3.1 Systems (体系 ) versus Control Volumes (控制体 )
System,an arbitrary quantity of mass of fixed
identity,Everything external to this system is denoted by
the term surroundings,and the system is separated from
its surroundings by it‘s boundaries through which no mass
across,(Lagrange 拉格朗日 )
Chapter 3 Integral Relations（ 积分关系式 ）
for a Control Volume in One-dimensional Steady Flows
Control Volume (CV),In the neighborhood of our
product the fluid forms the environment whose effect on
our product we wish to know,This specific region is called
control volume,with open boundaries through which mass,
momentum and energy are allowed to across,(Euler 欧拉 )
Fixed CV,moving CV,deforming CV
3.2 Basic Physical Laws of Fluid Mechanics
All the laws of mechanics are written for a system,which
state what happens when there is an interaction between
the system and it’s surroundings,
If m is the mass of the system
Conservation of
mass(质量守恒 ) ? 0dmm c o n s t o r
dt??
Newton’s
second law ? F m a?v v dVm
dt?
r
()d mVdt? r
Angular
momentum ? dHM
dt?
vv
()H r V m? ? ?? rv v
First law of
thermodynamic ? d Q d W d E
d t d t d t??
It is rare that we wish to follow the ultimate
path of a specific particle of fluid,Instead it is
likely that the fluid forms the environment
whose effect on our product we wish to know,
such as how an airplane is affected by the
surrounding air,how a ship is affected by the
surrounding water,This requires that the basic
laws be rewritten to apply to a specific region in
the neighbored of our product namely a control
volume ( CV),The boundary of the CV is called control
surface(CS)
?Basic Laws for system for CV
3.3 The Reynolds Transport Theorem (RTT)
雷诺输运定理
?
1122 is CV,
1*1*2*2* is system
which occupies the
CV at instant t.
d
dm?
??,The amount of per unit mass ?
CV cv d??? ? cv dm?? ?
The total amount of in the CV is, ?
t+dtt+dt tt
s
?, any property of fluid (,,,)m mV H Err
1 [ ( ) ( ) ]
C V C Vt d t tdt? ? ???
11[ ( ) ( ) ( ) ] ( )
o u t i n ss t d t d d td t d t? ? ? ? ? ? ? ??
11[ ( ) ( ) ] [ ( ) ( ) ]
s o u t i ns t d t t d dd t d t? ? ? ? ? ? ? ??
1 [ ( ) ( ) ]s
o u t i n
d dd
d t d t ?
?? ? ? ?
1 [ ( ) ( ) ]s c v
o u t i n
dd dd
d t d t d t
?? ? ? ? ? ?
()CVddt ??
t+dtt+dt tt
s
( ) ( ) ( ) ( )in in in ind d m A d s A V d t? ? ? ? ?? ? ? ?
In the like manner
( ) ( )o u t o u td A V d t????
1 [ ( ) ( ) ]s c v
o u t i n
dd dd
d t d t d t
?? ? ? ? ? ?
[ ( ) ( ) ]cv o u t i nd A V A Vdt ? ? ? ??? ? ?
s
1-D flow, is only the function of s, ()s? ? ??
For steady
flow, 0
cvd
dt
? ? ( ) ( )s
o u t ind A V A Vdt ? ? ? ?? ???
t+dtt+dt tt
ds
R T T
If there are several one-D inlets and outlets,
( ) ( )s outi i i i i i i i in
ii
d A V A V
dt ? ? ? ?
? ????
Steady,1-D only in inlets and outlets,no matter
how the flow is within the CV,
3.3 Conservation of mass (质量守恒 )
(Continuity Equation)
f=m ??dm/dm=1
( ) ( ) 0s o u t i niiiiii
ii
dm VVAA
dt ??? ? ???
( ) ( )o u t i niiii ii
ii
A VV A? ???? ( ) ( )i i n i o u t
ii
mm???&&
Mass flux (质量流量 )m&
For incompressible flow:
( ) ( )o u t i niiii
ii
VVAA??? i ii V o l u m e f l u xQ VA?
体积流量
-------Leonardo da Vinci in 15001212VVAA?
If only one inlet and one outlet
壶口瀑布是我国著名的第二大瀑布。两百多米宽的黄河河面，突然紧缩
为 50米左右，跌入 30多米的壶形峡谷。入壶之水，奔腾咆哮，势如奔马，浪
声震天，声闻十里。, 黄河之水天上来, 之惊心动魄的景观。
Example:
A jet engine working at design condition,At the inlet of the nozzle
At the outlet
Please find the mass flux and velocity at the outlet.
Given gas constant
521 2.05 10 /p N m?? T1 =865K,V1=288 m/s,A1=0.19㎡ ；
522 1,14 3 10 /p N m?? T2 =766K,A2=0.1538㎡
R=287.4 J/kg.K。
Solution
1 1 1
1
45.1 /p A V k g sRT??m A V??& p AVRT?
1 1 1 2 2 2
12
p A V p A V
R T R T?
According to the conservation of mass
1 1 1 2 2 2m A V A V????& ?
1 1 2
21
2 2 1
565,1 /A p TV V m sA p T???
Homework,P185 P3.12,P189P3.36
mV?? r ()l i n e a r m o m e n t u m?
dm V V
dm? ??
r v m o m e n t u m p e r u n i t m a s s
() ( ) ( )s
o u t iniiiiiiii
ii
d m V VVAAVV
dt ??????
r vv
3.4 The Linear Momentum Equation (动量方程 )
（ Newton’s Second Law ）
( ) ( )o u t inii
ii
mmVV????vv&&
Newton’s
second law () sd m VF dt??
rr ( ) ( )
o u t iniiii
ii
mmVV????vv&&
F?r,Net force on the system or CV (体系或控制体受到的合外力 )
iimVv&
,Momentum flux (动量流量 )
( ) ( )s o u t iniiiiii
ii
d VVAA
dt ????
? ????
1-D in & out
steady RTT
? flux
inoutm m m??& & &
For only one inlet and one outlet
According to continuity
)
() (s
o u t in
d m VF m V V
dt? ? ??
rr vv
&
2－ out,1－ in
Example,A fixed control volume of a streamtube in steady
flow has a uniform inlet (?1,A1,V1 )and a uniform exit
(?2,A2,V2), Find the net force on the control volume.
?
1Vv 1
2Vv2
o x
y
21()x x xF m V V??&
21()y y yF m V V??&
21()z z zF m V V??&?
o x
y
z
21()Fm VV? ? ?
v vv&Solution:
21()x xxm V VF? ? ?& 21( c o s )m VV ???&
21()y yym V VF? ? ?& 1
s i nm V ??? &
121212m VVAA????&
G iven 5 212 4.19 10 Npp m? ? ?
