , u
s s , s
= == +
= + −
=
= +
= +
=
= +
= +
=
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11MEC 451 – THEM!"#N$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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12MEC 451 – THEM!"#N$M%C&
( )
( )
2 1
,!,5 3,3,!42 2441!
2!43
Q m u u
"#
= −
= −=
$mount of heat su..lied 8
The change in entro.y 9s
2 1
!
-!2-, '!2,4
1!,-5 "# "g K
s s s∆ = −= −=
F l f M h i l E i i UiTM
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13MEC 451 – THEM!"#N$M%C&
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
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14MEC 451 – THEM!"#N$M%C&
( )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!//
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&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!$lso show the .rocess on a T?D diagram with res.ect to saturation
lines!
A 4211o- B
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1'MEC 451 – THEM!"#N$M%C&
&ule(entary -ro6le(s 1
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! $n insulated tan is divided into two .arts "y a .artition! ne .art of
the tan contains ' g of an ideal gas at 5,:C and /,, a while the
other .art is evacuated! The .artition is now removed and the gas
e;.ands to fill the entire tan! etermine the final tem.erature and the
.ressure in the tan!
A07-, 800 "aB
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1-MEC 451 – THEM!"#N$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
− =
=
∆
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1/MEC 451 – THEM!"#N$M%C&
*oun)ary +ors
2
4
5
1
P
V
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1MEC 451 – THEM!"#N$M%C&
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*
$ccording to a law of constant=nV
2
2
1
1
.
.
=
2
2
1
1
. V
. V =
2211 V V =
1
2
1
1
2
2
1−
=
= n
nn
.
.
V
V
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2,MEC 451 – THEM!"#N$M%C&
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
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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 !
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22MEC 451 – THEM!"#N$M%C&
Solution:
rocess "y .rocess analysis
( )
( )
( )
12 2 1
2
3
2
1,!5
14, ,!,2/ 1,!5
1
,!1 3
2
,
W V V
V
+ection isobari
V m
c
= − =
=
=
−
−
The isothermal wor
( )
( )
( )
2 2 3 3
3
323 2 2
2
,!1,314, 515
,!,2/
ln
,!,2/14, ,!1,3 ln
,!1,3
1/!-/
2 3
V V
"a
V W V
+ection isothermal
V
"#
=
= = ÷
→ =
= ÷
−
−
=
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23MEC 451 – THEM!"#N$M%C&
The net wor
( )31
12 23 31
,
1,!5 1/!-/
/!2
1
/
3
net
W
W W
+ection isochoric
W W
"#
=∴ = + +
= −
= −
−
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Faculty of Mechanical Engineering, UiTM
24MEC 451 – THEM!"#N$M%C&
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!
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Faculty of Mechanical Engineering, UiTM
25MEC 451 – THEM!"#N$M%C&
Solution:
rocess "y .rocess analysis
( )
( )
( )
1 1 2 2
1
3
212 1 1
1
115,!22 ,!12
415
,!,,-32
ln
,!,2'4415 ,!,,-32 l
1
n,!,,-32
3!/ 5
2
+ection isothe
V V
V
m
V W V
V
"
rm l
#
a
=
= ÷
=
=
=
=
−
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2'MEC 451 – THEM!"#N$M%C&
( )
( )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
"#
= + + +=
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2-MEC 451 – THEM!"#N$M%C&
&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! $lso determine the value of n
+inde;* and the "oundary wor done during this .rocess! A1289,19 " B
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acu y o ec a ca g ee g, U
2/MEC 451 – THEM!"#N$M%C&
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
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y g g,
2MEC 451 – THEM!"#N$M%C&
FIRS! LA( OF !HERMOD)NAMI*S
MASS & ENERGY ANALYSISOF CONTROL VOLUME
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y g g,
3,MEC 451 – THEM!"#N$M%C&
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!
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y g g
31MEC 451 – THEM!"#N$M%C&
Mass an) 8olu(e Flo'ates
Mass flow through a cross?sectional area .er unit time is called themass flow rate! ote the dot over the mass sym"ol indicates a time
rate of change! (t is e;.ressed as
∫ = d)V m ! ρ
(f the fluid density and velocity are constant over the flow cross?
sectional area the mass flow rate is
'oulme s(ecificcalled is
%here
)V )V m
ν
ρ ν
ν
ρ
1=
==
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32MEC 451 – THEM!"#N$M%C&
-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
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$s the fluid u.stream .ushes mass across the control volume wor
done on that unit of mass is
flo%
flo%
flo%
)W F d* F d* dV ' m
)W
% 'm
δ δ
δ δ
δ
= = = =
= =
Flo' +or 9 The Energy of a Flo'ingFlui)
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34MEC 451 – THEM!"#N$M%C&
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)
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35MEC 451 – THEM!"#N$M%C&
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
−+−+−=−
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&tea)y;o' Engineering "e7ices
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No
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3/MEC 451 – THEM!"#N$M%C&
Energy "alance +noGGle O diffuser*>
∑∑
++++=
++++
out
out out
out out out out
in
inin
inininin g$V
hmW Q g$V
hmW Q22
2!!!
2!!!
+=
+
22
2!2!out
out out in
inin
V hm
V hm
+=
+22
2
22
2
11
V hV h
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3MEC 451 – THEM!"#N$M%C&
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
= −
=
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4,MEC 451 – THEM!"#N$M%C&
E0a(le 23?
