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Momentum Transport（Summary）
动量传输（摘要）

2．Momentum Transport and Laminar Flow of Newtonian Fluids
2．牛顿流体的动量传输和层流
（1）Explanation of terms  （1）术语解释
Laminar flow：Flow is regarded as being the unidirectional movement of lamellae of fluid sliding over one another，with no macroscopic mixing or intermingling of the fluid in the radial direction．
层流：流动被视为流体层在相互滑动时的单向运动，且在径向上没有宏观混合或交融
Turbulent flow：In contrast to laminar flow，the fluid undergoes chaotic fluctuations，leading to intense mixing both in the axial direction and in the radial direction．
湍流：与层流相反，流体经历混乱的波动，导致在轴向和径向上都发生强烈的混合
Newton＇s law of viscosity：
牛顿粘度定律：
τ = μ d v x d y τ y x = − μ d v x d y τ = μ d v x d y τ y x = − μ d v x d y tau=mu(dv_(x))/(dy)quadtau_(yx)=-mu(dv_(x))/(dy)\tau=\mu \frac{d v_{x}}{d y} \quad \tau_{y x}=-\mu \frac{d v_{x}}{d y}
τ τ tau\tau－－－－－viscous shear or momentum flux， Pa
τ τ tau\tau －－－－－粘性剪切或动量通量，Pa
μ μ mu\mu－－－－－dynamic viscosity，Pa＇s
μ μ mu\mu －－－－－动态粘度，Pa·s

c．iviomentum Transport and Laminar Flow of Newtonian Fluids
c．牛顿流体的动量传输和层流

1）Explanation of terms  1）术语解释

Newton＇s law of viscosity： → τ y x = − μ d v x d y ⟹ Momentum transport is driven by velocity gradient． → τ y x = − μ d v x d y ⟹  Momentum transport is driven by   velocity gradient． rarr"tau_(yx)=-mu(dv_(x))/(dy)"Longrightarrow{:[" Momentum transport is driven by "],[" velocity gradient．"]:}\xrightarrow{\tau_{y x}=-\mu \frac{d v_{x}}{d y}} \Longrightarrow \begin{aligned} & \text { Momentum transport is driven by } \\ & \text { velocity gradient．}\end{aligned}．
牛顿粘度定律： → τ y x = − μ d v x d y ⟹ Momentum transport is driven by velocity gradient． → τ y x = − μ d v x d y ⟹  Momentum transport is driven by   velocity gradient． rarr"tau_(yx)=-mu(dv_(x))/(dy)"Longrightarrow{:[" Momentum transport is driven by "],[" velocity gradient．"]:}\xrightarrow{\tau_{y x}=-\mu \frac{d v_{x}}{d y}} \Longrightarrow \begin{aligned} & \text { Momentum transport is driven by } \\ & \text { velocity gradient．}\end{aligned}．
Reynolds number：  雷诺数：
Re = D × V 2 x 8 × ρ η = ( characteristic length ) × ( average velocity ) × ( density ) ( viscosity ) F g F η = ρ L 2 V x z 2 η V vzz L = ρ L V x z ― η = Re = Inertial force Viscous force Re = D × V 2 x 8 × ρ η = (  characteristic length  ) × (  average velocity  ) × (  density  ) (  viscosity  ) F g F η = ρ L 2 V x z 2 η V vzz L = ρ L V x z ¯ η = Re =  Inertial force   Viscous force  {:[Re=(D xxV_(2x8)xx rho)/(eta)],[=((" characteristic length ")xx(" average velocity ")xx(" density "))/((" viscosity "))],[(F_(g))/(F_(eta))=(rhoL^(2)V_(xz)^(2))/(etaV_(vzz)L)=(rho LV_( bar(xz)))/(eta)=Re=(" Inertial force ")/(" Viscous force ")]:}\begin{aligned} & \operatorname{Re}=\frac{D \times V_{2 x 8} \times \rho}{\eta} \\ & =\frac{(\text { characteristic length }) \times(\text { average velocity }) \times(\text { density })}{(\text { viscosity })} \\ & \frac{F_{\mathrm{g}}}{F_{\eta}}=\frac{\rho L^{2} V_{x z}^{2}}{\eta V_{\mathrm{vzz}} L}=\frac{\rho L V_{\overline{x z}}}{\eta}=\mathrm{Re}=\frac{\text { Inertial force }}{\text { Viscous force }} \end{aligned}
Viscosity of gases： μ = 2.6693 × 10 − 5 M T σ 2 Ω μ = 2.6693 × 10 − 5 M T σ 2 Ω mu=2.6693 xx10^(-5)(sqrt(MT))/(sigma^(2)Omega)\mu=2.6693 \times 10^{-5} \frac{\sqrt{M T}}{\sigma^{2} \Omega}
气体的粘度： μ = 2.6693 × 10 − 5 M T σ 2 Ω μ = 2.6693 × 10 − 5 M T σ 2 Ω mu=2.6693 xx10^(-5)(sqrt(MT))/(sigma^(2)Omega)\mu=2.6693 \times 10^{-5} \frac{\sqrt{M T}}{\sigma^{2} \Omega}
The viscosity of a gas is independent of the pressure of the gas and is a linear function of the square root of temperature．
气体的粘度与气体的压力无关，并且是温度平方根的线性函数

