Transcript Solution

Chapter 3 Integral Relations（积分关系式）
for a Control Volume in One-dimensional Steady Flows
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)
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)

Newton’s
second law

Angular
momentum

First law of
thermodynamic

m  const
or
dm
0
dt
dV d
 ( mV )
F  ma  m
dt dt
dH
M
H   (r  V )  m
dt
dQ dW dE


dt
dt
dt
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 .
s
1*1*2*2* is system
which occupies the
CV at instant t.
t
t+dt
 : any property of fluid (m, mV , H , E )
d

dm
:The amount of  per unit mass
The total amount of

in the CV is :
CV  cv d   cv  dm
t
t+dt
d
( CV ) 
dt
1
 [ CV (t  dt )   CV (t )]
dt
s
t
t+dt
1
1
 [ s (t  dt )  (d )out  (d )in]  s(t )
dt
dt
1
1
 [ s (t  dt )  s (t )]  [(d  ) out  (d  )in]
dt
dt
d s 1

 [(d  )out  (d  )in]
dt dt
d s d cv 1

 [(d  )out  (d  )in]
dt
dt
dt
t
t+dt
1-D flow : 
is only the function of s .    ( s )
(d )in  (  dm)in  (  Ads)in  (  AVdt )in
In the like manner
s
(d )out  (  AVdt )out
ds
t
t
t+dt
t+dt
d s d cv 1

 [( d  ) out  ( d  )in]
dt
dt
dt
d cv

 [(  AV )out  (  AV )in]
dt
For steady
flow :
d cv
0
dt

RTT
ds
 (  AV ) out  (  AV )in
dt
If there are several one-D inlets and outlets :
d s
  ( i i AV
i i ) out   (  i i AV
i i )in
dt
i
i
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
dms
 (  i AiV i)out  (  i AiV i)in  0
dt
i
i
( 
i
i
AiV i ) out  (  i AiV i )in
i
Mass flux (质量流量
m)
(m )
i in
i
 (m i )out
i
For incompressible flow:
( A V )
i
i
i out
 ( AiV i )in
Qi  AiV i Volume flux
体积流量
i
If only one inlet and one outlet
A1V 1  A2V 2
-------Leonardo da Vinci in 1500
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
p1  2.05 105 N / m 2
T1 =865K，V1=288 m/s，A1=0.19㎡；
At the outlet p2  1.143 105 N / m 2 T2 =766K，A2=0.1538㎡
Please find the mass flux and velocity at the outlet.
Given
R=287.4 J/kg.K。
Solution
gas constant
p1 AV
p
1 1
 45.1 kg / s
AV 
m   AV 
RT
RT1
According to the conservation of mass
p1 AV
p2 A2V2
1 1
m  1 AV


A
V


1 1
2 2 2
RT1
RT2
A1 p1 T2
 V2  V1 A p T  565.1 m / s
2
2
1
Homework: P185 P3.12, P189P3.36
3.4 The Linear Momentum Equation (动量方程)
（ Newton’s Second Law ）
ds
  (   i AiV i )out   (   i AiV i )in
dt
i
i
  mV
dmV

V
dm
1-D in & out
steady RTT
(linear  momentum)
 flux
momentum perunit mass
d (mV ) s
 (  i AiV iV i )out  (  i AiV iV i )in  (mi V i )out  (mi V i )in
i
i
dt
i
i
Newton’s
second law
d (mV ) s  (m V i )out  (m V i )in
i i
 F  dt i i
mi V i ：Momentum flux (动量流量)
F
:Net force on the system or CV (体系或控制体受到的合外力)
For only one inlet and one outlet
According to continuity
d (mV ) s
F

 m(Vout  Vin )

dt
2－out, 1－ in
mout  min  m

Fx  m(V 2 x  V 1x )
y
Fy  m(V 2 y  V 1 y )
o
Fz  m(V 2 z  V 1z )
z
x
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.
Solution:
F  m(V 2  V 1)
m   1 A1V 1   2 A2V 2
 F x  m(V2 x  V1x )  m(V 2  V 1 cos  )
2
 F y  m(V2 y  V1 y )  mV 1 sin 
V1
1

