Misplaced Pages

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100-432: The Navier–Stokes equations mathematically express momentum balance for Newtonian fluids and make use of conservation of mass . They are sometimes accompanied by an equation of state relating pressure , temperature and density . They arise from applying Isaac Newton's second law to fluid motion , together with the assumption that the stress in the fluid is the sum of a diffusing viscous term (proportional to

200-466: A {\displaystyle \rho {\frac {\mathrm {D} \mathbf {u} }{\mathrm {D} t}}=-\nabla p+\nabla \cdot {\boldsymbol {\tau }}+\rho \,\mathbf {a} } where In this form, it is apparent that in the assumption of an inviscid fluid – no deviatoric stress – Cauchy equations reduce to the Euler equations . Assuming conservation of mass , with the known properties of divergence and gradient we can use

300-644: A . {\displaystyle \left({\frac {\partial }{\partial t}}+\mathbf {u} \cdot \nabla -\nu \,\nabla ^{2}-({\tfrac {1}{3}}\nu +\xi )\,\nabla (\nabla \cdot )\right)\mathbf {u} =-{\frac {1}{\rho }}\nabla p+\mathbf {a} .} The convective acceleration term can also be written as u ⋅ ∇ u = ( ∇ × u ) × u + 1 2 ∇ u 2 , {\displaystyle \mathbf {u} \cdot \nabla \mathbf {u} =(\nabla \times \mathbf {u} )\times \mathbf {u} +{\tfrac {1}{2}}\nabla \mathbf {u} ^{2},} where

400-446: A . {\displaystyle \rho {\frac {\mathrm {D} \mathbf {u} }{\mathrm {D} t}}=\rho \left({\frac {\partial \mathbf {u} }{\partial t}}+(\mathbf {u} \cdot \nabla )\mathbf {u} \right)=-\nabla p+\nabla \cdot \left\{\mu \left[\nabla \mathbf {u} +(\nabla \mathbf {u} )^{\mathrm {T} }-{\tfrac {2}{3}}(\nabla \cdot \mathbf {u} )\mathbf {I} \right]\right\}+\nabla [\zeta (\nabla \cdot \mathbf {u} )]+\rho \mathbf {a} .} in index notation,

500-520: A . {\displaystyle \rho {\frac {\mathrm {D} \mathbf {u} }{\mathrm {D} t}}=\rho \left({\frac {\partial \mathbf {u} }{\partial t}}+(\mathbf {u} \cdot \nabla )\mathbf {u} \right)=-\nabla p+\nabla \cdot \left\{\mu \left[\nabla \mathbf {u} +(\nabla \mathbf {u} )^{\mathrm {T} }-{\tfrac {2}{3}}(\nabla \cdot \mathbf {u} )\mathbf {I} \right]\right\}+\rho \mathbf {a} .} If the dynamic μ and bulk ζ {\displaystyle \zeta } viscosities are assumed to be uniform in space,

600-459: A . {\displaystyle {\frac {D\mathbf {u} }{Dt}}=-{\frac {1}{\rho }}\nabla p+\nu \,\nabla ^{2}\mathbf {u} +({\tfrac {1}{3}}\nu +\xi )\,\nabla (\nabla \cdot \mathbf {u} )+\mathbf {a} .} where D D t {\textstyle {\frac {\mathrm {D} }{\mathrm {D} t}}} is the material derivative . ν = μ ρ {\displaystyle \nu ={\frac {\mu }{\rho }}}

700-409: A . {\displaystyle {\frac {\partial }{\partial t}}(\rho \mathbf {u} )+\nabla \cdot \left(\rho \mathbf {u} \otimes \mathbf {u} +[p-\zeta (\nabla \cdot \mathbf {u} )]\mathbf {I} -\mu \left[\nabla \mathbf {u} +(\nabla \mathbf {u} )^{\mathrm {T} }-{\tfrac {2}{3}}(\nabla \cdot \mathbf {u} )\mathbf {I} \right]\right)=\rho \mathbf {a} .} Apart from its dependence of pressure and temperature,

800-601: A i . {\displaystyle \rho \left({\frac {\partial u_{i}}{\partial t}}+u_{k}{\frac {\partial u_{i}}{\partial x_{k}}}\right)=-{\frac {\partial p}{\partial x_{i}}}+{\frac {\partial }{\partial x_{k}}}\left[\mu \left({\frac {\partial u_{i}}{\partial x_{k}}}+{\frac {\partial u_{k}}{\partial x_{i}}}-{\frac {2}{3}}\delta _{ik}{\frac {\partial u_{l}}{\partial x_{l}}}\right)\right]+{\frac {\partial }{\partial x_{i}}}\left(\zeta {\frac {\partial u_{l}}{\partial x_{l}}}\right)+\rho a_{i}.} The corresponding equation in conservation form can be obtained by considering that, given

900-402: A n t . {\displaystyle m_{A}v_{A}+m_{B}v_{B}+m_{C}v_{C}+...=constant.} This conservation law applies to all interactions, including collisions (both elastic and inelastic ) and separations caused by explosive forces. It can also be generalized to situations where Newton's laws do not hold, for example in the theory of relativity and in electrodynamics . Momentum

1000-418: A Galilean transformation . If a particle is moving at speed ⁠ d x / d t ⁠ = v in the first frame of reference, in the second, it is moving at speed v ′ = d x ′ d t = v − u . {\displaystyle v'={\frac {{\text{d}}x'}{{\text{d}}t}}=v-u\,.} Since u does not change,

1100-585: A US$ 1 million prize for a solution or a counterexample. The solution of the equations is a flow velocity . It is a vector field —to every point in a fluid, at any moment in a time interval, it gives a vector whose direction and magnitude are those of the velocity of the fluid at that point in space and at that moment in time. It is usually studied in three spatial dimensions and one time dimension, although two (spatial) dimensional and steady-state cases are often used as models, and higher-dimensional analogues are studied in both pure and applied mathematics. Once

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1200-402: A collision. For example, suppose there are two bodies of equal mass m , one stationary and one approaching the other at a speed v (as in the figure). The center of mass is moving at speed ⁠ v / 2 ⁠ and both bodies are moving towards it at speed ⁠ v / 2 ⁠ . Because of the symmetry, after the collision both must be moving away from the center of mass at