vivmentum Transport（Summary）
vivmentum Transport（摘要）

2．Momentum Transport and Laminar Flow of Newtonian Fluids
2．牛顿流体的动量传输和层流
（2）Analytical solutions  （2）解析解
Momentum balance based on a control volume（steady－state，constant density and viscosity）
基于控制体的动量平衡（稳态，恒定密度和粘度）
R Rate = = == Flux × × xx\times Area
R 速率 = = == 通量 × × xx\times 面积
◻ ◻ ◻\square Momentum includes convective momentum and viscous momentum
◻ ◻ ◻\square 动量包括对流动量和粘性动量
◻ ◻ ◻\square Forces include pressure and gravity
◻ ◻ ◻\square 力包括压力和重力
－Couette flow  －库埃特流
－Fluid flow between two parallel flat plates
－两个平行平板之间的流体流动
－Fluid flow down an inclined plane
－流体在倾斜平面上的流动
－Fluid flow in a cylindrical tube
－圆柱管中的流体流动

Momentum Transport（Summary）
动量传输（摘要）

2．Momentum Transport and Laminar Flow of Newtonian Fluids
2．牛顿流体的动量传输和层流
2）Analytical solutions  2）解析解
◻ ◻ ◻\square Couette flow   ◻ ◻ ◻\square 库埃特流
v x = V Y y τ z = − μ d v z d y } τ y x = − μ V Y v x = V Y y τ z = − μ d v z d y τ y x = − μ V Y {:[v_(x)=(V)/(Y)y],[tau_(z)=-mu(dv_(z))/(dy)]}tau_(yx)=-mu(V)/(Y)\left.\begin{array}{l} v_{x}=\frac{V}{Y} y \\ \tau_{z}=-\mu \frac{d v_{z}}{d y} \end{array}\right\} \tau_{y x}=-\mu \frac{V}{Y}
Application：viscometer  应用：粘度计

L．Momentum Transport and Laminar Flow of Newtonian Fluids
L．牛顿流体的动量传输和层流

2）Analytical solutions  2）解析解
I Fluid flow between two parallel flat plates
我在两个平行平板之间的流体流动
v x = Δ P 2 μ L ( δ 2 − y 2 ) τ y = − μ d v x d y } τ y = Δ P L y v z = m = v x | , − ∞ = Δ P 2 μ L δ 2 v x = 2 3 v x = Δ P 3 μ δ v x = Δ P 2 μ L δ 2 − y 2 τ y = − μ d v x d y τ y = Δ P L y      v z = m = v x , − ∞ = Δ P 2 μ L δ 2 v x = 2 3 v x = Δ P 3 μ δ {:[v_(x)=(Delta P)/(2mu L)(delta^(2)-y^(2))],[tau_(y)=-mu(dv_(x))/(dy)]}{:[tau_(y)=(Delta P)/(L)y,v_(z=m)=v_(x)|_(,-oo)=(Delta P)/(2mu L)delta^(2)],[v_(x)=(2)/(3)v_(x)=(Delta P)/(3mu)delta]:}\left.\begin{array}{ll} v_{x}=\frac{\Delta P}{2 \mu L}\left(\delta^{2}-y^{2}\right) \\ \tau_{y}=-\mu \frac{d v_{x}}{d y} \end{array}\right\} \begin{array}{ll} \tau_{y}=\frac{\Delta P}{L} y & v_{z=m}=\left.v_{x}\right|_{,-\infty}=\frac{\Delta P}{2 \mu L} \delta^{2} \\ v_{x}=\frac{2}{3} v_{x}=\frac{\Delta P}{3 \mu} \delta \end{array}
c．iviumentum Transport and Laminar Flow of Newtonian Fluids
c．iviumentum 纽顿流体的运输和层流
2）Analytical solutions  2）解析解
－Fluid flow down an inclined plane
－流体在倾斜平面上的流动
v x = ρ g cos ⁡ θ 2 μ ( δ 2 − y 2 ) τ y x = − μ d v x d y } τ y z = ρ g y cos ⁡ θ v x = ρ g cos ⁡ θ 2 μ δ 2 − y 2 τ y x = − μ d v x d y τ y z = ρ g y cos ⁡ θ {:[v_(x)=(rho g cos theta)/(2mu)(delta^(2)-y^(2))],[tau_(yx)=-mu(dv_(x))/(dy)]}tau_(yz)=rho gy cos theta\left.\begin{array}{l} v_{x}=\frac{\rho g \cos \theta}{2 \mu}\left(\delta^{2}-y^{2}\right) \\ \tau_{y x}=-\mu \frac{d v_{x}}{d y} \end{array}\right\} \tau_{y z}=\rho g y \cos \theta
v x , m a x = v x | y − 0 = ρ g cos ⁡ θ 2 μ δ 2 v x , m a x = v x y − 0 = ρ g cos ⁡ θ 2 μ δ 2 v_(x,max)=v_(x)|_(y-0)=(rho g cos theta)/(2mu)delta^(2)v_{x, m a x}=\left.v_{x}\right|_{y-0}=\frac{\rho g \cos \theta}{2 \mu} \delta^{2}
v ¯ x = 2 3 v x , m a x = ρ g cos ⁡ θ 3 μ δ 2 v ¯ x = 2 3 v x , m a x = ρ g cos ⁡ θ 3 μ δ 2 bar(v)_(x)=(2)/(3)v_(x,max)=(rho g cos theta)/(3mu)delta^(2)\bar{v}_{x}=\frac{2}{3} v_{x, m a x}=\frac{\rho g \cos \theta}{3 \mu} \delta^{2}

