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#유체역학2, #유체역학의이해, #베르누이방정식, # Navier-Stokes Equation, Navier-Stokes, Navier-Stokes 방정식, #권오붕,
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나비에 스토크스 방정식 유도 (Navier-Stokes equations) 이해하기
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Navier-Stokes 방정식 – 1
Navier-Stokes 방정식은 뉴톤 제2법칙으로부터 유도될 수 있다. 공기를 비롯한 유체는 고체와 달리 정해진 모양이 없기 때문에 뉴톤 제2법칙을 적용 …
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주제와 관련된 이미지 navier stokes equation 유도
주제와 관련된 더 많은 사진을 참조하십시오 유체역학2 – 유체역학의이해 3 12 Navier-Stokes Equation. 댓글에서 더 많은 관련 이미지를 보거나 필요한 경우 더 많은 관련 기사를 볼 수 있습니다.
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나비에 스토크스 방정식 유도 (Navier-Stokes equations) 이해하기
송도방랑객
나비에 스토크스 방정식에 대해 이해한 바를 정리하고자 합니다.
그 전에 기본적으로 알아야 할 사항입니다.
나비에 스토크스 방정식을 한 문장으로 표현하면 아래와 같습니다.
‘점성을 가진 유체에 작용하는 힘과 운동량의 변화를 기술하는 비선형 편미분 방정식’
‘점탄성이 없는 유체(뉴턴 유체)에 대한 일반적인 운동 방정식’
(점탄성과 점성은 다릅니다)
나비에 스토크스 방정식은
유도 과정이 복잡하고
그 중간중간 단순화를 가정하는 것이 많아
공식이 여러가지가 있지만
가장 보기 편한 것이 아래의 식 같습니다.
위 식을 이해하기 위해서
먼저 알아야 할 것이 있습니다.
수학적인 내용입니다.
——————————————————-
미소 유체 시스템에 Newton의 제 2법칙(F=ma)을 적용하기 위해
유동의 가속도 벡터장 a 를 계산하여야 합니다.
속도 벡터 V의 전시간 미분을 계산하면 아래와 같습니다.
(일반적으로 d는 전미분, ∂는 편미분이라고 생각하면 된다.)
여기서 u, v, w 는 시간 t에 영향을 받는 함수이므로, 아래와 같이 표기 됩니다.
이를 전미분이라 하며, 편미분처럼 다른 변수를 상수로 취급하여 단순 미분해버리면 안됩니다.
즉, x축 속도벡터인 u를 보면, 아래와 같습니다.
위에서 x축에 대한 가속도를 구했지만
유체는 3차원 공간에서 흐르니 x, y, z 다 필요하겠죠.
따라서 이를 x, y, z축으로 확장시키면 아래와 같습니다.
이게 바로 나비에 스토크스 방정식의 좌변이 2가지로 쓰이는 이유입니다.
이제 본격적으로 들어갑니다.
미소 유체 시스템에 가해지는 힘은 중력에 의한 힘과, 미소체적의 x, y, z 표면에 가해지는 힘 두 가지가 있습니다.
이를 표현하면 아래와 같습니다.
즉,
그렇다면, 저 각각의 힘을 어떻게 구할까요?
중력에 의한 힘 Fgrav 는 그래도 간단합니다.
미소구간 표면에 가해지는 외력을 구하는게 조금 어려운데,
아래를 잘 살펴보세요
이제 나비에 스토크스 방정식을 완성하기 위해
점성력 까지 고려해 줍니다.
좀 더 볼까요?
이를 바탕으로, x축에 대한 표면력을 먼저 구하면
이를 또 3차원으로 확장하면
이제 위에서 봤던 식
에 위에서 구한것들을 넣어줍시다.
이를 다시 표현하면
최종적으로 처음 봤던 식이 다시 나오져
의미를 살펴보면 아래와 같다고 할 수 있습니다.
응력 텀에 점성력까지 반영 돼 있고
체적력이 뭐 중력에 의한 힘이겠죠?
여기까지가 뉴턴유체, 비압축성 유체에 대한 나비에 스토크스 방정식
유도라고 할 수 있겠고, 비점성이라면 좀 더 단순해지구요
압축성을 고려하면 좀 더 복잡해지게 됩니다.
Navier Stokes Equation 유도 | [박재우] 유체역학 1 – Navier-Stokes 방정식 20 개의 정답
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Navier-Stokes 방정식 – 1
Navier-Stokes 방정식은 뉴톤 제2법칙으로부터 유도될 수 있다. 공기를 비롯한 유체는 고체와 달리 정해진 모양이 없기 때문에 뉴톤 제2법칙을 적용하기 위해서는 특별한 아이디어가 필요하다.
공기와 같은 속도로 움직이는 미소 유체요소(infinitesimal fluid element)를 생각해보자. 이 유체요소는 일정한 질량을 가지고 있으며, 질량을 유지하기 위해서 부피는 변할 수 있다고 가정한다. 이 유체요소를 질점으로 보면 뉴톤 제2법칙을 적용할 수 있다.
이 유체요소에 작용하는 힘은 체적력(body force), 압력, 그리고 점성력(viscous force)이 있다. 먼저 체적력에는 대표적으로 중력이 있으며 이밖에 관성력과 전자기력 등이 있다. 체적력 \(d\mathbf{F}_b\) 를 단위 질량당 체적력인 \(\mathbf{f}\) 를 이용해 표현하면 다음과 같다.
\[ d\mathbf{F}_b=dxdydz \ (\rho f_x \ \mathbf{i} + \rho f_y \ \mathbf{j} + \rho f_z \ \mathbf{k} ) \tag{1} \]
가 된다. 여기서 \(dx, dy, dz\) 는 유체요소의 각 변의 길이이며 \(\rho\) 는 밀도, \(f_x, f_y, f_z\) 는 단위질량장 체적력의 각 축 성분이다.
압력은 유체요소의 표면에 작용하는 표면력(surface force)으로서 표면에 수직으로 작용한다. 먼저 그림에 표시된 대로 \(x\) 축 방향의 압력을 계산해 보자.
\[ \begin{align} dF_{px} &= p \ dydz – \left( p+ \frac{\partial p}{\partial x} dx \right) dydz \\ \\ &= – \frac{\partial p}{\partial x} dxdydz \end{align} \]
\(y, z\) 축 방향도 같은 방법으로 계산하면 유체요소에 작용하는 총 압력 \(d \mathbf{F}_p\) 는 다음과 같다.
\[ d \mathbf{F}_p= -dxdydz \left( \frac{\partial p}{\partial x} \ \mathbf{i} +\frac{\partial p}{\partial y} \ \mathbf{j} + \frac{\partial p}{\partial z} \ \mathbf{k} \right) \tag{2} \]
점성력도 유체요소의 표면에 작용하는 표면력으로서 표면에 수직으로 작용하는 수직응력(normal stress)과 평행하게 작용하는 전단응력(shear stress)로 나눠진다.
먼저 그림에 표시된 대로 \(x\) 축 방향의 점성력을 계산해 보자.
\[ \begin{align} dF_{vx} &= \left( \tau_{xx} + \frac{\partial \tau_{xx}}{\partial x} dx \right) dydz – \tau_{xx} dydz \\ \\ & \ \ \ + \left( \tau_{yx} + \frac{\partial \tau_{yx}}{\partial y} dy \right) dxdz – \tau_{yx} dxdz \\ \\ & \ \ \ + \left( \tau_{zx} + \frac{\partial \tau_{zx}}{\partial z} dz \right) dxdy – \tau_{zx} dxdy \\ \\ &= dxdydz \ \left( \frac{\partial \tau_{xx}}{\partial x} +\frac{\partial \tau_{yx}}{\partial y}+ \frac{\partial \tau_{zx}}{\partial z} \right) \end{align} \]
\(y, z\) 축 방향도 같은 방법으로 계산하면 유체요소에 작용하는 총 점성력 \(d\mathbf{F}_v\) 는 다음과 같다.
\[ \begin{align} d\mathbf{F}_v &= dxdydz \ \left( \frac{\partial \tau_{xx}}{\partial x} +\frac{\partial \tau_{yx}}{\partial y}+ \frac{\partial \tau_{zx}}{\partial z} \right) \ \mathbf{i} \tag{3} \\ \\ & \ \ \ + dxdydz \ \left( \frac{\partial \tau_{xy}}{\partial x} +\frac{\partial \tau_{yy}}{\partial y}+ \frac{\partial \tau_{zy}}{\partial z} \right) \ \mathbf{j} \\ \\ & \ \ \ + dxdydz \ \left( \frac{\partial \tau_{xz}}{\partial x} +\frac{\partial \tau_{yz}}{\partial y}+ \frac{\partial \tau_{zz}}{\partial z} \right) \ \mathbf{k} \end{align} \]
한편 유체요소의 가속도를 속도를 미분해서 계산할 수 있다. 먼저 \(x\) 축 방향의 속도 \(u(x, y, z, t)\) 를 미분해 보자.
\[ \begin{align} \frac{du}{dt} &= \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt}+ \frac{\partial u}{\partial z} \frac{dz}{dt} + \frac{\partial u}{\partial t} \\ \\ &= u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y}+ w \frac{\partial u}{\partial z} + \frac{\partial u}{\partial t} \end{align} \]
\(y, z\) 축 방향도 같은 방법으로 계산하면 유체요소의 가속도는 다음과 같이 계산된다.
\[ \begin{align} \frac{d \mathbf{V}}{dt} &= \left( u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y}+ w \frac{\partial u} {\partial z} + \frac{\partial u}{\partial t} \right) \ \mathbf{i} \tag{4} \\ \\ & \ \ + \left( u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y}+ w \frac{\partial v} {\partial z} + \frac{\partial v}{\partial t} \right) \ \mathbf{j} \\ \\ & \ \ + \left( u \frac{\partial w}{\partial x} + v \frac{\partial w}{\partial y}+ w \frac{\partial w} {\partial z} + \frac{\partial w}{\partial t} \right) \ \mathbf{k} \end{align} \]
위 식은 속도벡터 \(\mathbf{V}\) 와 연산자 \(
abla\) 를 사용하면 다음과 같이 고쳐 쓸 수 있다.
\[ \begin{align} \frac{d\mathbf{V}}{dt} &= \left( \frac{\partial u}{\partial t} + \mathbf{V} \cdot
abla u \right) \mathbf{i} +\left( \frac{\partial v}{\partial t} + \mathbf{V} \cdot
abla v \right) \mathbf{j} + \left( \frac{\partial w}{\partial t} + \mathbf{V} \cdot
abla w \right) \mathbf{k} \tag{5} \\ \\ &= \frac{\partial \mathbf{V}}{\partial t}+ ( \mathbf{V} \cdot
abla ) \mathbf{V} \end{align} \]
유체요소의 질량 \(dm=\rho \ dxdydz\) 는 일정하다고 가정했으므로 뉴톤의 제2법칙에 의하면
\[ \rho \ dxdydz \frac{d\mathbf{V}}{dt} = d \mathbf{F}_b + d \mathbf{F}_p + d \mathbf{F}_v \tag{6} \]
이 성립한다. 식 (1), (2), (3), (5)를 식 (6)에 대입하고 \(x, y, z\) 축 성분으로 나누어서 식을 쓰면 다음과 같이 된다.
\[ \begin{align} & \rho \left( \frac{\partial u}{\partial t} + \mathbf{V} \cdot
abla u \right) = -\frac{\partial p}{\partial x} + \frac{\partial \tau_{xx}}{\partial x} + \frac{\partial \tau_{yx}}{\partial y} + \frac{\partial \tau_{zx}}{\partial z} + \rho f_x \tag{7} \\ \\ & \rho \left( \frac{\partial v}{\partial t} + \mathbf{V} \cdot
abla v \right) = -\frac{\partial p}{\partial y} + \frac{\partial \tau_{xy}}{\partial x} + \frac{\partial \tau_{yy}}{\partial y} + \frac{\partial \tau_{zy}}{\partial z} + \rho f_y \\ \\ & \rho \left( \frac{\partial w}{\partial t} + \mathbf{V} \cdot
abla w \right) = -\frac{\partial p}{\partial z} + \frac{\partial \tau_{xz}}{\partial x} + \frac{\partial \tau_{yz}}{\partial y} + \frac{\partial \tau_{zz}}{\partial z} + \rho f_z \end{align} \]
위 식에 의하면 미지수가 \(u, v, w, p, \rho, \tau_{xx}, …\) 등 14개에 달한다. 반면 식은 식 (7)로 주어지는 3개와 연속 방정식 1개 등 4개 밖에 안된다. 따라서 더 많은 식을 찾아내야 한다.