1300-410: A flow with respect to space. While individual fluid particles indeed experience time-dependent acceleration, the convective acceleration of the flow field is a spatial effect, one example being fluid speeding up in a nozzle. Remark: here, the deviatoric stress tensor is denoted τ {\textstyle {\boldsymbol {\tau }}} as it was in the general continuum equations and in

1400-656: A momentum of 1 kg⋅m/s due north measured with reference to the ground. The momentum of a system of particles is the vector sum of their momenta. If two particles have respective masses m 1 and m 2 , and velocities v 1 and v 2 , the total momentum is p = p 1 + p 2 = m 1 v 1 + m 2 v 2 . {\displaystyle {\begin{aligned}p&=p_{1}+p_{2}\\&=m_{1}v_{1}+m_{2}v_{2}\,.\end{aligned}}} The momenta of more than two particles can be added more generally with

1500-514: A purely mathematical sense. Despite their wide range of practical uses, it has not yet been proven whether smooth solutions always exist in three dimensions—i.e., whether they are infinitely differentiable (or even just bounded) at all points in the domain . This is called the Navier–Stokes existence and smoothness problem. The Clay Mathematics Institute has called this one of the seven most important open problems in mathematics and has offered

1600-502: A vector quantity), then the object's momentum p (from Latin pellere "push, drive") is: p = m v . {\displaystyle \mathbf {p} =m\mathbf {v} .} In the International System of Units (SI), the unit of measurement of momentum is the kilogram metre per second (kg⋅m/s), which is dimensionally equivalent to the newton-second . Newton's second law of motion states that

1700-524: Is ∇ ( ∇ ⋅ u ) {\textstyle \nabla \left(\nabla \cdot \mathbf {u} \right)} , one finally arrives to the compressible Navier–Stokes momentum equation: D u D t = − 1 ρ ∇ p + ν ∇ 2 u + ( 1 3 ν + ξ ) ∇ ( ∇ ⋅ u ) +

1800-413: Is a good example of an almost totally elastic collision, due to their high rigidity , but when bodies come in contact there is always some dissipation . A head-on elastic collision between two bodies can be represented by velocities in one dimension, along a line passing through the bodies. If the velocities are v A1 and v B1 before the collision and v A2 and v B2 after,

1900-603: Is a measurable quantity, and the measurement depends on the frame of reference . For example: if an aircraft of mass 1000 kg is flying through the air at a speed of 50 m/s its momentum can be calculated to be 50,000 kg.m/s. If the aircraft is flying into a headwind of 5 m/s its speed relative to the surface of the Earth is only 45 m/s and its momentum can be calculated to be 45,000 kg.m/s. Both calculations are equally correct. In both frames of reference, any change in momentum will be found to be consistent with

2000-472: Is a negative unit vector. Then, we calculated the stress vector by definition s → = n → ⋅ σ = [ − σ x x , − σ x y , − σ x z ] {\displaystyle {\vec {s}}={\vec {n}}\cdot {\boldsymbol {\sigma }}=[-\sigma _{xx},-\sigma _{xy},-\sigma _{xz}]} , thus

2100-466: Is an inelastic collision . An elastic collision is one in which no kinetic energy is transformed into heat or some other form of energy. Perfectly elastic collisions can occur when the objects do not touch each other, as for example in atomic or nuclear scattering where electric repulsion keeps the objects apart. A slingshot maneuver of a satellite around a planet can also be viewed as a perfectly elastic collision. A collision between two pool balls

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2200-452: Is an expression of one of the fundamental symmetries of space and time: translational symmetry . Advanced formulations of classical mechanics, Lagrangian and Hamiltonian mechanics , allow one to choose coordinate systems that incorporate symmetries and constraints. In these systems the conserved quantity is generalized momentum , and in general this is different from the kinetic momentum defined above. The concept of generalized momentum

2300-511: Is carried over into quantum mechanics, where it becomes an operator on a wave function . The momentum and position operators are related by the Heisenberg uncertainty principle . In continuous systems such as electromagnetic fields , fluid dynamics and deformable bodies , a momentum density can be defined as momentum per volume (a volume-specific quantity ). A continuum version of the conservation of momentum leads to equations such as

2400-605: Is equal to the instantaneous force F acting on it, F = d p d t . {\displaystyle F={\frac {{\text{d}}p}{{\text{d}}t}}.} If the net force experienced by a particle changes as a function of time, F ( t ) , the change in momentum (or impulse J ) between times t 1 and t 2 is Δ p = J = ∫ t 1 t 2 F ( t ) d t . {\displaystyle \Delta p=J=\int _{t_{1}}^{t_{2}}F(t)\,{\text{d}}t\,.} Impulse

2500-407: Is known as Euler's first law . If the net force F applied to a particle is constant, and is applied for a time interval Δ t , the momentum of the particle changes by an amount Δ p = F Δ t . {\displaystyle \Delta p=F\Delta t\,.} In differential form, this is Newton's second law ; the rate of change of the momentum of a particle

2600-468: Is measured in the derived units of the newton second (1 N⋅s = 1 kg⋅m/s) or dyne second (1 dyne⋅s = 1 g⋅cm/s) Under the assumption of constant mass m , it is equivalent to write F = d ( m v ) d t = m d v d t = m a , {\displaystyle F={\frac {{\text{d}}(mv)}{{\text{d}}t}}=m{\frac {{\text{d}}v}{{\text{d}}t}}=ma,} hence

2700-651: Is momentum at time t , and F ¯ → {\displaystyle {\vec {\bar {F}}}} is force averaged over Δ t {\displaystyle \Delta t} . After dividing by Δ t {\displaystyle \Delta t} and passing to the limit Δ t → 0 {\displaystyle \Delta t\to 0} we get ( derivative ): d p → d t = F → {\displaystyle {\frac {d{\vec {p}}}{dt}}={\vec {F}}} Let us analyse each side of

2800-466: Is numerically equivalent to 3 newtons. In a closed system (one that does not exchange any matter with its surroundings and is not acted on by external forces) the total momentum remains constant. This fact, known as the law of conservation of momentum , is implied by Newton's laws of motion . Suppose, for example, that two particles interact. As explained by the third law, the forces between them are equal in magnitude but opposite in direction. If

2900-748: Is proportional to the shear viscosity: σ ′ = τ = μ [ ∇ u + ( ∇ u ) T − 2 3 ( ∇ ⋅ u ) I ] {\displaystyle {\boldsymbol {\sigma }}'={\boldsymbol {\tau }}=\mu \left[\nabla \mathbf {u} +(\nabla \mathbf {u} )^{\mathrm {T} }-{\tfrac {2}{3}}(\nabla \cdot \mathbf {u} )\mathbf {I} \right]} Both bulk viscosity ζ {\textstyle \zeta } and dynamic viscosity μ {\textstyle \mu } need not be constant – in general, they depend on two thermodynamics variables if