Nomentum Transport（Summary）
Nomentum Transport（摘要）

2．Momentum Transport and Laminar Flow of Newtonian Fluids
2．牛顿流体的动量传输和层流
（2）Analytical solutions  （2）解析解
Fluid flow in a cylindrical tube（fully developed flow）
圆柱管中的流体流动（完全发展流动）
2）Analytical solutions  2）解析解

Couette
flow  库埃特流动

Fluid flow between two parallel flat plates
两个平行平板之间的流体流动
v x = V Y y τ μ = − μ V Y v x , m x = V v z = 0 = Δ P 2 μ L δ 2 v ¯ x = 2 3 v x = v ¯ z = 1 2 v z = x ¯ ν x = Δ P 2 μ L ( δ 2 − y 2 ) v z = ρ G cos ⁡ θ 2 μ ( δ 2 − y 2 ) v z = ( Δ P L ± ρ g cos ⁡ θ ) R 2 − r 2 4 μ τ y z = Δ P L y τ y z = ρ g y cos ⁡ θ τ r = ( Δ P L ± ρ g cos ⁡ θ ) r 2 v x = π = ρ g cos ⁡ θ 2 μ δ 2 v z = = ( Δ P L ± ρ g cos ⁡ θ ) R 2 4 μ v ¯ x = 2 3 v z = v ¯ z = 1 2 v z = v x = V Y y τ μ = − μ V Y v x , m x = V v z = 0 = Δ P 2 μ L δ 2 v ¯ x = 2 3 v x = v ¯ z = 1 2 v z = x ¯ ν x = Δ P 2 μ L δ 2 − y 2 v z = ρ G cos ⁡ θ 2 μ δ 2 − y 2 v z = Δ P L ± ρ g cos ⁡ θ R 2 − r 2 4 μ τ y z = Δ P L y τ y z = ρ g y cos ⁡ θ τ r = Δ P L ± ρ g cos ⁡ θ r 2 v x = π = ρ g cos ⁡ θ 2 μ δ 2 v z = = Δ P L ± ρ g cos ⁡ θ R 2 4 μ v ¯ x = 2 3 v z = v ¯ z = 1 2 v z = {:[v_(x)=(V)/(Y)y],[tau_(mu)=-mu(V)/(Y)],[v_(x,mx)=V quadv_(z=0)=(Delta P)/(2mu L)delta^(2)],[ bar(v)_(x)=(2)/(3)v_(x)=],[ bar(v)_(z)=(1)/(2)v_(z)= bar(x)],[nu_(x)=(Delta P)/(2mu L)(delta^(2)-y^(2))],[v_(z)=(rho G cos theta)/(2mu)(delta^(2)-y^(2))],[v_(z)=((Delta P)/(L)+-rho g cos theta)((R^(2)-r^(2))/(4mu))],[tau_(yz)=(Delta P)/(L)y],[tau_(yz)=rho gy cos theta],[tau_(r)=((Delta P)/(L)+-rho g cos theta)((r)/(2))],[v_(x=pi)=(rho g cos theta)/(2mu)delta^(2)],[v_(z=)=((Delta P)/(L)+-rho g cos theta)((R^(2))/(4mu))],[ bar(v)_(x)=(2)/(3)v_(z)=],[ bar(v)_(z)=(1)/(2)v_(z=)]:}\begin{aligned} & v_{x}=\frac{V}{Y} y \\ & \tau_{\mu}=-\mu \frac{V}{Y} \\ & v_{x, m x}=V \quad v_{z=0}=\frac{\Delta P}{2 \mu L} \delta^{2} \\ & \bar{v}_{\mathrm{x}}=\frac{2}{3} v_{\mathrm{x}}= \\ & \bar{v}_{\mathrm{z}}=\frac{1}{2} v_{\mathrm{z}}=\bar{x} \\ & \nu_{x}=\frac{\Delta P}{2 \mu L}\left(\delta^{2}-y^{2}\right) \\ & v_{z}=\frac{\rho G \cos \theta}{2 \mu}\left(\delta^{2}-y^{2}\right) \\ & v_{z}=\left(\frac{\Delta P}{L} \pm \rho g \cos \theta\right) \frac{R^{2}-r^{2}}{4 \mu} \\ & \tau_{y z}=\frac{\Delta P}{L} y \\ & \tau_{y z}=\rho g y \cos \theta \\ & \tau_{r}=\left(\frac{\Delta P}{L} \pm \rho g \cos \theta\right) \frac{r}{2} \\ & v_{x=\pi}=\frac{\rho g \cos \theta}{2 \mu} \delta^{2} \\ & v_{z=}=\left(\frac{\Delta P}{L} \pm \rho g \cos \theta\right) \frac{R^{2}}{4 \mu} \\ & \bar{v}_{x}=\frac{2}{3} v_{z}= \\ & \bar{v}_{z}=\frac{1}{2} v_{z=} \end{aligned}