7 8,5 Kgm s?& 121 0,8c m c mdd??
3998 Kg m? ?
Neglect the weight of the fluid,Find the force on the
water by the elbow pipe.
Example:
1 2
1
2
Solution:
x
y
o
select coordinate,control volume
21()Fm VV? ? ?v vv&
21()x xxm VVF
?? ? ?
2m V
?
?
2 22sx mp VFA
???
2 22sx mp VFA
???
22
2
22
2
478.5
998 4
dp
d ??
??? 3696 N?
22 5934sxs s yF F NF? ? ?
1 111 ( 0 )sy mmp VVFA
??? ? ? ? ?
-4642N?1 11sy mp VFA ?? ? ?
In the like manner
1 sy
sx
Ftg
F?
?? 141.47?
o
Find the force to fix the elbow.
Solution,coordinate,CV
Net force on the control volume:
22 2()x L R e xaap p pF A A A A F? ? ? ? ? ?
Where Fex is the force on the CV by pipe,( on elbow)
2 )(LR L exa a ap p p p FAA A? ? ? ? ?
1
2
x
y
o
22 )(x e x appF F A? ? ? ?
I n l i k e m a n n e r
11 )(y e y appF F A? ? ? ?
Fex
Surface force,(1) Forces exposed by cutting though solid bodies
which protrude into the surface.(2)Pressure,viscous stress.
A fixed vane turns a water jet of
area A through an angle ? without
changing its velocity magnitude,
The flow is steady,pressure pa is
everywhere,and friction on the
vane is negligible,Find the force F
applied to vane.
V
?
F
Vv
( c o s 1 )x mVF ???& s i ny mVF ?? &
A water jet of velocity Vj impinges
normal to a flat plate which moves
to the right at velocity Vc,Find the
force required to keep the plate
moving at constant velocity and the
power delivered to the cart if the
jet density is 1000kg/m3
the jet area is 3cm2,and
Vj=20m/s,Vc=15m/s
jV cV
x
Neglect the weight of the jet and plate,and
assume steady flow with respect to the moving
plate with the jet splitting into an equal upward
and downward half-jet.
7.5N?
Home work:
P190-p3.46
P191-p3.50
P192-p3.54
P192-p3.58
Derive the thrust(推力 ) equation for the jet engine,
air drag is neglect
:inlet,au n ifo r m Vp outlet
outer V ap
inner eV ep
Solution:
( ) ( )o u t i nF m V m V? ? ?&&
eem V m V??&& ()
efm m V mV? ? ?& & &
fm&
,mass flux of fuel
o?
o?
e?
e?
e
e
m&
o
o
V ap
V ap
xF? right 'eeeaeppAA??left ''ooaAp
Balance with thrust
e e e eaeFx ppAA?? ?
Fx
eV
ep
em&
V
ap
Fx
)(ee eeeaF x m V m V pp A? ? ? ?&&
[ ) ](ee eeeaR F x m V m V pp A? ? ? ? ? ? ?&&
[ ( ) ) ](e eeeaR m V V pp A? ? ? ? ?&
)( eeaeFx pp A?? eem V m V??&&
'''oo e e eea aF x Ap p pAA? ? ?
F??
Coordinate,CV
emm?&&
0,0 2fmm?&&
oom?& eem?&
oom?& eem ?? &
Example,In a ground test of a jet engine,
pa=1.0133× 105N/m2,Ae=0.1543m2,Pe=1.141× 105N/m2,
Ve=542m/s,.4 3,4 /m K g s?& Find the thrust force.
eV
ep
em&
ap
Solution:
[ ( ) ]
25493
e e a eR mV P P A
N
? ? ? ? ?
??
&
0V ?
F16 R=65.38KN
[ ( ) ) ](e eeeaR m V V pp A? ? ? ? ?&
x
coordinate
A rocket moving straight up,Let
the initial mass be M0,and assume a
steady exhaust mass flow and exhaust
velocity ve relative to the rocket,If
the flow pattern within the rocket
motor is steady and air drag is neglect,
Derive the differential equation
of vertical rocket motion v(t) and
integrate using the initial condition v=0
at t=0,
Example:
ev
,eep A
fm&
()vt
Solution:
The CV enclose the rocket,cuts through
the exit jet,and accelerates upward at
rocket speed v(t).
coordinate
z
v(t)
Z-momentum equation:
( ) ( )o u t i nF m v m v??? &&
ef
dvm g m m v
dt? ? ? ? &
0() fm t M m t?? &f emdv v dt gd tm??
&
0 0 00
v t t
ef
f
dtdv m v g dt
M m t?? ?? ? ?& &
0
( ) l n( 1 )fe
mtv t v gt
M? ? ? ?
&
Am&
Am&
( ) ( )f e A Am v m v m v? ? ? ? ?& & &femv??&
aPA? ()aeP A A?? eePA? mg? dvm
dt?
F??
()e a e dvP P A m g m dt? ? ? ?
aeif P P?
ev
,eep A
fm&
()vt
v(t)
z
A
3.5 The Angular-Momentum Equation
( ) ( )s o u t in
d A V A V
dt ? ? ? ?
? ?? RTT
d rv
dm?
?? ? ?vv
2121()|
s
z
dH m vvrr
dt ? ? ? ?
v
vv&
2121()zMm vvrr? ? ? ? ?
v vv&
F o r t u r b o m a c h i n e s
(,,) c o l u m n c o o r d i n a t e r z?
0zi f M ?? 2121vvrr???
2121()zMm vvrr??? ? ?&
??
r zo
rv
zv
v?
()zH r v m? ? ? ?v vv (Angular-Momentum)
zM?
v, Net moment(合力矩 )
Example:Centrifugal (离心 )pump
The velocity of the fluid is
changed from v1 to v2 and
its pressure from p1 to p2.
Find (a).an expression for
the torque T0 which must be
applied those blades to
maintain this flow,(b).the
power supplied to the pump.
0 2121( ) ( )o u t inmm vvM rr? ? ???vv&&
blade
2vv
2rv
2v?
1vv 1rv
1v?
w
1p
2p
o
121212r o u t rin V m V mm AA?? ? ? ?? &&&
mQ??&For incompressible flow
()rff? 1-D
Continuity,
Solution,The CV is chosen,
0 2121()m vvM rr??? ? ?&
0 2121()m vvT rr????&
blade
2vv
2rv
2v?
1vv 1rv
1v?
w
1p
2p
o
Pressure has no contribution
to the torque
0PTw?