유체의 점성은 유체의 변형에 대항하는 저항력과 같은 역할을 하는데 전단응력이 유체의 점성과 관련이 있다. 아래 그림과 같이 물체의 표면 주위를 흐르는 유체는 점성 때문에 표면에서는 속도가 \(0\) 이고 표면으로부터의 거리에 따라 유속이 달라진다.
속도가 느린 유동층은 속도가 빠른 유동층에 힘을 가하여 속도가 느려지게 하고, 반대로 속도가 빠른 층은 속도가 느린 층에 반대의 힘을 가하여 속도가 느려지게 하는데 이와 같은 속도의 차이 때문에 전단응력이 발생하게 된다.
뉴톤유체(Newtonian fluid)는 전단응력 \(\tau\) 와 속도 변화율 사이에 비례관계가 성립하는 유체로서 다음과 같은 관계식을 갖는다.
\[ \tau = \mu \frac{du}{dh} \tag{8} \]
여기서 비례상수 \(\mu\) 를 점성계수(viscosity coefficient)라고 한다.
식 (7)의 유체가 뉴톤유체라는 가정을 하면 다음 식이 추가로 얻어진다.
\[ \begin{align} & \tau_{xy}= \tau_{yx} =\mu \left( \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} \right) \tag{9} \\ \\ & \tau_{xz}= \tau_{zx} =\mu \left( \frac{\partial w}{\partial x} + \frac{\partial u}{\partial z} \right) \\ \\ & \tau_{yz}= \tau_{zy} =\mu \left( \frac{\partial w}{\partial y} + \frac{\partial v}{\partial z} \right) \end{align} \]
한편, 수직응력은 점성력뿐만 아니라 유체요소가 일정한 질량을 유지하기 위한 팽창과 수축 과정에서의 저항력도 고려해야 하므로 다음과 같은 관계식을 갖는다.
\[ \begin{align} & \tau_{xx} = \lambda (
abla \cdot \mathbf{V} ) + 2 \mu \frac{\partial u}{\partial x} \tag{10} \\ \\ & \tau_{yy} = \lambda (
abla \cdot \mathbf{V} ) + 2 \mu \frac{\partial v}{\partial y} \\ \\ & \tau_{zz} = \lambda (
abla \cdot \mathbf{V} ) + 2 \mu \frac{\partial w}{\partial z} \\ \\ \end{align} \]
여기서 \(\lambda\) 를 제2의 점성계수라고 하는데 Stokes는 이 값을 \(\lambda= -\frac{2}{3} \mu\) 일 것으로 생각하였다. Stokes의 가설이 맞는지 여부는 아직 명확히 확인되지 않았다고 한다. 식 (7), (9), (10)을 Navier-Stokes 방정식이라고 한다.
Navier Stokes Equation 유도 | [박재우] 유체역학 1 – Navier-Stokes 방정식 빠른 답변
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Navier-Stokes 방정식은 뉴톤 제2법칙으로부터 유도될 수 있다. 공기를 비롯한 유체는 고체와 달리 정해진 모양이 없기 때문에 뉴톤 제2법칙을 적용 …
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주제와 관련된 더 많은 사진을 참조하십시오 [박재우] 유체역학 1 – Navier-Stokes 방정식. 댓글에서 더 많은 관련 이미지를 보거나 필요한 경우 더 많은 관련 기사를 볼 수 있습니다.
The intent of this article is to highlight the important points of the derivation of the Navier–Stokes equations as well as its application and formulation for different families of fluids.
Basic assumptions [ edit ]
The Navier–Stokes equations are based on the assumption that the fluid, at the scale of interest, is a continuum – a continuous substance rather than discrete particles. Another necessary assumption is that all the fields of interest including pressure, flow velocity, density, and temperature are at least weakly differentiable.
The equations are derived from the basic principles of continuity of mass, momentum, and energy. Sometimes it is necessary to consider a finite arbitrary volume, called a control volume, over which these principles can be applied. This finite volume is denoted by Ω and its bounding surface ∂Ω. The control volume can remain fixed in space or can move with the fluid.
The material derivative [ edit ]
Changes in properties of a moving fluid can be measured in two different ways. One can measure a given property by either carrying out the measurement on a fixed point in space as particles of the fluid pass by, or by following a parcel of fluid along its streamline. The derivative of a field with respect to a fixed position in space is called the Eulerian derivative, while the derivative following a moving parcel is called the advective or material (or Lagrangian[1]) derivative.
The material derivative is defined as the nonlinear operator:
D D t = d e f ∂ ∂ t + u ⋅ ∇ {\displaystyle {\frac {D}{Dt}}\ {\stackrel {\mathrm {def} }{=}}\ {\frac {\partial }{\partial t}}+\mathbf {u} \cdot
abla }
where u is the flow velocity. The first term on the right-hand side of the equation is the ordinary Eulerian derivative (the derivative on a fixed reference frame, representing changes at a point with respect to time) whereas the second term represents changes of a quantity with respect to position (see advection). This “special” derivative is in fact the ordinary derivative of a function of many variables along a path following the fluid motion; it may be derived through application of the chain rule in which all independent variables are checked for change along the path (which is to say, the total derivative).
For example, the measurement of changes in wind velocity in the atmosphere can be obtained with the help of an anemometer in a weather station or by observing the movement of a weather balloon. The anemometer in the first case is measuring the velocity of all the moving particles passing through a fixed point in space, whereas in the second case the instrument is measuring changes in velocity as it moves with the flow.
Continuity equations [ edit ]
The Navier–Stokes equation is a special continuity equation. A continuity equation may be derived from conservation principles of:
A continuity equation (or conservation law) is an integral relation stating that the rate of change of some integrated property φ defined over a control volume Ω must be equal to what amount is lost or gained through the boundaries Γ of the volume plus what is created or consumed by sources and sinks inside the volume. This is expressed by the following integral continuity equation:
d d t ∫ Ω φ d Ω = − ∫ Γ φ u ⋅ n d Γ − ∫ Ω s d Ω {\displaystyle {\frac {d}{dt}}\int _{\Omega }\varphi \ d\Omega =-\int _{\Gamma }\varphi \mathbf {u\cdot n} \ d\Gamma -\int _{\Omega }s\ d\Omega }
where u is the flow velocity of the fluid, n is the outward-pointing unit normal vector, and s represents the sources and sinks in the flow, taking the sinks as positive.
The divergence theorem may be applied to the surface integral, changing it into a volume integral:
d d t ∫ Ω φ d Ω = − ∫ Ω ∇ ⋅ ( φ u ) d Ω − ∫ Ω s d Ω . {\displaystyle {\frac {d}{dt}}\int _{\Omega }\varphi \ d\Omega =-\int _{\Omega }
abla \cdot (\varphi \mathbf {u} )\ d\Omega -\int _{\Omega }s\ d\Omega .}
Applying the Reynolds transport theorem to the integral on the left and then combining all of the integrals:
∫ Ω ∂ φ ∂ t d Ω = − ∫ Ω ∇ ⋅ ( φ u ) d Ω − ∫ Ω s d Ω ⇒ ∫ Ω ( ∂ φ ∂ t + ∇ ⋅ ( φ u ) + s ) d Ω = 0. {\displaystyle \int _{\Omega }{\frac {\partial \varphi }{\partial t}}\ d\Omega =-\int _{\Omega }
abla \cdot (\varphi \mathbf {u} )\ d\Omega -\int _{\Omega }s\ d\Omega \quad \Rightarrow \quad \int _{\Omega }\left({\frac {\partial \varphi }{\partial t}}+
abla \cdot (\varphi \mathbf {u} )+s\right)d\Omega =0.}
The integral must be zero for any control volume; this can only be true if the integrand itself is zero, so that:
∂ φ ∂ t + ∇ ⋅ ( φ u ) + s = 0. {\displaystyle {\frac {\partial \varphi }{\partial t}}+
abla \cdot (\varphi \mathbf {u} )+s=0.}
From this valuable relation (a very generic continuity equation), three important concepts may be concisely written: conservation of mass, conservation of momentum, and conservation of energy. Validity is retained if φ is a vector, in which case the vector-vector product in the second term will be a dyad.
Conservation of momentum [ edit ]
A general momentum equation is obtained when the conservation relation is applied to momentum. When the intensive property φ is considered as the mass flux (also momentum density), that is, the product of mass density and flow velocity ρu, by substitution into the general continuum equation:
∂ ∂ t ( ρ u ) + ∇ ⋅ ( ρ u ⊗ u ) = s {\displaystyle {\frac {\partial }{\partial t}}(\rho \mathbf {u} )+
abla \cdot (\rho \mathbf {u} \otimes \mathbf {u} )=\mathbf {s} }
where u ⊗ u is a dyad, a special case of tensor product, which results in a second rank tensor; the divergence of a second rank tensor is again a vector (a first-rank tensor).[2]
Using the formula for the divergence of a dyad,
∇ ⋅ ( a ⊗ b ) = ( ∇ ⋅ a ) b + a ⋅ ∇ b {\displaystyle
abla \cdot (\mathbf {a} \otimes \mathbf {b} )=(
abla \cdot \mathbf {a} )\mathbf {b} +\mathbf {a} \cdot
abla \mathbf {b} }
we then have
u ∂ ρ ∂ t + ρ ∂ u ∂ t + u u ⋅ ∇ ρ + ρ u ⋅ ∇ u + ρ u ∇ ⋅ u = s {\displaystyle \mathbf {u} {\frac {\partial \rho }{\partial t}}+\rho {\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \mathbf {u} \cdot
abla \rho +\rho \mathbf {u} \cdot
abla \mathbf {u} +\rho \mathbf {u}
abla \cdot \mathbf {u} =\mathbf {s} }
Note that the gradient of a vector is a special case of the covariant derivative, the operation results in second rank tensors;[2] except in Cartesian coordinates, it is important to understand that this is not simply an element by element gradient. Rearranging and recognizing that u ⋅ ∇ρ + ρ∇ ⋅ u = ∇ ⋅ (ρu):
u ( ∂ ρ ∂ t + u ⋅ ∇ ρ + ρ ∇ ⋅ u ) + ρ ( ∂ u ∂ t + u ⋅ ∇ u ) = s u ( ∂ ρ ∂ t + ∇ ⋅ ( ρ u ) ) + ρ ( ∂ u ∂ t + u ⋅ ∇ u ) = s {\displaystyle {\begin{aligned}\mathbf {u} \left({\frac {\partial \rho }{\partial t}}+\mathbf {u} \cdot
abla \rho +\rho
abla \cdot \mathbf {u} \right)+\rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot
abla \mathbf {u} \right)&=\mathbf {s} \\[6px]\mathbf {u} \left({\frac {\partial \rho }{\partial t}}+
abla \cdot (\rho \mathbf {u} )\right)+\rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot
abla \mathbf {u} \right)&=\mathbf {s} \end{aligned}}}
The leftmost expression enclosed in parentheses is, by mass continuity (shown in a moment), equal to zero. Noting that what remains on the left side of the equation is the material derivative of flow velocity:
ρ D u D t = ρ ( ∂ u ∂ t + u ⋅ ∇ u ) = s {\displaystyle \rho {\frac {D\mathbf {u} }{Dt}}=\rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot
abla \mathbf {u} \right)=\mathbf {s} }
This appears to simply be an expression of Newton’s second law (F = ma) in terms of body forces instead of point forces. Each term in any case of the Navier–Stokes equations is a body force. A shorter though less rigorous way to arrive at this result would be the application of the chain rule to acceleration:
ρ d d t ( u ( x , y , z , t ) ) = s ⇒ ρ ( ∂ u ∂ t + ∂ u ∂ x d x d t + ∂ u ∂ y d y d t + ∂ u ∂ z d z d t ) = s ⇒ ρ ( ∂ u ∂ t + u ∂ u ∂ x + v ∂ u ∂ y + w ∂ u ∂ z ) = s ⇒ ρ ( ∂ u ∂ t + u ⋅ ∇ u ) = s {\displaystyle {\begin{aligned}\rho {\frac {d}{dt}}{\bigl (}\mathbf {u} (x,y,z,t){\bigr )}=\mathbf {s} \quad &\Rightarrow &\rho \left({\frac {\partial \mathbf {u} }{\partial t}}+{\frac {\partial \mathbf {u} }{\partial x}}{\frac {dx}{dt}}+{\frac {\partial \mathbf {u} }{\partial y}}{\frac {dy}{dt}}+{\frac {\partial \mathbf {u} }{\partial z}}{\frac {dz}{dt}}\right)&=\mathbf {s} \\\quad &\Rightarrow &\rho \left({\frac {\partial \mathbf {u} }{\partial t}}+u{\frac {\partial \mathbf {u} }{\partial x}}+v{\frac {\partial \mathbf {u} }{\partial y}}+w{\frac {\partial \mathbf {u} }{\partial z}}\right)&=\mathbf {s} \\\quad &\Rightarrow &\rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot
abla \mathbf {u} \right)&=\mathbf {s} \end{aligned}}}
where u = (u, v, w). The reason why this is “less rigorous” is that we haven’t shown that the choice of
u = ( d x d t , d y d t , d z d t ) {\displaystyle \mathbf {u} =\left({\frac {dx}{dt}},{\frac {dy}{dt}},{\frac {dz}{dt}}\right)}
is correct; however it does make sense since with that choice of path the derivative is “following” a fluid “particle”, and in order for Newton’s second law to work, forces must be summed following a particle. For this reason the convective derivative is also known as the particle derivative.
