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Chapter 8 Differential Analysis
8.1 Introduction
The integral analysis has a limited accuracy, which leads to a different approach of differential analysis.
The differential analysis allows the flow field investigation in greater detail.
In differential analysis, the emphasis is on infinitesimal scale and thus the analysis provides
This analysis leads to partial differential equations which are referred to as the Navier–Stokes equations.
These equations are named after Claude–Louis Navier–Marie and George Gabriel Stokes.
Like many equations they were independently derived by several people.
First these equations were derived by Claude–Louis–Marie Navier as it is known in 1827.
As usual Simeon–Denis Poisson independently, as he done to many other equations or conditions,
derived these equations in 1831 for the same arguments as Navier.
The foundations for their arguments or motivations are based on a molecular view of how stresses are exerted
between fluid layers.
Barre de Saint Venant (1843) and George Gabriel Stokes (1845) derived
these equation based on the relationship between stress and rate–of–strain (this approach is presented in this book).
Navier–Stokes equations are non–linear and there are more than one possible solution in many cases
(if not most cases) e.g. the solution is not unique.
A discussion about the ``regular'' solution is present and a brief discussion about limitations when the solution
is applicable.
Later in the Chapters on Real Fluid and Turbulence, with a presentation of the ``non–regular'' solutions will
be presented with the associated issues of stability.
However even for the ``regular'' solution the mathematics is very complex.
One of the approaches is to reduce the equations by eliminating the viscosity effects.
The equations without the viscosity effects are referred to as
the ideal flow equations (Euler Equations) which will be discussed in the next chapter.
The concepts of the Add Mass and the Add Force, which are easier to discuss when the viscosity is ignored,
and will be presented in the Ideal Flow chapter.
It has to be pointed out that the Add Mass and Add Force appear regardless to the viscosity.
Historically, complexity of the equations, on one hand, leads to approximations and consequently
to the ideal flow approximation (equations) and on the other hand experimental solutions of Navier–Stokes equations.
The connection between these two ideas or fields was done via introduction of the boundary layer theory
by Prandtl which will be discussed as well.
Even for simple situations, there are cases when complying with the boundary conditions leads to
a discontinuity (shock or choked flow).
These equations cannot satisfy the boundary conditions in other cases
and in way the fluid pushes the boundary condition(s) further downstream (choked flow).
These issues are discussed in Open Channel Flow and Compressible Flow chapters.
Sometimes, the boundary conditions create instability which alters the boundary conditions itself
which is known as Interfacial instability.
The choked flow is associated with a single phase flow (even the double choked flow)
while the Interfacial instability associated with the Multi–Phase flow.
This phenomenon is presented in Multi–phase chapter and briefly discussed in this chapter.
8.2 Mass Conservation
Fig. 8.1 The mass balance on the infinitesimal control volume.
Fluid flows into and from a three dimensional infinitesimal control volume
depicted in Figure 8.1.
At a specific time this control volume can be viewed as a system.
The mass conservation for this infinitesimal small system is zero thus
\begin{align}
\label{dif:eq:RTT}
\dfrac{D}{Dt} \int_V \rho dV = 0
\end{align}
However for a control volume using Reynolds Transport Theorem (RTT), the following can be written
\begin{align}
\label{dif:eq:RTTe}
\dfrac{D}{Dt} \int_V \rho dV = \dfrac{d}{dt} \int_V \rho dV + \int_A U_{rn} \, \rho\, dA = 0
\end{align}
For a constant control volume, the derivative can enter into the integral (see also for the divergence theorem
in the appendix A.1.2) on the right hand side and hence
\begin{align}
\label{dif:eq:RTTcvM}
\overbrace{\int_V \dfrac{d\rho}{dt} \, dV}^{\dfrac{d\rho}{dt}\,dV}
+ \int_A U_{rn} \, \rho\, dA = 0
\end{align}
The first term in equation \eqref{dif:eq:RTTcvM} for the infinitesimal volume is expressed, neglecting higher order
derivatives, as
\begin{align}
\label{dif:eq:controlVolRho}
\int_V \dfrac{d\rho}{dt} \, dV = \dfrac{d\rho}{dt} \,\overbrace{dx\,dy\,dz}^{\scriptsize dV} +
\overbrace{f\left( \dfrac{d^2\rho}{dt^2\dfrac{}{}} \right) + \cdots}^{\sim 0}
\end{align}
The second term in the LHS of equation \eqref{dif:eq:RTTe} is expressed
as
\begin{multline}
\label{dif:eq:mass2}
\displaystyle \int_A U_{rn} \, \rho\, dA =
\overbrace{dy\,dz}^{dA_{yz}}\,\left[ \left.\dfrac{}{\dfrac{}{}} (\rho\, U_x)\right_x  \left.(\rho\,U_x)\right_{x+dx} \right] + \\
\displaystyle \overbrace{dx\,dz}^{dA_{xz}}\left[\left.\dfrac{}{\dfrac{}{}}(\rho\,U_y)\right_y \left.(\rho\,U_y)\right_{y+dy} \right] + \\
\displaystyle \overbrace{dx\,dy}^{dA_{xz}}\,\left[\left.\dfrac{}{\dfrac{}{}}(\rho\, U_z)\right_z  \left.(\rho\,U_z)\right_{z+dz} \right]
\end{multline}
The difference between point $x$ and $x+dx$ can be obtained by developing Taylor series as
\begin{align}
\label{dif:eq:mass3}
\left.(\rho\,U_x)\right_{x+dx} = \left. (\rho\, U_x)\right_x + \left.\dfrac{\partial \left(\rho\, U_x\right) }
{\partial x}\right_x dx
\end{align}
The same can be said for the $y$ and $z$ coordinates.
