Return to Answer

Let's take the general conservative compressible form of the mass continuity \eqref{1} and momentum equations \eqref{2}

$\frac{\partial \rho}{\partial t} + \sum\limits_{j \in \mathcal{D}} \frac{\partial (\rho u_j )}{\partial x_j }=0 \tag{1}\label{1}$

$\frac{\partial (\rho u_i )}{\partial t} + \sum\limits_{j \in \mathcal{D}}\frac{\partial (\rho u_i u_j )}{\partial x_j} = \sum\limits_{j \in \mathcal{D}} \frac{\partial \sigma_{ij}}{\partial x_j } + \rho g_i \tag{2}\label{2}$

where $\mathcal{D} = \{ x, y, z \}$ and assume a Newtonian fluid with the Stokes' hypothesis $\sigma_{ij} = - p \delta_{ij} + 2 \mu S_{ij} \tag{3}\label{3} - \frac{2}{3} \mu \sum\limits_{k \in \mathcal{D}} S_{kk} \delta_{ij}$

where the rate of strain tensor is given by

$$S_{ij} := \frac{1}{2} \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right).$$

Now let's go through what simplification have been made to arrive at your form.

First insert equation \eqref{3} into \eqref{2} and apply the chain rule

$$\underbrace{u_i \frac{\partial \rho}{\partial t} + u_i \sum\limits_{j \in \mathcal{D}}\frac{\partial (\rho u_j)}{\partial x_j}}_{u_i \cdot (1) = 0} + \rho \frac{\partial u_i}{\partial t} + \sum\limits_{j \in \mathcal{D}} (\rho u_j) \frac{\partial u_i}{\partial x_j} = - \frac{\partial p}{\partial x_i } + \sum\limits_{j \in \mathcal{D}} \frac{\partial}{\partial x_j } \Big[ \mu \big( \frac{\partial u_i}{\partial x_j} + \underbrace{ \frac{\partial u_j}{\partial x_i} }_{II}\big) \underbrace{ - \frac{2}{3} \mu \sum\limits_{k \in \mathcal{D}} \frac{\partial u_k}{\partial x_k} \delta_{ij} }_{I} \Big] + \rho g_i.$$

The first term cancels out due to continuity equation \eqref{1} and we are left with the non-conservative form of the momentum equation. Your equation looks fairly similar but we instantly see that the shear viscosity $\mu$ is in your formula replaced by the kinematic viscosity $\nu := \frac{\mu}{\rho}$ and it stands outside the derivative - it is assumed constant.

Furthermore the term $- \frac{2 \mu}{3} \sum\limits_{k \in \mathcal{D}} \frac{\partial u_k}{\partial x_k}$ corresponding to the volumetric dilatation is not there. This means a fluid parcel is not compressed along its way on a stream line

$$\frac{D \rho}{D t} = \frac{\partial \rho}{\partial t} + \sum\limits_{j \in \mathcal{D}} u_j \frac{\partial \rho}{\partial x_j} = 0,$$

something termed incompressible flow (it could also be slightly more strict and be an incompressible fluid). With equation \eqref{1} we can find - applying the chain rule - that in such a case the velocity field is divergence free

$$\sum\limits_{j \in \mathcal{D}} \frac{\partial u_j}{\partial x_j} = 0.$$

This is the reason why the two terms I and II are equivalent to zero and not present in your equation.

Summing up, these equations are only valid for incompressible flow of a Newtonian fluid with a constant viscosity.

Let's take the general conservative compressible form of the mass continuity \eqref{1} and momentum equations \eqref{2}

$\frac{\partial \rho}{\partial t} + \sum\limits_{j \in \mathcal{D}} \frac{\partial (\rho u_j )}{\partial x_j }=0 \tag{1}\label{1}$

$\frac{\partial (\rho u_i )}{\partial t} + \sum\limits_{j \in \mathcal{D}}\frac{\partial (\rho u_i u_j )}{\partial x_j} = \sum\limits_{j \in \mathcal{D}} \frac{\partial \sigma_{ij}}{\partial x_j } + \rho g_i \tag{2}\label{2}$

where $\mathcal{D} = \{ x, y, z \}$ and assume a Newtonian fluid with the Stokes' hypothesis $\sigma_{ij} = - p \delta_{ij} + 2 \mu S_{ij} \tag{3}\label{3} - \frac{2}{3} \mu \sum\limits_{k \in \mathcal{D}} S_{kk} \delta_{ij}$

where the rate of strain tensor is given by

$$S_{ij} := \frac{1}{2} \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right). \tag{4}\label{4}$$

Now let's go through what simplification have been made to arrive at your form.