2121()m vvrr?? w??&
2 2 1 1()ttm v V v V????&
are blade rotational speeds
12ttVV
pw
m? & 2 2 1 1tt
v V v V???? Work on per unit mass
Homework,P192-p3.55; P194-p3.68,p3.78 ; P200-
p3.114,p3.116
Brief Review
? Basic Physical Laws of Fluid Mechanics:
0?dtdm )( VmdtdF ?? ?
inoutCVs y s AVAVdt
d
dt
d )()()()( ???? ?????
inout VmVmF )()(
?? ???
inout rVmrVmM )()( ??
?? ???
)( rVmdtdM ??? ?? dtdWdtdQdtdE ??
? The Reynolds Transport Theorem:
? The Linear Momentum Equation:
? The Angular-Momentum Theorem:
? Conservation of Mass:
?? ? outin AVAV )()( ??
Review of Fluid Statics
? Especially,
）zZyYxXz
z
py
y
px
x
pp ddd(dddd ???
?
??
?
??
?
?? ?
Cpzpz ＝?? 00 ???
Question
When fluid
flowing…
Bernoulli(1700~1782)
What relations are there in
velocity,height and pressure?
Several Tragedies in History:
? A little railway
station in 19th Russia.
? The ‘Olimpic’ shipwreck in the Pacific
? The bumping accident of B-52 bomber of
the U.S,air force in 1960s.
3.6 Frictionless Flow:
The Bernoulli Equation
1.Differential Form of Linear Momentum Equation
Elemental fixed streamtube CV of variable area
A(s),and length ds.
ds
Al
?? d?
dAA?
dVV ?
dpp?
2
dpp ?
?
?
p
V
A
??
z
s
Linear momentum relation in the
streamwise direction:
ino u ts VmVmF )()( ?? ???
dVmVVmR i g h t inout
?
?
? ?? )( AVm ???
b o d ys u rs FFL e f tF ????
321 ssss u r FFFF ???
?s in)2/(3 ls AdppF ??
?? s ins in)2/( dAdpp ??
dAdpp )2/( ??
ds
Al
?? d?
dAA?
dVV?
dpp?
2dpp?
?
?
p
V
A
??
z
s
p d Ap d AAd p ????
A dp??
dAdppdAAdpppA )2/())(( ??????
321 ssss u r FFFF ???
?? c o sAd sF b o d y ??
A dz???
AV d VAd pAd z ?? ??? 0??? Ad pAd zAV d V ??
0??? V d Vgdzdp?
one-D,steady,frictionless flow
ds
Al
?? d?
dAA?
dVV?
dpp?
2dpp?
?
?
p
V
A
??
z
s
0??? V d Vgdzdp?
For incompressible flow,?=const.
Integral between any points 1 and 2 on the streamline:
0)()(21 12212212 ?????? zzgVVpp ?
czgVpzgVp ?????? 2
2
22
1
2
11
22 ??
f o re q u a t io n B e r n o u lli
ssfr ic t io n les t e a d y
f lo w ib lein c o m p r e s s
.s t r e a m lin e a a lo n g
A Question:
Is the Bernoulli
equation a
momentum or
energy equation?
Hydraulic and energy grade lines for frictionless flow in a duct.
Example 1:
Find a relation between nozzle discharge
velocity and tank free-surface height h,
Assume steady frictionless flow.
1,2 maximum information is known or desired.
h
1
2
V2
Solution:
h
1
2
V2
Continuity,VAVA 2211 ?
Bernoulli,z
g
Vpz
g
Vp
2
2
22
1
2
11
22 ????? ??
ppp a?? 11?
)(2 212122 zzgVV ??? gh2?
AA
ghV
2122
2
2 /1
2
??
AA 21 ???
ghV 22 ? Torricelli 1644
According to the Bernoulli
equation,the velocity of a
fluid flowing through a hole in
the side of an open tank or
reservoir is proportional to
the square root of the depth
of fluid above the hole.
The velocity of a jet of water from an open pop
bottle containing four holes is clearly related to
the depth of water above the hole,The greater
the depth,the higher the velocity,
Review of
Bernoulli equation
The dimensions of above three items
are the same of length!
cz
g
Vp
z
g
Vp
????
??
2
2
22
1
2
11
2
2
?
?
0??? V d Vgdzdp?
Example 1:
Find a relation between nozzle discharge velocity
and tank free-surface height h,
Assume steady frictionless flow.
V2
h
1
2 ghV 22 ?
Example 2:
Find velocity in
the right tube,h
h?
AB
g
Vp
g
Vp AABB
22
22
??? ?? hgV B ?? 2
In like manner:
?
??
h
V
?
?? )(2 ??? ghV
Example 3,Find velocity in the Venturi tube.
h?
'?
1 2
22
1 1 2 2
1222
p V p V
gg??? ? ?
1 1 2 2A V A V?
2V??
2 2
2
(1 )
pV
??
??
? 12p p p? ? ?
)β1β
ρρhg
2?
???
(
)(2 '
2
1
2
1 VA
AV ?
As a fluid flows through a Venturi tube,the
pressure is reduced in accordance with the
continuity and Bernoulli equations.
Example 4,Estimate required to keep
the plate in a balance state.
(Assume the flow is steady and frictionless.)
1h
?
1 h
?
2h
V
A A
Solution:
For plate,
?
1 h
A A
V
?
2h
122 g h g h? 12
1
2hh?
12 ghV ?
AghF r i g h t ?? 2?
VAVF l e f t ?? ?by lineal momentum equation,
by Bernoulli equation,
Example 5,Fire hose,Q=1.5m3/min
Find the force on the bolts.
10cm? 3cm? ap
1
1
2
2
Solution:
1..F l m e p B e r n o u l l i c o n t i n u i t y? ? ? ?
By continuity:
1 1 2 2A V A V Q?? 1
1
QV
A? 2 2
QV
A?
By Bernoulli,22
1 1 2 2
22
p V p V
gg??? ? ?
22
1 2 1()2ap p V V
?? ? ?
10cm? 3cm? ap
1
1
2
2
By momentum,
21()F m V V
?? ? ?
1 1 2 1()gp A F m V V
?? ? ?
2 1 1 1() gF m V V p A
?? ? ?
4067 N??
4067b o l tF F N? ? ?
Example 6,Find the aero-force on the blade
(cascade).
A
B
D
C
S
S
1xV
1yV
1V
2xV
2yV
2V
Solution:
21()x x xF m V V
?? ? ?
21()y y yF m V V
?? ? ?
.
1 2 2 1()x x xp s p s F m V V? ? ? ?
.