3000-455: Is represented by the nonlinear quantity u ⋅ ∇ u , which may be interpreted either as ( u ⋅ ∇) u or as u ⋅ (∇ u ) , with ∇ u the tensor derivative of the velocity vector u . Both interpretations give the same result. The convective acceleration ( u ⋅ ∇) u can be thought of as the advection operator u ⋅ ∇ acting on the velocity field u . This contrasts with the expression in terms of tensor derivative ∇ u , which

3100-422: Is the center of mass frame – one that is moving with the center of mass. In this frame, the total momentum is zero. If two particles, each of known momentum, collide and coalesce, the law of conservation of momentum can be used to determine the momentum of the coalesced body. If the outcome of the collision is that the two particles separate, the law is not sufficient to determine the momentum of each particle. If

Navier–Stokes equations - Misplaced Pages Continue

3200-704: Is the identity tensor , and tr ⁡ ( ε ) {\textstyle \operatorname {tr} ({\boldsymbol {\varepsilon }})} is the trace of the rate-of-strain tensor. So this decomposition can be explicitly defined as: σ = − p I + λ ( ∇ ⋅ u ) I + μ ( ∇ u + ( ∇ u ) T ) . {\displaystyle {\boldsymbol {\sigma }}=-p\mathbf {I} +\lambda (\nabla \cdot \mathbf {u} )\mathbf {I} +\mu \left(\nabla \mathbf {u} +(\nabla \mathbf {u} )^{\mathrm {T} }\right).} Since

3300-801: Is the momentum density at a given space-time point, F is the flux associated to the momentum density, and s contains all of the body forces per unit volume. Let us start with the generalized momentum conservation principle which can be written as follows: "The change in system momentum is proportional to the resulting force acting on this system". It is expressed by the formula: p → ( t + Δ t ) − p → ( t ) = Δ t F ¯ → {\displaystyle {\vec {p}}(t+\Delta t)-{\vec {p}}(t)=\Delta t{\vec {\bar {F}}}} where p → ( t ) {\displaystyle {\vec {p}}(t)}

3400-421: Is the momentum density at the point considered in the continuum (for which the continuity equation holds), F is the flux associated to the momentum density, and s contains all of the body forces per unit volume. u ⊗ u is the dyad of the velocity. Here j and s have same number of dimensions N as the flow speed and the body acceleration, while F , being a tensor , has N . In

3500-415: Is the outer product of the flow velocity ( u {\displaystyle \mathbf {u} } ): u ⊗ u = u u T {\displaystyle \mathbf {u} \otimes \mathbf {u} =\mathbf {u} \mathbf {u} ^{\mathrm {T} }} The left side of the equation describes acceleration, and may be composed of time-dependent and convective components (also

3600-1571: Is the approximation of the values of stress acting on the walls opposite the walls covering the axes. Let us focus on the right wall where the stress is an approximation of stress σ x x {\displaystyle \sigma _{xx}} from the left wall at points with coordinates x + d x {\displaystyle x+dx} and it is equal to σ x x + ∂ σ x x ∂ x d x {\displaystyle \sigma _{xx}+{\frac {\partial \sigma _{xx}}{\partial x}}dx} . This approximation suffices since, as d x {\displaystyle dx} goes to zero, σ x x ( x + d x ) − ( σ x x ( x ) + d x ∂ σ x x ( x ) ∂ x ) d x = σ x x ( x + d x ) − σ x x ( x ) d x − ∂ σ x x ( x ) ∂ x {\displaystyle {\frac {\sigma _{xx}(x+dx)-(\sigma _{xx}(x)+dx{\frac {\partial \sigma _{xx}(x)}{\partial x}})}{dx}}={\frac {\sigma _{xx}(x+dx)-\sigma _{xx}(x)}{dx}}-{\frac {\partial \sigma _{xx}(x)}{\partial x}}} goes to zero as well, by definition of partial derivative. A more intuitive representation of

3700-628: Is the component-wise derivative of the velocity vector defined by [∇ u ] mi = ∂ m v i , so that [ u ⋅ ( ∇ u ) ] i = ∑ m v m ∂ m v i = [ ( u ⋅ ∇ ) u ] i . {\displaystyle \left[\mathbf {u} \cdot \left(\nabla \mathbf {u} \right)\right]_{i}=\sum _{m}v_{m}\partial _{m}v_{i}=\left[(\mathbf {u} \cdot \nabla )\mathbf {u} \right]_{i}\,.} The vector calculus identity of

3800-784: Is the shear kinematic viscosity and ξ = ζ ρ {\displaystyle \xi ={\frac {\zeta }{\rho }}} is the bulk kinematic viscosity. The left-hand side changes in the conservation form of the Navier–Stokes momentum equation. By bringing the operator on the flow velocity on the left side, on also has: ( ∂ ∂ t + u ⋅ ∇ − ν ∇ 2 − ( 1 3 ν + ξ ) ∇ ( ∇ ⋅ ) ) u = − 1 ρ ∇ p +

3900-535: Is three: tr ⁡ ( I ) = 3. {\displaystyle \operatorname {tr} ({\boldsymbol {I}})=3.} the trace of the stress tensor in three dimensions becomes: tr ⁡ ( σ ) = − 3 p + ( 3 λ + 2 μ ) ∇ ⋅ u . {\displaystyle \operatorname {tr} ({\boldsymbol {\sigma }})=-3p+(3\lambda +2\mu )\nabla \cdot \mathbf {u} .} So by alternatively decomposing

4000-455: Is unchanged. Forces such as Newtonian gravity, which depend only on the scalar distance between objects, satisfy this criterion. This independence of reference frame is called Newtonian relativity or Galilean invariance . A change of reference frame, can, often, simplify calculations of motion. For example, in a collision of two particles, a reference frame can be chosen, where, one particle begins at rest. Another, commonly used reference frame,

4100-399: Is usually neglected most of the time (that is whenever we are not dealing with processes such as sound absorption and attenuation of shock waves, where second viscosity coefficient becomes important) by explicitly assuming ζ = 0 {\textstyle \zeta =0} . The assumption of setting ζ = 0 {\textstyle \zeta =0} is called as