Conservation of mass [ edit ]
Mass may be considered also. When the intensive property φ is considered as the mass, by substitution into the general continuum equation, and taking s = 0 (no sources or sinks of mass):
∂ ρ ∂ t + ∇ ⋅ ( ρ u ) = 0 {\displaystyle {\frac {\partial \rho }{\partial t}}+
abla \cdot (\rho \mathbf {u} )=0}
where ρ is the mass density (mass per unit volume), and u is the flow velocity. This equation is called the mass continuity equation, or simply the continuity equation. This equation generally accompanies the Navier–Stokes equation.
In the case of an incompressible fluid, Dρ/Dt = 0 (the density following the path of a fluid element is constant) and the equation reduces to:
∇ ⋅ u = 0 {\displaystyle
abla \cdot \mathbf {u} =0}
which is in fact a statement of the conservation of volume.
Cauchy momentum equation [ edit ]
The generic density of the momentum source s seen previously is made specific first by breaking it up into two new terms, one to describe internal stresses and one for external forces, such as gravity. By examining the forces acting on a small cube in a fluid, it may be shown that
ρ D u D t = ∇ ⋅ σ + f {\displaystyle \rho {\frac {D\mathbf {u} }{Dt}}=
abla \cdot {\boldsymbol {\sigma }}+\mathbf {f} }
where σ is the Cauchy stress tensor, and f accounts for body forces present. This equation is called the Cauchy momentum equation and describes the non-relativistic momentum conservation of any continuum that conserves mass. σ is a rank two symmetric tensor given by its covariant components. In orthogonal coordinates in three dimensions it is represented as the 3 × 3 matrix:
σ i j = ( σ x x τ x y τ x z τ y x σ y y τ y z τ z x τ z y σ z z ) {\displaystyle \sigma _{ij}={\begin{pmatrix}\sigma _{xx}&\tau _{xy}&\tau _{xz}\\\tau _{yx}&\sigma _{yy}&\tau _{yz}\\\tau _{zx}&\tau _{zy}&\sigma _{zz}\end{pmatrix}}}
where the σ are normal stresses and τ shear stresses. This matrix is split up into two terms:
σ i j = ( σ x x τ x y τ x z τ y x σ y y τ y z τ z x τ z y σ z z ) = − ( p 0 0 0 p 0 0 0 p ) + ( σ x x + p τ x y τ x z τ y x σ y y + p τ y z τ z x τ z y σ z z + p ) = − p I + τ {\displaystyle \sigma _{ij}={\begin{pmatrix}\sigma _{xx}&\tau _{xy}&\tau _{xz}\\\tau _{yx}&\sigma _{yy}&\tau _{yz}\\\tau _{zx}&\tau _{zy}&\sigma _{zz}\end{pmatrix}}=-{\begin{pmatrix}p&0&0\\0&p&0\\0&0&p\end{pmatrix}}+{\begin{pmatrix}\sigma _{xx}+p&\tau _{xy}&\tau _{xz}\\\tau _{yx}&\sigma _{yy}+p&\tau _{yz}\\\tau _{zx}&\tau _{zy}&\sigma _{zz}+p\end{pmatrix}}=-p\mathbf {I} +{\boldsymbol {\tau }}}
where I is the 3 × 3 identity matrix and τ is the deviatoric stress tensor. Note that the mechanical pressure p is equal to the negative of the mean normal stress:
p = − 1 3 ( σ x x + σ y y + σ z z ) . {\displaystyle p=-{\tfrac {1}{3}}\left(\sigma _{xx}+\sigma _{yy}+\sigma _{zz}\right).}
The motivation for doing this is that pressure is typically a variable of interest, and also this simplifies application to specific fluid families later on since the rightmost tensor τ in the equation above must be zero for a fluid at rest. Note that τ is traceless. The Cauchy equation may now be written in another more explicit form:
ρ D u D t = − ∇ p + ∇ ⋅ τ + f {\displaystyle \rho {\frac {D\mathbf {u} }{Dt}}=-
abla p+
abla \cdot {\boldsymbol {\tau }}+\mathbf {f} }
This equation is still incomplete. For completion, one must make hypotheses on the forms of τ and p, that is, one needs a constitutive law for the stress tensor which can be obtained for specific fluid families and on the pressure. Some of these hypotheses lead to the Euler equations (fluid dynamics), other ones lead to the Navier–Stokes equations. Additionally, if the flow is assumed compressible an equation of state will be required, which will likely further require a conservation of energy formulation.
Application to different fluids [ edit ]
The general form of the equations of motion is not “ready for use”, the stress tensor is still unknown so that more information is needed; this information is normally some knowledge of the viscous behavior of the fluid. For different types of fluid flow this results in specific forms of the Navier–Stokes equations.
Newtonian fluid [ edit ]
Compressible Newtonian fluid [ edit ]
The formulation for Newtonian fluids stems from an observation made by Newton that, for most fluids,
τ ∝ ∂ u ∂ y {\displaystyle \tau \propto {\frac {\partial u}{\partial y}}}
In order to apply this to the Navier–Stokes equations, three assumptions were made by Stokes:
The stress tensor is a linear function of the strain rate tensor or equivalently the velocity gradient.
The fluid is isotropic.
For a fluid at rest, ∇ ⋅ τ must be zero (so that hydrostatic pressure results).
The above list states the classic argument[4] that the shear strain rate tensor (the (symmetric) shear part of the velocity gradient) is a pure shear tensor and does not include any inflow/outflow part (any compression/expansion part). This means that its trace is zero, and this is achieved by subtracting ∇ ⋅ u in a symmetric way from the diagonal elements of the tensor. The compressional contribution to viscous stress is added as a separate diagonal tensor.
Applying these assumptions will lead to :
τ = μ ( ∇ u + ( ∇ u ) T ) + λ ( ∇ ⋅ u ) I {\displaystyle \tau =\mu \left(
abla \mathbf {u} +\left(
abla \mathbf {u} \right)^{\mathsf {T}}\right)+\lambda \left(
abla \cdot \mathbf {u} \right)\mathbf {I} }
or in tensor form
τ i j = μ ( ∂ u i ∂ x j + ∂ u j ∂ x i ) + δ i j λ ∂ u k ∂ x k {\displaystyle \tau _{ij}=\mu \left({\frac {\partial u_{i}}{\partial x_{j}}}+{\frac {\partial u_{j}}{\partial x_{i}}}\right)+\delta _{ij}\lambda {\frac {\partial u_{k}}{\partial x_{k}}}}
That is, the deviatoric of the deformation rate tensor is identified to the deviatoric of the stress tensor, up to a factor μ.[5]
δ ij is the Kronecker delta. μ and λ are proportionality constants associated with the assumption that stress depends on strain linearly; μ is called the first coefficient of viscosity or shear viscosity (usually just called “viscosity”) and λ is the second coefficient of viscosity or volume viscosity (and it is related to bulk viscosity). The value of λ, which produces a viscous effect associated with volume change, is very difficult to determine, not even its sign is known with absolute certainty. Even in compressible flows, the term involving λ is often negligible; however it can occasionally be important even in nearly incompressible flows and is a matter of controversy. When taken nonzero, the most common approximation is λ ≈ − 2/3μ.
A straightforward substitution of τ ij into the momentum conservation equation will yield the Navier–Stokes equations, describing a compressible Newtonian fluid:
ρ ( ∂ u ∂ t + u ⋅ ∇ u ) = − ∇ p + ∇ ⋅ [ μ ( ∇ u + ( ∇ u ) T ) ] + ∇ ⋅ [ λ ( ∇ ⋅ u ) I ] + ρ g {\displaystyle \rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot
abla \mathbf {u} \right)=-
abla p+
abla \cdot \left[\mu \left(
abla \mathbf {u} +\left(
abla \mathbf {u} \right)^{\mathsf {T}}\right)\right]+
abla \cdot \left[\lambda \left(
abla \cdot \mathbf {u} \right)\mathbf {I} \right]+\rho \mathbf {g} }
The body force has been decomposed into density and external acceleration, that is, f = ρg. The associated mass continuity equation is:
∂ ρ ∂ t + ∇ ⋅ ( ρ u ) = 0 {\displaystyle {\frac {\partial \rho }{\partial t}}+
abla \cdot (\rho \mathbf {u} )=0}
In addition to this equation, an equation of state and an equation for the conservation of energy is needed. The equation of state to use depends on context (often the ideal gas law), the conservation of energy will read:
ρ D h D t = D p D t + ∇ ⋅ ( k ∇ T ) + Φ {\displaystyle \rho {\frac {Dh}{Dt}}={\frac {Dp}{Dt}}+
abla \cdot (k
abla T)+\Phi }
Here, h is the specific enthalpy, T is the temperature, and Φ is a function representing the dissipation of energy due to viscous effects:
Φ = μ ( 2 ( ∂ u ∂ x ) 2 + 2 ( ∂ v ∂ y ) 2 + 2 ( ∂ w ∂ z ) 2 + ( ∂ v ∂ x + ∂ u ∂ y ) 2 + ( ∂ w ∂ y + ∂ v ∂ z ) 2 + ( ∂ u ∂ z + ∂ w ∂ x ) 2 ) + λ ( ∇ ⋅ u ) 2 . {\displaystyle \Phi =\mu \left(2\left({\frac {\partial u}{\partial x}}\right)^{2}+2\left({\frac {\partial v}{\partial y}}\right)^{2}+2\left({\frac {\partial w}{\partial z}}\right)^{2}+\left({\frac {\partial v}{\partial x}}+{\frac {\partial u}{\partial y}}\right)^{2}+\left({\frac {\partial w}{\partial y}}+{\frac {\partial v}{\partial z}}\right)^{2}+\left({\frac {\partial u}{\partial z}}+{\frac {\partial w}{\partial x}}\right)^{2}\right)+\lambda (
abla \cdot \mathbf {u} )^{2}.}
With a good equation of state and good functions for the dependence of parameters (such as viscosity) on the variables, this system of equations seems to properly model the dynamics of all known gases and most liquids.
