3.1 Systems versus Control Volumes

stickshrivelΜηχανική

24 Οκτ 2013 (πριν από 3 χρόνια και 5 μήνες)

66 εμφανίσεις

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
)

Newton’s
second law

Angular
momentum

First law of
thermodynamic




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
.

:
The amount of per unit mass


The total amount of in the CV is
:

t+dt

t+dt

t

t

s

:
any property of fluid

t+dt

t+dt

t

t

s

In the like manner


s

1
-
D flow

:

is only the function of s

.

For steady
flow
:

t+dt

t+dt

t

t

ds

R T T

If there are several one
-
D inlets and outlets
:

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

Mass flux (
质量流量

)

For incompressible flow:

体积流量

-------
Leonardo da Vinci in 1500

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

T1
=865K

V1=288 m/s

A1=0.19




T2
=766K

A2=0.1538



R=287.4 J/kg.K



Solution

According to the conservation of mass

Homework: P185 P3.12, P189P3.36


3.4 The Linear Momentum Equation
(
动量方程
)



Newton

s Second Law




Newton’s
second law

:
Net force on the system or CV (
体系或控制体受到的合外力
)


Momentum flux (
动量流量
)

1
-
D in & out

steady RTT


flux

For only one inlet and one outlet

According to continuity

2

out, 1


in

Example: A fixed
control volume

of a streamtube in steady
flow has a uniform inlet (

1
,A1
,V1
aa楦ore楴i
(

2
,A2
,
V2
) . Find the net force on the control volume.

Solution
:

Neglect the weight of the fluid. Find the force on the
water by the elbow pipe.

Example:

1

2

1

2

Solution:

select coordinate

,
control volume

In the like manner

Find the force to fix the elbow.

Solution:

coordinate,
CV

Net force on the control volume:

Where F
ex

is the force on the CV by pipe,( on elbow)

1

2

F
ex

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.

A water jet of velocity V
j

impinges
normal to a flat plate which moves
to the right at velocity V
c.
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

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.

Home work:

P190
-
p3.46

P191
-
p3.50

P192
-
p3.54

P192
-
p3.58

Derive the thrust(
推力
) equa瑩潮 for 瑨e 橥琠en杩ge

a楲 dra朠楳 ne杬ect

Solu瑩tn:

:

mass flux of fuel

x

Balance with thrust

Coordinate, CV


Example:
In a ground test of a jet engine,
p
a
=1.0133
×
10
5
N/m
2 ,
Ae=0.1543m2,Pe=1.141
×
105N/m2,
Ve=542m/s
, .

Find the thrust force.

Solution:

F16 R=65.38KN

x

coordinate


A rocket moving straight up. Let
the initial mass be M
0
,and assume a
steady exhaust mass flow and exhaust
velocity v
e

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:

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:

v(t)

z

3.5 The Angular
-
Momentum Equation

(Angular
-
Momentum)


Net moment(
合力矩
)

Example:
Centrifugal (
离心
)pump

The veloc楴i of 瑨e flu楤 楳
changed from v
1

to v
2

and
its pressure from p
1

to p
2
.

Find (a).an expression for
the torque T
0

which must be
applied those blades to
maintain this flow. (b).the
power supplied to the pump.


blade

w

For incompressible flow

1
-
D

Continuity
:

Solution: The CV is chosen .

blade

w


Pressure has no contribution
to the torque

are blade rotational speeds

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:


The Reynolds Transport Theorem:


The Linear Momentum Equation:


The Angular
-
Momentum Theorem:


Conservation of Mass:

Review of Fluid Statics


Especially :


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 19
th

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.

Linear momentum relation in the
streamwise direction:

one
-
D,steady,frictionless flow

For incompressible flow,

=const.

Integral between any points 1 and 2 on the 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.


h

1

2

V
2

Solution:

h

1

2

V
2

Continuity:

Bernoulli:

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!

Example 1:

Find a relation between nozzle discharge velocity
and tank free
-
surface height h.
Assume steady frictionless flow.

V
2

h

1

2



Example 2:

Find velocity in
the right tube.

h

A

B

In like manner:

V

Example 3:

Find velocity in the
Venturi tube.

1

2

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.)

Solution:

For plate,

by lineal momentum equation,

by Bernoulli equation,

Example 5:
Fire hose,Q=1.5m
3
/min
Find the force on the bolts.

1

1

2

2

Solution:

By continuity:

By Bernoulli:

1

1

2

2

By momentum :

Example 6:
Find the aero
-
force on the blade


(cascade).

A

B

D

C

S

S

Solution:

A

B

D

C

S

S

By continuity,

叶片越弯,做功量越大。

A

B

D

C

S

S

By Bernoulli,

Bernoulli Equation for compressible flow

Specific
-
heat ratio

For isentropic flow:

Gas Weight neglect

For nozzle:

For diffuser:

Extended Bernoulli Equation

For compressor

多变压缩功

For turbine

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


Review of examples:

V

1

2


Analysis


Choose your control volumn


Body force and Surface force


Solution

1

1

2

2

x

Find the aero
-
force on the blade

(cascade).

叶片越弯,做功量越大。

A

B

D

C

S

S

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.


Energy Per Unit Mass

1

1

2

2

e

First Laws of
Thermodynamics



Conservation of Energy

1

1

2

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:

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

CV

(1)

(2)

(3)

110KW

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.

Solution:

Mass conservation:

By energy equation:

CV

(1)

(2)

(3)

110K
W