Navier–Stokes equations - Misplaced Pages Continue

4200-763: The Cartesian coordinate system ) above for clarity, but the equation is written using physical components (which are neither covariants ("column") nor contravariants ("row") ). However, if we chose a non-orthogonal curvilinear coordinate system , then we should calculate and write equations in covariant ("row vectors") or contravariant ("column vectors") form. After an appropriate change of variables, it can also be written in conservation form : ∂ j ∂ t + ∇ ⋅ F = s {\displaystyle {\frac {\partial \mathbf {j} }{\partial t}}+\nabla \cdot \mathbf {F} =\mathbf {s} } where j

4300-595: The Cauchy stress tensor σ {\textstyle {\boldsymbol {\sigma }}} to be the sum of a viscosity term τ {\textstyle {\boldsymbol {\tau }}} (the deviatoric stress ) and a pressure term − p I {\textstyle -p\mathbf {I} } (volumetric stress), we arrive at: ρ D u D t = − ∇ p + ∇ ⋅ τ + ρ

4400-459: The Franck–Hertz experiment ); and particle accelerators in which the kinetic energy is converted into mass in the form of new particles. In a perfectly inelastic collision (such as a bug hitting a windshield), both bodies have the same motion afterwards. A head-on inelastic collision between two bodies can be represented by velocities in one dimension, along a line passing through the bodies. If

4500-429: The Navier–Stokes equations for fluids or the Cauchy momentum equation for deformable solids or fluids. Momentum is a vector quantity : it has both magnitude and direction. Since momentum has a direction, it can be used to predict the resulting direction and speed of motion of objects after they collide. Below, the basic properties of momentum are described in one dimension. The vector equations are almost identical to

4600-905: The Stokes hypothesis . The validity of Stokes hypothesis can be demonstrated for monoatomic gas both experimentally and from the kinetic theory; for other gases and liquids, Stokes hypothesis is generally incorrect. With the Stokes hypothesis, the Navier–Stokes equations become ρ D u D t = ρ ( ∂ u ∂ t + ( u ⋅ ∇ ) u ) = − ∇ p + ∇ ⋅ { μ [ ∇ u + ( ∇ u ) T − 2 3 ( ∇ ⋅ u ) I ] } + ρ

4700-954: The bulk viscosity ζ {\textstyle \zeta } , ζ ≡ λ + 2 3 μ , {\displaystyle \zeta \equiv \lambda +{\tfrac {2}{3}}\mu ,} we arrive to the linear constitutive equation in the form usually employed in thermal hydraulics : σ = − [ p − ζ ( ∇ ⋅ u ) ] I + μ [ ∇ u + ( ∇ u ) T − 2 3 ( ∇ ⋅ u ) I ] {\displaystyle {\boldsymbol {\sigma }}=-[p-\zeta (\nabla \cdot \mathbf {u} )]\mathbf {I} +\mu \left[\nabla \mathbf {u} +(\nabla \mathbf {u} )^{\mathrm {T} }-{\tfrac {2}{3}}(\nabla \cdot \mathbf {u} )\mathbf {I} \right]} which can also be arranged in

4800-417: The deviatoric stress tensor σ ′ {\displaystyle {\boldsymbol {\sigma }}'} is still coincident with the shear stress tensor τ {\displaystyle {\boldsymbol {\tau }}} (i.e. the deviatoric stress in a Newtonian fluid has no normal stress components), and it has a compressibility term in addition to the incompressible case, which

4900-588: The gradient of velocity) and a pressure term—hence describing viscous flow . The difference between them and the closely related Euler equations is that Navier–Stokes equations take viscosity into account while the Euler equations model only inviscid flow . As a result, the Navier–Stokes are a parabolic equation and therefore have better analytic properties, at the expense of having less mathematical structure (e.g. they are never completely integrable ). The Navier–Stokes equations are useful because they describe

5000-667: The incompressible flow section . The compressible momentum Navier–Stokes equation results from the following assumptions on the Cauchy stress tensor: σ ( ε ) = − p I + λ tr ⁡ ( ε ) I + 2 μ ε {\displaystyle {\boldsymbol {\sigma }}({\boldsymbol {\varepsilon }})=-p\mathbf {I} +\lambda \operatorname {tr} ({\boldsymbol {\varepsilon }})\mathbf {I} +2\mu {\boldsymbol {\varepsilon }}} where I {\textstyle \mathbf {I} }

5100-768: The streamlines of a vector field, interpreted as flow velocity, are the paths along which a massless fluid particle would travel. These paths are the integral curves whose derivative at each point is equal to the vector field, and they can represent visually the behavior of the vector field at a point in time. The Navier–Stokes momentum equation can be derived as a particular form of the Cauchy momentum equation , whose general convective form is: D u D t = 1 ρ ∇ ⋅ σ + f . {\displaystyle {\frac {\mathrm {D} \mathbf {u} }{\mathrm {D} t}}={\frac {1}{\rho }}\nabla \cdot {\boldsymbol {\sigma }}+\mathbf {f} .} By setting

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5200-410: The trace of the rate-of-strain tensor in three dimensions is the divergence (i.e. rate of expansion) of the flow: tr ⁡ ( ε ) = ∇ ⋅ u . {\displaystyle \operatorname {tr} ({\boldsymbol {\varepsilon }})=\nabla \cdot \mathbf {u} .} Given this relation, and since the trace of the identity tensor in three dimensions

5300-552: The Cauchy momentum equation is written as: D u D t = 1 ρ ∇ ⋅ σ + f {\displaystyle {\frac {D\mathbf {u} }{Dt}}={\frac {1}{\rho }}\nabla \cdot {\boldsymbol {\sigma }}+\mathbf {f} } where Commonly used SI units are given in parentheses although the equations are general in nature and other units can be entered into them or units can be removed at all by nondimensionalization . Note that only we use column vectors (in

5400-474: The Eulerian forms it is apparent that the assumption of no deviatoric stress brings Cauchy equations to the Euler equations . A significant feature of the Navier–Stokes equations is the presence of convective acceleration: the effect of time-independent acceleration of a flow with respect to space. While individual continuum particles indeed experience time dependent acceleration, the convective acceleration of

5500-446: The X component of this vector is s x = − σ x x {\displaystyle s_{x}=-\sigma _{xx}} (we use similar reasoning for stresses acting on the bottom and back walls, i.e.: − σ y x , − σ z x {\displaystyle -\sigma _{yx},-\sigma _{zx}} ). The second element requiring explanation