Viomentum Transport（Summary）
Viomentum Transport（摘要）

3．Equations of Continuity and Conservation of Momentum and Fluid Flow Past Submerged Objects
3．连续性方程、动量守恒方程及流体流过沉没物体的流动
1）Explanation of terms  1）术语解释
Incompressible fluid：A fluid that has constant density（ ρ = ρ = rho=\rho= const．）
不可压缩流体：一种具有恒定密度（ ρ = ρ = rho=\rho= const.）的流体
Entry length at entrance to a pipe：The distance a flow travels after entering a pipe before the flow becomes fully developed．
管道入口的入口长度：流体进入管道后在流动完全发展之前流动的距离
2）Equations  2）方程式

Equation of continuily  连续性方程

∂ ρ ∂ t + ∇ ⋅ ( ρ v ) = 0 ∇ ⋅ v = 0 ∂ ρ ∂ t + ∇ ⋅ ( ρ v ) = 0 ∇ ⋅ v = 0 (del rho)/(del t)+grad*(rhov)=0quad grad*v=0\frac{\partial \rho}{\partial t}+\nabla \cdot(\rho \mathrm{v})=0 \quad \nabla \cdot \mathrm{v}=0
Equation of momentum conservation
动量守恒方程
∂ ∂ t ρ v = − ∇ ⋅ τ ¯ ― − ∇ ⋅ ρ v v − ∇ P + ρ g ∂ ∂ t ρ v = − ∇ ⋅ τ ¯ ¯ − ∇ ⋅ ρ v v − ∇ P + ρ g (del)/(del t)rhov=-grad* bar(bar(tau))-grad*rhovv-grad P+rhog\frac{\partial}{\partial t} \rho \mathbf{v}=-\nabla \cdot \overline{\bar{\tau}}-\nabla \cdot \rho \mathbf{v} \mathbf{v}-\nabla P+\rho \mathbf{g}
◻ ◻ ◻\square Accumulation   ◻ ◻ ◻\square 积累
◻ ◻ ◻\square Viscous   ◻ ◻ ◻\square 粘性
－Convective  －对流
◻ ◻ ◻\square Surface force   ◻ ◻ ◻\square 表面力
◻ ◻ ◻\square Body force   ◻ ◻ ◻\square 体力
3．Equations of Continuity and Conservation of Momentum and Fluid Flow Past Submerged Objects
3．连续性方程、动量守恒方程及流体流过沉没物体的流动
2）Equations  2）方程式
ρ D v D t = η ∇ 2 v − ∇ P + ρ g { ρ D v z D t = η ( ∂ 2 v z ∂ x 2 + σ ^ 2 v z ∂ y 2 + σ ^ 2 v z σ ^ 2 ) − σ ^ P ∂ x + ρ g z ρ D v y D t = η ( σ ^ 2 v y ∂ x 2 + σ ^ 2 v y ∂ y 2 + σ ^ 2 v y a ^ 2 ) − σ ^ P ∂ ^ y + ρ g y ρ D v z D t = η ( ∂ 2 v z ∂ x ^ x 2 + σ ^ 2 v z ∂ y 2 + σ ^ 2 v z z ^ 2 ) − ∂ ^ P c ^ z + ρ g ρ D v D t = η ∇ 2 v − ∇ P + ρ g ρ D v z D t = η ∂ 2 v z ∂ x 2 + σ ^ 2 v z ∂ y 2 + σ ^ 2 v z σ ^ 2 − σ ^ P ∂ x + ρ g z ρ D v y D t = η σ ^ 2 v y ∂ x 2 + σ ^ 2 v y ∂ y 2 + σ ^ 2 v y a ^ 2 − σ ^ P ∂ ^ y + ρ g y ρ D v z D t = η ∂ 2 v z ∂ x ^ x 2 + σ ^ 2 v z ∂ y 2 + σ ^ 2 v z z ^ 2 − ∂ ^ P c ^ z + ρ g rho(Dv)/(Dt)=etagrad^(2)v-grad P+rhog{[rho(Dv_(z))/(Dt)=eta((del^(2)v_(z))/(delx^(2))+( hat(sigma)^(2)v_(z))/(dely^(2))+( hat(sigma)^(2)v_(z))/( hat(sigma)^(2)))-(( hat(sigma))P)/(del