It also can be noticed that, for example, the operation, in the $x$ coordinate, produces additional $dx$ thus
a infinitesimal volume element $dV$ is obtained for all directions.
The combination can be divided by $dx\,dy\,dz$ and simplified by using the definition of the partial
derivative in the regular process to be
\begin{align}
\label{dif:eq:mass4}
{\displaystyle \int_A U_{rn} \, \rho\, dA =  \left[
\dfrac{\partial (\rho\, U_x) }{\partial x} +
\dfrac{\partial (\rho\, U_y) }{\partial y} +
\dfrac{\partial (\rho\, U_z) }{\partial z} \right] }
\end{align}
Combining the first term with the second term results in
the continuity equation in Cartesian coordinates as
Continuity in Cartesian Coordinates
\begin{align}
\label{dif:eq:continuityCart}
\dfrac{\partial \rho }{\partial t} +
\dfrac{\partial \rho\,U_x }{\partial x} +
\dfrac{\partial \rho\,U_y }{\partial y} +
\dfrac{\partial \rho\,U_z }{\partial z} = 0
\end{align}
Cylindrical Coordinates
Fig. 8.2 The mass conservation in cylindrical coordinates.
The same equation can be derived in cylindrical coordinates.
The net mass change, as depicted in Figure 8.2, in the control volume is
\begin{align}
\label{dif:eq:cyl:dRhodt}
d\,\dot{m} = \dfrac{\partial \rho}{\partial t} \overbrace{dr\,dz\,r\,d\theta}^{\scriptsize dv}
\end{align}
The net mass flow out or in the $\mathbf{\widehat{r}}$ direction has an additional term which is the area
change compared to the Cartesian coordinates.
This change creates a different differential equation with additional complications.
The change is
\begin{align}
\label{dif:eq:cly:massR}
\left( \, \dfrac{}{\dfrac{}{}} \text{flux in $\pmb{\mathbf{r}}$ direction} \right)=
d\theta\,dz\,\left( r\,\rho\, U_r
 \left(r\,\rho\, U_r + \dfrac{\partial \rho\,U_r\,r}{ \partial r \dfrac{}{} } dr \right)
\right)
\end{align}
The net flux in the $r$ direction is then
\begin{align}
\label{dif:eq:cly:massRnet}
\text{ net flux in the $\pmb{\mathbf{r}}$ direction} =
d\theta\,dz\,\dfrac{ \partial \rho\,U_r\,r}{ \partial r} dr
\end{align}
Note
that the $r$ is still inside the derivative since it is a function of $r$, e.g. the change of $r$ with $r$.
In a similar fashion, the net flux in the $z$ coordinate be written as
\begin{align}
\label{dif:eq:cly:massZnet}
\mbox{net flux in $z$ direction} =
r\,d\theta\,dr\,\dfrac{ \partial \left( \rho\,U_z\right)}{ \partial z} dz
\end{align}
The net change in the $\theta$ direction is then
\begin{align}
\label{dif:eq:cly:massThetanet}
\mbox{net flux in $\theta$ direction} =
dr\,dz\,\dfrac{\partial\rho\,U_{\theta}}{\partial \theta} d\theta
\end{align}
Combining equations \eqref{dif:eq:cly:massRnet}–qref{dif:eq:cly:massThetanet} and dividing by infinitesimal
control volume, $dr\;r\,d\theta\,dz$, results in
\begin{align}
\label{dif:eq:cly:massCombP}
\text{total net flux } =
\left(
\dfrac{1}{r}\dfrac{ \partial \left( \rho\,U_r\,r \right) }{ \partial r\dfrac{}{}} + \dfrac{ \partial \rho\,U_z\,r}{ \partial z}
+\dfrac{\partial\rho\,U_{\theta}}{\partial \theta}
\right)
\end{align}
Combining equation \eqref{dif:eq:cly:massCombP} with the change in the control volume qref{dif:eq:cyl:dRhodt}
divided by infinitesimal control volume, $dr\;r\,d\theta\,dz$ yields
Continuity in Cylindrical Coordinates
\begin{align}
\label{dif:eq:continuityCyl}
\dfrac{\partial \rho }{\partial t} +
\dfrac{1}{r}\dfrac{\partial \left(r\,\rho\,U_r\right) }{\partial r} +
\dfrac{1}{r} \dfrac{\partial \rho\,U_{\theta }}{\partial \theta} +
\dfrac{\partial \rho\,U_z }{\partial z} = 0
\end{align}
Carrying similar operations for the spherical coordinates, the continuity equation becomes
Continuity in Spherical Coordinates
\begin{align}
\label{dif:eq:continuitySph}
\dfrac{\partial \rho }{\partial t} +
\dfrac{1}{r^2}\dfrac{\partial \left(r^2\,\rho\,U_r\right) }{\partial r} +
\dfrac{1}{r\,\sin\theta} \dfrac{\partial \left(\rho\,U_{\theta } \,\sin\theta\right)}
{\partial \theta} +
\dfrac{1}{r\,\sin\theta} \dfrac{\partial \rho\,U_{\phi} }{\partial z} = 0
\end{align}
The continuity equations \eqref{dif:eq:continuityCart}, qref{dif:eq:continuityCyl} and
qref{dif:eq:continuitySph} can be expressed in different coordinates.