5600-431: The compressible case the pressure is no more proportional to the isotropic stress term, since there is the additional bulk viscosity term: p = − 1 3 tr ⁡ ( σ ) + ζ ( ∇ ⋅ u ) {\displaystyle p=-{\frac {1}{3}}\operatorname {tr} ({\boldsymbol {\sigma }})+\zeta (\nabla \cdot \mathbf {u} )} and

5700-621: The conservation form of the equations of motion. This is often written: ∂ ∂ t ( ρ u ) + ∇ ⋅ ( ρ u ⊗ u ) = − ∇ p + ∇ ⋅ τ + ρ a {\displaystyle {\frac {\partial }{\partial t}}(\rho \,\mathbf {u} )+\nabla \cdot (\rho \,\mathbf {u} \otimes \mathbf {u} )=-\nabla p+\nabla \cdot {\boldsymbol {\tau }}+\rho \,\mathbf {a} } where ⊗ {\textstyle \otimes }

5800-1127: The control volume. Since this equation must hold for any control volume, it must be true that the integrand is zero, from this the Cauchy momentum equation follows. The main step (not done above) in deriving this equation is establishing that the derivative of the stress tensor is one of the forces that constitutes F i . The Cauchy momentum equation can also be put in the following form: ∂ j ∂ t + ∇ ⋅ F = s {\displaystyle {\frac {\partial \mathbf {j} }{\partial t}}+\nabla \cdot \mathbf {F} =\mathbf {s} } simply by defining: j = ρ u F = ρ u ⊗ u − σ s = ρ f {\displaystyle {\begin{aligned}{\mathbf {j} }&=\rho \mathbf {u} \\{\mathbf {F} }&=\rho \mathbf {u} \otimes \mathbf {u} -{\boldsymbol {\sigma }}\\{\mathbf {s} }&=\rho \mathbf {f} \end{aligned}}} where j

5900-1385: The cube walls, we get: F p x = ( σ x x + ∂ σ x x ∂ x d x ) d y d z − σ x x d y d z + ( σ y x + ∂ σ y x ∂ y d y ) d x d z − σ y x d x d z + ( σ z x + ∂ σ z x ∂ z d z ) d x d y − σ z x d x d y {\displaystyle F_{p}^{x}=\left(\sigma _{xx}+{\frac {\partial \sigma _{xx}}{\partial x}}dx\right)dy\,dz-\sigma _{xx}dy\,dz+\left(\sigma _{yx}+{\frac {\partial \sigma _{yx}}{\partial y}}dy\right)dx\,dz-\sigma _{yx}dx\,dz+\left(\sigma _{zx}+{\frac {\partial \sigma _{zx}}{\partial z}}dz\right)dx\,dy-\sigma _{zx}dx\,dy} After ordering F p x {\displaystyle F_{p}^{x}} and performing similar reasoning for components F p y , F p z {\displaystyle F_{p}^{y},F_{p}^{z}} (they have not been shown in

6000-2126: The cube: p → = u m = u ρ d x d y d z {\displaystyle {\vec {p}}=\mathbf {u} m=\mathbf {u} \rho \,dx\,dy\,dz} Because we assume that tested mass (cube) m = ρ d x d y d z {\displaystyle m=\rho \,dx\,dy\,dz} is constant in time, so d p → d t = d u d t ρ d x d y d z {\displaystyle {\frac {d{\vec {p}}}{dt}}={\frac {d\mathbf {u} }{dt}}\rho \,dx\,dy\,dz} We have d p → d t = F → {\displaystyle {\frac {d{\vec {p}}}{dt}}={\vec {F}}} then d p → d t = F → p + F → m {\displaystyle {\frac {d{\vec {p}}}{dt}}={\vec {F}}_{p}+{\vec {F}}_{m}} then d u d t ρ d x d y d z = ( ∇ ⋅ σ ) d x d y d z + f ρ d x d y d z {\displaystyle {\frac {d\mathbf {u} }{dt}}\rho \,dx\,dy\,dz=(\nabla \cdot {\boldsymbol {\sigma }})dx\,dy\,dz+\mathbf {f} \rho \,dx\,dy\,dz} Divide both sides by ρ d x d y d z {\displaystyle \rho \,dx\,dy\,dz} , and because d u d t = D u D t {\textstyle {\frac {d\mathbf {u} }{dt}}={\frac {D\mathbf {u} }{Dt}}} we get: D u D t = 1 ρ ∇ ⋅ σ + f {\displaystyle {\frac {D\mathbf {u} }{Dt}}={\frac {1}{\rho }}\nabla \cdot {\boldsymbol {\sigma }}+\mathbf {f} } which finishes

6100-595: The cubic fluid element. For each wall, the X component of these forces was marked in the figure with a cubic element (in the form of a product of stress and surface area e.g. − σ x x d y d z {\displaystyle -\sigma _{xx}\,dy\,dz} with units P a ⋅ m ⋅ m = N m 2 ⋅ m 2 = N {\textstyle \mathrm {Pa\cdot m\cdot m={\frac {N}{m^{2}}}\cdot m^{2}=N} } ). It requires some explanation why stress applied to

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6200-1869: The derivation. Applying Newton's second law ( i th component) to a control volume in the continuum being modeled gives: m a i = F i {\displaystyle ma_{i}=F_{i}} Then, based on the Reynolds transport theorem and using material derivative notation, one can write ∫ Ω ρ D u i D t d V = ∫ Ω ∇ j σ i j d V + ∫ Ω ρ f i d V ∫ Ω ( ρ D u i D t − ∇ j σ i j − ρ f i ) d V = 0 ρ D u i D t − ∇ j σ i j − ρ f i = 0 D u i D t − ∇ j σ i j ρ − f i = 0 {\displaystyle {\begin{aligned}\int _{\Omega }\rho {\frac {Du_{i}}{Dt}}\,dV&=\int _{\Omega }\nabla _{j}\sigma _{i}^{j}\,dV+\int _{\Omega }\rho f_{i}\,dV\\\int _{\Omega }\left(\rho {\frac {Du_{i}}{Dt}}-\nabla _{j}\sigma _{i}^{j}-\rho f_{i}\right)\,dV&=0\\\rho {\frac {Du_{i}}{Dt}}-\nabla _{j}\sigma _{i}^{j}-\rho f_{i}&=0\\{\frac {Du_{i}}{Dt}}-{\frac {\nabla _{j}\sigma _{i}^{j}}{\rho }}-f_{i}&=0\end{aligned}}} where Ω represents