x)+rhog_(z)],[rho(Dv_(y))/(Dt)=eta(( hat(sigma)^(2)v_(y))/(delx^(2))+( hat(sigma)^(2)v_(y))/(dely^(2))+( hat(sigma)^(2)v_(y))/( hat(a)^(2)))-(( hat(sigma))P)/(( hat(del))y)+rhog_(y)],[rho(Dv_(z))/(Dt)=eta((del^(2)v_(z))/(( hat(del x))x^(2))+( hat(sigma)^(2)v_(z))/(dely^(2))+( hat(sigma)^(2)v_(z))/( hat(z)^(2)))-(( hat(del))P)/(( hat(c))z)+rho g]:}\rho \frac{D \mathbf{v}}{D t}=\eta \nabla^{2} \mathbf{v}-\nabla P+\rho \mathbf{g}\left\{\begin{array}{l} \rho \frac{D v_{z}}{D t}=\eta\left(\frac{\partial^{2} v_{z}}{\partial x^{2}}+\frac{\hat{\sigma}^{2} v_{z}}{\partial y^{2}}+\frac{\hat{\sigma}^{2} v_{z}}{\hat{\sigma}^{2}}\right)-\frac{\hat{\sigma} P}{\partial x}+\rho g_{z} \\ \rho \frac{D v_{y}}{D t}=\eta\left(\frac{\hat{\sigma}^{2} v_{y}}{\partial x^{2}}+\frac{\hat{\sigma}^{2} v_{y}}{\partial y^{2}}+\frac{\hat{\sigma}^{2} v_{y}}{\hat{a}^{2}}\right)-\frac{\hat{\sigma} P}{\hat{\partial} y}+\rho g_{y} \\ \rho \frac{D v_{z}}{D t}=\eta\left(\frac{\partial^{2} v_{z}}{\hat{\partial x} x^{2}}+\frac{\hat{\sigma}^{2} v_{z}}{\partial y^{2}}+\frac{\hat{\sigma}^{2} v_{z}}{\hat{z}^{2}}\right)-\frac{\hat{\partial} P}{\hat{c} z}+\rho g \end{array}\right.
Prandtl boundary layer equation：
普朗特边界层方程：
∂ ^ v x ∂ x + ∂ ^ v y ∂ y ^ = 0 v x = v y = 0 ( a y = 0 v x ∂ ^ v z ∂ x + v y ∂ v x ∂ y ∂ y = v ∂ ^ 2 v x ∂ y 2 v x = 0.99 v x ( a y = δ ∂ ^ v x ∂ x + ∂ ^ v y ∂ y ^ = 0      v x = v y = 0      ( a y = 0 v x ∂ ^ v z ∂ x + v y ∂ v x ∂ y ∂ y = v ∂ ^ 2 v x ∂ y 2      v x = 0.99 v x      ( a y = δ {:[(( hat(del))v_(x))/(del x)+(( hat(del))v_(y))/(( hat(del y)))=0,v_(x)=v_(y)=0,(ay=0],[v_(x)(( hat(del))v_(z))/(del x)+v_(y)(delv_(x))/((del y)/(del y)=v( hat(del)^(2)v_(x))/(dely^(2))),v_(x)=0.99v_(x),(ay=delta]:}\begin{array}{lll} \frac{\hat{\partial} v_{x}}{\partial x}+\frac{\hat{\partial} v_{y}}{\hat{\partial y}}=0 & v_{x}=v_{y}=0 & (a y=0 \\ v_{x} \frac{\hat{\partial} v_{z}}{\partial x}+v_{y} \frac{\partial v_{x}}{\frac{\partial y}{\partial y}=v \frac{\hat{\partial}^{2} v_{x}}{\partial y^{2}}} & v_{x}=0.99 v_{x} & (a y=\delta \end{array}
3．Equations of Continuity and Conservation of Momentum and Fluid Flow Past Submerged Objects
3．连续性方程、动量守恒方程及流体流过沉没物体的流动
2）Equations  2）方程式