V2
y
o
x
Example:
Given
p1  p 2  4.19 105 N
m  78.5 Kg
  998 Kg
s
m
2
m
d 1  10cm, d 2  8cm
2
1
3
Neglect the weight of the fluid. Find the force on the
water by the elbow pipe.
Solution: select coordinate ,control volume
F  m(V 2  V 1)
2


 F x  m(V 2 x  V 1x )  m V 2

1
F sx  p 2 A2  mV 2

2
2

4
78.5
d
2
F sx  p 2 A2  mV 2 

  3696N
p
2
2
998 d 2
4
y
o
x
In the like manner


 V 1)   mV 1
F sy  p1 A1  m(0

F sy   p1 A1  mV 1  -4642N
Find the force to fix the elbow.
Fs 
2
2
F sx  Fsy  5934 N
F
  tg 1 sy  141.47
F sx
F
ex
Solution: coordinate, CV
Net force on the control volume:
F x  p a AL  p a( AR  A2)  p 2 A2  F ex
2
1
 p a AL  p a AR  ( p 2  p a) AL  F ex
Where Fex is the force on the CV by pipe,( on elbow)
y
o
F x  F ex  ( p 2  p a) A2
In like manner
F y  F ey  ( p1  p a) A1
Surface force: (1) Forces exposed by cutting though solid bodies
which protrude into the surface.(2)Pressure,viscous stress.
x
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.
F x  mV (cos   1)
V
V
F y  mV sin 

F
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
V
j
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
Vc
x
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
Fx
Solution: Coordinate, CV
e
o
outer V p a
inlet : uniform V , p a outlet
F  (mV )out  (mV )in
inner V e p e
V
moo  mee
o
pa
o
 meVe  mV  (m  m f )Ve  mV
m f : mass flux of fuel
F
left p a Ao'o'
right  p a Aee'  p e Ae o
Fx Balance with thrust
e
Ve
m
moo
mee
x
pe
e
e
me
V pa
F  Fx  p a Ao'o'  p a Aee'  pe Aee  Fx  p a Aee p e Aee
Fx  ( p a  pe) Aee  meVe  mV
Fx  meVe  mV  ( pe  p a) Aee
R  Fx  [meVe  mV  ( pe  p a) Aee]
R  [m(Ve  V )  ( pe  p a) Aee]
V pa
m f  0.02m
me  m
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, m  43.4 Kg / s . Find the thrust force.
Solution:
coordinate
R  [m(Ve  V )  ( pe  p a) Aee]
Ve
pe
pa
me
V 0
x
 R  [mVe  ( Pe  Pa ) Ae]
 25493 N
F16
R=65.38KN
Example:
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 .
Solution: coordinate
The CV enclose the rocket,cuts through
the exit jet,and accelerates upward at
rocket speed v(t).
v(t)
v (t )
z
mf
ve
p e, Ae
Z-momentum equation:
v(t) A
F  (mv)out  (mv)in
 (m f ve  mAv)  (mAv)   m f ve
F   Pa A  Pa ( A  Ae )  Pe Ae  mg m dv
dt
dv
 ( Pe  Pa ) Ae  mg  m
dt
dv
if Pa  Pe
mg  m   m f ve
dt
mf
dv 
vedt  gdt m(t )  M 0  m f t
m
v
t
t
dt
0 dv  m f ve 0 M 0  m f t  g 0 dt
v(t )  ve ln(1 
mf t
M0
)  gt
v (t )
mA
z
mf
mA
ve
p e, Ae
3.5 The Angular-Momentum Equation
d s
 (  AV )out  (  AV )in
dt
  H z  (r  v )m
RTT
(Angular-Momentum)
dHs
 m(r 2  v 2  r 1  v 1)
|
z
dt
d