6300-407: The deviatoric (shear) stress tensor in terms of viscosity and the fluid velocity gradient, and assuming constant viscosity, the above Cauchy equations will lead to the Navier–Stokes equations below. A significant feature of the Cauchy equation and consequently all other continuum equations (including Euler and Navier–Stokes) is the presence of convective acceleration: the effect of acceleration of

6400-704: The effect of the volume viscosity ζ {\textstyle \zeta } is that the mechanical pressure is not equivalent to the thermodynamic pressure : as demonstrated below. ∇ ⋅ ( ∇ ⋅ u ) I = ∇ ( ∇ ⋅ u ) , {\displaystyle \nabla \cdot (\nabla \cdot \mathbf {u} )\mathbf {I} =\nabla (\nabla \cdot \mathbf {u} ),} p ¯ ≡ p − ζ ∇ ⋅ u , {\displaystyle {\bar {p}}\equiv p-\zeta \,\nabla \cdot \mathbf {u} ,} However, this difference

6500-418: The effects of non-inertial coordinates if present). The right side of the equation is in effect a summation of hydrostatic effects, the divergence of deviatoric stress and body forces (such as gravity). All non-relativistic balance equations, such as the Navier–Stokes equations, can be derived by beginning with the Cauchy equations and specifying the stress tensor through a constitutive relation . By expressing

6600-516: The equation above. We split the forces into body forces F → m {\displaystyle {\vec {F}}_{m}} and surface forces F → p {\displaystyle {\vec {F}}_{p}} F → = F → p + F → m {\displaystyle {\vec {F}}={\vec {F}}_{p}+{\vec {F}}_{m}} Surface forces act on walls of

6700-918: The equation can be written as ρ ( ∂ u i ∂ t + u k ∂ u i ∂ x k ) = − ∂ p ∂ x i + ∂ ∂ x k [ μ ( ∂ u i ∂ x k + ∂ u k ∂ x i − 2 3 δ i k ∂ u l ∂ x l ) ] + ∂ ∂ x i ( ζ ∂ u l ∂ x l ) + ρ

6800-914: The equations expressing conservation of momentum and kinetic energy are: m A v A 1 + m B v B 1 = m A v A 2 + m B v B 2 1 2 m A v A 1 2 + 1 2 m B v B 1 2 = 1 2 m A v A 2 2 + 1 2 m B v B 2 2 . {\displaystyle {\begin{aligned}m_{A}v_{A1}+m_{B}v_{B1}&=m_{A}v_{A2}+m_{B}v_{B2}\\{\tfrac {1}{2}}m_{A}v_{A1}^{2}+{\tfrac {1}{2}}m_{B}v_{B1}^{2}&={\tfrac {1}{2}}m_{A}v_{A2}^{2}+{\tfrac {1}{2}}m_{B}v_{B2}^{2}\,.\end{aligned}}} A change of reference frame can simplify analysis of

6900-478: The equations in convective form can be simplified further. By computing the divergence of the stress tensor, since the divergence of tensor ∇ u {\textstyle \nabla \mathbf {u} } is ∇ 2 u {\textstyle \nabla ^{2}\mathbf {u} } and the divergence of tensor ( ∇ u ) T {\textstyle \left(\nabla \mathbf {u} \right)^{\mathrm {T} }}

7000-2032: The figure, but these would be vectors parallel to the Y and Z axes, respectively) we get: F p x = ∂ σ x x ∂ x d x d y d z + ∂ σ y x ∂ y d y d x d z + ∂ σ z x ∂ z d z d x d y F p y = ∂ σ x y ∂ x d x d y d z + ∂ σ y y ∂ y d y d x d z + ∂ σ z y ∂ z d z d x d y F p z = ∂ σ x z ∂ x d x d y d z + ∂ σ y z ∂ y d y d x d z + ∂ σ z z ∂ z d z d x d y {\displaystyle {\begin{aligned}F_{p}^{x}&={\frac {\partial \sigma _{xx}}{\partial x}}\,dx\,dy\,dz+{\frac {\partial \sigma _{yx}}{\partial y}}\,dy\,dx\,dz+{\frac {\partial \sigma _{zx}}{\partial z}}\,dz\,dx\,dy\\[6pt]F_{p}^{y}&={\frac {\partial \sigma _{xy}}{\partial x}}\,dx\,dy\,dz+{\frac {\partial \sigma _{yy}}{\partial y}}\,dy\,dx\,dz+{\frac {\partial \sigma _{zy}}{\partial z}}\,dz\,dx\,dy\\[6pt]F_{p}^{z}&={\frac {\partial \sigma _{xz}}{\partial x}}\,dx\,dy\,dz+{\frac {\partial \sigma _{yz}}{\partial y}}\,dy\,dx\,dz+{\frac {\partial \sigma _{zz}}{\partial z}}\,dz\,dx\,dy{\vphantom {\begin{matrix}\\\\\end{matrix}}}\end{aligned}}} We can then write it in

7100-408: The flow field is a spatial effect, one example being fluid speeding up in a nozzle. Regardless of what kind of continuum is being dealt with, convective acceleration is a nonlinear effect. Convective acceleration is present in most flows (exceptions include one-dimensional incompressible flow), but its dynamic effect is disregarded in creeping flow (also called Stokes flow). Convective acceleration

7200-887: The fluid contains a single chemical species, say for example, pressure and temperature. Any equation that makes explicit one of these transport coefficient in the conservation variables is called an equation of state . The most general of the Navier–Stokes equations become ρ D u D t = ρ ( ∂ u ∂ t + ( u ⋅ ∇ ) u ) = − ∇ p + ∇ ⋅ { μ [ ∇ u + ( ∇ u ) T − 2 3 ( ∇ ⋅ u ) I ] } + ∇ [ ζ ( ∇ ⋅ u ) ] + ρ

7300-404: The following assumptions on the Cauchy stress tensor: Momentum In Newtonian mechanics , momentum ( pl. : momenta or momentums ; more specifically linear momentum or translational momentum ) is the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction. If m is an object's mass and v is its velocity (also