Blasius＇s exact solution：
Blasius 的精确解：

δ x = 5.0 Re x δ x = 5.0 Re x (delta )/(x)=(5.0)/(sqrt(Re_(x)))\frac{\delta}{x}=\frac{5.0}{\sqrt{\mathrm{Re}_{x}}}
C x = 0.664 Re x C x = 0.664 Re x C_(x)=(0.664)/(sqrt(Re_(x)))C_{x}=\frac{0.664}{\sqrt{\mathrm{Re}_{\mathrm{x}}}}
C ¯ L = 1.328 Re L C ¯ L = 1.328 Re L bar(C)_(L)=(1.328)/(sqrt(Re_(L)))\bar{C}_{L}=\frac{1.328}{\sqrt{\operatorname{Re}_{L}}}
Approximate solution：  近似解：
δ x = 4.641 Re x δ x = 4.641 Re x (delta )/(x)=(4.641)/(sqrt(Re_(x)))\frac{\delta}{x}=\frac{4.641}{\sqrt{\mathrm{Re}_{x}}}
C x = 0.646 Re x C x = 0.646 Re x C_(x)=(0.646)/(sqrt(Re_(x)))C_{x}=\frac{0.646}{\sqrt{\operatorname{Re}_{x}}}
C ¯ L = 1.292 Re L C ¯ L = 1.292 Re L bar(C)_(L)=(1.292)/(sqrt(Re_(L)))\bar{C}_{L}=\frac{1.292}{\sqrt{\mathrm{Re}_{L}}}
v x = 3 2 v x δ y − 1 2 v x δ 3 y 3 τ xall = − τ y x | y = 0 = μ ∂ v x ∂ y | y = 0 ∂ v x ∂ y | y = 0 = 0 F = C ¯ L 1 2 ρ v x 2 W L v x = 3 2 v x δ y − 1 2 v x δ 3 y 3 τ xall = − τ y x y = 0 = μ ∂ v x ∂ y y = 0 ∂ v x ∂ y y = 0 = 0 F = C ¯ L 1 2 ρ v x 2 W L {:[v_(x)=(3)/(2)(v_(x))/(delta)y-(1)/(2)(v_(x))/(delta^(3))y^(3)],[tau_(xall)=-tau_(yx)|_(y=0)= mu(delv_(x))/(del y)|_(y=0)],[(delv_(x))/(del y)|_(y=0)=0],[F= bar(C)_(L)(1)/(2)rhov_(x)^(2)WL]:}\begin{aligned} & v_{x}=\frac{3}{2} \frac{v_{x}}{\delta} y-\frac{1}{2} \frac{v_{x}}{\delta^{3}} y^{3} \\ & \tau_{\mathrm{xall}}=-\left.\tau_{y x}\right|_{y=0}=\left.\mu \frac{\partial v_{x}}{\partial y}\right|_{y=0} \\ & \left.\frac{\partial v_{x}}{\partial y}\right|_{y=0}=0 \\ & F=\bar{C}_{L} \frac{1}{2} \rho v_{x}^{2} W L \end{aligned}

Stokes＇s law：  斯托克斯定律：

4 3 π R 3 ( ρ 2 − ρ ) g = 6 π R μ ν t Re = ρ D v t μ < 0.1 4 3 π R 3 ρ 2 − ρ g = 6 π R μ ν t Re = ρ D v t μ < 0.1 (4)/(3)piR^(3)(rho_(2)-rho)g=6pi R munu_(t)quadRe=(rho Dv_(t))/(mu) < 0.1\frac{4}{3} \pi R^{3}\left(\rho_{2}-\rho\right) g=6 \pi R \mu \nu_{\mathrm{t}} \quad \mathrm{Re}=\frac{\rho D v_{\mathrm{t}}}{\mu}<0.1