First insert equation \eqref{3} and \eqref{4} into \eqref{2} and apply the chain rule

$$\underbrace{u_i \frac{\partial \rho}{\partial t} + u_i \sum\limits_{j \in \mathcal{D}}\frac{\partial (\rho u_j)}{\partial x_j}}_{u_i \cdot (1) = 0} + \rho \frac{\partial u_i}{\partial t} + \sum\limits_{j \in \mathcal{D}} (\rho u_j) \frac{\partial u_i}{\partial x_j} = - \frac{\partial p}{\partial x_i } + \sum\limits_{j \in \mathcal{D}} \frac{\partial}{\partial x_j } \Big[ \mu \big( \frac{\partial u_i}{\partial x_j} + \underbrace{ \frac{\partial u_j}{\partial x_i} }_{II}\big) \underbrace{ - \frac{2}{3} \mu \sum\limits_{k \in \mathcal{D}} \frac{\partial u_k}{\partial x_k} \delta_{ij} }_{I} \Big] + \rho g_i.$$

The first term cancels out due to continuity equation \eqref{1} and we are left with the non-conservative form of the momentum equation. Your equation looks fairly similar but we instantly see that the shear viscosity $\mu$ is in your formula replaced by the kinematic viscosity $\nu := \frac{\mu}{\rho}$ and it stands outside the derivative - it is assumed constant.

Furthermore the term $- \frac{2 \mu}{3} \sum\limits_{k \in \mathcal{D}} \frac{\partial u_k}{\partial x_k}$ corresponding to the volumetric dilatation is not there. This means a fluid parcel is not compressed along its way on a stream line

$$\frac{D \rho}{D t} = \frac{\partial \rho}{\partial t} + \sum\limits_{j \in \mathcal{D}} u_j \frac{\partial \rho}{\partial x_j} = 0,$$

something termed incompressible flow (it could also be slightly more strict and be an incompressible fluid). With equation \eqref{1} we can find - applying the chain rule - that in such a case the velocity field is divergence free

$$\sum\limits_{j \in \mathcal{D}} \frac{\partial u_j}{\partial x_j} = 0.$$

This is the reason why the two terms I and II are equivalent to zero and not present in your equation.

Summing up, these equations are only valid for incompressible flow of a Newtonian fluid with a constant viscosity.

Let's take the general conservative compressible form of the mass continuity \eqref{1} and momentum equations \eqref{2}

$\frac{\partial \rho}{\partial t} + \sum\limits_{j \in \mathcal{D}} \frac{\partial (\rho u_j )}{\partial x_j }=0 \tag{1}\label{1}$

$\frac{\partial (\rho u_i )}{\partial t} + \sum\limits_{j \in \mathcal{D}}\frac{\partial (\rho u_i u_j )}{\partial x_j} = \sum\limits_{j \in \mathcal{D}} \frac{\partial \sigma_{ij}}{\partial x_j } + \rho g_i \tag{2}\label{2}$

where $\mathcal{D} = \{ x, y, z \}$ and assume a Newtonian fluid with the Stokes' hypothesis $\sigma_{ij} = - p \delta_{ij} - \frac{2}{3} \mu \sum\limits_{k \in \mathcal{D}} S_{kk} \delta_{ij} + 2 \mu S_{ij} \tag{3}\label{3}$

where the rate of strain tensor is given by

$$S_{ij} := \frac{1}{2} \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right).$$

Now let's go through what simplification have been made to arrive at your form.