7400-765: The following: p = ∑ i m i v i . {\displaystyle p=\sum _{i}m_{i}v_{i}.} A system of particles has a center of mass , a point determined by the weighted sum of their positions: r cm = m 1 r 1 + m 2 r 2 + ⋯ m 1 + m 2 + ⋯ = ∑ i m i r i ∑ i m i . {\displaystyle r_{\text{cm}}={\frac {m_{1}r_{1}+m_{2}r_{2}+\cdots }{m_{1}+m_{2}+\cdots }}={\frac {\sum _{i}m_{i}r_{i}}{\sum _{i}m_{i}}}.} If one or more of

7500-456: The force is between particles. Similarly, if there are several particles, the momentum exchanged between each pair of particles adds to zero, so the total change in momentum is zero. The conservation of the total momentum of a number of interacting particles can be expressed as m A v A + m B v B + m C v C + . . . = c o n s t

7600-1047: The initial velocities are known, the final velocities are given by v A 2 = ( m A − m B m A + m B ) v A 1 + ( 2 m B m A + m B ) v B 1 v B 2 = ( m B − m A m A + m B ) v B 1 + ( 2 m A m A + m B ) v A 1 . {\displaystyle {\begin{aligned}v_{A2}&=\left({\frac {m_{A}-m_{B}}{m_{A}+m_{B}}}\right)v_{A1}+\left({\frac {2m_{B}}{m_{A}+m_{B}}}\right)v_{B1}\\v_{B2}&=\left({\frac {m_{B}-m_{A}}{m_{A}+m_{B}}}\right)v_{B1}+\left({\frac {2m_{A}}{m_{A}+m_{B}}}\right)v_{A1}\,.\end{aligned}}} If one body has much greater mass than

7700-1012: The mass continuity equation , the left side is equivalent to: ρ D u D t = ∂ ∂ t ( ρ u ) + ∇ ⋅ ( ρ u ⊗ u ) {\displaystyle \rho {\frac {\mathrm {D} \mathbf {u} }{\mathrm {D} t}}={\frac {\partial }{\partial t}}(\rho \mathbf {u} )+\nabla \cdot (\rho \mathbf {u} \otimes \mathbf {u} )} To give finally: ∂ ∂ t ( ρ u ) + ∇ ⋅ ( ρ u ⊗ u + [ p − ζ ( ∇ ⋅ u ) ] I − μ [ ∇ u + ( ∇ u ) T − 2 3 ( ∇ ⋅ u ) I ] ) = ρ

7800-1573: The mass continuity equation, which represents the mass per unit volume of a homogenous fluid with respect to space and time (i.e., material derivative D D t {\displaystyle {\frac {\mathbf {D} }{\mathbf {Dt} }}} ) of any finite volume ( V ) to represent the change of velocity in fluid media: D m D t = ∭ V ( D ρ D t + ρ ( ∇ ⋅ u ) ) d V D ρ D t + ρ ( ∇ ⋅ u ) = ∂ ρ ∂ t + ( ∇ ρ ) ⋅ u + ρ ( ∇ ⋅ u ) = ∂ ρ ∂ t + ∇ ⋅ ( ρ u ) = 0 {\displaystyle {\begin{aligned}{\frac {\mathbf {D} m}{\mathbf {Dt} }}&={\iiint \limits _{V}}\left({{\frac {\mathbf {D} \rho }{\mathbf {Dt} }}+\rho (\nabla \cdot \mathbf {u} )}\right)dV\\{\frac {\mathbf {D} \rho }{\mathbf {Dt} }}+\rho (\nabla \cdot {\mathbf {u} })&={\frac {\partial \rho }{\partial t}}+({\nabla \rho })\cdot {\mathbf {u} }+{\rho }(\nabla \cdot \mathbf {u} )={\frac {\partial \rho }{\partial t}}+\nabla \cdot ({\rho \mathbf {u} })=0\end{aligned}}} where to arrive at

7900-437: The mass is in kilograms and the velocity is in meters per second then the momentum is in kilogram meters per second (kg⋅m/s). In cgs units , if the mass is in grams and the velocity in centimeters per second, then the momentum is in gram centimeters per second (g⋅cm/s). Being a vector, momentum has magnitude and direction. For example, a 1 kg model airplane, traveling due north at 1 m/s in straight and level flight, has

8000-411: The momentum of one particle after the collision is known, the law can be used to determine the momentum of the other particle. Alternatively if the combined kinetic energy after the collision is known, the law can be used to determine the momentum of each particle after the collision. Kinetic energy is usually not conserved. If it is conserved, the collision is called an elastic collision ; if not, it

8100-742: The negative sign indicating that the forces oppose. Equivalently, d d t ( p 1 + p 2 ) = 0. {\displaystyle {\frac {\text{d}}{{\text{d}}t}}\left(p_{1}+p_{2}\right)=0.} If the velocities of the particles are v A1 and v B1 before the interaction, and afterwards they are v A2 and v B2 , then m A v A 1 + m B v B 1 = m A v A 2 + m B v B 2 . {\displaystyle m_{A}v_{A1}+m_{B}v_{B1}=m_{A}v_{A2}+m_{B}v_{B2}.} This law holds no matter how complicated

8200-400: The net force is equal to the mass of the particle times its acceleration . Example : A model airplane of mass 1 kg accelerates from rest to a velocity of 6 m/s due north in 2 s. The net force required to produce this acceleration is 3  newtons due north. The change in momentum is 6 kg⋅m/s due north. The rate of change of momentum is 3 (kg⋅m/s)/s due north which

8300-566: The other usual form: σ = − p I + μ ( ∇ u + ( ∇ u ) T ) + ( ζ − 2 3 μ ) ( ∇ ⋅ u ) I . {\displaystyle {\boldsymbol {\sigma }}=-p\mathbf {I} +\mu \left(\nabla \mathbf {u} +(\nabla \mathbf {u} )^{\mathrm {T} }\right)+\left(\zeta -{\frac {2}{3}}\mu \right)(\nabla \cdot \mathbf {u} )\mathbf {I} .} Note that in

8400-447: The other, its velocity will be little affected by a collision while the other body will experience a large change. In an inelastic collision, some of the kinetic energy of the colliding bodies is converted into other forms of energy (such as heat or sound ). Examples include traffic collisions , in which the effect of loss of kinetic energy can be seen in the damage to the vehicles; electrons losing some of their energy to atoms (as in

8500-456: The particles are numbered 1 and 2, the second law states that F 1 = ⁠ d p 1 / d t ⁠ and F 2 = ⁠ d p 2 / d t ⁠ . Therefore, d p 1 d t = − d p 2 d t , {\displaystyle {\frac {{\text{d}}p_{1}}{{\text{d}}t}}=-{\frac {{\text{d}}p_{2}}{{\text{d}}t}},} with