21()y y yF m V V??
.
2 1 2 1( ) ( )x x xF m V V p p s? ? ? ?21()p p s??
.
12xxm sV sV???? 12x x xV V V??
A
B
D
C
S
S
1xV
1yV
1V
2xV
2yV
2V
By continuity,
22
21
12 ()2
VVpp
g?
??? 2 2 2 2
2 2 1 1[ ( ) ( ) ]2 x y x yV V V V
?? ? ? ?22
21()2 yyVV
???
22
21()2x y yF V V s
???
21()y x y yF s V V V???
叶片越弯，做功量越大。
A
B
D
C
S
S
1xV
1yV
1V
2xV
2yV
2V
By Bernoulli,
Bernoulli Equation for compressible flow
222
21
1
02VVdp? ????
Specific-heat ratio 1,4 p
v
Ck
C??For isentropic flow:
12
12
k k k
ppp C
? ? ?? ? ?
Gas Weight neglect
1 1
1
22 12
11 11[ ( ) 1 ]1
kkk
k
ppd p d p kC
kpp??
?? ? ?
???
1
22
2 2 1
1
1
[ ( ) 1 ] 012 k kp V Vk RTkp ? ?? ? ?? 1
22
2 1 2
1
1
[ 1 ( ) ]21 k kV V pk RTkp ?? ???
For nozzle:
21VV? 21pp?
For diffuser,21VV? 21pp?
k
k
k
k
p
C
p
p
1
1
1
1
1
1 11 ???
??
Extended Bernoulli Equation
222
21
1 2 sf
VVdp ww
?
?? ? ??
sw
m a ch in e
fw l o s s f r i c t i o n
For compressor 多变压缩功
For turbine 多变膨胀功
222
21
1
02VVdp? ????
0?sw
0?sw
?p ?V
?p ?V
Home work!
? Page 206,P3.158,P3.161
? Page 207,P3.164,P3.165
?, 气体动力学, 第二章习题第一
部分,Page 20 33题
Review of examples:
?
??
h
V
?
?? )(2 ??? ghV
h?
'?
1 2
)1(
)(2
2??
??
?
???? hgV
hg
ppp
???
???
)'(
21
??
)( 11
'
12
hhgp
hgghp
????
???
?
??
h?
'?
?
1p 2p
1h
2h
?
1h
A A
V
?
2h
12
1
2hh?
)(
)(
12
.
11
VVm
FAppFFF b o l tar i g h tl e f t
??
??????
b o l te l s eal e f t FApApF ??? 11
)( 1 e l s ear i g h t AApF ???
NApp a 4 8 7 2)( 11 ??
NF b o l t 4067?
?Analysis
?Choose your control volumn
?Body force and Surface force
?Solution
NVVm 8 0 5)( 12,??
10cm? 3cm? ap
1
1
2
2
x
Find the aero-force on the blade (cascade).
22
21
12 ()2
VVpp
g?
??? 2 2 2 2
2 2 1 1[ ( ) ( ) ]2 x y x yV V V V
?? ? ? ?22
21()2 yyVV
???
22
21()2x y yF V V s
???
21()y x y yF s V V V???
叶片越弯，做功量越大。
A
B
D
C
S
S
1xV
1yV
1V
2xV
2yV
2V
By Bernoulli,
3.7 The Energy Equation
? Conservation of Energy
Various types of energy occur in flowing fluids.
Work must be done on the
device shown to turn it
over because the system
gains potential energy as
the heavy(dark) liquid is
raised above the
light(clear) liquid.
This potential energy is converted into
kinetic energy which is either dissipated due
to friction as the fluid flows down ramp or
is converted into power by the turbine and
dissipated by friction.
The fluid finally becomes stationary again.
The initial work done in turning it over
eventually results in a very slight increase in
the system temperature.
s
o u t in
d A V A V
dt ? ? ? ?
? ??
,()
o u t inm ????
,dEEedm?? ? ? ?
Energy Per Unit Mass
1
1
2
2
e
First Laws of
Thermodynamics
d E d Q d W
d t d t d t??
..QW??
? Conservation of Energy
i n t e r n a l k i n e t i c p o t e n t i a l o t h e re e e e e? ? ? ?
2
2
vu g z?? ? ?
22.
[ ( ) ( ) ]o u t i nd E v vm u g z u g zdt ??? ? ? ? ? ?
1
1
2
2
)(,inout eemdtdE ??
.W ()a,
shaftW
.()
SW
()b,
pressW
.()
pW
o n l y 1 - 1,2 - 2
.
2 2 2 1 1 1pW p A V p A V??
,21
21
()ppm ????
.
v i s c o u s s t r e s sW()c
0cV ?Q
0vW??
22.,,,
21
21
21
[ ( ) ] [ ( ) ( ) ]22s pp VVQ W m m u g z u g z??
??
? ? ? ? ? ? ? ? ? ?
22.,,
21[ ( ) ( ) ]22 s
V p V pQ m u g z u g z W
??
??
? ? ? ? ? ? ? ? ?
22
2 1 2 1 2 1
1 ( ) ( ) ( )
2sq w V V g z z h h? ? ? ? ? ? ?
puh
?
?
?? ph C T?
The energy equation!
Example,A steady flow machine takes in air
at section 1 and discharged it at section 2 and
3.The properties at each section are as follows:
section A,Q,T,P,Pa Z,m
1 0.04 2.8 21 1000 0.3
2 0.09 1.1 38 1440 1.2
3 0.02 1.4 100? 0.4
2m 3/ms Co
CV
(1)
(2)
(3)
110KW,?Q?
Work is provided to the machine at the rate of 110kw,
Find the pressure (abs) and the heat transfer,
Assume that air is a perfect gas with R=287,Cp=1005.
3p
.Q
Solution:
3 3 3P R T?? 3?? ?
Mass conservation,1 1 2 2 3 3Q Q Q? ? ???
32 0, 0 1 6 1 /K g m? ?3
1 0, 0 1 1 9 /K g m? ?
31 1 2 2
3
3
0, 0 1 1 2 /QQ K g mQ??? ??? 3 3 3 1199P R T P a???
1
1
1
70 /QV m sA?? 2 1 2,2 /V m s? 3 5 5 /V m s?
By energy equation:
22.,,,
[ ( ) ] [ ( ) ]s o u t i nVVQ W m h g z m h g z? ? ? ? ? ? ? ? ?
22.,,
3223
2 2 3 3( ) ( )22p p
VVQ m C T g z m C T g z? ? ? ? ? ? ?
.
2.
11
11() 2p
s
Vm C T gz W? ? ? ?