First insert equation \eqref{3} into \eqref{2} and apply the chain rule

$$\underbrace{u_i \frac{\partial \rho}{\partial t} + u_i \sum\limits_{j \in \mathcal{D}}\frac{\partial (\rho u_j)}{\partial x_j}}_{u_i \cdot (1) = 0} + \rho \frac{\partial u_i}{\partial t} + \sum\limits_{j \in \mathcal{D}} (\rho u_j) \frac{\partial u_i}{\partial x_j} = - \frac{\partial p}{\partial x_i } + \sum\limits_{j \in \mathcal{D}} \frac{\partial}{\partial x_j } \Big[ \mu \big( \frac{\partial u_i}{\partial x_j} + \underbrace{ \frac{\partial u_j}{\partial x_i} }_{I}\big) \underbrace{ - \frac{2 \mu}{3} \sum\limits_{k \in \mathcal{D}} \frac{\partial u_k}{\partial x_k} }_{II} \Big] + \rho g_i.$$

The first term cancels out due to continuity equation \eqref{1} and we are left with the non-conservative form of the momentum equation. Your equation looks fairly similar but we instantly see that the shear viscosity $\mu$ is in your formula replaced by the kinematic viscosity $\nu := \frac{\mu}{\rho}$ and it stands outside the derivative - it is assumed constant.

Furthermore the term $- \frac{2 \mu}{3} \frac{\partial u_k}{\partial x_k}$ corresponding to the volumetric dilatation is not there. This means a fluid parcel is not compressed along its way on a stream line

$$\frac{D \rho}{D t} = \frac{\partial \rho}{\partial t} + \sum\limits_{j \in \mathcal{D}} u_j \frac{\partial \rho}{\partial x_j} = 0,$$

something termed incompressible flow (it could also be slightly more strict and be an incompressible fluid). With equation \eqref{1} we can find - applying the chain rule - that in such a case the velocity field is divergence free

$$\sum\limits_{j \in \mathcal{D}} \frac{\partial u_j}{\partial x_j} = 0.$$

This is the reason why the two terms I and II are equivalent to zero and not present in your equation.

Summing up, these equations are only valid for incompressible flow of a Newtonian fluid with a constant viscosity.

Let's take the general conservative compressible form of the mass continuity \eqref{1} and momentum equations \eqref{2}

$\frac{\partial \rho}{\partial t} + \sum\limits_{j \in \mathcal{D}} \frac{\partial (\rho u_j )}{\partial x_j }=0 \tag{1}\label{1}$

$\frac{\partial (\rho u_i )}{\partial t} + \sum\limits_{j \in \mathcal{D}}\frac{\partial (\rho u_i u_j )}{\partial x_j} = \sum\limits_{j \in \mathcal{D}} \frac{\partial \sigma_{ij}}{\partial x_j } + \rho g_i \tag{2}\label{2}$

where $\mathcal{D} = \{ x, y, z \}$ and assume a Newtonian fluid with the Stokes' hypothesis $\sigma_{ij} = - p \delta_{ij} + 2 \mu S_{ij} \tag{3}\label{3} - \frac{2}{3} \mu \sum\limits_{k \in \mathcal{D}} S_{kk} \delta_{ij}$

where the rate of strain tensor is given by

$$S_{ij} := \frac{1}{2} \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right).$$

Now let's go through what simplification have been made to arrive at your form.

First insert equation \eqref{3} into \eqref{2} and apply the chain rule

$$\underbrace{u_i \frac{\partial \rho}{\partial t} + u_i \sum\limits_{j \in \mathcal{D}}\frac{\partial (\rho u_j)}{\partial x_j}}_{u_i \cdot (1) = 0} + \rho \frac{\partial u_i}{\partial t} + \sum\limits_{j \in \mathcal{D}} (\rho u_j) \frac{\partial u_i}{\partial x_j} = - \frac{\partial p}{\partial x_i } + \sum\limits_{j \in \mathcal{D}} \frac{\partial}{\partial x_j } \Big[ \mu \big( \frac{\partial u_i}{\partial x_j} + \underbrace{ \frac{\partial u_j}{\partial x_i} }_{II}\big) \underbrace{ - \frac{2}{3} \mu \sum\limits_{k \in \mathcal{D}} \frac{\partial u_k}{\partial x_k} \delta_{ij} }_{I} \Big] + \rho g_i.$$

The first term cancels out due to continuity equation \eqref{1} and we are left with the non-conservative form of the momentum equation. Your equation looks fairly similar but we instantly see that the shear viscosity $\mu$ is in your formula replaced by the kinematic viscosity $\nu := \frac{\mu}{\rho}$ and it stands outside the derivative - it is assumed constant.