4．Turbulent Flow  4．湍流

（1）Friction factor（flow within pipes）
（1）摩擦因子（管道内流动）
Δ P = ∫ 1 2 p v ¯ − 2 4 L D Δ P = ∫ 1 2 p v ¯ − 2 4 L D Delta P=int(1)/(2)p bar(v)^(-2)(4L)/(D)\Delta P=\int \frac{1}{2} p \bar{v}^{-2} \frac{4 L}{D}
v ¯ v max = 0.817 v ¯ v max = 0.817 (( bar(v)))/(v_(max))=0.817\frac{\bar{v}}{v_{\max }}=0.817
Smooth－walled circular pipes：
光滑壁圆形管：
As for laminar flows ( Re < 2100 ) f = 16 Re ( Re < 2100 ) f = 16 Re (Re < 2100)quad f=(16)/(Re)(\operatorname{Re}<2100) \quad f=\frac{16}{\mathrm{Re}}
关于层流 ( Re < 2100 ) f = 16 Re ( Re < 2100 ) f = 16 Re (Re < 2100)quad f=(16)/(Re)(\operatorname{Re}<2100) \quad f=\frac{16}{\mathrm{Re}}
Blasius one－seventh power law
布拉修斯一七次方定律
As for turbulent flows ( 2100 < Re < 10 5 ) f = 0.0791 Re 025 2100 < Re < 10 5 f = 0.0791 Re 025 (2100 < Re < 10^(5))quad f=(0.0791)/(Re^(025))\left(2100<\operatorname{Re}<10^{5}\right) \quad f=\frac{0.0791}{\operatorname{Re}^{025}}
关于湍流 ( 2100 < Re < 10 5 ) f = 0.0791 Re 025 2100 < Re < 10 5 f = 0.0791 Re 025 (2100 < Re < 10^(5))quad f=(0.0791)/(Re^(025))\left(2100<\operatorname{Re}<10^{5}\right) \quad f=\frac{0.0791}{\operatorname{Re}^{025}}
As for turbulent flows ( 3 × 10 3 < Re < 10 8 ) 1 f = 4 log ⁡ ( 2 Re f ) − 1.6 3 × 10 3 < Re < 10 8 1 f = 4 log ⁡ ( 2 Re f ) − 1.6 (3xx10^(3) < Re < 10^(8))(1)/(sqrtf)=4log(2Resqrtf)-1.6\left(3 \times 10^{3}<\operatorname{Re}<10^{8}\right) \frac{1}{\sqrt{f}}=4 \log (2 \operatorname{Re} \sqrt{f})-1.6
关于湍流 ( 3 × 10 3 < Re < 10 8 ) 1 f = 4 log ⁡ ( 2 Re f ) − 1.6 3 × 10 3 < Re < 10 8 1 f = 4 log ⁡ ( 2 Re f ) − 1.6 (3xx10^(3) < Re < 10^(8))(1)/(sqrtf)=4log(2Resqrtf)-1.6\left(3 \times 10^{3}<\operatorname{Re}<10^{8}\right) \frac{1}{\sqrt{f}}=4 \log (2 \operatorname{Re} \sqrt{f})-1.6
Rough－walled circular pipes：
粗壁圆形管道：
As for laminar flows ( Re < 2100 ) : f = 16 Re ( Re < 2100 ) : f = 16 Re (Re < 2100):quad f=(16 )/(Re)(\operatorname{Re}<2100): \quad f=\frac{16}{\operatorname{Re}}
关于层流 ( Re < 2100 ) : f = 16 Re ( Re < 2100 ) : f = 16 Re (Re < 2100):quad f=(16 )/(Re)(\operatorname{Re}<2100): \quad f=\frac{16}{\operatorname{Re}}
As for turbulent flows ( Re > 2100 ) : 1 f = − 4 log ⁡ [ 1.26 Re f + ε 3.7 D ] ( Re > 2100 ) : 1 f = − 4 log ⁡ 1.26 Re f + ε 3.7 D (Re > 2100):(1)/(sqrtf)=-4log[(1.26)/(Resqrtf)+(epsi)/(3.7 D)](\operatorname{Re}>2100): \frac{1}{\sqrt{f}}=-4 \log \left[\frac{1.26}{\operatorname{Re} \sqrt{f}}+\frac{\varepsilon}{3.7 D}\right] or 1 f = − 3.6 log ⁡ [ 6.9 Re + ( ε 3.7 D ) LII ] 2 : 1 f = − 3.6 log ⁡ 6.9 Re + ε 3.7 D LII 2 : (1)/(sqrtf)=-3.6 log [(6.9 )/(Re)+((epsi)/(3.7 D))^(LII)]_(2:)\frac{1}{\sqrt{f}}=-3.6 \log \left[\frac{6.9}{\operatorname{Re}}+\left(\frac{\varepsilon}{3.7 D}\right)^{\mathrm{LII}}\right]_{2:}
至于湍流 ( Re > 2100 ) : 1 f = − 4 log ⁡ [ 1.26 Re f + ε 3.7 D ] ( Re > 2100 ) : 1 f = − 4 log ⁡ 1.26 Re f + ε 3.7 D (Re > 2100):(1)/(sqrtf)=-4log[(1.26)/(Resqrtf)+(epsi)/(3.7 D)](\operatorname{Re}>2100): \frac{1}{\sqrt{f}}=-4 \log \left[\frac{1.26}{\operatorname{Re} \sqrt{f}}+\frac{\varepsilon}{3.7 D}\right] 或 1 f = − 3.6 log ⁡ [ 6.9 Re + ( ε 3.7 D ) LII ] 2 : 1 f = − 3.6 log ⁡ 6.9 Re + ε 3.7 D LII 2 : (1)/(sqrtf)=-3.6 log [(6.9 )/(Re)+((epsi)/(3.7 D))^(LII)]_(2:)\frac{1}{\sqrt{f}}=-3.6 \log \left[\frac{6.9}{\operatorname{Re}}+\left(\frac{\varepsilon}{3.7 D}\right)^{\mathrm{LII}}\right]_{2:}