CV
(1)
(2)
(3)
110K
W
,?Q?
System,an arbitrary quantity of mass of fixed
identity,Everything external to this system is denoted by
the term surroundings,and the system is separated from
its surroundings by it‘s boundaries through which no mass
across,(Lagrange 拉格朗日 )
Chapter 3 Integral Relations（ 积分关系式 ）
for a Control Volume in One-dimensional Steady Flows
Control Volume (CV),In the neighborhood of our
product the fluid forms the environment whose effect on
our product we wish to know,This specific region is called
control volume,with open boundaries through which mass,
momentum and energy are allowed to across,(Euler 欧拉 )
Fixed CV,moving CV,deforming CV
3.2 Basic Physical Laws of Fluid Mechanics
All the laws of mechanics are written for a system,which
state what happens when there is an interaction between
the system and it’s surroundings,
If m is the mass of the system
Conservation of
mass(质量守恒 ) ? 0dmm c o n s t o r
dt??
Newton’s
second law ? F m a?v v dVm
dt?
r
()d mVdt? r
Angular
momentum ? dHM
dt?
vv
()H r V m? ? ?? rv v
First law of
thermodynamic ? d Q d W d E
d t d t d t??
It is rare that we wish to follow the ultimate
path of a specific particle of fluid,Instead it is
likely that the fluid forms the environment
whose effect on our product we wish to know,
such as how an airplane is affected by the
surrounding air,how a ship is affected by the
surrounding water,This requires that the basic
laws be rewritten to apply to a specific region in
the neighbored of our product namely a control
volume ( CV),The boundary of the CV is called control
surface(CS)
?Basic Laws for system for CV
3.3 The Reynolds Transport Theorem (RTT)
雷诺输运定理
?
1122 is CV,
1*1*2*2* is system
which occupies the
CV at instant t.
d
dm?
??,The amount of per unit mass ?
CV cv d??? ? cv dm?? ?
The total amount of in the CV is, ?
t+dtt+dt tt
s
?, any property of fluid (,,,)m mV H Err
1 [ ( ) ( ) ]
C V C Vt d t tdt? ? ???
11[ ( ) ( ) ( ) ] ( )
o u t i n ss t d t d d td t d t? ? ? ? ? ? ? ??
11[ ( ) ( ) ] [ ( ) ( ) ]
s o u t i ns t d t t d dd t d t? ? ? ? ? ? ? ??
1 [ ( ) ( ) ]s
o u t i n
d dd
d t d t ?
?? ? ? ?
1 [ ( ) ( ) ]s c v
o u t i n
dd dd
d t d t d t
?? ? ? ? ? ?
()CVddt ??
t+dtt+dt tt
s
( ) ( ) ( ) ( )in in in ind d m A d s A V d t? ? ? ? ?? ? ? ?
In the like manner
( ) ( )o u t o u td A V d t????
1 [ ( ) ( ) ]s c v
o u t i n
dd dd
d t d t d t
?? ? ? ? ? ?
[ ( ) ( ) ]cv o u t i nd A V A Vdt ? ? ? ??? ? ?
s
1-D flow, is only the function of s, ()s? ? ??
For steady
flow, 0
cvd
dt
? ? ( ) ( )s
o u t ind A V A Vdt ? ? ? ?? ???
t+dtt+dt tt
ds
R T T
If there are several one-D inlets and outlets,
( ) ( )s outi i i i i i i i in
ii
d A V A V
dt ? ? ? ?
? ????
Steady,1-D only in inlets and outlets,no matter
how the flow is within the CV,
3.3 Conservation of mass (质量守恒 )
(Continuity Equation)
f=m ??dm/dm=1
( ) ( ) 0s o u t i niiiiii
ii
dm VVAA
dt ??? ? ???
( ) ( )o u t i niiii ii
ii
A VV A? ???? ( ) ( )i i n i o u t
ii
mm???&&
Mass flux (质量流量 )m&
For incompressible flow:
( ) ( )o u t i niiii
ii
VVAA??? i ii V o l u m e f l u xQ VA?
体积流量
-------Leonardo da Vinci in 15001212VVAA?
If only one inlet and one outlet
壶口瀑布是我国著名的第二大瀑布。两百多米宽的黄河河面，突然紧缩
为 50米左右，跌入 30多米的壶形峡谷。入壶之水，奔腾咆哮，势如奔马，浪
声震天，声闻十里。, 黄河之水天上来, 之惊心动魄的景观。
Example:
A jet engine working at design condition,At the inlet of the nozzle
At the outlet
Please find the mass flux and velocity at the outlet.
Given gas constant
521 2.05 10 /p N m?? T1 =865K,V1=288 m/s,A1=0.19㎡ ；
522 1,14 3 10 /p N m?? T2 =766K,A2=0.1538㎡
R=287.4 J/kg.K。
Solution
1 1 1
1
45.1 /p A V k g sRT??m A V??& p AVRT?
1 1 1 2 2 2
12
p A V p A V
R T R T?
According to the conservation of mass
1 1 1 2 2 2m A V A V????& ?
1 1 2
21
2 2 1
565,1 /A p TV V m sA p T???
Homework,P185 P3.12,P189P3.36
mV?? r ()l i n e a r m o m e n t u m?
dm V V
dm? ??
r v m o m e n t u m p e r u n i t m a s s
() ( ) ( )s
o u t iniiiiiiii
ii
d m V VVAAVV
dt ??????
r vv
3.4 The Linear Momentum Equation (动量方程 )
（ Newton’s Second Law ）
( ) ( )o u t inii
ii
mmVV????vv&&
Newton’s
second law () sd m VF dt??
rr ( ) ( )
o u t iniiii
ii
mmVV????vv&&
F?r,Net force on the system or CV (体系或控制体受到的合外力 )
iimVv&
,Momentum flux (动量流量 )
( ) ( )s o u t iniiiiii
ii
d VVAA
dt ????
? ????
1-D in & out
steady RTT
? flux
inoutm m m??& & &
For only one inlet and one outlet
According to continuity
)
() (s
o u t in
d m VF m V V
dt? ? ??
rr vv
&
2－ out,1－ in
Example,A fixed control volume of a streamtube in steady
flow has a uniform inlet (?1,A1,V1 )and a uniform exit
(?2,A2,V2), Find the net force on the control volume.
?
1Vv 1
2Vv2
o x
y
21()x x xF m V V??&
21()y y yF m V V??&
21()z z zF m V V??&?
o x
y
z
21()Fm VV? ? ?
v vv&Solution:
21()x xxm V VF? ? ?& 21( c o s )m VV ???&
21()y yym V VF? ? ?& 1
s i nm V ??? &
121212m VVAA????&
G iven 5 212 4.19 10 Npp m? ? ?