Incompressible Newtonian fluid [ edit ]
For the special (but very common) case of incompressible flow, the momentum equations simplify significantly. Using the following assumptions:
Viscosity μ will now be a constant
will now be a constant The second viscosity effect λ = 0
The simplified mass continuity equation ∇ ⋅ u = 0
This gives incompressible Navier-Stokes equations, describing incompressible Newtonian fluid:
ρ ( ∂ u ∂ t + u ⋅ ∇ u ) = − ∇ p + ∇ ⋅ [ μ ( ∇ u + ( ∇ u ) T ) ] + ρ g {\displaystyle \rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot
abla \mathbf {u} \right)=-
abla p+
abla \cdot \left[\mu \left(
abla \mathbf {u} +\left(
abla \mathbf {u} \right)^{\mathsf {T}}\right)\right]+\rho \mathbf {g} }
then looking at the viscous terms of the x momentum equation for example we have:
∂ ∂ x ( 2 μ ∂ u ∂ x ) + ∂ ∂ y ( μ ( ∂ u ∂ y + ∂ v ∂ x ) ) + ∂ ∂ z ( μ ( ∂ u ∂ z + ∂ w ∂ x ) ) = 2 μ ∂ 2 u ∂ x 2 + μ ∂ 2 u ∂ y 2 + μ ∂ 2 v ∂ y ∂ x + μ ∂ 2 u ∂ z 2 + μ ∂ 2 w ∂ z ∂ x = μ ∂ 2 u ∂ x 2 + μ ∂ 2 u ∂ y 2 + μ ∂ 2 u ∂ z 2 + μ ∂ 2 u ∂ x 2 + μ ∂ 2 v ∂ y ∂ x + μ ∂ 2 w ∂ z ∂ x = μ ∇ 2 u + μ ∂ ∂ x ( ∂ u ∂ x + ∂ v ∂ y + ∂ w ∂ z ) 0 = μ ∇ 2 u {\displaystyle {\begin{aligned}&{\frac {\partial }{\partial x}}\left(2\mu {\frac {\partial u}{\partial x}}\right)+{\frac {\partial }{\partial y}}\left(\mu \left({\frac {\partial u}{\partial y}}+{\frac {\partial v}{\partial x}}\right)\right)+{\frac {\partial }{\partial z}}\left(\mu \left({\frac {\partial u}{\partial z}}+{\frac {\partial w}{\partial x}}\right)\right)\\[8px]&\qquad =2\mu {\frac {\partial ^{2}u}{\partial x^{2}}}+\mu {\frac {\partial ^{2}u}{\partial y^{2}}}+\mu {\frac {\partial ^{2}v}{\partial y\,\partial x}}+\mu {\frac {\partial ^{2}u}{\partial z^{2}}}+\mu {\frac {\partial ^{2}w}{\partial z\,\partial x}}\\[8px]&\qquad =\mu {\frac {\partial ^{2}u}{\partial x^{2}}}+\mu {\frac {\partial ^{2}u}{\partial y^{2}}}+\mu {\frac {\partial ^{2}u}{\partial z^{2}}}+\mu {\frac {\partial ^{2}u}{\partial x^{2}}}+\mu {\frac {\partial ^{2}v}{\partial y\,\partial x}}+\mu {\frac {\partial ^{2}w}{\partial z\,\partial x}}\\[8px]&\qquad =\mu
abla ^{2}u+\mu {\frac {\partial }{\partial x}}{\cancelto {0}{\left({\frac {\partial u}{\partial x}}+{\frac {\partial v}{\partial y}}+{\frac {\partial w}{\partial z}}\right)}}\\[8px]&\qquad =\mu
abla ^{2}u\end{aligned}}\,}
Similarly for the y and z momentum directions we have μ∇2v and μ∇2w.
The above solution is key to deriving Navier–Stokes equations from the equation of motion in fluid dynamics when density and viscosity are constant.
Non-Newtonian fluids [ edit ]
A non-Newtonian fluid is a fluid whose flow properties differ in any way from those of Newtonian fluids. Most commonly the viscosity of non-Newtonian fluids is a function of shear rate or shear rate history. However, there are some non-Newtonian fluids with shear-independent viscosity, that nonetheless exhibit normal stress-differences or other non-Newtonian behaviour. Many salt solutions and molten polymers are non-Newtonian fluids, as are many commonly found substances such as ketchup, custard, toothpaste, starch suspensions, paint, blood, and shampoo. In a Newtonian fluid, the relation between the shear stress and the shear rate is linear, passing through the origin, the constant of proportionality being the coefficient of viscosity. In a non-Newtonian fluid, the relation between the shear stress and the shear rate is different, and can even be time-dependent. The study of the non-Newtonian fluids is usually called rheology. A few examples are given here.
Bingham fluid [ edit ]
In Bingham fluids, the situation is slightly different:
∂ u ∂ y = { 0 , τ < τ 0 τ − τ 0 μ , τ ≥ τ 0 {\displaystyle {\frac {\partial u}{\partial y}}={\begin{cases}0,&\tau 나비어-스토크스 방정식 (Naview stokes equation) 소개와 유도 클레이 수학 연구소가 밀레니엄 난제로 지목한 7개 문제중 하나에 들어가는 나이어스토크스 관계식은 사실 수십년전부터 기계공학을 비롯한 각종 공학에서는 유용한 지배방정식으로 이미 활용되고 있었다. 단, 수학적 일반해를 구해서 이용한 것이 아니라 수치해석적으로 접근하여 공학적으로 이용한것에 불과하다. 이 방정식의 일반해를 구하는 것은 앞으로도 오랫동안 영구미제로 남을수도 있을 것이다. 비선형 방정식은 아주 오래전부터 지독하게도 어려운 방정식으로 정평이 나 있었기도 하다… 이미 공학을 접었으나 여전히 수학교육이라는 다분히 이공학적 마인드와 관계가 있는 분야에 종사하고 있는 관계로 과거의 악몽이었었던 나비어스토크스 관계식은 여전히 나의 가슴에 영원한 관심사로 남아있기도 하다. 수학교육과 의학의 도상에서 잠시 공학의 기억을 되살려보고자한다. 이동현상은 크게 세가지의 전달현상을 다룬다. 운동량과 열 그리고 물질전달이 이 세가지이다. 이들의 각각 이동현상은 전달매체가 다를 뿐만 아니라 물리적인 이동과정도 매우 상이하다. 그러나 이들의 현상을 설명하는 수학적 체계는 매우 단순하리만큼 일치성을 보이고 있다는 것은 놀라운 점이다. 여기서 내가 다루고자 하는 분야는 바로 운동량 전달이다. 이름대로 운동량의 전달현상을 목적으로 하기에 당연히 운동량의 변화를 설명할 수 있는 관계식으로부터 현상을 이해하여야 할 것이다. 에서 유체 흐름의 운동관계식을 적분형으로 다루어 볼수도 있으나, 적분형 식은 여러 현상을 적용하기에 한계가 있기에 이를 더욱 유용한 미분형으로 바꾸어 일반식을 만들어 놓는 것이 여러 가지 이유로 필요할 것이다. 따라서 여기서는 운동량 관계식을 미분형으로 유도하는 과정을 설명하고, 이에 대한 여러 관점을 함께 논하고자 한다. 일단 나비어 스토크스 방정식을 이해하기 위한 기본 개념을 짚고 넘어가자. 질량보존의 법칙은 자연의 기본 원리로서 잘 알려진 법칙 중 하나이다. 즉 어떠한 물리적 과정을 거치게 될 때, 과정 전과 후의 총 질량이 변하지 않다는 원리이다. 이 법칙을 유체 흐름에 적용시키면 매우 유용하게 사용할 수 있는 관계식을 얻을 수 있게 된다. 우리가 관심을 가지는 대상부피에 대한 질량보존의 개략적 표현은 다음과 같이 나타낼 수 있다. (1) 이 관계를 잘 살펴보면 매우 당연한 표현이다. 즉, 질량의 유출과 유입의 차이가 바로 대상부피 내의 질량 축적량을 의미하는 것이다. 이 관계를 수식으로 표현하기 위하여 다음의 간단한 모형을 생각해 볼 수 있다. 이 모형은 어느 특정 시간에 형성되는 유선을 표현하였다. dA는 대상부피 표면의 미세면적이고, v는 유선방향의 속도벡터, n는 dA의 법선벡터이다. 당연히 θ는 v와 n의 차이각이다. 따라서 dA로 통과되는 질량의 유출속도는 가 된다. 여기서 를 질량플럭스(mass flux)라고 하며 단위시간당 및 단위면적을 통과하는 질량을 의미한다. 이러한 질량의 유출속도를 벡터 표현으로 바꾸면 다음과 같이 나타낼 수 있다. 미세면적의 질량 유출속도 = (2) 이 값을 전체 대상표면에 대하여 적분을 하면 총 대상면적을 통과하는 질량의 순유출량을 알 수 있다. 이 값은 윗 식을 적분하면 얻게 된다. (3) 이 값이 양수이면 질량의 순유출이 일어나며, 반대로 음수이면 순유입이 된다. 또한 0이면 대상부피 내의 질량은 항상 정량을 유지하게 된다. 한편, 질량 축적속도는 결국 시간에 대한 밀도변화이므로 다음과 같이 표현할 수 있다. (4) 결과적으로 질량보존의 최종 관계식은 식(3)과 (4)를 식(1)에 적용시켜서 다음의 적분형 물질수지식을 얻게 된다. (5) 유체흐름의 선운동량 식은 질량보존의 관계식과 매우 비슷하게 유도될 수 있다. 아래 그림에서 운동량의 유출속도는 질량유출속에 속도를 곱하는 것으로 생각할 수 있다. 자세한 관계식 유도는 질량보존의 관계식을 참고하도록 하고, 여기서는 최종 결과식만을 제시하고자 한다. (1) 1. 유체흐름의 운동량 관계식 모형 유체 운동량의 적분형 관계식을 다음과 같이 유도됨을 전 절에서 다루었다. (1) 이 관계식을 풀어서 설명하면 다음과 같은 개념관계로 도시할 수 있다. (2) 이 관계식을 쉽게 설명하기 위하여 식(1)을 대상부피로 나눈 다음 극한값을 취하면 다음과 같이 쓸 수 있다. (3) 이제 이 식을 하나하나의 항으로 분류하여 풀어보도록 하자. 2. 외부 힘의 합 대상부피에 작용하는 외부 힘의 영향으로는 수직응력과 전단응력, 그리고 중력과 외부로부터 받는 압력의 영향을 적용할 수 있다. 먼저 x 방향으로 작용하는 힘만을 고려한다면 다음과 같이 쓸 수 있다. (4) 식(3)의 형식을 맞추기 위하여 식(4)를 미소부피( )로 나누어 극한을 취하고, 동일한 방법으로 y 방향과 z 방향을 적용하면 다음과 같이 종합할 수 있다. (5) 3. 대상부피를 통과하는 선운동량의 유출속도 위 그림은 대상 미소부피를 통과하는 운동량 flux를 표현한 것이다. 이 그림을 참조하여 식(2)와 (3)의 선운동량의 순 유출속도항을 플어서 쓰면 다음과 같이 표현할 수 있다. (6) 마지막 식은 연속방정식을 적용시킨 결과이다. 4. 대상부피내의 운동량 변화 속도 (7) 5. 종합된 결과와 Navier-Stokes 방정식 지금까지 얻어진 결과들을 종합하면 다음과 같이 정리할 수 있다. (8) 식(8)의 결과식을 식(3)에 대입하고 성분별로 분리하여 쓰면 다음과 같다. (9) 여기서 D/Dt는 전미분을 의미한다. 식(9)는 응력-변형률 관계에 상관없이 모든 유체에 적용할 수 있으며, Stokes의 점도법칙으로부터 다음과 같이 나타낼 수 있다. (10) 이렇게 유도된 식(10)을 Navier-Stokes 방정식이라고 부른다. 물론 이 식은 직교좌표의 경우이다. 6. ρ와 μ가 일정한 비압축성 흐름에서의 Navier-Stokes 방정식 일반적인 경우는 이러한 구속조건을 가지는 경우가 많이 발생한다. 비압축성 흐름의 경우는 이므로, 이 경우에는 다음과 같이 전 좌표의 상황을 하나의 식으로 표현할 수 있다. (11) 특히 더욱 특수한 경우로서 비점성인 경우 즉, 점도가 ‘0’인 경우의 식을 Euler의 식이라고 불리며, 다음과 같이 나타낸다. (12) 7. 역학적 에너지관계식의 표현 Navier-Stokes 방정식을 조금 관점을 달리 하여, 흐르는 유체상에서 에너지 관계성이 어떠한지에 대하여 알아보고자 한다. 식(9)를 벡터형식으로 통합하여 다음과 같이 나타낼 수 있다. (13) 식(13)을 에너지 rate형식으로 나타내기 위하여 양 변에 지역속도를 곱하면 다음과 같이 쓸 수 있다. (14) 식(14)를 각각 의미를 가지는 항으로 풀어서 정리하면 다음처럼 종합할 수 있다. : net rate of input of kinetic energy by virtue of bulk flow (15) : rate of increase in kinetic energy per unit volume : rate of work done by pressure od surroundings on volume element : rate of reversible conversion to internal energy : rate of work done by viscous force on volume element : rate of irreversible conversion to internal energy : rate of work done by gravity force on volume element 여기서 항은 Newtonian 흐름에서 항상 (+)값을 가지며, 다음과 같은 의미를 지닌다. 키워드에 대한 정보 navier stokes equation 유도 다음은 Bing에서 navier stokes equation 유도 주제에 대한 검색 결과입니다. 필요한 경우 더 읽을 수 있습니다. 이 기사는 인터넷의 다양한 출처에서 편집되었습니다. 이 기사가 유용했기를 바랍니다. 이 기사가 유용하다고 생각되면 공유하십시오. 매우 감사합니다! 사람들이 주제에 대해 자주 검색하는 키워드 [박재우] 유체역학 1 – Navier-Stokes 방정식 박재우 징짤마 xpertian 유체역학 전공인강 기계공학 YouTube에서 navier stokes equation 유도 주제의 다른 동영상 보기 주제에 대한 기사를 시청해 주셔서 감사합니다 [박재우] 유체역학 1 – Navier-Stokes 방정식 | navier stokes equation 유도, 이 기사가 유용하다고 생각되면 공유하십시오, 매우 감사합니다.