Iviomentum Transport（Summary）
Iviomentum Transport（摘要）

4．Turbulent Flow  4．湍流
2）Friction factor（flow over flat plates）
2）摩擦因子（平板流动）
3）Friction factor（flow across spheres／cylinders）
3）摩擦因子（流经球体／圆柱体）
F = f A 1 2 ρ v 2 2 = f π 2 2 1 2 ρ t 2 F = f A 1 2 ρ v 2 2 = f π 2 2 1 2 ρ t 2 F=fA(1)/(2)rhov_(2)^(2)=fpi_(2)^(2)(1)/(2)rho_(t)^(2)F=f A \frac{1}{2} \rho v_{2}^{2}=f \pi_{2}^{2} \frac{1}{2} \rho_{t}^{2}
f = 24 Re f = 24 Re f=(24 )/(Re)f=\frac{24}{\operatorname{Re}}
f = 18.5 Re 0.6 f = 18.5 Re 0.6 f=(18.5)/(Re^(0.6))f=\frac{18.5}{\operatorname{Re}^{0.6}}
f = 0.44 f = 0.44 f=0.44f=0.44

Momentum Transport（Summary）
动量传输（摘要）

5．Mechanical Energy Balance and Its Application to Fluid Flow
5．机械能平衡及其在流体流动中的应用
＂1）Modified Bernoulli equation
1）修正的伯努利方程
{ P 1 ρ + 1 2 β v ¯ 1 2 + g z 1 + α ^ − E f = P 2 ρ + 1 2 β v ¯ 2 2 + g z 2 β = { 0.5 Laminar 1 Turbulent P 1 ρ + 1 2 β v ¯ 1 2 + g z 1 + α ^ − E f = P 2 ρ + 1 2 β v ¯ 2 2 + g z 2 β = 0.5  Laminar  1  Turbulent  {[(P_(1))/(rho)+(1)/(2beta) bar(v)_(1)^(2)+gz_(1)+ hat(alpha)-E_(f)=(P_(2))/(rho)+(1)/(2beta) bar(v)_(2)^(2)+gz_(2)],[beta={[0.5," Laminar "],[1," Turbulent "]:}]:}\left\{\begin{array}{l} \frac{P_{1}}{\rho}+\frac{1}{2 \beta} \bar{v}_{1}^{2}+g z_{1}+\hat{\alpha}-E_{\mathrm{f}}=\frac{P_{2}}{\rho}+\frac{1}{2 \beta} \bar{v}_{2}^{2}+g z_{2} \\ \beta= \begin{cases}0.5 & \text { Laminar } \\ 1 & \text { Turbulent }\end{cases} \end{array}\right.
Friction loss { wall shear E f = f 4 L D 1 2 v ¯ 2 bends and fittings E f = f 4 L e D 1 2 v ¯ 2 changes in radius E f = e f 1 2 v ¯ s 2  wall shear       E f = f 4 L D 1 2 v ¯ 2  bends and fittings       E f = f 4 L e D 1 2 v ¯ 2  changes in radius       E f = e f 1 2 v ¯ s 2 {[" wall shear ",E_(f)=f((4L)/(D))(1)/(2) bar(v)^(2)],[" bends and fittings ",E_(f)=f((4L_(e))/(D))(1)/(2) bar(v)^(2)],[" changes in radius ",E_(f)=e_(f)(1)/(2) bar(v)_(s)^(2)]:}\begin{cases}\text { wall shear } & E_{\mathrm{f}}=f \frac{4 L}{D} \frac{1}{2} \bar{v}^{2} \\ \text { bends and fittings } & E_{\mathrm{f}}=f \frac{4 L_{\mathrm{e}}}{D} \frac{1}{2} \bar{v}^{2} \\ \text { changes in radius } & E_{\mathrm{f}}=e_{\mathrm{f}} \frac{1}{2} \bar{v}_{\mathrm{s}}^{2}\end{cases}  摩擦损失 { wall shear E f = f 4 L D 1 2 v ¯ 2 bends and fittings E f = f 4 L e D 1 2 v ¯ 2 changes in radius E f = e f 1 2 v ¯ s 2  wall shear       E f = f 4 L D 1 2 v ¯ 2  bends and fittings       E f = f 4 L e D 1 2 v ¯ 2  changes in radius       E f = e f 1 2 v ¯ s 2 {[" wall shear ",E_(f)=f((4L)/(D))(1)/(2) bar(v)^(2)],[" bends and fittings ",E_(f)=f((4L_(e))/(D))(1)/(2) bar(v)^(2)],[" changes in radius ",E_(f)=e_(f)(1)/(2) bar(v)_(s)^(2)]:}\begin{cases}\text { wall shear } & E_{\mathrm{f}}=f \frac{4 L}{D} \frac{1}{2} \bar{v}^{2} \\ \text { bends and fittings } & E_{\mathrm{f}}=f \frac{4 L_{\mathrm{e}}}{D} \frac{1}{2} \bar{v}^{2} \\ \text { changes in radius } & E_{\mathrm{f}}=e_{\mathrm{f}} \frac{1}{2} \bar{v}_{\mathrm{s}}^{2}\end{cases}
set 限制解除