7 8,5 Kgm s?& 121 0,8c m c mdd??
3998 Kg m? ?
Neglect the weight of the fluid,Find the force on the
water by the elbow pipe.
Example:
1 2
1
2
Solution:
x
y
o
select coordinate,control volume
21()Fm VV? ? ?v vv&
21()x xxm VVF
?? ? ?
2m V
?
?
2 22sx mp VFA
???
2 22sx mp VFA
???
22
2
22
2
478.5
998 4
dp
d ??
??? 3696 N?
22 5934sxs s yF F NF? ? ?
1 111 ( 0 )sy mmp VVFA
??? ? ? ? ?
-4642N?1 11sy mp VFA ?? ? ?
In the like manner
1 sy
sx
Ftg
F?
?? 141.47?
o
Find the force to fix the elbow.
Solution,coordinate,CV
Net force on the control volume:
22 2()x L R e xaap p pF A A A A F? ? ? ? ? ?
Where Fex is the force on the CV by pipe,( on elbow)
2 )(LR L exa a ap p p p FAA A? ? ? ? ?
1
2
x
y
o
22 )(x e x appF F A? ? ? ?
I n l i k e m a n n e r
11 )(y e y appF F A? ? ? ?
Fex
Surface force,(1) Forces exposed by cutting though solid bodies
which protrude into the surface.(2)Pressure,viscous stress.
A fixed vane turns a water jet of
area A through an angle ? without
changing its velocity magnitude,
The flow is steady,pressure pa is
everywhere,and friction on the
vane is negligible,Find the force F
applied to vane.
V
?
F
Vv
( c o s 1 )x mVF ???& s i ny mVF ?? &
A water jet of velocity Vj impinges
normal to a flat plate which moves
to the right at velocity Vc,Find the
force required to keep the plate
moving at constant velocity and the
power delivered to the cart if the
jet density is 1000kg/m3
the jet area is 3cm2,and
Vj=20m/s,Vc=15m/s
jV cV
x
Neglect the weight of the jet and plate,and
assume steady flow with respect to the moving
plate with the jet splitting into an equal upward
and downward half-jet.
7.5N?
Home work:
P190-p3.46
P191-p3.50
P192-p3.54
P192-p3.58
Derive the thrust(推力 ) equation for the jet engine,
air drag is neglect
:inlet,au n ifo r m Vp outlet
outer V ap
inner eV ep
Solution:
( ) ( )o u t i nF m V m V? ? ?&&
eem V m V??&& ()
efm m V mV? ? ?& & &
fm&
,mass flux of fuel
o?
o?
e?
e?
e
e
m&
o
o
V ap
V ap
xF? right 'eeeaeppAA??left ''ooaAp
Balance with thrust
e e e eaeFx ppAA?? ?
Fx
eV
ep
em&
V
ap
Fx
)(ee eeeaF x m V m V pp A? ? ? ?&&
[ ) ](ee eeeaR F x m V m V pp A? ? ? ? ? ? ?&&
[ ( ) ) ](e eeeaR m V V pp A? ? ? ? ?&
)( eeaeFx pp A?? eem V m V??&&
'''oo e e eea aF x Ap p pAA? ? ?
F??
Coordinate,CV
emm?&&
0,0 2fmm?&&
oom?& eem?&
oom?& eem ?? &
Example,In a ground test of a jet engine,
pa=1.0133× 105N/m2,Ae=0.1543m2,Pe=1.141× 105N/m2,
Ve=542m/s,.4 3,4 /m K g s?& Find the thrust force.
eV
ep
em&
ap
Solution:
[ ( ) ]
25493
e e a eR mV P P A
N
? ? ? ? ?
??
&
0V ?
F16 R=65.38KN
[ ( ) ) ](e eeeaR m V V pp A? ? ? ? ?&
x
coordinate
A rocket moving straight up,Let
the initial mass be M0,and assume a
steady exhaust mass flow and exhaust
velocity ve relative to the rocket,If
the flow pattern within the rocket
motor is steady and air drag is neglect,
Derive the differential equation
of vertical rocket motion v(t) and
integrate using the initial condition v=0
at t=0,
Example:
ev
,eep A
fm&
()vt
Solution:
The CV enclose the rocket,cuts through
the exit jet,and accelerates upward at
rocket speed v(t).
coordinate
z
v(t)
Z-momentum equation:
( ) ( )o u t i nF m v m v??? &&
ef
dvm g m m v
dt? ? ? ? &
0() fm t M m t?? &f emdv v dt gd tm??
&
0 0 00
v t t
ef
f
dtdv m v g dt
M m t?? ?? ? ?& &
0
( ) l n( 1 )fe
mtv t v gt
M? ? ? ?
&
Am&
Am&
( ) ( )f e A Am v m v m v? ? ? ? ?& & &femv??&
aPA? ()aeP A A?? eePA? mg? dvm
dt?
F??
()e a e dvP P A m g m dt? ? ? ?
aeif P P?
ev
,eep A
fm&
()vt
v(t)
z
A
3.5 The Angular-Momentum Equation
( ) ( )s o u t in
d A V A V
dt ? ? ? ?
? ?? RTT
d rv
dm?
?? ? ?vv
2121()|
s
z
dH m vvrr
dt ? ? ? ?
v
vv&
2121()zMm vvrr? ? ? ? ?
v vv&
F o r t u r b o m a c h i n e s
(,,) c o l u m n c o o r d i n a t e r z?
0zi f M ?? 2121vvrr???
2121()zMm vvrr??? ? ?&
??
r zo
rv
zv
v?
()zH r v m? ? ? ?v vv (Angular-Momentum)
zM?
v, Net moment(合力矩 )
Example:Centrifugal (离心 )pump
The velocity of the fluid is
changed from v1 to v2 and
its pressure from p1 to p2.
Find (a).an expression for
the torque T0 which must be
applied those blades to
maintain this flow,(b).the
power supplied to the pump.
0 2121( ) ( )o u t inmm vvM rr? ? ???vv&&
blade
2vv
2rv
2v?
1vv 1rv
1v?
w
1p
2p
o
121212r o u t rin V m V mm AA?? ? ? ?? &&&
mQ??&For incompressible flow
()rff? 1-D
Continuity,
Solution,The CV is chosen,
0 2121()m vvM rr??? ? ?&
0 2121()m vvT rr????&
blade
2vv
2rv
2v?
1vv 1rv
1v?
w
1p
2p
o
Pressure has no contribution
to the torque
0PTw?
