Chapter 8 Thermodynamics of High-Speed Gas Flow (第8章气体和蒸气的流动) 8.1声速与马赫数 Velocity of Sound and Mach Number 8.2一维定熵稳定流动 One dimensional Isentropic Steady Flow °8.3喷管出口流速和流量的计算 Outlet velocity and Flow rate Calculation for Nozzles °8.4具有摩擦的绝热稳定流动 Adiabatic steady flow with friction 85绝热节流 Adiabatic throttling
Chapter 8 Thermodynamics of High-Speed Gas Flow (第8章 气体和蒸气的流动) ⚫ 8.1 声速与马赫数 Velocity of Sound and Mach Number ⚫ 8.2 一维定熵稳定流动 One dimensional Isentropic Steady Flow ⚫ 8.3 喷管出口流速和流量的计算 Outlet Velocity and Flow rate Calculation for Nozzles ⚫ 8.4具有摩擦的绝热稳定流动 Adiabatic steady flow with friction ⚫ 8.5 绝热节流 Adiabatic throttling
8.1声速与马赫数 Velocity of Sound and Mach Number 1. Velocity of Sound( or Sonic velocity)(声速) pressure wave travels through a mediuhmall It is the velocity at which infinitesimally si 0o For adiabatic process For ideal gas〔对于理想气体) 2s、kp=vkRT
8.1 声速与马赫数 Velocity of Sound and Mach Number 1. Velocity of Sound (or Sonic Velocity) (声速) It is the velocity at which infinitesimally small pressure wave travels through a medium. For adiabatic process For Ideal Gas (对于理想气体) s s v p v p a ( ) ( ) 2 = − = a = kpv = kRT
2. Mach number(马赫数) 定义:流体某一点的运动速度和该点当地声速之比,以M表示 Definition: the mach number m is the ratio of the flow speed c to the velocity of sound in the same fluid at the same state. it is denoted as m Varieties of flow(流动的种类): M1 supersonic flow(超声速流) M>>1 hypersonic
2. Mach number (马赫数) 定义：流体某一点的运动速度和该点当地声速之比, 以M 表示 Definition: The Mach number, M, is the ratio of the flow speed, c, to the velocity of sound in the same fluid at the same state. It is denoted as M. Varieties of flow (流动的种类): M1 supersonic flow (超声速流) M>>1 hypersonic a c M = M 1
8.2一维定熵稳定流动) One dimensional Isentropic Steady Flow 1. Physical Problem(物理问题) (1 Gas steady flow 气体的稳定流动 (2)The flow in short duct with variable cross- sectional area 变截面短管中的流动 (3)The process is isentropic, that is reversible adiabatic process 可逆绝热的流动过程,即定熵流动
8.2一维定熵稳定流动） One dimensional Isentropic Steady Flow 1. Physical Problem （物理问题） (1) Gas steady flow 气体的稳定流动 (2)The flow in short duct with variable crosssectional area 变截面短管中的流动 (3)The process is isentropic, that is, reversible adiabatic process 可逆绝热的流动过程，即定熵流动
2. Mathematica| Model(数学模型) For flow in duct with variable cross-sectional area it is necessary to use differential equations to reveal the relationships between 1.C ◆ Conservation Equation of Mass〔质量守恒方程) ◆ Conservation Equation of Energy(能量守恒方程) Equation of State(状态方程) For ideal as(对理想气体) PeRT 4y-=0fp,=0 For real gas(对实际气体) ◆Pr。 cess Equation(过程方程) ◆ Equation of Entr。py(熵方程)
2. Mathematical Model (数学模型） For flow in duct with variable cross-sectional area, it is necessary to use differential equations to reveal the relationships between Conservation Equation of Mass (质量守恒方程） Conservation Equation of Energy (能量守恒方程） Equation of State (状态方程) For Ideal Gas (对理想气体） For Real Gas (对实际气体) ◆ Process Equation (过程方程） Equation of Entropy(熵方程） p,v,T,m ,c, f
Given m,p→)fc,v,T设计计算 Given p,f→mn,c,v,T校核计算 (1) Continuity Equation(连续性方程) C-m v f·de+c.d=m·dhy ac C If y c then c must be adopted K then df<o c must be adopted
Given p, f m,c, v,T m,p f, c, v,T 校核计算 设计计算 → Given → (1) Continuity Equation (连续性方程） If then , c , must be adopted; If then , c , must be adopted; f c = m v f dc + c df = m dv c dc v dv f df = − c dc v dv df 0 c dc v dv df 0 (A)
For incompressible fluid 0 df dc f' C C (2). Energy Equation(能量方程) 0 0 0 8gdh+ dc2+g.dit Ws =-dh 2 c2=2(h1-h2) c2=V2(hn-h2)+c2
For incompressible fluid , (2). Energy Equation (能量方程) = 0 v dv c dc f df = − f c f , c q dh dc g dz +ws = + + 2 2 1 0 0 0 dh dc = − 2 2 2( ) 1 2 2 1 2 c2 − c = h − h 2 2 1 2 1 c = 2(h − h ) + c
For reversible process(可逆过程) dh=-w,=vdp C 1 2 Cac =-vap(B) If C thenp;如果C变大(C>0,则p必减少(dp0 (3)Process Equation k k= For ideal gas(对理想气体)c For real gas, k is an empirical constant.〔对实际气体来二 说k是经验常数)
For reversible process (可逆过程) If then ; If then ; (3) Process Equation For ideal gas(对理想气体) For real gas, k is an empirical constant.(对实际气体来 说,k是经验常数) dh = −wt = vdp vdp dc = − 2 2 cdc = −vdp c p c p pv C k = v p c c k = (B) 如果 c 变大（ dc >0),则p必减少(dp0)
k h1+1 p=0 vdp= kdv k (C) Eq (BX kpv cac= C C M p kp From Eq (c) du
0 1 + = − kpv dv v dp k k − vdp = kpdv p dp v dv k = − (C ) Eq. (B)× 2 c kp dp c kpv cdc c k p 2 2 = − a = kpv 2 dp M dc c k p 2 1 = − dv v k p From Eq.(C) dp = −
dc dy M ( D) Substitute Eq. D) into Eq (A) (M2-1 l) M 1> Supersonic region dc>0, then df>0 Mo, then df<o
v dv c dc M = 2 (D) Substitute Eq.(D) into Eq.(A) c dc M f df ( 1) 2 = − M 1 If dc 0, then df 0 Supersonic region M 1 Subsonic region If dc 0, then df 0