It can be noticed that the second part of these equations is the divergence (see the Appendix
A.1.2 page \pageref{math:sec:difOperator}).
Hence, the continuity equation can be written in a general vector form as
Continuity Equation
\begin{align}
\label{dif:eq:continuityV}
\dfrac{\partial \rho }{\partial t} + \boldsymbol{\nabla} \cdot \left( \rho \,\pmb{U}\right) = 0
\end{align}
Advance Material
The mass equation can be written in index notation for Cartesian coordinates.
The index notation really does not add much to the scientific understanding.
However, this writing reduce the amount of writing and potentially can help
the thinking about the problem or situation in more conceptional way.
The mass equation (see in the appendix for more information on the index notation) written as
\begin{align}
\label{dif:eq:massIndexNotation}
\dfrac{\partial \rho }{\partial t} + \dfrac{\partial \left(\rho\,U\right)_i }{\partial x_i} = 0
\end{align}
Where $i$ is is of the $i$, $j$, and $k$
.
Compare to equation \eqref{dif:eq:continuityCart}.
Again remember that the meaning of repeated index is summation.
End Advance Material
The use of these equations is normally combined with other equations (momentum and or energy equations).
There are very few cases where this equation is used on its own merit.
For academic purposes, several examples are constructed here.
8.2.1 Mass Conservation Examples
A layer of liquid has an initial height of $H_0$ with an uniform temperature of $T_0$.
At time, $t_0$, the upper surface is exposed to temperature $T_1$ (see Figure 8.3).
Assume that
Fig. 8.3 Mass flow due to temperature difference for example
the actual temperature is exponentially approaches to a linear temperature profile
as depicted in Figure 8.3.
The density is a function of the temperature according to
\begin{align}
\label{massT:Trho}
\dfrac{TT_0}{T_1T_0} = \alpha\, \left( \dfrac{\rho\rho_0}{\rho_1\rho_0} \right)
\end{align}
where $\rho_1$ is the density at the surface and
where $\rho_0$ is the density at the bottom.
Assume that the velocity is only a function of the $y$ coordinate.
Calculates the velocity of the liquid.
Assume that the velocity at the lower boundary is zero at all times.
Neglect the mutual dependency of the temperature and the height.
Solution
The situation is unsteady state thus the unsteady state and one dimensional continuity equation has to be used
which is
\begin{align}
\label{massT:gov}
\dfrac{\partial \rho}{\partial t} + \dfrac{\partial \left(\rho U_y\right)}{\partial y } = 0
\end{align}
with the boundary condition of zero velocity at the lower surface $U_y(y=0)=0$.
The expression that connects the temperature with the space for the final temperature as
\begin{align}
\label{massT:connectT:Ty}
\dfrac{TT_0}{T_1T_0} = \alpha \, \dfrac{H_0y}{H_0}
\end{align}
The exponential decay is $\left(1e^{\beta\,t}\right)$ and thus the combination (with
equation \eqref{massT:Trho}) is
\begin{align}
\label{massT:connectY:Ts}
\dfrac{\rho \rho_0}{\rho_1\rho_0} = \alpha\, \dfrac{H_0y}{H_0} \left( 1  e^{\beta\,t} \right)
\end{align}
Equation \eqref{massT:connectY:Ts} relates the temperature with the time and the location was given
in the question (it is not the solution of any model).
It can be noticed that the height $H_0$ is a function of time.
For this question, it is treated as a constant.
