Chapter3 Integral Relations(积分关系式) for a control volume in One-dimensional steady Flows 31 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 ts 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 Efi) Fixed Cv, moving Cv, deforming Cv
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 ma5质量守→ m= const or dm=0 Newtons second aw →F=m=mm= Angular dH momentum →M H=∑(×1)·m dt First law of do dw dE thermodynamic
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) 0 dm m const or dt = = Newton’s second law F ma = dV m dt = ( ) d mV dt = Angular momentum dH M dt = H r V m = ( ) First law of thermodynamic 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
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 112>2 is system which occupies the cv at instant t 2 tt+at dp: any property of fluid(m, mv,H,e) dΦ B dm The amount of per unit mass The total amount of g in the cv is db=「B Cv
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+dt t+dt t t s : any property of fluid ( , , , ) m mV H E
(④cp) [Φcp(t+l)-Φc(t) t+at 2 tt+at s (t+dt)-(dq)out +(do)in]-aps(t dig(t+dt)-ops(t]-[(da )out -(dd )in] dt dΦs1 [dop) out-(ddp)in dt dt dΦde1 +,[(d)am-(d)m dt dt
1 [ ( ) ( )] CV CV t dt t dt = + − 1 1 [ ( ) ( ) ( ) ] ( ) out in s s t dt d d t dt dt = + − + − 1 1 [ ( ) ( )] [( ) ( ) ] s out in s t dt t d d dt dt = + − − − 1 [( ) ( ) ] s out in d d d dt dt − = − 1 [( ) ( ) ] s cv out in d d d d dt dt dt = + − ( ) CV d dt = t+dt t+dt t t s
1-Dfow:< Tis only the function of s.Φ=Φ(s) (d)mn=(Bdm)m=(pAds)im=(BpAVdt) In the like manner (d)out =(Bpavdtout 2 tt+at dd +-,[(dd)o-(a)m dt -+lBpADout-BpAl)in RTT For stead docy dd y (BpADout-BpADin flow dt
( ) ( ) ( ) ( ) d dm Ads AVdt = = = in in in in In the like manner ( ) ( ) d AVdt = out out 1 [( ) ( ) ] s cv out in d d d d dt dt dt = + − [( ) ( ) ] cv out in d AV AV dt = + − s 1-D flow : is only the function of s . = ( )s For steady flow : 0 d cv dt = ( ) ( ) s out in d AV AV dt = − t+dt t+dt t t ds R T T
If there are several one-D inlets and outlets: dd ∑(月41)m-∑(月pA Steady, 1-D only in inlets and outlets, no matter how the flow is within the Cv 33 Conservation of mass(质量守恒) (Continuity Equation) O=m B=dm/dm ∑(A1om-(OA)m=0 Σ(P,A1)=E(P,A)m∑(m)m=∑(m2) out Mass fluⅹ(质量流量m)
If there are several one-D inlets and outlets : ( ) ( ) s out i i i i i i i i in i i d AV AV 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 ( ) ( ) 0 s out in i i i i i i i i dm A A V V dt = − = ( ) ( ) out in i i i i i i i i AV V = A ( ) ( ) i in i out i i m m = Mass flux (质量流量 m )
For incompressible flow ∑(A)m=∑(A)mQ=A1 Volume flu体积流量 If only one inlet and one outlet A11=A22 -Leonardo da vinci in 1500 妻口瀑布是我国著名的第二大瀑布。两百多來宽的黄炣河面,突然紧缩 为50米左右,跌入30多米的形蛱谷。入妻之水,奔腾咆,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
For incompressible flow: ( ) ( ) out in i i i i i i A A V V = i i Qi = AV Volume flux 体积流量 A A 1 2 V V 1 2 = -------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
Exampl l Ajet engine working at design condition at the inlet of the nozzle p1=205×10Nm2T1=865K,Ⅵ1=288m/s,A1=0.19m3; At the outlet p,=1.143 x10N/m T2=766K, A2=0.1538m Please find the mass flux and velocity at the outlet Given R=287. 4 J/k, K gas constant Solution rm= pAv Rn AV-P,AV 45.1kg/s R71 According to the conservation of mass m=14V1=P2A12→P41 RT RT 2=565.1m/s Homework P185P3 12 P189P336
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 5 2 1 p N m = 2.05 10 / T1 =865K，V1=288 m/s，A1=0.19㎡； 5 2 2 p N m = 1.143 10 / T2 =766K，A2=0.1538㎡ R=287.4 J/kg.K。 Solution 1 1 1 1 45.1 / p AV kg s RT m AV = = = p AV RT = 1 1 1 2 2 2 1 2 p AV p A V RT RT = According to the conservation of mass m AV A V = = 1 1 1 2 2 2 1 1 2 2 1 2 2 1 565.1 / A p T V V m s A p T = = Homework: P185 P3.12, P189P3.36
34 The Linear Momentum Equation(动量方程) Newton's Second Law dop dt ∑(BpA)m-∑(B,A) steady RtT g-my (linear -momentum) g flux dmk B v momentum perunit mass d(mv) ∑(p,A1)m-∑(pA)m=∑(mV)-∑(mT)m Newton's d(my) ∑(mV1)om-2(mv)m second law ∑ m;: 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
= mV ( ) linear momentum − dmV V dm = = momentum perunit mass ( ) ( ) ( ) s out in i i i i i i i i i i d mV A A V V dt = − 3.4 The Linear Momentum Equation (动量方程) （ Newton’s Second Law ） ( ) ( ) out in i i i i i i = − m m V V Newton’s second law d mV ( )s F dt = ( ) ( ) out in i i i i i i = − m m V V F :Net force on the system or CV (体系或控制体受到的合外力) miV i ：Momentum flux (动量流量) ( ) ( ) s out in i i i i i i i i d A A V V dt = − 1-D in & out steady RTT flux