Derivation of the Navier–Stokes equations
The intent of this article is to highlight the important points of the derivation of the Navier–Stokes equations as well as its application and formulation for different families of fluids.
Basic assumptions [ edit ]
The Navier–Stokes equations are based on the assumption that the fluid, at the scale of interest, is a continuum – a continuous substance rather than discrete particles. Another necessary assumption is that all the fields of interest including pressure, flow velocity, density, and temperature are at least weakly differentiable.
The equations are derived from the basic principles of continuity of mass, momentum, and energy. Sometimes it is necessary to consider a finite arbitrary volume, called a control volume, over which these principles can be applied. This finite volume is denoted by Ω and its bounding surface ∂Ω. The control volume can remain fixed in space or can move with the fluid.
The material derivative [ edit ]
Changes in properties of a moving fluid can be measured in two different ways. One can measure a given property by either carrying out the measurement on a fixed point in space as particles of the fluid pass by, or by following a parcel of fluid along its streamline. The derivative of a field with respect to a fixed position in space is called the Eulerian derivative, while the derivative following a moving parcel is called the advective or material (or Lagrangian[1]) derivative.
The material derivative is defined as the nonlinear operator:
D D t = d e f ∂ ∂ t + u ⋅ ∇ {\displaystyle {\frac {D}{Dt}}\ {\stackrel {\mathrm {def} }{=}}\ {\frac {\partial }{\partial t}}+\mathbf {u} \cdot
abla }
where u is the flow velocity. The first term on the right-hand side of the equation is the ordinary Eulerian derivative (the derivative on a fixed reference frame, representing changes at a point with respect to time) whereas the second term represents changes of a quantity with respect to position (see advection). This “special” derivative is in fact the ordinary derivative of a function of many variables along a path following the fluid motion; it may be derived through application of the chain rule in which all independent variables are checked for change along the path (which is to say, the total derivative).
For example, the measurement of changes in wind velocity in the atmosphere can be obtained with the help of an anemometer in a weather station or by observing the movement of a weather balloon. The anemometer in the first case is measuring the velocity of all the moving particles passing through a fixed point in space, whereas in the second case the instrument is measuring changes in velocity as it moves with the flow.
Continuity equations [ edit ]
The Navier–Stokes equation is a special continuity equation. A continuity equation may be derived from conservation principles of:
A continuity equation (or conservation law) is an integral relation stating that the rate of change of some integrated property φ defined over a control volume Ω must be equal to what amount is lost or gained through the boundaries Γ of the volume plus what is created or consumed by sources and sinks inside the volume. This is expressed by the following integral continuity equation:
d d t ∫ Ω φ d Ω = − ∫ Γ φ u ⋅ n d Γ − ∫ Ω s d Ω {\displaystyle {\frac {d}{dt}}\int _{\Omega }\varphi \ d\Omega =-\int _{\Gamma }\varphi \mathbf {u\cdot n} \ d\Gamma -\int _{\Omega }s\ d\Omega }
where u is the flow velocity of the fluid, n is the outward-pointing unit normal vector, and s represents the sources and sinks in the flow, taking the sinks as positive.
The divergence theorem may be applied to the surface integral, changing it into a volume integral:
d d t ∫ Ω φ d Ω = − ∫ Ω ∇ ⋅ ( φ u ) d Ω − ∫ Ω s d Ω . {\displaystyle {\frac {d}{dt}}\int _{\Omega }\varphi \ d\Omega =-\int _{\Omega }
abla \cdot (\varphi \mathbf {u} )\ d\Omega -\int _{\Omega }s\ d\Omega .}
Applying the Reynolds transport theorem to the integral on the left and then combining all of the integrals:
∫ Ω ∂ φ ∂ t d Ω = − ∫ Ω ∇ ⋅ ( φ u ) d Ω − ∫ Ω s d Ω ⇒ ∫ Ω ( ∂ φ ∂ t + ∇ ⋅ ( φ u ) + s ) d Ω = 0. {\displaystyle \int _{\Omega }{\frac {\partial \varphi }{\partial t}}\ d\Omega =-\int _{\Omega }
abla \cdot (\varphi \mathbf {u} )\ d\Omega -\int _{\Omega }s\ d\Omega \quad \Rightarrow \quad \int _{\Omega }\left({\frac {\partial \varphi }{\partial t}}+
abla \cdot (\varphi \mathbf {u} )+s\right)d\Omega =0.}
The integral must be zero for any control volume; this can only be true if the integrand itself is zero, so that:
∂ φ ∂ t + ∇ ⋅ ( φ u ) + s = 0. {\displaystyle {\frac {\partial \varphi }{\partial t}}+
abla \cdot (\varphi \mathbf {u} )+s=0.}
From this valuable relation (a very generic continuity equation), three important concepts may be concisely written: conservation of mass, conservation of momentum, and conservation of energy. Validity is retained if φ is a vector, in which case the vector-vector product in the second term will be a dyad.
Conservation of momentum [ edit ]
A general momentum equation is obtained when the conservation relation is applied to momentum. When the intensive property φ is considered as the mass flux (also momentum density), that is, the product of mass density and flow velocity ρu, by substitution into the general continuum equation:
∂ ∂ t ( ρ u ) + ∇ ⋅ ( ρ u ⊗ u ) = s {\displaystyle {\frac {\partial }{\partial t}}(\rho \mathbf {u} )+
abla \cdot (\rho \mathbf {u} \otimes \mathbf {u} )=\mathbf {s} }
where u ⊗ u is a dyad, a special case of tensor product, which results in a second rank tensor; the divergence of a second rank tensor is again a vector (a first-rank tensor).[2]
Using the formula for the divergence of a dyad,
∇ ⋅ ( a ⊗ b ) = ( ∇ ⋅ a ) b + a ⋅ ∇ b {\displaystyle
abla \cdot (\mathbf {a} \otimes \mathbf {b} )=(
abla \cdot \mathbf {a} )\mathbf {b} +\mathbf {a} \cdot
abla \mathbf {b} }
we then have
u ∂ ρ ∂ t + ρ ∂ u ∂ t + u u ⋅ ∇ ρ + ρ u ⋅ ∇ u + ρ u ∇ ⋅ u = s {\displaystyle \mathbf {u} {\frac {\partial \rho }{\partial t}}+\rho {\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \mathbf {u} \cdot
abla \rho +\rho \mathbf {u} \cdot
abla \mathbf {u} +\rho \mathbf {u}
abla \cdot \mathbf {u} =\mathbf {s} }
Note that the gradient of a vector is a special case of the covariant derivative, the operation results in second rank tensors;[2] except in Cartesian coordinates, it is important to understand that this is not simply an element by element gradient. Rearranging and recognizing that u ⋅ ∇ρ + ρ∇ ⋅ u = ∇ ⋅ (ρu):
u ( ∂ ρ ∂ t + u ⋅ ∇ ρ + ρ ∇ ⋅ u ) + ρ ( ∂ u ∂ t + u ⋅ ∇ u ) = s u ( ∂ ρ ∂ t + ∇ ⋅ ( ρ u ) ) + ρ ( ∂ u ∂ t + u ⋅ ∇ u ) = s {\displaystyle {\begin{aligned}\mathbf {u} \left({\frac {\partial \rho }{\partial t}}+\mathbf {u} \cdot
abla \rho +\rho
abla \cdot \mathbf {u} \right)+\rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot
abla \mathbf {u} \right)&=\mathbf {s} \\[6px]\mathbf {u} \left({\frac {\partial \rho }{\partial t}}+
abla \cdot (\rho \mathbf {u} )\right)+\rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot
abla \mathbf {u} \right)&=\mathbf {s} \end{aligned}}}
The leftmost expression enclosed in parentheses is, by mass continuity (shown in a moment), equal to zero. Noting that what remains on the left side of the equation is the material derivative of flow velocity:
ρ D u D t = ρ ( ∂ u ∂ t + u ⋅ ∇ u ) = s {\displaystyle \rho {\frac {D\mathbf {u} }{Dt}}=\rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot
abla \mathbf {u} \right)=\mathbf {s} }
This appears to simply be an expression of Newton’s second law (F = ma) in terms of body forces instead of point forces. Each term in any case of the Navier–Stokes equations is a body force. A shorter though less rigorous way to arrive at this result would be the application of the chain rule to acceleration:
ρ d d t ( u ( x , y , z , t ) ) = s ⇒ ρ ( ∂ u ∂ t + ∂ u ∂ x d x d t + ∂ u ∂ y d y d t + ∂ u ∂ z d z d t ) = s ⇒ ρ ( ∂ u ∂ t + u ∂ u ∂ x + v ∂ u ∂ y + w ∂ u ∂ z ) = s ⇒ ρ ( ∂ u ∂ t + u ⋅ ∇ u ) = s {\displaystyle {\begin{aligned}\rho {\frac {d}{dt}}{\bigl (}\mathbf {u} (x,y,z,t){\bigr )}=\mathbf {s} \quad &\Rightarrow &\rho \left({\frac {\partial \mathbf {u} }{\partial t}}+{\frac {\partial \mathbf {u} }{\partial x}}{\frac {dx}{dt}}+{\frac {\partial \mathbf {u} }{\partial y}}{\frac {dy}{dt}}+{\frac {\partial \mathbf {u} }{\partial z}}{\frac {dz}{dt}}\right)&=\mathbf {s} \\\quad &\Rightarrow &\rho \left({\frac {\partial \mathbf {u} }{\partial t}}+u{\frac {\partial \mathbf {u} }{\partial x}}+v{\frac {\partial \mathbf {u} }{\partial y}}+w{\frac {\partial \mathbf {u} }{\partial z}}\right)&=\mathbf {s} \\\quad &\Rightarrow &\rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot
abla \mathbf {u} \right)&=\mathbf {s} \end{aligned}}}
where u = (u, v, w). The reason why this is “less rigorous” is that we haven’t shown that the choice of
u = ( d x d t , d y d t , d z d t ) {\displaystyle \mathbf {u} =\left({\frac {dx}{dt}},{\frac {dy}{dt}},{\frac {dz}{dt}}\right)}
is correct; however it does make sense since with that choice of path the derivative is “following” a fluid “particle”, and in order for Newton’s second law to work, forces must be summed following a particle. For this reason the convective derivative is also known as the particle derivative.
Conservation of mass [ edit ]
Mass may be considered also. When the intensive property φ is considered as the mass, by substitution into the general continuum equation, and taking s = 0 (no sources or sinks of mass):
∂ ρ ∂ t + ∇ ⋅ ( ρ u ) = 0 {\displaystyle {\frac {\partial \rho }{\partial t}}+
abla \cdot (\rho \mathbf {u} )=0}
where ρ is the mass density (mass per unit volume), and u is the flow velocity. This equation is called the mass continuity equation, or simply the continuity equation. This equation generally accompanies the Navier–Stokes equation.
In the case of an incompressible fluid, Dρ/Dt = 0 (the density following the path of a fluid element is constant) and the equation reduces to:
∇ ⋅ u = 0 {\displaystyle
abla \cdot \mathbf {u} =0}
which is in fact a statement of the conservation of volume.