Substituting the density, $\rho$, as a function of time into the governing equation \eqref{massT:gov} results in
\begin{align}
\label{massT:rhoGov1}
\overbrace{\alpha\,\beta\,\left(\dfrac{ H_0  y}{H_0}\right) e^{\beta\,t}}^{\dfrac{\partial \rho}{\partial t}} +
\overbrace{\dfrac{\partial \left( U_y\, \alpha\, \dfrac{H_0y}{H_0} \left( 1  e^{\beta\,t} \right) \right) }{\partial y}}
^{\dfrac{\partial\rho\, U_y}{\partial y}} = 0
\end{align}
Equation \eqref{massT:rhoGov1} is first order ODE with the boundary condition $U_y(y=0)=0$
which can be arranged as
\begin{align}
\label{massT:rhoGov}
{\dfrac{\partial \left( U_y\, \alpha\, \dfrac{H_0y}{H_0} \left( 1  e^{\beta\,t} \right) \right) }{\partial y}}
=  {\alpha\,\beta\,\left(\dfrac{ H_0  y}{H_0}\right) e^{\beta\,t}}
\end{align}
$U_y$ is a function of the time but not $y$.
Equation \eqref{massT:rhoGov} holds for any time and thus,
it can be treated for the solution of equation \eqref{massT:rhoGov} as a constant
Hence, the integration with respect to $y$ yields
\begin{align}
\label{massT:rhoSol3}
\left( U_y\, \alpha\, \dfrac{H_0y}{H_0} \left( 1  e^{\beta\,t} \right) \right)
=  {\alpha\,\beta\,\left(\dfrac{ 2\,H_0  y}{2\,H_0}\right) e^{\beta\,t}} y + c
\end{align}
Utilizing the boundary condition $U_y(y=0)=0$ yields
\begin{align}
\label{massT:rhoSol2}
\left( U_y\, \alpha\, \dfrac{H_0y}{H_0} \left( 1  e^{\beta\,t} \right) \right)
=  {\alpha\,\beta\,\left(\dfrac{2\, H_0  y}{2\,H_0}\right) e^{\beta\,t}} \left( y  1 \right)
\end{align}
or the velocity is
\begin{align}
\label{massT:rhoSol1}
U_y =
{\beta\,\left(\dfrac{2\, H_0  y}{2\,\left(H_0y\right)}\right)
\dfrac{e^{\beta\,t}} { \left( 1  e^{\beta\,t} \right)} } \left( 1  y \right)
\end{align}
It can be noticed that indeed the velocity is a function of the time and space $y$.
Simplified Continuity Equation
A simplified equation can be obtained for a steady state in which the transient term is eliminated as
(in a vector form)
\begin{align}
\label{dif:eq:massSS}
\boldsymbol{\nabla} \cdot \left( \rho \,\pmb{U}\right) = 0
\end{align}
If the fluid is incompressible then the governing equation is a volume conservation as
\begin{align}
\label{dif:eq:massSSRho}
\boldsymbol{\nabla} \cdot \pmb{U} = 0
\end{align}
Note that this equation appropriate only for a single phase case.
In many coating processes a thin film is created by a continuous process in which liquid
injected into a moving belt which carries the material out
as \hbox{exhibited in~Figure~.}
Fig. 8.4 Mass flow in coating process for example .
The temperature and mass transfer taking place which reduces (or increases) the thickness of the film.
For this example, assume that no mass transfer occurs or can be neglected and the main mechanism
is heat transfer.
Assume that the film temperature is only a function of the distance from the extraction point.
Calculate the film velocity field if the density is a function of the temperature.
The relationship between the density and the temperature is linear as
\begin{align}
\label{coating:rhoT}
\dfrac{\rho  \rho_{\infty}} {\rho_0  \rho_{\infty}} = \alpha \left(\dfrac{T  T_{\infty}} {T_{0}  T_{\infty}}\right)
\end{align}
State your assumptions.
Solution
This problem is somewhat similar to Example 8.1
however it can be considered as steady state.
At any point the governing equation in coordinate system that moving with the belt is
\begin{align}
\label{coating:gov}
\dfrac{\partial \left( \rho\,U_x \right)}{\partial x} + \dfrac{\partial \left( \rho\,U_y \right) }{\partial y} = 0
\end{align}
At first, it can be assumed that the material moves with the belt in the $x$ direction
in the same velocity.
This assumption is consistent with the first solution (no stability issues).
If the frame of reference was moving with the belt then there is only velocity
component in the $y$ direction
.
Hence equation \eqref{coating:gov} can be written as
\begin{align}
\label{coating:govTrans}
U_x\,\dfrac{\partial \rho }{\partial x} =  \dfrac{\partial \left( \rho\,U_y \right) }{\partial y}
\end{align}
Where $U_x$ is the belt velocity.
See the resembles to equation \eqref{massT:gov}.