8600-421: The particles is moving, the center of mass of the system will generally be moving as well (unless the system is in pure rotation around it). If the total mass of the particles is m {\displaystyle m} , and the center of mass is moving at velocity v cm , the momentum of the system is: p = m v cm . {\displaystyle p=mv_{\text{cm}}.} This

8700-564: The physics of many phenomena of scientific and engineering interest. They may be used to model the weather, ocean currents , water flow in a pipe and air flow around a wing . The Navier–Stokes equations, in their full and simplified forms, help with the design of aircraft and cars, the study of blood flow , the design of power stations , the analysis of pollution , and many other problems. Coupled with Maxwell's equations , they can be used to model and study magnetohydrodynamics . The Navier–Stokes equations are also of great interest in

8800-486: The rate of change of a body's momentum is equal to the net force acting on it. Momentum depends on the frame of reference , but in any inertial frame it is a conserved quantity, meaning that if a closed system is not affected by external forces, its total momentum does not change. Momentum is also conserved in special relativity (with a modified formula) and, in a modified form, in electrodynamics , quantum mechanics , quantum field theory , and general relativity . It

8900-402: The relevant laws of physics. Suppose x is a position in an inertial frame of reference. From the point of view of another frame of reference, moving at a constant speed u relative to the other, the position (represented by a primed coordinate) changes with time as x ′ = x − u t . {\displaystyle x'=x-ut\,.} This is called

9000-644: The same speed. Adding the speed of the center of mass to both, we find that the body that was moving is now stopped and the other is moving away at speed v . The bodies have exchanged their velocities. Regardless of the velocities of the bodies, a switch to the center of mass frame leads us to the same conclusion. Therefore, the final velocities are given by v A 2 = v B 1 v B 2 = v A 1 . {\displaystyle {\begin{aligned}v_{A2}&=v_{B1}\\v_{B2}&=v_{A1}\,.\end{aligned}}} In general, when

9100-414: The scalar equations (see multiple dimensions ). The momentum of a particle is conventionally represented by the letter p . It is the product of two quantities, the particle's mass (represented by the letter m ) and its velocity ( v ): p = m v . {\displaystyle p=mv.} The unit of momentum is the product of the units of mass and velocity. In SI units , if

9200-418: The second reference frame is also an inertial frame and the accelerations are the same: a ′ = d v ′ d t = a . {\displaystyle a'={\frac {{\text{d}}v'}{{\text{d}}t}}=a\,.} Thus, momentum is conserved in both reference frames. Moreover, as long as the force has the same form, in both frames, Newton's second law

9300-521: The second viscosity coefficient also depends on the process, that is to say, the second viscosity coefficient is not just a material property. Example: in the case of a sound wave with a definitive frequency that alternatively compresses and expands a fluid element, the second viscosity coefficient depends on the frequency of the wave. This dependence is called the dispersion . In some cases, the second viscosity ζ {\textstyle \zeta } can be assumed to be constant in which case,

9400-842: The stress tensor into isotropic and deviatoric parts, as usual in fluid dynamics: σ = − [ p − ( λ + 2 3 μ ) ( ∇ ⋅ u ) ] I + μ ( ∇ u + ( ∇ u ) T − 2 3 ( ∇ ⋅ u ) I ) {\displaystyle {\boldsymbol {\sigma }}=-\left[p-\left(\lambda +{\tfrac {2}{3}}\mu \right)\left(\nabla \cdot \mathbf {u} \right)\right]\mathbf {I} +\mu \left(\nabla \mathbf {u} +\left(\nabla \mathbf {u} \right)^{\mathrm {T} }-{\tfrac {2}{3}}\left(\nabla \cdot \mathbf {u} \right)\mathbf {I} \right)} Introducing

9500-705: The symbolic operational form: F → p = ( ∇ ⋅ σ ) d x d y d z {\displaystyle {\vec {F}}_{p}=(\nabla \cdot {\boldsymbol {\sigma }})\,dx\,dy\,dz} There are mass forces acting on the inside of the control volume. We can write them using the acceleration field f {\displaystyle \mathbf {f} } (e.g. gravitational acceleration): F → m = f ρ d x d y d z {\displaystyle {\vec {F}}_{m}=\mathbf {f} \rho \,dx\,dy\,dz} Let us calculate momentum of

9600-477: The value of approximation σ x x {\displaystyle \sigma _{xx}} in point x + d x {\displaystyle x+dx} has been shown in the figure below the cube. We proceed with similar reasoning for stress approximations σ y x , σ z x {\displaystyle \sigma _{yx},\sigma _{zx}} . Adding forces (their X components) acting on each of

9700-620: The vector ( ∇ × u ) × u {\textstyle (\nabla \times \mathbf {u} )\times \mathbf {u} } is known as the Lamb vector . For the special case of an incompressible flow , the pressure constrains the flow so that the volume of fluid elements is constant: isochoric flow resulting in a solenoidal velocity field with ∇ ⋅ u = 0 {\textstyle \nabla \cdot \mathbf {u} =0} . The incompressible momentum Navier–Stokes equation results from

9800-489: The velocities are v A1 and v B1 before the collision then in a perfectly inelastic collision both bodies will be travelling with velocity v 2 after the collision. The equation expressing conservation of momentum is: Cauchy momentum equation The Cauchy momentum equation is a vector partial differential equation put forth by Cauchy that describes the non-relativistic momentum transport in any continuum . In convective (or Lagrangian ) form

9900-486: The velocity field is calculated, other quantities of interest such as pressure or temperature may be found using dynamical equations and relations. This is different from what one normally sees in classical mechanics , where solutions are typically trajectories of position of a particle or deflection of a continuum . Studying velocity instead of position makes more sense for a fluid, although for visualization purposes one can compute various trajectories . In particular,

10000-625: The walls covering the coordinate axes takes a minus sign (e.g. for the left wall we have − σ x x {\displaystyle -\sigma _{xx}} ). For simplicity, let us focus on the left wall with tension − σ x x {\displaystyle -\sigma _{xx}} . The minus sign is due to the fact that a vector normal to this wall n → = [ − 1 , 0 , 0 ] = − e → x {\displaystyle {\vec {n}}=[-1,0,0]=-{\vec {e}}_{x}}

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