Cauchy momentum equation [ edit ]
The generic density of the momentum source s seen previously is made specific first by breaking it up into two new terms, one to describe internal stresses and one for external forces, such as gravity. By examining the forces acting on a small cube in a fluid, it may be shown that
ρ D u D t = ∇ ⋅ σ + f {\displaystyle \rho {\frac {D\mathbf {u} }{Dt}}=
abla \cdot {\boldsymbol {\sigma }}+\mathbf {f} }
where σ is the Cauchy stress tensor, and f accounts for body forces present. This equation is called the Cauchy momentum equation and describes the non-relativistic momentum conservation of any continuum that conserves mass. σ is a rank two symmetric tensor given by its covariant components. In orthogonal coordinates in three dimensions it is represented as the 3 × 3 matrix:
σ i j = ( σ x x τ x y τ x z τ y x σ y y τ y z τ z x τ z y σ z z ) {\displaystyle \sigma _{ij}={\begin{pmatrix}\sigma _{xx}&\tau _{xy}&\tau _{xz}\\\tau _{yx}&\sigma _{yy}&\tau _{yz}\\\tau _{zx}&\tau _{zy}&\sigma _{zz}\end{pmatrix}}}
where the σ are normal stresses and τ shear stresses. This matrix is split up into two terms:
σ i j = ( σ x x τ x y τ x z τ y x σ y y τ y z τ z x τ z y σ z z ) = − ( p 0 0 0 p 0 0 0 p ) + ( σ x x + p τ x y τ x z τ y x σ y y + p τ y z τ z x τ z y σ z z + p ) = − p I + τ {\displaystyle \sigma _{ij}={\begin{pmatrix}\sigma _{xx}&\tau _{xy}&\tau _{xz}\\\tau _{yx}&\sigma _{yy}&\tau _{yz}\\\tau _{zx}&\tau _{zy}&\sigma _{zz}\end{pmatrix}}=-{\begin{pmatrix}p&0&0\\0&p&0\\0&0&p\end{pmatrix}}+{\begin{pmatrix}\sigma _{xx}+p&\tau _{xy}&\tau _{xz}\\\tau _{yx}&\sigma _{yy}+p&\tau _{yz}\\\tau _{zx}&\tau _{zy}&\sigma _{zz}+p\end{pmatrix}}=-p\mathbf {I} +{\boldsymbol {\tau }}}
where I is the 3 × 3 identity matrix and τ is the deviatoric stress tensor. Note that the mechanical pressure p is equal to the negative of the mean normal stress:
p = − 1 3 ( σ x x + σ y y + σ z z ) . {\displaystyle p=-{\tfrac {1}{3}}\left(\sigma _{xx}+\sigma _{yy}+\sigma _{zz}\right).}
The motivation for doing this is that pressure is typically a variable of interest, and also this simplifies application to specific fluid families later on since the rightmost tensor τ in the equation above must be zero for a fluid at rest. Note that τ is traceless. The Cauchy equation may now be written in another more explicit form:
ρ D u D t = − ∇ p + ∇ ⋅ τ + f {\displaystyle \rho {\frac {D\mathbf {u} }{Dt}}=-
abla p+
abla \cdot {\boldsymbol {\tau }}+\mathbf {f} }
This equation is still incomplete. For completion, one must make hypotheses on the forms of τ and p, that is, one needs a constitutive law for the stress tensor which can be obtained for specific fluid families and on the pressure. Some of these hypotheses lead to the Euler equations (fluid dynamics), other ones lead to the Navier–Stokes equations. Additionally, if the flow is assumed compressible an equation of state will be required, which will likely further require a conservation of energy formulation.
Application to different fluids [ edit ]
The general form of the equations of motion is not “ready for use”, the stress tensor is still unknown so that more information is needed; this information is normally some knowledge of the viscous behavior of the fluid. For different types of fluid flow this results in specific forms of the Navier–Stokes equations.
Newtonian fluid [ edit ]
Compressible Newtonian fluid [ edit ]
The formulation for Newtonian fluids stems from an observation made by Newton that, for most fluids,
τ ∝ ∂ u ∂ y {\displaystyle \tau \propto {\frac {\partial u}{\partial y}}}
In order to apply this to the Navier–Stokes equations, three assumptions were made by Stokes:
The stress tensor is a linear function of the strain rate tensor or equivalently the velocity gradient.
The fluid is isotropic.
For a fluid at rest, ∇ ⋅ τ must be zero (so that hydrostatic pressure results).
The above list states the classic argument[4] that the shear strain rate tensor (the (symmetric) shear part of the velocity gradient) is a pure shear tensor and does not include any inflow/outflow part (any compression/expansion part). This means that its trace is zero, and this is achieved by subtracting ∇ ⋅ u in a symmetric way from the diagonal elements of the tensor. The compressional contribution to viscous stress is added as a separate diagonal tensor.
Applying these assumptions will lead to :
τ = μ ( ∇ u + ( ∇ u ) T ) + λ ( ∇ ⋅ u ) I {\displaystyle \tau =\mu \left(
abla \mathbf {u} +\left(
abla \mathbf {u} \right)^{\mathsf {T}}\right)+\lambda \left(
abla \cdot \mathbf {u} \right)\mathbf {I} }
or in tensor form
τ i j = μ ( ∂ u i ∂ x j + ∂ u j ∂ x i ) + δ i j λ ∂ u k ∂ x k {\displaystyle \tau _{ij}=\mu \left({\frac {\partial u_{i}}{\partial x_{j}}}+{\frac {\partial u_{j}}{\partial x_{i}}}\right)+\delta _{ij}\lambda {\frac {\partial u_{k}}{\partial x_{k}}}}
That is, the deviatoric of the deformation rate tensor is identified to the deviatoric of the stress tensor, up to a factor μ.[5]
δ ij is the Kronecker delta. μ and λ are proportionality constants associated with the assumption that stress depends on strain linearly; μ is called the first coefficient of viscosity or shear viscosity (usually just called “viscosity”) and λ is the second coefficient of viscosity or volume viscosity (and it is related to bulk viscosity). The value of λ, which produces a viscous effect associated with volume change, is very difficult to determine, not even its sign is known with absolute certainty. Even in compressible flows, the term involving λ is often negligible; however it can occasionally be important even in nearly incompressible flows and is a matter of controversy. When taken nonzero, the most common approximation is λ ≈ − 2/3μ.
A straightforward substitution of τ ij into the momentum conservation equation will yield the Navier–Stokes equations, describing a compressible Newtonian fluid:
ρ ( ∂ u ∂ t + u ⋅ ∇ u ) = − ∇ p + ∇ ⋅ [ μ ( ∇ u + ( ∇ u ) T ) ] + ∇ ⋅ [ λ ( ∇ ⋅ u ) I ] + ρ g {\displaystyle \rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot
abla \mathbf {u} \right)=-
abla p+
abla \cdot \left[\mu \left(
abla \mathbf {u} +\left(
abla \mathbf {u} \right)^{\mathsf {T}}\right)\right]+
abla \cdot \left[\lambda \left(
abla \cdot \mathbf {u} \right)\mathbf {I} \right]+\rho \mathbf {g} }
The body force has been decomposed into density and external acceleration, that is, f = ρg. The associated mass continuity equation is:
∂ ρ ∂ t + ∇ ⋅ ( ρ u ) = 0 {\displaystyle {\frac {\partial \rho }{\partial t}}+
abla \cdot (\rho \mathbf {u} )=0}
In addition to this equation, an equation of state and an equation for the conservation of energy is needed. The equation of state to use depends on context (often the ideal gas law), the conservation of energy will read:
ρ D h D t = D p D t + ∇ ⋅ ( k ∇ T ) + Φ {\displaystyle \rho {\frac {Dh}{Dt}}={\frac {Dp}{Dt}}+
abla \cdot (k
abla T)+\Phi }
Here, h is the specific enthalpy, T is the temperature, and Φ is a function representing the dissipation of energy due to viscous effects:
Φ = μ ( 2 ( ∂ u ∂ x ) 2 + 2 ( ∂ v ∂ y ) 2 + 2 ( ∂ w ∂ z ) 2 + ( ∂ v ∂ x + ∂ u ∂ y ) 2 + ( ∂ w ∂ y + ∂ v ∂ z ) 2 + ( ∂ u ∂ z + ∂ w ∂ x ) 2 ) + λ ( ∇ ⋅ u ) 2 . {\displaystyle \Phi =\mu \left(2\left({\frac {\partial u}{\partial x}}\right)^{2}+2\left({\frac {\partial v}{\partial y}}\right)^{2}+2\left({\frac {\partial w}{\partial z}}\right)^{2}+\left({\frac {\partial v}{\partial x}}+{\frac {\partial u}{\partial y}}\right)^{2}+\left({\frac {\partial w}{\partial y}}+{\frac {\partial v}{\partial z}}\right)^{2}+\left({\frac {\partial u}{\partial z}}+{\frac {\partial w}{\partial x}}\right)^{2}\right)+\lambda (
abla \cdot \mathbf {u} )^{2}.}
With a good equation of state and good functions for the dependence of parameters (such as viscosity) on the variables, this system of equations seems to properly model the dynamics of all known gases and most liquids.
Incompressible Newtonian fluid [ edit ]
For the special (but very common) case of incompressible flow, the momentum equations simplify significantly. Using the following assumptions:
Viscosity μ will now be a constant
will now be a constant The second viscosity effect λ = 0
The simplified mass continuity equation ∇ ⋅ u = 0
This gives incompressible Navier-Stokes equations, describing incompressible Newtonian fluid:
ρ ( ∂ u ∂ t + u ⋅ ∇ u ) = − ∇ p + ∇ ⋅ [ μ ( ∇ u + ( ∇ u ) T ) ] + ρ g {\displaystyle \rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot
abla \mathbf {u} \right)=-
abla p+
abla \cdot \left[\mu \left(
abla \mathbf {u} +\left(
abla \mathbf {u} \right)^{\mathsf {T}}\right)\right]+\rho \mathbf {g} }
then looking at the viscous terms of the x momentum equation for example we have:
∂ ∂ x ( 2 μ ∂ u ∂ x ) + ∂ ∂ y ( μ ( ∂ u ∂ y + ∂ v ∂ x ) ) + ∂ ∂ z ( μ ( ∂ u ∂ z + ∂ w ∂ x ) ) = 2 μ ∂ 2 u ∂ x 2 + μ ∂ 2 u ∂ y 2 + μ ∂ 2 v ∂ y ∂ x + μ ∂ 2 u ∂ z 2 + μ ∂ 2 w ∂ z ∂ x = μ ∂ 2 u ∂ x 2 + μ ∂ 2 u ∂ y 2 + μ ∂ 2 u ∂ z 2 + μ ∂ 2 u ∂ x 2 + μ ∂ 2 v ∂ y ∂ x + μ ∂ 2 w ∂ z ∂ x = μ ∇ 2 u + μ ∂ ∂ x ( ∂ u ∂ x + ∂ v ∂ y + ∂ w ∂ z ) 0 = μ ∇ 2 u {\displaystyle {\begin{aligned}&{\frac {\partial }{\partial x}}\left(2\mu {\frac {\partial u}{\partial x}}\right)+{\frac {\partial }{\partial y}}\left(\mu \left({\frac {\partial u}{\partial y}}+{\frac {\partial v}{\partial x}}\right)\right)+{\frac {\partial }{\partial z}}\left(\mu \left({\frac {\partial u}{\partial z}}+{\frac {\partial w}{\partial x}}\right)\right)\\[8px]&\qquad =2\mu {\frac {\partial ^{2}u}{\partial x^{2}}}+\mu {\frac {\partial ^{2}u}{\partial y^{2}}}+\mu {\frac {\partial ^{2}v}{\partial y\,\partial x}}+\mu {\frac {\partial ^{2}u}{\partial z^{2}}}+\mu {\frac {\partial ^{2}w}{\partial z\,\partial x}}\\[8px]&\qquad =\mu {\frac {\partial ^{2}u}{\partial x^{2}}}+\mu {\frac {\partial ^{2}u}{\partial y^{2}}}+\mu {\frac {\partial ^{2}u}{\partial z^{2}}}+\mu {\frac {\partial ^{2}u}{\partial x^{2}}}+\mu {\frac {\partial ^{2}v}{\partial y\,\partial x}}+\mu {\frac {\partial ^{2}w}{\partial z\,\partial x}}\\[8px]&\qquad =\mu
abla ^{2}u+\mu {\frac {\partial }{\partial x}}{\cancelto {0}{\left({\frac {\partial u}{\partial x}}+{\frac {\partial v}{\partial y}}+{\frac {\partial w}{\partial z}}\right)}}\\[8px]&\qquad =\mu
abla ^{2}u\end{aligned}}\,}
Similarly for the y and z momentum directions we have μ∇2v and μ∇2w.
