工程流体力学第三章1

dΦs = ( β ρ A V ) o ut − ( β ρ A V ) in dt
If there are several one-D inlets and outlets :
dΦs = dt
∑ (β ρ AV )
i i i i
i out
− ∑ ( β i ρ i AiVi ) in
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)
φ=m
β=dm/dm=1
dms = Σ ( ρ i AiV i ) out − Σ ( ρ i AiV i )in = 0 dt i i
Σ (ρ
i
i
A iV i ) o u t =
Σ (ρ
i
i
A iV i ) i n
Solu = m(V 2 − V 1)
& m = ρ 1 A1V 1 = ρ 2 A2V 2
2
v V1
v V2
& Σ F x = m(V2 x − V1x ) = m(V 2 − V 1 cos θ ) &
& & Σ F y = m(V2 y − V1 y ) = − mV 1 sin θ
∑
r r d (m V )s v v & = m (V out − V in ) F = dt
2－out, 1－ in
{
& F x = m (V 2 x − V 1 x )
y
& F y = m (V 2 y − V 1 y )
& F z = m (V 2 z − V 1 z )
o
z
x
Example: A fixed control volume of a streamtube in steady flow has a uniform inlet (ρ1,Α1,V1 )and a uniform exit ρ (ρ2,Α2 V2) . Find the net force on the control volume. ρ ,Α2,
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
Homework: P185 P3.12, P189P3.36
3.4 The Linear Momentum Equation (动量方程 动量方程) （ Newton’s Second Law ）
r Φ = mV
r v dm V β = =V dm
dΦs = dt
∑ (β ρ
i
i
A iV i ) out − ∑ ( β ρ i A iV i ) in
⇒
⇒
⇒
⇒
or r v r v dV d = (mV ) F = ma = m dt dt v v dH v v r M= H = ∑ (r × V ) ⋅ m dt
dQ dW dE − = dt dt dt
m = const
dm =0 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
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
& Σ (m
i
i in
) =
& Σ (m
i
i out
)
Mass flux (质量流量
& m )
For incompressible flow:
Σ( A V )
i i
i out
= Σ( AiV i )in
i
Qi = AiV i Volume flux
体积流量
If only one inlet and one outlet
s
t t+dt
t
t+dt
r r Φ : any property of fluid (m, mV , H , E )
dΦ β= dm
:The amount of Φ per unit mass
The total amount of
Φ
in the CV is :
Φ CV = ∫cv d Φ = ∫cv β dm
1
θ
y
o
x
Example:
G iv e n
m = 10 cm , d 2 = 8 cm & m = 78.5 K g s d1
ρ = 998 K g
m
3
p 1 = p 2 = 4.19 × 10 5 N
2
1
2
Neglect the weight of the fluid. Find the force on the water by the elbow pipe. Solution: select coordinate , ,control volume
surface(CS)
Basic Laws for system
⇑ 3.3 The Reynolds Transport Theorem (RTT) 雷诺输运定理
⇒ for CV
1122 is CV . 1*1*2*2* is system which occupies the CV at instant t.
d (Φ CV ) = dt
s
t t+dt
1 = [Φ CV (t + dt ) − Φ CV (t )] dt
t
t+dt
1 1 = [Φ s (t + dt ) − (d Φ )out + (d Φ )in] − Φs (t ) dt dt 1 1 = [Φ s (t + dt ) − Φs (t )] − [(d Φ )out − (d Φ )in ] dt dt d Φs 1 = − [(d Φ )out − (d Φ )in] dt dt d Φs d Φcv 1 = + [(d Φ )out − (d Φ )in] dt dt dt
r ∑ F :Net force on the system or CV (体系或控制体受到的合外力)
v & i V i ：Momentum flux (动量流量) m
For only one inlet and one outlet According to continuity
& & & mout = min = m
Chapter 3 Integral Relations（积分关系式 积分关系式） 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 体系) 控制体) 体系 控制体
1-D flow : Φ is only the function of s .
Φ = Φ (s)
( d Φ ) in = ( β d m ) in = ( β ρ A d s ) in = ( β ρ A V d t ) in
In the like manner
( d Φ ) out = ( β ρ A V dt ) out
i
1-D in & out steady RTT
( linear − momentum )
momentum perunit mass
Φ flux