The solution is similar to the previous Example 8.1 for a general function $T=F(x)$.
\begin{align}
\label{coating:drhodx}
\dfrac{\partial \rho}{\partial x} = \dfrac{ \alpha}{U_x}\,
\dfrac{\partial F(x) }{\partial x} \left( \rho_0  \rho_{\infty} \right)
\end{align}
Substituting this relationship in equation \eqref{coating:drhodx} into the governing equation results in
\begin{align}
\label{coating:govAdd}
\dfrac{\partial U_y\,\rho}{\partial y} = \dfrac{ \alpha}{U_x}\,
\dfrac{\partial F(x) }{\partial x} \left( \rho_0  \rho_{\infty} \right)
\end{align}
The density is expressed by equation \eqref{coating:rhoT} and thus
\begin{align}
\label{coating:Uy}
U_y = \dfrac{ \alpha}{\rho\, U_x}\, \dfrac{\partial F(x) }{\partial x} \left( \rho_0  \rho_{\infty} \right)\,y + c
\end{align}
Notice that $\rho$ could ``come'' out of the derivative (why?) and move into the RHS.
Applying the boundary condition $U_y(t=0) =0 $ results in
\begin{align}
\label{coating:Uy2}
U_y = \dfrac{ \alpha}{\rho(x)\, U_x}\, \dfrac{\partial F(x) }{\partial x} \left( \rho_0  \rho_{\infty} \right)\, y
\end{align}
The velocity in a two dimensional field is assumed to be in a steady state.
Assume that the density is constant and calculate the vertical velocity ($y$ component)
for the following $x$ velocity component.
\begin{align}
\label{massFxy:Ux}
U_x = a\,x^2 + b\,y^2
\end{align}
Next, assume the density is also a function of the location in the form of
\begin{align}
\label{massFxy:rho}
\rho = m\,e^{x+y}
\end{align}
Where $m$ is constant.
Calculate the velocity field in this case.
Solution
The flow field must comply with the mass conservation \eqref{dif:eq:massSSRho} thus
\begin{align}
\label{massFxy:Rhocomply}
2\,a\, x + \dfrac{\partial U_y }{\partial y} = 0
\end{align}
Equation \eqref{massFxy:Rhocomply} is an ODE with constant coefficients.
It can be noted that $x$ should be treated as a constant parameter for the $y$ coordinate
integration.
Thus,
\begin{align}
\label{massFxy:U_y}
U_y =  \int 2\,a\, x + f(x) = 2\,x\,y + f(x)
\end{align}
The integration constant in this case is not really a constant but rather an arbitrary function of $x$.
Notice the symmetry of the situation.
The velocity, $U_x$ has also arbitrary function in the $y$ component.
For the second part equation \eqref{dif:eq:massSS} is applicable and used as
\begin{align}
\label{massFxy:Govfull}
\dfrac{\partial\, \left( a\,x^2 + b\,y^2 \right) \left( m\,e^{x+y}\right) }{\partial x} +
\dfrac{\partial\,U_y\, \left( m\,e^{x+y}\right) }{\partial y} = 0
\end{align}
Taking the derivative of the first term while moving the second part to the other
side results in
\begin{align}
\label{massFxy:GovExplnation1}
a\,\left( 2\,x+ x^2 + \dfrac{b}{a\dfrac{}{}}\,y^2 \right) \,e^{x+y} = 
\left( e^{x+y} \right)\, \left( \dfrac{\partial\,U_y\, }{\partial y \dfrac{}{}} + U_y \right)
\end{align}
The exponent can be canceled to further simplify the equation \eqref{massFxy:GovExplnation1}
and switching sides to be
\begin{align}
\label{massFxy:GovExplnation}
\left( \dfrac{\partial\,U_y\, }{\partial y\dfrac{}{}} + U_y \right)
=  a\,\left( 2\,x+ x^2 + \dfrac{b}{a\dfrac{}{}}\,y^2 \right)
\end{align}
Equation \eqref{massFxy:GovExplnation} is a first order ODE that can be solved by combination of the homogeneous
solution with the private solution (see for an explanation in the Appendix).
The homogeneous equation is
\begin{align}
\label{massFxy:GovHomogeneous}
\dfrac{\partial\,U_y\, }{\partial y} + U_y = 0
\end{align}
The solution for \eqref{massFxy:GovHomogeneous} is $U_y= c\,e^{y}$ (see for an explanation in the appendix).
The private solution is
\begin{align}
\label{massFxy:GovPrivate}
\left.U_y\right_{private} = \left( b\,\left( {y}^{2}2\,y+2\right) a\,{x}^{2}2\,a\,x\right)
\end{align}
The total solution is
\begin{align}
\label{massFxy:GovF}
U_y = c\,e^{y} + \left( b\,\left( {y}^{2}2\,y+2\right) a\,{x}^{2}2\,a\,x\right)
\end{align}
Can the following velocities coexist
\begin{align}
\label{canUbe:Uxyx}
\begin{array}{lcccr}
U_x = \left(x\,t\right) ^2 \, z &&
U_y = \left(x\,t\right) + \left( y\,t\right) + \left( z\,t\right) &&
U_z = \left(x\,t\right) + \left( y\,t\right) + \left( z\,t\right)
\end{array}
\end{align}
in the flow field.