2121()m vvrr?? w??&
2 2 1 1()ttm v V v V????&
are blade rotational speeds
12ttVV
pw
m? & 2 2 1 1tt
v V v V???? Work on per unit mass
Homework,P192-p3.55; P194-p3.68,p3.78 ; P200-
p3.114,p3.116
Brief Review
? Basic Physical Laws of Fluid Mechanics:
0?dtdm )( VmdtdF ?? ?
inoutCVs y s AVAVdt
d
dt
d )()()()( ???? ?????
inout VmVmF )()(
?? ???
inout rVmrVmM )()( ??
?? ???
)( rVmdtdM ??? ?? dtdWdtdQdtdE ??
? The Reynolds Transport Theorem:
? The Linear Momentum Equation:
? The Angular-Momentum Theorem:
? Conservation of Mass:
?? ? outin AVAV )()( ??
Review of Fluid Statics
? Especially,
）zZyYxXz
z
py
y
px
x
pp ddd(dddd ???
?
??
?
??
?
?? ?
Cpzpz ＝?? 00 ???
Question
When fluid
flowing…
Bernoulli(1700~1782)
What relations are there in
velocity,height and pressure?
Several Tragedies in History:
? A little railway
station in 19th Russia.
? The ‘Olimpic’ shipwreck in the Pacific
? The bumping accident of B-52 bomber of
the U.S,air force in 1960s.
3.6 Frictionless Flow:
The Bernoulli Equation
1.Differential Form of Linear Momentum Equation
Elemental fixed streamtube CV of variable area
A(s),and length ds.
ds
Al
?? d?
dAA?
dVV ?
dpp?
2
dpp ?
?
?
p
V
A
??
z
s
Linear momentum relation in the
streamwise direction:
ino u ts VmVmF )()( ?? ???
dVmVVmR i g h t inout
?
?
? ?? )( AVm ???
b o d ys u rs FFL e f tF ????
321 ssss u r FFFF ???
?s in)2/(3 ls AdppF ??
?? s ins in)2/( dAdpp ??
dAdpp )2/( ??
ds
Al
?? d?
dAA?
dVV?
dpp?
2dpp?
?
?
p
V
A
??
z
s
p d Ap d AAd p ????
A dp??
dAdppdAAdpppA )2/())(( ??????
321 ssss u r FFFF ???
?? c o sAd sF b o d y ??
A dz???
AV d VAd pAd z ?? ??? 0??? Ad pAd zAV d V ??
0??? V d Vgdzdp?
one-D,steady,frictionless flow
ds
Al
?? d?
dAA?
dVV?
dpp?
2dpp?
?
?
p
V
A
??
z
s
0??? V d Vgdzdp?
For incompressible flow,?=const.
Integral between any points 1 and 2 on the streamline:
0)()(21 12212212 ?????? zzgVVpp ?
czgVpzgVp ?????? 2
2
22
1
2
11
22 ??
f o re q u a t io n B e r n o u lli
ssfr ic t io n les t e a d y
f lo w ib lein c o m p r e s s
.s t r e a m lin e a a lo n g
A Question:
Is the Bernoulli
equation a
momentum or
energy equation?
Hydraulic and energy grade lines for frictionless flow in a duct.
Example 1:
Find a relation between nozzle discharge
velocity and tank free-surface height h,
Assume steady frictionless flow.
1,2 maximum information is known or desired.
h
1
2
V2
Solution:
h
1
2
V2
Continuity,VAVA 2211 ?
Bernoulli,z
g
Vpz
g
Vp
2
2
22
1
2
11
22 ????? ??
ppp a?? 11?
)(2 212122 zzgVV ??? gh2?
AA
ghV
2122
2
2 /1
2
??
AA 21 ???
ghV 22 ? Torricelli 1644
According to the Bernoulli
equation,the velocity of a
fluid flowing through a hole in
the side of an open tank or
reservoir is proportional to
the square root of the depth
of fluid above the hole.
The velocity of a jet of water from an open pop
bottle containing four holes is clearly related to
the depth of water above the hole,The greater
the depth,the higher the velocity,
Review of
Bernoulli equation
The dimensions of above three items
are the same of length!
cz
g
Vp
z
g
Vp
????
??
2
2
22
1
2
11
2
2
?
?
0??? V d Vgdzdp?
Example 1:
Find a relation between nozzle discharge velocity
and tank free-surface height h,
Assume steady frictionless flow.
V2
h
1
2 ghV 22 ?
Example 2:
Find velocity in
the right tube,h
h?
AB
g
Vp
g
Vp AABB
22
22
??? ?? hgV B ?? 2
In like manner:
?
??
h
V
?
?? )(2 ??? ghV
Example 3,Find velocity in the Venturi tube.
h?
'?
1 2
22
1 1 2 2
1222
p V p V
gg??? ? ?
1 1 2 2A V A V?
2V??
2 2
2
(1 )
pV
??
??
? 12p p p? ? ?
)β1β
ρρhg
2?
???
(
)(2 '
2
1
2
1 VA
AV ?
As a fluid flows through a Venturi tube,the
pressure is reduced in accordance with the
continuity and Bernoulli equations.
Example 4,Estimate required to keep
the plate in a balance state.
(Assume the flow is steady and frictionless.)
1h
?
1 h
?
2h
V
A A
Solution:
For plate,
?
1 h
A A
V
?
2h
122 g h g h? 12
1
2hh?
12 ghV ?
AghF r i g h t ?? 2?
VAVF l e f t ?? ?by lineal momentum equation,
by Bernoulli equation,
Example 5,Fire hose,Q=1.5m3/min
Find the force on the bolts.
10cm? 3cm? ap
1
1
2
2
Solution:
1..F l m e p B e r n o u l l i c o n t i n u i t y? ? ? ?
By continuity:
1 1 2 2A V A V Q?? 1
1
QV
A? 2 2
QV
A?
By Bernoulli,22
1 1 2 2
22
p V p V
gg??? ? ?
22
1 2 1()2ap p V V
?? ? ?
10cm? 3cm? ap
1
1
2
2
By momentum,
21()F m V V
?? ? ?
1 1 2 1()gp A F m V V
?? ? ?
2 1 1 1() gF m V V p A
?? ? ?
4067 N??
4067b o l tF F N? ? ?
Example 6,Find the aero-force on the blade
(cascade).
A
B
D
C
S
S
1xV
1yV
1V
2xV
2yV
2V
Solution:
21()x x xF m V V
?? ? ?
21()y y yF m V V
?? ? ?
.
1 2 2 1()x x xp s p s F m V V? ? ? ?
.
21()y y yF m V V??
.