$ir at 1,:C and /, a enters the
diffuser of a Net engine steadilywith a velocity of 2,, m#s! The
inlet area of the diffuser is ,!4 m2!
The air leaves the diffuser with a
velocity that is very small
com.ared with the inlet velocity!
etermine +a* the mass flow rate
of the air and +"* the tem.erature
of the air leaving the diffuser!
1 2
1
1
2
1
/, ,
1
1
,
2,, #
,!
2
4
o
"a V
. -
V m s
+t
)
ate +tate
m
=
==
=
5
Solution:
2221
1 22
V V h h
+ = + ÷
,
2
÷ ÷
6im.lified energy "alance>
From (deal Ras Law>
311
1
1!,15 m"g
:. '
= =
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41MEC 451 – THEM!"#N$M%C&
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
= +
= +
=
=
=
=
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42MEC 451 – THEM!"#N$M%C&
Tur6ine 9 Co(ressor
Tur"ine S a wor .roducing device through the e;.ansion of a
fluid!
Com.ressor +as well as .um. and fan* ? device used to increase
.ressure of a fluid and involves wor in.ut!
8 % , +well insulated* 9E % , 97E % , +very small com.are
to 9enthal.y*!
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43MEC 451 – THEM!"#N$M%C&
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 −=
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44MEC 451 – THEM!"#N$M%C&
Energy "alance> for com.ressor .um. and fan
∑∑
++++=
++++
out
out out
out out out out
in
inin
inininin g$
V hmW Q g$
V hmW Q
22
2!!!
2!!!
( ) ( )out out ininin hmhmW !!!
=+
( )12!!
hhmW in −=
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45MEC 451 – THEM!"#N$M%C&
E0a(le 23@
The .ower out.ut of an adia"atic steam tur"ine is 5 M! Com.arethe magnitudes of 9h 9e and 9.e! Then determine the wor done
.er unit mass of the steam flowing through the tur"ine and calculate
the mass flow rate of the steam!
ata > (nlet + % 2 Ma T % 4,,PCv % 5, m#s G % 1, m*
E;it + % 15 a ; % ,Q v % 1/, m#s G % 'm*
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4'MEC 451 – THEM!"#N$M%C&
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
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4-MEC 451 – THEM!"#N$M%C&
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
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4/MEC 451 – THEM!"#N$M%C&
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! $ssuming the
changes in inetic and
.otential energies are
negligi"le determine thenecessary .ower in.ut to the
com.ressor!
Solution:
sim.lified energy "alance>
( )( )
2 1
2 1
in out
out
W m h h Q
m h h m&= − += − +
99 9
9 9
1
11
2
22
1
1,,
2/,!132/,
2
',,
4,,!/4,,
"# "g
"# "g
+tate
air "a
h. K
+tate
air "a
h. K
= ==
= ==
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4MEC 451 – THEM!"#N$M%C&
Thus
( ),!,2 4,,!/ 2/,!13 1'2!-4
inW
"W = − +
=9
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5,MEC 451 – THEM!"#N$M%C&
Throttling 8al7e
Flow?restricting devices that
cause a significant .ressure dro.
in the fluid!
6ome familiar e;am.les are
ordinary adNusta"le valves and
ca.illary tu"es!
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51MEC 451 – THEM!"#N$M%C&
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
= ==
= =
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52MEC 451 – THEM!"#N$M%C&
2 2,1!3o
sat . . - = =
$t state 2 the region is sat!
mi;ture
Retting the )uality at state 2
2 2
2
2 2
2-/'!5 /5-!4
2-2!'/ /5-!4
,!-
f
g f
h h ,h h
−=−
−=
−=
6.ecific volume at state 2
( )3
2 2 2 2
,!,,115//
,!- ,!1244,2 ,!,,115//
,!124,
f fg
m
"g
' ' , '= += +
−
=
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53
MEC 451 – THEM!"#N$M%C&
The section where the mi;ing .rocess
taes .lace!
$n ordinary T?el"ow or a ?el"ow in
a shower for e;am.le serves as the
mi;ing cham"er for the cold? and
hot?water streams!
Mi0ing Cha(6er
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54
MEC 451 – THEM!"#N$M%C&
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
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55
MEC 451 – THEM!"#N$M%C&
evices where two moving fluid
streams e;change heat without
mi;ing!
eat e;changers ty.ically involve
no wor interactions +w % ,* and
negligi"le inetic and .otentialenergy changes for each fluid
stream!
Heat E0changer
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5'
MEC 451 – THEM!"#N$M%C&
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! $lso calculate the
rate of heat transfer!
E0a(le 2311 Solution:
sim.lified energy "alance>
( ) ( )
( ) ( )
1 1 2 2
1 2 2 1
1 2 2 1
s s % % s s % %
s s s % % %
s ( s s s % % %
m h m h m h m h
m h h m h h
m - . . m h h
+ = +− = −
− = −
9 9 9 9
9 9
9 9
1
11
2
2
1>
!5,,,
//!'12,
2 >
5,,,
2-4!2
o "# % "g
"# % "g
+tate %ater com( li&uid "a
h. -
+tate %ater
"a
h
= ==
==
Assume a sat.vapor state toobtain the max.allowable exitingenthalpy.
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5-
MEC 451 – THEM!"#N$M%C&
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
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5/
MEC 451 – THEM!"#N$M%C&
&ule(entary -ro6le(s 2
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!
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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!