Furthermore the term $- \frac{2 \mu}{3} \sum\limits_{k \in \mathcal{D}} \frac{\partial u_k}{\partial x_k}$ corresponding to the volumetric dilatation is not there. This means a fluid parcel is not compressed along its way on a stream line

$$\frac{D \rho}{D t} = \frac{\partial \rho}{\partial t} + \sum\limits_{j \in \mathcal{D}} u_j \frac{\partial \rho}{\partial x_j} = 0,$$

something termed incompressible flow (it could also be slightly more strict and be an incompressible fluid). With equation \eqref{1} we can find - applying the chain rule - that in such a case the velocity field is divergence free

$$\sum\limits_{j \in \mathcal{D}} \frac{\partial u_j}{\partial x_j} = 0.$$

This is the reason why the two terms I and II are equivalent to zero and not present in your equation.

Summing up, these equations are only valid for incompressible flow of a Newtonian fluid with a constant viscosity.

Let's take the general conservative compressible form of the mass continuity \eqref{1} and momentum equations \eqref{2}

$\frac{\partial \rho}{\partial t} + \sum\limits_{j \in \mathcal{D}} \frac{\partial (\rho u_j )}{\partial x_j }=0 \tag{1}\label{1}$

$\frac{\partial (\rho u_i )}{\partial t} + \sum\limits_{j \in \mathcal{D}}\frac{\partial (\rho u_i u_j )}{\partial x_j} = \sum\limits_{j \in \mathcal{D}} \frac{\partial \sigma_{ij}}{\partial x_j } + \rho g_i \tag{2}\label{2}$

where $\mathcal{D} = \{ x, y, z \}$ and assume a Newtonian fluid with the Stokes' hypothesis $\sigma_{ij} = - p \delta_{ij} - \frac{2}{3} \mu \sum\limits_{k \in \mathcal{D}} S_{kk} \delta_{ij} + 2 \mu S_{ij} \tag{3}\label{3}$

where the rate of strain tensor is given by

$$S_{ij} := \frac{1}{2} \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right).$$

Now let's go through what simplification have been made to arrive at your form.

First insert equation \eqref{3} into \eqref{2} and apply the chain rule

$$\underbrace{u_i \frac{\partial \rho}{\partial t} + u_i \sum\limits_{j \in \mathcal{D}}\frac{\partial (\rho u_j)}{\partial x_j}}_{u_i \cdot (1) = 0} + \rho \frac{\partial u_i}{\partial t} + \sum\limits_{j \in \mathcal{D}} (\rho u_j) \frac{\partial u_i}{\partial x_j} = - \frac{\partial p}{\partial x_i } + \sum\limits_{j \in \mathcal{D}} \frac{\partial}{\partial x_j } \Big[ \mu \big( \frac{\partial u_i}{\partial x_j} + \underbrace{ \frac{\partial u_j}{\partial x_i} }_{I}\big) \underbrace{ - \frac{2 \mu}{3} \frac{\partial u_k}{\partial x_k} }_{II} \Big] + \rho g_i.$$

The first term cancels out due to continuity equation \eqref{1} and we are left with the non-conservative form of the momentum equation. Your equation looks fairly similar but we instantly see that the shear viscosity $\mu$ is in your formula replaced by the kinematic viscosity $\nu := \frac{\mu}{\rho}$ and it stands outside the derivative - it is assumed constant.