2 1 2 1( ) ( )x x xF m V V p p s? ? ? ?21()p p s??
.
12xxm sV sV???? 12x x xV V V??
A
B
D
C
S
S
1xV
1yV
1V
2xV
2yV
2V
By continuity,
22
21
12 ()2
VVpp
g?
??? 2 2 2 2
2 2 1 1[ ( ) ( ) ]2 x y x yV V V V
?? ? ? ?22
21()2 yyVV
???
22
21()2x y yF V V s
???
21()y x y yF s V V V???
叶片越弯，做功量越大。
A
B
D
C
S
S
1xV
1yV
1V
2xV
2yV
2V
By Bernoulli,
Bernoulli Equation for compressible flow
222
21
1
02VVdp? ????
Specific-heat ratio 1,4 p
v
Ck
C??For isentropic flow:
12
12
k k k
ppp C
? ? ?? ? ?
Gas Weight neglect
1 1
1
22 12
11 11[ ( ) 1 ]1
kkk
k
ppd p d p kC
kpp??
?? ? ?
???
1
22
2 2 1
1
1
[ ( ) 1 ] 012 k kp V Vk RTkp ? ?? ? ?? 1
22
2 1 2
1
1
[ 1 ( ) ]21 k kV V pk RTkp ?? ???
For nozzle:
21VV? 21pp?
For diffuser,21VV? 21pp?
k
k
k
k
p
C
p
p
1
1
1
1
1
1 11 ???
??
Extended Bernoulli Equation
222
21
1 2 sf
VVdp ww
?
?? ? ??
sw
m a ch in e
fw l o s s f r i c t i o n
For compressor 多变压缩功
For turbine 多变膨胀功
222
21
1
02VVdp? ????
0?sw
0?sw
?p ?V
?p ?V
Home work!
? Page 206,P3.158,P3.161
? Page 207,P3.164,P3.165
?, 气体动力学, 第二章习题第一
部分,Page 20 33题
Review of examples:
?
??
h
V
?
?? )(2 ??? ghV
h?
'?
1 2
)1(
)(2
2??
??
?
???? hgV
hg
ppp
???
???
)'(
21
??
)( 11
'
12
hhgp
hgghp
????
???
?
??
h?
'?
?
1p 2p
1h
2h
?
1h
A A
V
?
2h
12
1
2hh?
)(
)(
12
.
11
VVm
FAppFFF b o l tar i g h tl e f t
??
??????
b o l te l s eal e f t FApApF ??? 11
)( 1 e l s ear i g h t AApF ???
NApp a 4 8 7 2)( 11 ??
NF b o l t 4067?
?Analysis
?Choose your control volumn
?Body force and Surface force
?Solution
NVVm 8 0 5)( 12,??
10cm? 3cm? ap
1
1
2
2
x
Find the aero-force on the blade (cascade).
22
21
12 ()2
VVpp
g?
??? 2 2 2 2
2 2 1 1[ ( ) ( ) ]2 x y x yV V V V
?? ? ? ?22
21()2 yyVV
???
22
21()2x y yF V V s
???
21()y x y yF s V V V???
叶片越弯，做功量越大。
A
B
D
C
S
S
1xV
1yV
1V
2xV
2yV
2V
By Bernoulli,
3.7 The Energy Equation
? Conservation of Energy
Various types of energy occur in flowing fluids.
Work must be done on the
device shown to turn it
over because the system
gains potential energy as
the heavy(dark) liquid is
raised above the
light(clear) liquid.
This potential energy is converted into
kinetic energy which is either dissipated due
to friction as the fluid flows down ramp or
is converted into power by the turbine and
dissipated by friction.
The fluid finally becomes stationary again.
The initial work done in turning it over
eventually results in a very slight increase in
the system temperature.
s
o u t in
d A V A V
dt ? ? ? ?
? ??
,()
o u t inm ????
,dEEedm?? ? ? ?
Energy Per Unit Mass
1
1
2
2
e
First Laws of
Thermodynamics
d E d Q d W
d t d t d t??
..QW??
? Conservation of Energy
i n t e r n a l k i n e t i c p o t e n t i a l o t h e re e e e e? ? ? ?
2
2
vu g z?? ? ?
22.
[ ( ) ( ) ]o u t i nd E v vm u g z u g zdt ??? ? ? ? ? ?
1
1
2
2
)(,inout eemdtdE ??
.W ()a,
shaftW
.()
SW
()b,
pressW
.()
pW
o n l y 1 - 1,2 - 2
.
2 2 2 1 1 1pW p A V p A V??
,21
21
()ppm ????
.
v i s c o u s s t r e s sW()c
0cV ?Q
0vW??
22.,,,
21
21
21
[ ( ) ] [ ( ) ( ) ]22s pp VVQ W m m u g z u g z??
??
? ? ? ? ? ? ? ? ? ?
22.,,
21[ ( ) ( ) ]22 s
V p V pQ m u g z u g z W
??
??
? ? ? ? ? ? ? ? ?
22
2 1 2 1 2 1
1 ( ) ( ) ( )
2sq w V V g z z h h? ? ? ? ? ? ?
puh
?
?
?? ph C T?
The energy equation!
Example,A steady flow machine takes in air
at section 1 and discharged it at section 2 and
3.The properties at each section are as follows:
section A,Q,T,P,Pa Z,m
1 0.04 2.8 21 1000 0.3
2 0.09 1.1 38 1440 1.2
3 0.02 1.4 100? 0.4
2m 3/ms Co
CV
(1)
(2)
(3)
110KW,?Q?
Work is provided to the machine at the rate of 110kw,
Find the pressure (abs) and the heat transfer,
Assume that air is a perfect gas with R=287,Cp=1005.
3p
.Q
Solution:
3 3 3P R T?? 3?? ?
Mass conservation,1 1 2 2 3 3Q Q Q? ? ???
32 0, 0 1 6 1 /K g m? ?3
1 0, 0 1 1 9 /K g m? ?
31 1 2 2
3
3
0, 0 1 1 2 /QQ K g mQ??? ??? 3 3 3 1199P R T P a???
1
1
1
70 /QV m sA?? 2 1 2,2 /V m s? 3 5 5 /V m s?
By energy equation:
22.,,,
[ ( ) ] [ ( ) ]s o u t i nVVQ W m h g z m h g z? ? ? ? ? ? ? ? ?
22.,,
3223
2 2 3 3( ) ( )22p p
VVQ m C T g z m C T g z? ? ? ? ? ? ?
.
2.
11
11() 2p
s
Vm C T gz W? ? ? ?
CV
(1)
(2)
(3)
110K
W
,?Q?