Is the flow is incompressible? Is the flow in a steady state condition?
Solution
Whether the solution is in a steady state or not can be observed from whether the
velocity contains time component.
Thus, this flow field is not steady state since it contains time component.
This continuity equation is checked if the flow incompressible (constant density).
The derivative of each component are
\begin{align}
\label{canUbe:parUxParX}
\begin{array}{lcccr}
\dfrac{\partial U_x}{\partial x} = t^2\,z &&
\dfrac{\partial U_y}{\partial y} = t &&
\dfrac{\partial U_z}{\partial z} = t
\end{array}
\end{align}
Hence the gradient or the combination of these derivatives is
\begin{align}
\label{canUbe:divergence}
\nabla \pmb{U} = t^2\,z + 2\,t
\end{align}
The divergence isn't zero thus this flow, if it exist, must be compressible flow.
This flow can exist only for a limit time since over time the divergence is unbounded
(a source must exist).
Find the density as a function of the time for a given one dimensional flow
with $U_x = x \,e^{5\,\alpha\,y} \,\left( \cos\left(\alpha\,t \right) \right)$.
The initial density is $\rho(t=0)= \rho_0$.
Solution
This problem is one dimensional unsteady state and for a compressible substance.
Hence, the mass conservation is reduced only for one dimensional form as
\begin{align}
\label{massWhatRho:gov}
\dfrac{\partial \rho}{\partial t} + \dfrac{\partial \left(U_x\, \rho\right) }{\partial x} = 0
\end{align}
Mathematically speaking, this kind of presentation is possible.
However physically there are velocity components in $y$ and $z$ directions.
In this problem, these physical components are ignored for academic reasons.
Equation \eqref{massWhatRho:gov} is first order partial differential equation which can be converted to
an ordinary differential equations when the velocity component, $U_x$, is substituted.
Using,
\begin{align}
\label{massWhatRho:Ux}
\dfrac{\partial U_x}{\partial x} = e^{5\,\alpha\,y} \,\left( \cos\left(\alpha\,t \right) \right)
\end{align}
Substituting equation \eqref{massWhatRho:Ux} into equation qref{massWhatRho:gov} and
noticing that the density, $\rho$, is a function of $x$ results of
\begin{align}
\label{massWhatRho:dRho}
\dfrac{\partial \rho}{\partial t} =  \rho\, x \,e^{5\,\alpha\,y} \,\left( \cos\left(\alpha\,t \right) \right)
 \dfrac{\partial \rho} {\partial x} \, e^{5\,\alpha\,y} \,\left( \cos\left(\alpha\,t \right) \right)
\end{align}
Equation \eqref{massWhatRho:dRho} can be separated to yield
\begin{align}
\label{massWhatRho:RhoofY}
\overbrace{\dfrac{1}{\cos\left(\alpha\,t \right) } \dfrac{\partial \rho}{\partial t}}
^{f(t)} =
\overbrace{ \rho\, x \,e^{5\,\alpha\,y}  \dfrac{\partial \rho} {\partial x} \, e^{5\,\alpha\,y}}
^{f(y)}
\end{align}
A possible solution is when the left and the right hand sides are equal to a constant.
In that case the left hand side is
\begin{align}
\label{massWhatRho:rhoEgov}
\dfrac{1}{\cos\left(\alpha\,t \right) } \dfrac{\partial \rho}{\partial t} = c_1
\end{align}
The solution of equation \eqref{massWhatRho:rhoEgov} is reduced to ODE and its solution is
\begin{align}
\label{massWhatRho:rhoEsol}
\rho =\dfrac{c_1\,sin\left( \alpha\,t\right) }{\alpha}+c_2
\end{align}
The same can be done for the right hand side as
\begin{align}
\label{massWhatRho:UxEgovtmp}
\rho\, x \,e^{5\,\alpha\,y} + \dfrac{\partial \rho} {\partial x} \, e^{5\,\alpha\,y} = c_1
\end{align}
The term $\,e^{5\,\alpha\,y}$ is always positive, real value, and independent of $y$ thus
equation \eqref{massWhatRho:UxEgovtmp} becomes
\begin{align}
\label{massWhatRho:UxEgov}
\rho\, x + \dfrac{\partial \rho} {\partial x} = \dfrac{c_1}{e^{5\,\alpha\,y}} = c_3
\end{align}
Equation \eqref{massWhatRho:UxEgov} is a constant coefficients first order ODE which its solution discussed
extensively in the appendix.