The above solution is key to deriving Navier–Stokes equations from the equation of motion in fluid dynamics when density and viscosity are constant.
Non-Newtonian fluids [ edit ]
A non-Newtonian fluid is a fluid whose flow properties differ in any way from those of Newtonian fluids. Most commonly the viscosity of non-Newtonian fluids is a function of shear rate or shear rate history. However, there are some non-Newtonian fluids with shear-independent viscosity, that nonetheless exhibit normal stress-differences or other non-Newtonian behaviour. Many salt solutions and molten polymers are non-Newtonian fluids, as are many commonly found substances such as ketchup, custard, toothpaste, starch suspensions, paint, blood, and shampoo. In a Newtonian fluid, the relation between the shear stress and the shear rate is linear, passing through the origin, the constant of proportionality being the coefficient of viscosity. In a non-Newtonian fluid, the relation between the shear stress and the shear rate is different, and can even be time-dependent. The study of the non-Newtonian fluids is usually called rheology. A few examples are given here.
Bingham fluid [ edit ]
In Bingham fluids, the situation is slightly different:
∂ u ∂ y = { 0 , τ < τ 0 τ − τ 0 μ , τ ≥ τ 0 {\displaystyle {\frac {\partial u}{\partial y}}={\begin{cases}0,&\tau <\tau _{0}\\[5px]{\dfrac {\tau -\tau _{0}}{\mu }},&\tau \geq \tau _{0}\end{cases}}} These are fluids capable of bearing some shear before they start flowing. Some common examples are toothpaste and clay. Power-law fluid [ edit ] A power law fluid is an idealised fluid for which the shear stress, τ, is given by τ = K ( ∂ u ∂ y ) n {\displaystyle \tau =K\left({\frac {\partial u}{\partial y}}\right)^{n}} This form is useful for approximating all sorts of general fluids, including shear thinning (such as latex paint) and shear thickening (such as corn starch water mixture). Stream function formulation [ edit ] In the analysis of a flow, it is often desirable to reduce the number of equations and/or the number of variables. The incompressible Navier–Stokes equation with mass continuity (four equations in four unknowns) can be reduced to a single equation with a single dependent variable in 2D, or one vector equation in 3D. This is enabled by two vector calculus identities: ∇ × ( ∇ ϕ ) = 0 ∇ ⋅ ( ∇ × A ) = 0 {\displaystyle {\begin{aligned} abla \times ( abla \phi )&=0\\ abla \cdot ( abla \times \mathbf {A} )&=0\end{aligned}}} for any differentiable scalar φ and vector A. The first identity implies that any term in the Navier–Stokes equation that may be represented as the gradient of a scalar will disappear when the curl of the equation is taken. Commonly, pressure p and external acceleration g will be eliminated, resulting in (this is true in 2D as well as 3D): ∇ × ( ∂ u ∂ t + u ⋅ ∇ u ) = ν ∇ × ( ∇ 2 u ) {\displaystyle abla \times \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot abla \mathbf {u} \right)= u abla \times \left( abla ^{2}\mathbf {u} \right)} where it is assumed that all body forces are describable as gradients (for example it is true for gravity), and density has been divided so that viscosity becomes kinematic viscosity. The second vector calculus identity above states that the divergence of the curl of a vector field is zero. Since the (incompressible) mass continuity equation specifies the divergence of flow velocity being zero, we can replace the flow velocity with the curl of some vector ψ so that mass continuity is always satisfied: ∇ ⋅ u = 0 ⇒ ∇ ⋅ ( ∇ × ψ ) = 0 ⇒ 0 = 0 {\displaystyle abla \cdot \mathbf {u} =0\quad \Rightarrow \quad abla \cdot ( abla \times {\boldsymbol {\psi }})=0\quad \Rightarrow \quad 0=0} So, as long as flow velocity is represented through u = ∇ × ψ, mass continuity is unconditionally satisfied. With this new dependent vector variable, the Navier–Stokes equation (with curl taken as above) becomes a single fourth order vector equation, no longer containing the unknown pressure variable and no longer dependent on a separate mass continuity equation: ∇ × ( ∂ ∂ t ( ∇ × ψ ) + ( ∇ × ψ ) ⋅ ∇ ( ∇ × ψ ) ) = ν ∇ × ( ∇ 2 ( ∇ × ψ ) ) {\displaystyle abla \times \left({\frac {\partial }{\partial t}}( abla \times {\boldsymbol {\psi }})+( abla \times {\boldsymbol {\psi }})\cdot abla ( abla \times {\boldsymbol {\psi }})\right)= u abla \times \left( abla ^{2}( abla \times {\boldsymbol {\psi }})\right)} Apart from containing fourth order derivatives, this equation is fairly complicated, and is thus uncommon. Note that if the cross differentiation is left out, the result is a third order vector equation containing an unknown vector field (the gradient of pressure) that may be determined from the same boundary conditions that one would apply to the fourth order equation above. 2D flow in orthogonal coordinates [ edit ] The true utility of this formulation is seen when the flow is two dimensional in nature and the equation is written in a general orthogonal coordinate system, in other words a system where the basis vectors are orthogonal. Note that this by no means limits application to Cartesian coordinates, in fact most of the common coordinates systems are orthogonal, including familiar ones like cylindrical and obscure ones like toroidal. The 3D flow velocity is expressed as (note that the discussion not used coordinates so far): u = u 1 e 1 + u 2 e 2 + u 3 e 3 {\displaystyle \mathbf {u} =u_{1}\mathbf {e} _{1}+u_{2}\mathbf {e} _{2}+u_{3}\mathbf {e} _{3}} where e i are basis vectors, not necessarily constant and not necessarily normalized, and u i are flow velocity components; let also the coordinates of space be (x 1 , x 2 , x 3 ). Now suppose that the flow is 2D. This does not mean the flow is in a plane, rather it means that the component of flow velocity in one direction is zero and the remaining components are independent of the same direction. In that case (take component 3 to be zero): u = u 1 e 1 + u 2 e 2 ; ∂ u 1 ∂ x 3 = ∂ u 2 ∂ x 3 = 0 {\displaystyle \mathbf {u} =u_{1}\mathbf {e} _{1}+u_{2}\mathbf {e} _{2};\qquad {\frac {\partial u_{1}}{\partial x_{3}}}={\frac {\partial u_{2}}{\partial x_{3}}}=0} The vector function ψ is still defined via: u = ∇ × ψ {\displaystyle \mathbf {u} = abla \times {\boldsymbol {\psi }}} but this must simplify in some way also since the flow is assumed 2D. If orthogonal coordinates are assumed, the curl takes on a fairly simple form, and the equation above expanded becomes: u 1 e 1 + u 2 e 2 = e 1 h 2 h 3 [ ∂ ∂ x 2 ( h 3 ψ 3 ) − ∂ ∂ x 3 ( h 2 ψ 2 ) ] + {\displaystyle u_{1}\mathbf {e} _{1}+u_{2}\mathbf {e} _{2}={\frac {\mathbf {e} _{1}}{h_{2}h_{3}}}\left[{\frac {\partial }{\partial x_{2}}}\left(h_{3}\psi _{3}\right)-{\frac {\partial }{\partial x_{3}}}\left(h_{2}\psi _{2}\right)\right]+} + e 2 h 3 h 1 [ ∂ ∂ x 3 ( h 1 ψ 1 ) − ∂ ∂ x 1 ( h 3 ψ 3 ) ] + e 3 h 1 h 2 [ ∂ ∂ x 1 ( h 2 ψ 2 ) − ∂ ∂ x 2 ( h 1 ψ 1 ) ] {\displaystyle {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }+{\frac {\mathbf {e} _{2}}{h_{3}h_{1}}}\left[{\frac {\partial }{\partial x_{3}}}\left(h_{1}\psi _{1}\right)-{\frac {\partial }{\partial x_{1}}}\left(h_{3}\psi _{3}\right)\right]+{\frac {\mathbf {e} _{3}}{h_{1}h_{2}}}\left[{\frac {\partial }{\partial x_{1}}}\left(h_{2}\psi _{2}\right)-{\frac {\partial }{\partial x_{2}}}\left(h_{1}\psi _{1}\right)\right]} Examining this equation shows that we can set ψ 1 = ψ 2 = 0 and retain equality with no loss of generality, so that: u 1 e 1 + u 2 e 2 = e 1 h 2 h 3 ∂ ∂ x 2 ( h 3 ψ 3 ) − e 2 h 3 h 1 ∂ ∂ x 1 ( h 3 ψ 3 ) {\displaystyle u_{1}\mathbf {e} _{1}+u_{2}\mathbf {e} _{2}={\frac {\mathbf {e} _{1}}{h_{2}h_{3}}}{\frac {\partial }{\partial x_{2}}}\left(h_{3}\psi _{3}\right)-{\frac {\mathbf {e} _{2}}{h_{3}h_{1}}}{\frac {\partial }{\partial x_{1}}}\left(h_{3}\psi _{3}\right)} the significance here is that only one component of ψ remains, so that 2D flow becomes a problem with only one dependent variable. The cross differentiated Navier–Stokes equation becomes two 0 = 0 equations and one meaningful equation. The remaining component ψ 3 = ψ is called the stream function. The equation for ψ can simplify since a variety of quantities will now equal zero, for example: ∇ ⋅ ψ = 1 h 1 h 2 h 3 ∂ ∂ x 3 ( ψ h 1 h 2 ) = 0 {\displaystyle abla \cdot {\boldsymbol {\psi }}={\frac {1}{h_{1}h_{2}h_{3}}}{\frac {\partial }{\partial x_{3}}}\left(\psi h_{1}h_{2}\right)=0} if the scale factors h 1 and h 2 also are independent of x 3 . Also, from the definition of the vector Laplacian ∇ × ( ∇ × ψ ) = ∇ ( ∇ ⋅ ψ ) − ∇ 2 ψ = − ∇ 2 ψ {\displaystyle abla \times ( abla \times {\boldsymbol {\psi }})= abla ( abla \cdot {\boldsymbol {\psi }})- abla ^{2}{\boldsymbol {\psi }}=- abla ^{2}{\boldsymbol {\psi }}} Manipulating the cross differentiated Navier–Stokes equation using the above two equations and a variety of identities[7] will eventually yield the 1D scalar equation for the stream function: ∂ ∂ t ( ∇ 2 ψ ) + ( ∇ × ψ ) ⋅ ∇ ( ∇ 2 ψ ) = ν ∇ 4 ψ {\displaystyle {\frac {\partial }{\partial t}}\left( abla ^{2}\psi \right)+( abla \times {\boldsymbol {\psi }})\cdot abla \left( abla ^{2}\psi \right)= u abla ^{4}\psi } where ∇4 is the biharmonic operator. This is very useful because it is a single self-contained scalar equation that describes both momentum and mass conservation in 2D. The only other equations that this partial differential equation needs are initial and boundary conditions. Derivation of the scalar stream function equation Distributing the curl: ∂ ∂ t ( ∇ × ( ∇ × ψ ) ) + ∇ × ( ( ∇ × ψ ) ⋅ ∇ ( ∇ × ψ ) ) = ν ∇ × ( ∇ 2 ( ∇ × ψ ) ) {\displaystyle {\frac {\partial }{\partial t}}( abla \times ( abla \times {\boldsymbol {\psi }}))+ abla \times {\bigl (}( abla \times {\boldsymbol {\psi }})\cdot abla ( abla \times {\boldsymbol {\psi }}){\bigr )}= u abla \times \left( abla ^{2}( abla \times {\boldsymbol {\psi }})\right)} Replacing curl of the curl with the Laplacian and expanding convection and viscosity: − ∂ ∂ t ( ∇ 2 ψ ) + ∇ × ( ∇ ( ( ∇ × ψ ) ⋅ ( ∇ × ψ ) 2 ) + ( ∇ × ( ∇ × ψ ) ) × ( ∇ × ψ ) ) = ν ( ∇ 2 ( ∇ ( ∇ ⋅ ψ ) ) − ∇ 4 ψ ) {\displaystyle -{\frac {\partial }{\partial t}}\left( abla ^{2}{\boldsymbol {\psi }}\right)+ abla \times \left( abla \left({\frac {( abla \times {\boldsymbol {\psi }})\cdot ( abla \times {\boldsymbol {\psi }})}{2}}\right)+\left( abla \times ( abla \times {\boldsymbol {\psi }})\right)\times ( abla \times {\boldsymbol {\psi }})\right)= u \left( abla ^{2}( abla ( abla \cdot {\boldsymbol {\psi }}))- abla ^{4}{\boldsymbol {\psi }}\right)} Above, the curl of a gradient is zero, and the divergence of ψ is zero. Negating: ∂ ∂ t ( ∇ 2 ψ ) + ∇ × ( ∇ 2 ψ × ( ∇ × ψ ) ) = ν ∇ 4 ψ {\displaystyle {\frac {\partial }{\partial t}}\left( abla ^{2}{\boldsymbol {\psi }}\right)+ abla \times \left( abla ^{2}{\boldsymbol {\psi }}\times ( abla \times {\boldsymbol {\psi }})\right)= u abla ^{4}{\boldsymbol {\psi }}} Expanding the curl of the cross product into four terms: ∂ ∂ t ( ∇ 2 ψ ) + ( ∇ × ψ ) ⋅ ∇ ( ∇ 2 ψ ) − ( ∇ 2 ψ ) ⋅ ∇ ( ∇ × ψ ) + ( ∇ 2 ψ ) ( ∇ ⋅ ( ∇ × ψ ) ) − ( ∇ × ψ ) ( ∇ ⋅ ( ∇ 2 ψ ) ) = ν ∇ 4 ψ {\displaystyle {\frac {\partial }{\partial t}}\left( abla ^{2}{\boldsymbol {\psi }}\right)+( abla \times {\boldsymbol {\psi }})\cdot abla \left( abla ^{2}{\boldsymbol {\psi }}\right)-\left( abla ^{2}{\boldsymbol {\psi }}\right)\cdot abla ( abla \times {\boldsymbol {\psi }})+\left( abla ^{2}{\boldsymbol {\psi }}\right)( abla \cdot ( abla \times {\boldsymbol {\psi }}))-( abla \times {\boldsymbol {\psi }})\left( abla \cdot \left( abla ^{2}{\boldsymbol {\psi }}\right)\right)= u abla ^{4}{\boldsymbol {\psi }}} Only one of four terms of the expanded curl is nonzero. The second is zero because it is the dot product of orthogonal vectors, the third is zero because it contains the divergence of flow velocity, and the fourth is zero because the divergence of a vector with only component three is zero (since it is assumed that nothing (except maybe h 3 ) depends on component three). ∂ ∂ t ( ∇ 2 ψ ) + ( ∇ × ψ ) ⋅ ∇ ( ∇ 2 ψ ) = ν ∇ 4 ψ {\displaystyle {\frac {\partial }{\partial t}}\left( abla ^{2}{\boldsymbol {\psi }}\right)+( abla \times {\boldsymbol {\psi }})\cdot abla \left( abla ^{2}{\boldsymbol {\psi }}\right)= u abla ^{4}{\boldsymbol {\psi }}} This vector equation is one meaningful scalar equation and two 0 = 0 equations. The assumptions for the stream function equation are: The flow is incompressible and Newtonian. Coordinates are orthogonal. Flow is 2D: u 3 = ∂ u 1 / ∂ x 3 = ∂ u 2 / ∂ x 3 = 0 The first two scale factors of the coordinate system are independent of the last coordinate: ∂h 1 / ∂x 3 = ∂h 2 / ∂x 3 = 0 , otherwise extra terms appear. The stream function has some useful properties: Since −∇ 2 ψ = ∇ × (∇ × ψ ) = ∇ × u , the vorticity of the flow is just the negative of the Laplacian of the stream function. , the vorticity of the flow is just the negative of the Laplacian of the stream function. The level curves of the stream function are streamlines. The stress tensor [ edit ] The derivation of the Navier–Stokes equation involves the consideration of forces acting on fluid elements, so that a quantity called the stress tensor appears naturally in the Cauchy momentum equation. Since the divergence of this tensor is taken, it is customary to write out the equation fully simplified, so that the original appearance of the stress tensor is lost. However, the stress tensor still has some important uses, especially in formulating boundary conditions at fluid interfaces. Recalling that σ = −pI + τ, for a Newtonian fluid the stress tensor is: σ i j = − p δ i j + μ ( ∂ u i ∂ x j + ∂ u j ∂ x i ) + δ i j λ ∇ ⋅ u . {\displaystyle \sigma _{ij}=-p\delta _{ij}+\mu \left({\frac {\partial u_{i}}{\partial x_{j}}}+{\frac {\partial u_{j}}{\partial x_{i}}}\right)+\delta _{ij}\lambda abla \cdot \mathbf {u} .} If the fluid is assumed to be incompressible, the tensor simplifies significantly. In 3D cartesian coordinates for example: σ = − ( p 0 0 0 p 0 0 0 p ) + μ ( 2 ∂ u ∂ x ∂ u ∂ y + ∂ v ∂ x ∂ u ∂ z + ∂ w ∂ x ∂ v ∂ x + ∂ u ∂ y 2 ∂ v ∂ y ∂ v ∂ z + ∂ w ∂ y ∂ w ∂ x + ∂ u ∂ z ∂ w ∂ y + ∂ v ∂ z 2 ∂ w ∂ z ) = − p I + μ ( ∇ u + ( ∇ u ) T ) = − p I + 2 μ e {\displaystyle {\begin{aligned}{\boldsymbol {\sigma }}&=-{\begin{pmatrix}p&0&0\\0&p&0\\0&0&p\end{pmatrix}}+\mu {\begin{pmatrix}2\displaystyle {\frac {\partial u}{\partial x}}&\displaystyle {{\frac {\partial u}{\partial y}}+{\frac {\partial v}{\partial x}}}&\displaystyle {{\frac {\partial u}{\partial z}}+{\frac {\partial w}{\partial x}}}\\\displaystyle {{\frac {\partial v}{\partial x}}+{\frac {\partial u}{\partial y}}}&2\displaystyle {\frac {\partial v}{\partial y}}&\displaystyle {{\frac {\partial v}{\partial z}}+{\frac {\partial w}{\partial y}}}\\\displaystyle {{\frac {\partial w}{\partial x}}+{\frac {\partial u}{\partial z}}}&\displaystyle {{\frac {\partial w}{\partial y}}+{\frac {\partial v}{\partial z}}}&2\displaystyle {\frac {\partial w}{\partial z}}\end{pmatrix}}\\[6px]&=-p\mathbf {I} +\mu \left( abla \mathbf {u} +\left( abla \mathbf {u} \right)^{\mathsf {T}}\right)\\[6px]&=-p\mathbf {I} +2\mu \mathbf {e} \end{aligned}}} e is the strain rate tensor, by definition: e i j = 1 2 ( ∂ u i ∂ x j + ∂ u j ∂ x i ) . {\displaystyle e_{ij}={\frac {1}{2}}\left({\frac {\partial u_{i}}{\partial x_{j}}}+{\frac {\partial u_{j}}{\partial x_{i}}}\right).} See also [ edit ]
연속방정식, 나비에-스톡스 방정식 (Continuity Equation, Navier-Stokes Equation)
먼저 연속방정식에 대해 생각해보자.
연속 방정식은 질량보존법칙을 오일러 표현으로 나타내었을 때
어떤 모양이 되는지를 설명해주는 식이다.
라그랑지언 표현을 오일러 표현으로 바꾸어주는 RTT를 사용해
미소 면적을 표현해보자.
미소면적의 중앙에서 모든 Property의 값을 갖는다고 가정했을 때,
상하좌우에서 Property의 값은 테일러 급수를 통해 값을 근사해서
나타낼 수 있다.
(1계 미분까지만 표현한 것은 2계미분 항부터는 너무 크기가 작아 무시할 수 있기 때문에)
이렇게 나온 식을 연속 방정식이라 부른다.
그런데 여기서 이 항을 조금더 풀게 되면 물질도함수의 모양으로 정리가 가능하다.
다시 연속방정식을 이 모양으로 해석하게 되면,
특정 점에서 밀도의 시간에 따른 변화는 경계면에서의 속도의 발산(다이버전스)값에 밀도를 곱해준 것과 같다.
로 해석할 수 있다.
여기서 만약 시간에 따라 밀도가 일정하다면,
즉, 특정점에서 밀도 변화가 없다면,
(밀도 = 상수, 비압축성 유체)
속도의 발산 값이 0, 즉 경계면에서 나가고 들어오는 양의 총합이 0,
우리가 생각하는 결과와 일치한다.
나비에 스톡스 방정식은 그럼 무엇인가.
뉴턴 법칙을 오일러 관점에서 서술한 식이다.
net_F = m*a에서 RHS의 서술이 오일러 관점에서 어떻게 서술될 수 있는지
알 수 있다.
이를 바꿔주는 RTT에서 B = V를 대입해서 정리해보자.
마찬가지로 미소면적에 대해 정리하면,
여기서 비압축성 유체(밀도 = 상수)일 경우,
첫번째 항에서 밀도가 상수로 빠져나올 수 있게 된다.
그리고 비압축성 유체의 연속방정식으로 부터,
결론적으로 미소 부피의 Total F는 밀도 * 속도에 대한 물질도함수로 나타낼 수 있다.
그런데, 이 건 외력의 총합이 시스템에서 어떻게 작용하는 지에 대한,
m*a에 대한 정리이고,
실제로 특정 시스템을 잡았을 때, 그 곳에 작용 할 수 있는 외력의 종류가
무엇이 있는지 정리할 필요가 있다.
우리가 가하는 힘을 제외했을 때, 자연적으로 발생하는 힘은
Body force와 Surface force로 나눌 수 있다.
그중 Body force는 쉽게 구할 수 있다.
표면힘이 조금 까다로운데, 미소면적의 중앙에서 작용하는 응력을 기준으로
잡았을 때, 테일러 급수에 의해 각 면에 해당하는 응력을 구할 수 있다.
이 때, 응력에 의해 이 미소면적에 작용하는 알짜 힘은 각면에서의 응력차에
해당하는 힘이 그 역할을 하므로, 다음과 같이 구할 수 있을 것이다.
만약 3차원이라면 각포인트에서 3개의 응력항이 나오고, 이것을 행렬형태로
만든것을 응력 텐서라 한다. 이것에 대해서는 다른 포스팅에서 좀더 자세하게
다루도록 하겠다.
이것들을 한번에 나타내면,
외력은 라그랑지언 관점 과, 오일러 관점 에서 서술할 수 있다.
라그랑지언 관점 은 기존의 생각하는 방식과 동일하게 작용하는 외력항을
다 더해준 형태를 말한다.
오일러 관점 에서는 조금 변화되서 생각하는 것이 좋은데,
다른 구조물에 의한 외력을 제외했을 때, body force와 surface force로
생각한다. 이는 유체내에서 해석하기 좋게 하기 위해 식을 풀어 쓰는 것이다.
ma항도 역시 라그랑지언 관점 과, 오일러 관점 에서 서술할 수 있다.
라그랑지언 관점 은 ma로 해석하면 된다.
오일러 관점 은 물질도함수를 사용하여 나타내는데,
전역가속항과 국소가속항으로 나누어서 표현한다.
유체역학에서 ma항은 보통 오일러 관점으로 서술한다.
뉴튼 유체, 즉 incompressible(밀도 상수)인 유체에서
나비에 스톡스 방정식을 정리해보자.
밑 부분은 사실 벡터 형태로 정리할 경우 더 직관적인 사실을 알 수 있다.
지금은 그냥 넘어가고 다음 포스팅에서 이를 알아보자.
마지막으로 이 나비에 스톡스 방정식에서 베르누이 방정식을 도출할 수 있는지
알아보자. 베르누이 정리를 쓰기위한 4가지 가정을 이 식에 적용시켜야 한다.
Steady No friction Incompressible Along the Streamline
키워드에 대한 정보 navier stokes equation 유도
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