 r v
dm
M z  m(r 2  v 2  r 1  v 1)
M z ：Net moment(合力矩)
For turbomachines
column coordinate(r ,  , z )
Mz  m(v 2 r 2  v1 r 1)
if
M  0
z
v 2 r 2  v1 r1
vr

r o

v
vz
z
Example:Centrifugal (离心)pump
v 2r
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.
Solution: The CV is chosen .
 M 0  mout (r 2  v 2)  min (r 1  v 1)
v2
v 2
v 1 v1r
v1
o
p1
p2
blade
f  f (r )
1-D
Continuity : min   1 A1V1r  mout   2 A2V2 r  m
For incompressible flow
m  Q
w
v 2r
M 0  m(r 2v 2  r1v1 )
v2
Pressure has no contribution
to the torque
T 0  m(r 2v 2  r1v1 )
P  T 0w
 m(r 2v 2  r1v1 )w
v 2
v 1 v1r
v1
o
p1
p2
 m(v 2V 2t  v1V 1t )
blade
V1t V2t are blade rotational speeds
p
w
 v 2V 2t  v1V 1t
m
Work on per unit mass
Homework: P192-p3.55; P194-p3.68, p3.78 ; P200p3.114,p3.116
w
Brief Review
• Basic Physical Laws of Fluid Mechanics:
dm
0
dt
 d

F  (mV )
dt
 d
 
M  (mV  r )
dt
dE dQ dW


dt dt dt
• The Reynolds Transport Theorem:
d
d
( sys ) 
( CV )  ( AV ) out  ( AV ) in
dt
dt
• Conservation of Mass:
 ( AV )in   ( AV )out
• The Linear Momentum Equation:


 F  (m V )  (m V )
• The Angular-Momentum Theorem:
 M  (mV r )  (mV r )
out
in


out
in
Review of Fluid Statics
p
p
p
dp  dx  dy  dz   ( Xdx  Ydy  Zdz）
x
y
z
• Especially :
z
p

 z0 
p0

＝C
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.
p
dp
2


z
p
V

p  dp
V  dV
A dA
A

  d
Al
ds
s
Linear momentum relation in the
streamwise direction:


Fs  (mV )out  (mV )in



m  AV
Right  m(Vout  Vin)  m dV
Fs  Left  Fsur  Fbody
p
dp
2

Fsur  Fs1  Fs 2  Fs3
Fs 3  ( p  dp / 2) Al sin 
dA
 ( p  dp / 2)
sin 
sin 
 ( p  dp / 2)dA

z
p
V

p  dp
V  dV
A dA
A

  d
Al
ds
s
Fsur  Fs1  Fs 2  Fs3
 pA  ( p  dp)( A  dA)  ( p  dp / 2)dA
  Adp  pdA  pdA
p
  Adp
Fbody  Ads cos 
 Adz
Adz  Adp  AVdV
dp

 gdz  VdV  0
dp
2


z
p
V

p  dp
V  dV
A dA
A

  d
Al
ds
AVdV  Adz  Adp  0
one-D,steady,frictionless flow
s
dp

 gdz  VdV  0
For incompressible flow, =const.
Integral between any points 1 and 2 on the streamline:
p 2  p1 1 2
 (V 2  V 12)  g ( z 2  z1)  0

2
p1 V 12
p 2 V 22
  g z1    g z 2  c
 2
 2
Bernoulli equation for
steady frictionle ss
incompress ible flow
along a streamline .
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.
1
h
V2
2
Solution:
A1V 1  A2 V 2
Continuity:
Bernoulli:
p1 V 12
p 2 V 22

 z1 

 z2
 2g
 2g
 p1  p1  pa
1
2

V V 1  2 g ( z1  z 2)  2 gh
2
2
2 gh
V 
1  A22 / A12
2
2
V 2  2 gh
 A1  A2
Torricelli 1644
h
V2
2
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
dp

 gdz  VdV  0
p1


p2


2
1
V
 z1
2g

2
2
V
 z2  c
2g
The dimensions of above three items
are the same of length!
Example 1:
Find a relation between nozzle discharge velocity
and tank free-surface height h.
Assume steady frictionless flow.
1
h
V2
2
V 2  2 gh
Example 2:
Find velocity in
the right tube.
h
h
B
2
2
V B pA V A