The solution of \eqref{massWhatRho:UxEgov} is given by
\begin{align}
\label{massWhatRho:UxEsolG}
\rho ={e}^{\dfrac{{x}^{2}}{2}}\,
\left(
c  \overbrace{\dfrac{\sqrt{\pi}\,i\,c_3\,erf\left( \dfrac{i\,x}{\sqrt{2}} \right) } {\sqrt{2}}}
^{\text{impossible solution}}
\right)
\end{align}
which indicates that the solution is a complex number thus the constant, $c_3$, must be zero
and thus the constant, $c_1$ vanishes as well and the solution
contain only the homogeneous part and the private solution is dropped
\begin{align}
\label{massWhatRho:UxEsol}
\rho = c_2\, {e}^{\dfrac{{x}^{2}}{2}}
\end{align}
The solution is the multiplication of equation \eqref{massWhatRho:UxEsol} by qref{massWhatRho:rhoEsol}
transferred to
\begin{align}
\label{massWhatRho:RhoTotalSol}
\rho = c_2\, {e}^{\dfrac{{x}^{2}}{2}} \left( \dfrac{c_1\,sin\left( \alpha\,t\right) }{\alpha}+c_2 \right)
\end{align}
Where the constant, $c_2$, is an arbitrary function of the $y$ coordinate.
8.3 Conservation of General Quantity
8.3.1 Generalization of Mathematical Approach for Derivations
In this section a general approach for the derivations for conservation of any
quantity e.g. scalar, vector or tensor, are presented.
Suppose that the property $\phi$ is under a study which is a function of the time and location as $\phi(x,y,z,t)$.
The total amount of quantity that exist in arbitrary system is
\begin{align}
\label{dif:eq:math:phiG}
\Phi = \int_{sys} \phi\,\rho\,dV
\end{align}
Where $\Phi$ is the total quantity of the system which has a volume $V$ and a surface area of $A$ which
is a function of time.
A change with time is
\begin{align}
\label{dif:eq:math:DphiDt1}
\dfrac{D\Phi}{Dt} = \dfrac{D}{Dt} \int_{sys} \phi\,\rho\,dV
\end{align}
Using RTT to change the system to a control volume (see equation \eqref{mass:eq:RT}) yields
\begin{align}
\label{dif:eq:math:DphiDt}
\dfrac{D}{Dt} \int_{sys} \phi\,\rho\,dV =
\dfrac{d}{dt} \int_{cv} \phi\,\rho\,dV + \int_{A} \rho\,\phi\,\pmb{U}\cdot dA
\end{align}
The last term on the RHS can be converted using the divergence theorem (see the appendix
)
from a surface integral into a volume integral (alternatively, the volume integral
can be changed to the surface integral) as
\begin{align}
\label{dif:math:divergenceTheorem}
\int_{A} \rho\,\phi\,\pmb{U}\cdot dA = \int_{V} \nabla\cdot\left(\rho\,\phi\,\pmb{U} \right) dV
\end{align}
Substituting equation \eqref{dif:math:divergenceTheorem} into equation qref{dif:eq:math:DphiDt}
yields
\begin{align}
\label{dif:eq:RTTextended1}
\dfrac{D}{Dt} \int_{sys} \phi\,\rho\,dV = \dfrac{d}{dt} \int_{cv} \phi\,\rho\,dV +
\int_{cv} \nabla\cdot\left(\rho\,\phi\,\pmb{U} \right) dV
\end{align}
Since the volume of the control volume remains independent of the time, the derivative
can enter into the integral and thus combining the two integrals on the RHS results in
\begin{align}
\label{dif:eq:RTTextended}
\dfrac{D}{Dt} \int_{sys} \phi\,\rho\,dV = \int_{cv} \left( \dfrac{d \left(\phi\,\rho\right)}{dt} +
\nabla\cdot\left(\rho\,\phi\,\pmb{U} \right) \right) dV
\end{align}
The definition of equation \eqref{dif:eq:math:phiG} LHS can be changed to simply the derivative of $\Phi$.
The integral is carried over arbitrary system.
For an infinitesimal control volume the change is
\begin{align}
\label{dif:eq:math:infinitesimalChange}
\dfrac{D\,\Phi}{Dt} \cong \left( \dfrac{d \left(\phi\,\rho\right)}{dt} +
\nabla\cdot\left(\rho\,\phi\,\pmb{U} \right) \right) \overbrace{dx\,dy\,dz}^{dV}
\end{align}
8.3.2 Examples of Several Quantities
8.3.2.1 The General Mass Time Derivative
Using $\phi=1$ is the same as dealing with the mass conservation.