Furthermore the term $- \frac{2 \mu}{3} \frac{\partial u_k}{\partial x_k}$ corresponding to the volumetric dilatation is not there. This means a fluid parcel is not compressed along its way on a stream line

$$\frac{D \rho}{D t} = \frac{\partial \rho}{\partial t} + \sum\limits_{j \in \mathcal{D}} u_j \frac{\partial \rho}{\partial x_j} = 0,$$

something termed incompressible flow (it could also be slightly more strict and be an incompressible fluid). With equation \eqref{1} we can find - applying the chain rule - that in such a case the velocity field is divergence free

$$\sum\limits_{j \in \mathcal{D}} \frac{\partial u_j}{\partial x_j} = 0.$$

This is the reason why the two terms I and II are equivalent to zero and not present in your equation.

Summing up, these equations are only valid for incompressible flow of a Newtonian fluid with a constant viscosity.

Let's take the general conservative compressible form of the mass continuity \eqref{1} and momentum equations \eqref{2}

$\frac{\partial \rho}{\partial t} + \sum\limits_{j \in \mathcal{D}} \frac{\partial (\rho u_j )}{\partial x_j }=0 \tag{1}\label{1}$

$\frac{\partial (\rho u_i )}{\partial t} + \sum\limits_{j \in \mathcal{D}}\frac{\partial (\rho u_i u_j )}{\partial x_j} = \sum\limits_{j \in \mathcal{D}} \frac{\partial \sigma_{ij}}{\partial x_j } + \rho g_i \tag{2}\label{2}$

where $\mathcal{D} = \{ x, y, z \}$ and assume a Newtonian fluid with the Stokes' hypothesis $\sigma_{ij} = - p \delta_{ij} - \frac{2}{3} \mu \sum\limits_{k \in \mathcal{D}} S_{kk} \delta_{ij} + 2 \mu S_{ij} \tag{3}\label{3}$

where the rate of strain tensor is given by

$$S_{ij} := \frac{1}{2} \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right).$$

Now let's go through what simplification have been made to arrive at your form.

First insert equation \eqref{3} into \eqref{2} and apply the chain rule

$$\underbrace{u_i \frac{\partial \rho}{\partial t} + u_i \sum\limits_{j \in \mathcal{D}}\frac{\partial (\rho u_j)}{\partial x_j}}_{u_i \cdot (1) = 0} + \rho \frac{\partial u_i}{\partial t} + \sum\limits_{j \in \mathcal{D}} (\rho u_j) \frac{\partial u_i}{\partial x_j} = - \frac{\partial p}{\partial x_i } + \sum\limits_{j \in \mathcal{D}} \frac{\partial}{\partial x_j } \Big[ \mu \big( \frac{\partial u_i}{\partial x_j} + \underbrace{ \frac{\partial u_j}{\partial x_i} }_{I}\big) \underbrace{ - \frac{2 \mu}{3} \sum\limits_{k \in \mathcal{D}} \frac{\partial u_k}{\partial x_k} }_{II} \Big] + \rho g_i.$$

The first term cancels out due to continuity equation \eqref{1} and we are left with the non-conservative form of the momentum equation. Your equation looks fairly similar but we instantly see that the shear viscosity $\mu$ is in your formula replaced by the kinematic viscosity $\nu := \frac{\mu}{\rho}$ and it stands outside the derivative - it is assumed constant.

Furthermore the term $- \frac{2 \mu}{3} \frac{\partial u_k}{\partial x_k}$ corresponding to the volumetric dilatation is not there. This means a fluid parcel is not compressed along its way on a stream line

$$\frac{D \rho}{D t} = \frac{\partial \rho}{\partial t} + \sum\limits_{j \in \mathcal{D}} u_j \frac{\partial \rho}{\partial x_j} = 0,$$

something termed incompressible flow (it could also be slightly more strict and be an incompressible fluid). With equation \eqref{1} we can find - applying the chain rule - that in such a case the velocity field is divergence free

$$\sum\limits_{j \in \mathcal{D}} \frac{\partial u_j}{\partial x_j} = 0.$$

This is the reason why the two terms I and II are equivalent to zero and not present in your equation.

Summing up, these equations are only valid for incompressible flow of a Newtonian fluid with a constant viscosity.