 2g  2g
pB
A
V B  2 gh
In like manner:
V

h

V
2 gh(     )

Example 3: Find velocity in the Venturi tube.
1
V12 p2 V22



 1 2g  2 2g
p1
2
AV
1 1  A2V2
h
'
A2  V
V1  V2
2
A1
V2 
2p
 (1   2 )
p  p1  p2
2 gh(ρ '  ρ )

β (1  β 2 )
As a fluid flows through a Venturi tube, the
pressure is reduced in accordance with the
continuity and Bernoulli equations.
Example 4: Estimate h1 required to keep
the plate in a balance state.
(Assume the flow is steady and frictionless.)

h
h2
V
1
A

A
Solution:
For plate,
Fright  gh2  A

h
h2
V
1
A

A
by lineal momentum equation, Fleft  AV V
by Bernoulli equation,
2gh1  gh2
V  2gh1
1
h1  h2
2
Example 5: Fire hose,Q=1.5m3/min
Find the force on the bolts.
1
2
 3cm
 10cm
1
2
pa
Solution:
F  l.m.e  p1  Bernoulli  continuity
By continuity:
AV
1 1  A2V2  Q
Q
V1 
A1
V12 p2 V22



 2g  2g
By Bernoulli:
p1
By momentum :
V2 
p1  pa 

2
Q
A2
(V22  V12 )

F  m(V2  V1 )

p1g A1  F  m(V2  V1 )
1
2

F  m(V2  V1 )  p1g A1
 3cm
 10cm
 4067N
Fbolt   F  4067 N
1
2
pa
Example 6: Find the aero-force on the blade
(cascade).
A
V1y
V1x
D
S
V1
B
S
V2 x
V2 y
C
V2
Solution:
V1y
V1x

Fx  m(V2 x  V1x )
A
D
V1 S
B
SV
2y
.
p1s  p2 s  Fx  m(V2 x  V1x )
.
m   sV1x   sV2 x
By continuity,
.
Fx  m(V2 x  V1x )  ( p2  p1 ) s  ( p2  p1 )s

Fy  m(V2 y  V1 y )
.
Fy  m(V2 y  V1 y )
V2 x
C
V1x  V2 x  Vx
V2
By Bernoulli,
V22  V12


p1  p2   (
)  [(V22x  V22y )  (V12x  V12y )]  (V22y  V12y )
2g
2
2
Fx 

2
(V  V ) s
2
2y
2
1y
Fy   sVx (V2 y  V1 y )
V1y
V1x
A
V1 S
B
叶片越弯，做功量越大。
D
SV
V2 x
2y
C
V2
Bernoulli Equation for compressible flow
Gas Weight neglect
Cp

1
Specific-heat ratio k  1.4 
Cv
For isentropic flow:
p

k

2
1


dp

p1
k
1

C
1
k
p2
2

2
1
k
C
dp
p
1
k

V2  V1


dp V22  V12

0

2
p1
1
k
1

1
p
1
k

C
p
1
k
1
k
k p1 p2 k 1k
[( )  1]
k  1 1 p1
p2 k 1k
V22  V12
k
RT1[( )  1] 
0
k 1
p1
2
For nozzle:
1
2
p2  p1
V22  V12
p k 1
k

RT1[1  ( 2 ) k ]
2
k 1
p1
For diffuser: V2  V1 p2  p1
Extended Bernoulli Equation