In that case $\dfrac{D\,\Phi}{Dt} = \dfrac{D\,\rho}{Dt}$ which is equal to zero as
\begin{align}
\label{dif:eq:math:infinitesimalChangeRho1}
\int \left(
\dfrac{d \left(\overbrace{1}^{\phi}\,\rho\right)}{dt} +
\nabla\cdot\left(\rho\,\overbrace{1}^{\phi}\,\pmb{U} \right) \right) \overbrace{dx\,dy\,dz}^{dV} = 0
\end{align}
Using equation \eqref{dif:eq:math:phiG} leads to
\begin{align}
\label{dif:eq:math:infinitesimalChangeRho}
\dfrac{D\,\rho}{Dt} = 0 \longrightarrow \dfrac{\partial \,\rho}{\partial t} +
\nabla\cdot\left(\rho\,\pmb{U} \right) = 0
\end{align}
Equation \eqref{dif:eq:math:infinitesimalChangeRho} can be rearranged as
\begin{align}
\label{dif:eq:math:mass:start}
\dfrac{\partial \,\rho}{\partial t} + \pmb{U}\,\nabla\cdot\rho + \rho\,\nabla\cdot\pmb{U} = 0
\end{align}
Equation \eqref{dif:eq:math:mass:start} can be further rearranged so derivative of the density is
equal the divergence of velocity as
\begin{align}
\label{dif:eq:math:mass:startRe}
\dfrac{1}{\rho}
\left( \overbrace{\dfrac{\partial \,\rho}{\partial t} +
\pmb{U}\,\nabla\cdot\rho}
^{\text{substantial derivative}} \right)
=  \nabla\cdot\pmb{U}
\end{align}
Equation \eqref{dif:eq:math:mass:startRe} relates the density rate of change or the volumetric
change to the velocity divergence of the flow field.
The term in the bracket LHS is referred in the literature as substantial derivative.
The substantial derivative represents the change rate of the density at a point which moves with the fluid.
Acceleration Direct Derivations
One of the important points is to find the fluid particles acceleration.
A fluid particle velocity is a function of the location and time.
Therefore, it can be written that
\begin{align}
\label{diff:eq:locatonU}
\pmb{U}(x,y,z,t) = U_x(x,y,x,t)\,\widehat{i} +
U_y(x,y,z,t)\,\widehat{j} +
U_z(x,y,z,t)\,\widehat{k}
\end{align}
Therefor the acceleration will be
\begin{align}
\label{diff:eq:locatonA}
\dfrac{D\pmb{U}}{Dt} = \dfrac{d\,U_x}{dt}\,\widehat{i} +
\dfrac{d\,U_y}{dt} \,\widehat{j} + \dfrac{d\,U_z}{dt} \,\widehat{k}
\end{align}
The velocity components are a function of four variables, ($x$, $y$, $z$, and $t$), and hence
\begin{align}
\label{dif:eq:dUxdt}
\dfrac{D\,U_x}{Dt} = \dfrac{\partial \,U_x}{\partial t} \overbrace{\dfrac{d\, t}{d\,t}}^{=1} +
\dfrac{\partial \,U_x}{\partial x} \overbrace{\dfrac{d\, x}{d\,t}}^{U_x} +
\dfrac{\partial \,U_x}{\partial y} \overbrace{\dfrac{d\, y}{d\,t}}^{U_y} +
\dfrac{\partial \,U_x}{\partial z} \overbrace{\dfrac{d\, z}{d\,t}}^{U_z}
\end{align}
The acceleration in the $x$ can be written as
\begin{align}
\label{dif:eq:dUxdtC1}
\dfrac{D\,U_x}{Dt} = \dfrac{\partial \,U_x}{\partial t} +
{U_x} \dfrac{\partial \,U_x}{\partial x} +
{U_y} \dfrac{\partial \,U_x}{\partial y} +
{U_z} \dfrac{\partial \,U_x}{\partial z} = \dfrac{\partial \,U_x}{\partial t} +
\left( \pmb{U}\cdot\nabla\right)\,U_x
\end{align}
The same can be developed to the other two coordinates which can be combined (in a
vector form) as
\begin{align}
\label{dif:eq:dUdt}
\dfrac{d\,\pmb{U}}{dt} = \dfrac{\partial \,\pmb{U}}{\partial t} + \left( \pmb{U}\cdot\nabla\right)\,\pmb{U}
\end{align}
or in a more explicit form as
\begin{align}
\label{dif:eq:dUxdtC}
\dfrac{d\,\pmb{U}}{dt} = \overbrace{\dfrac{\partial \,\pmb{U}}{\partial t}}^{\text{ local acceleration } } +
\overbrace{\pmb{U} \dfrac{\partial \,\pmb{U} }{\partial x} +
\pmb{U} \dfrac{\partial \,\pmb{U} }{\partial y} +
\pmb{U} \dfrac{\partial \,\pmb{U} }{\partial z} }^{\text { convective acceleration } }
\end{align}
The time derivative referred in the literature as the local acceleration which vanishes when the flow
is in a steady state.
While the flow is in a steady state there is only convective acceleration of the flow.
The flow in a nozzle is an example to flow at steady state but yet has acceleration
which flow with a very low velocity can achieve a supersonic flow.
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