2
1
ws
dp V22  V12

0

2
machine

2
1
dp V22  V12

 ws  w f

2
wf
loss friction
For compressor 多变压缩功
ws  0
p V 
For turbine
ws  0
p V 
多变膨胀功
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:
h1
V
2 gh(     )


h

1
2
h

'
2 gh(     )
V
 (1   2 )
p2
p1
V
h2


h
'
p2  gh1   ' gh
 p1  g (h1  h)
p  p1  p2
 (  '  ) gh
1

1h
h2
V
A

 3cm
 10cm
A
•Analysis
•Choose your control volumn
•Body force and Surface force
pa
x
1
h1  h2
2
•Solution
2
1
2
Fleft  p1 A1  pa Aelse  Fbolt
Fright   pa ( A1  Aelse )
F  F
left
 Fright  ( p1  pa ) A1  Fbolt
.
 m(V2  V1 )
( p1  pa ) A1  4872 N
.
m(V2  V1 )  805N
Fbolt  4067 N
Find the aero-force on the blade (cascade).
By Bernoulli,
V22  V12


p1  p2   (
)  [(V22x  V22y )  (V12x  V12y )]  (V22y  V12y )
2g
2
2
Fx 

2
(V  V ) s
2
2y
2
1y
Fy   sVx (V2 y  V1 y )
V1y
V1x
A
V1 S
B
叶片越弯，做功量越大。
D
SV
V2 x
2y
C
V2
3.7 The Energy Equation
• Conservation of Energy
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.
Various types of energy occur in flowing fluids.
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.
• Conservation of Energy
2
1
2
First Laws of
Thermodynamics
1
dE dQ dW


dt
dt
dt
.
ds
  AVout   AVin
dt
.
 m(  out  in )
.
 Q W
dE
  E,  
e
dm
e
Energy Per Unit Mass
e  einternal  ekinetic  e potential  eother

v2
 u   gz
2
.
W
2
1
.
.
( a ) W shaft (W S )
.
.
(b) W press (W p )
only 1- 1,2 - 2
.
2
1
dE .
 m(eout  ein )
dt
.
( c ) W viscous stress

dE .  v
v
 m[(u   gz )out  (u   gz )in ]
dt
2
2
2
W p  p2 A2V2  p1 AV
1 1
.
p
p
 m( 2  1 )
 2 1
2
Vc  0
Wv  0
 V2
V2
 Q [W s  m(  )]  m[(u 
 gz ) 2  (u 
 gz )1 ]
 2 1
2
2
.
.
p2
.
p1
.

 V2
.
V2
p
p
Q  m[(u 
 gz  )2  (u 
 gz  )1 ]  W s
2

2

.
.


u
p

h
h  C pT
1 2
q  ws  (V2  V12 )  g ( z2  z1 )  (h2  h1 )
2
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:
.
(2)
110KW
(1)
Q?
section A, m 2
1
0.04
2
0.09
(3)
3
0.02
CV
Q,m3 / s T, C P, Pa Z,m
2.8
21
1000 0.3
1.1
38
1440 1.2
1.4
100
?
0.4
Work is provided to the machine at the rate of 110kw.
.
Find the pressure p3 (abs) and the heat transfer Q .
Assume that air is a perfect gas with R=287, Cp=1005.
.
Solution: P3  3 RT3
3  ?
Mass conservation: 1Q1  2Q2  3Q3
3
3


0.0161
Kg
/
m
1  0.0119 Kg / m
2
3 
1Q1   2Q2
Q3
 0.0112 Kg / m3
Q1
V1 
 70m / s
A1
(2) Q  ?
110K
W
(1)
P3  3 RT3  1199Pa
V2  12.2m / s
V3  55m / s
By energy equation:
.
V2
V2
Q W s   m[(h 
 gz )]out   m[(h 
 gz )]in
2
2
.
.
.
V32
V2 2
 Q  m2 (C p T2 
 gz2 )  m3 (C pT3 
 gz3 )
2
2.
.
V12
 m1 (C pT1 
 gz1 )  W
2
s
.
.
.
(3)
CV