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Chapter 5 The Control Volume and Mass Conservation
Introduction
This chapter presents a discussion on the control volume and will be focused on
the conservation of the mass.
When the fluid system moves or changes, one wants to find or predict the velocities
in the system.
The main target of such analysis is to find the value of certain variables.
This kind of analysis is reasonable and it referred to in the literature as the Lagrangian Analysis.
This name is in honored J. L. Langrange (1736–1813) who formulated the equations of
motion for the moving fluid particles.
Even though this system looks reasonable, the Lagrangian system turned out to be difficult to solve
and to analyze.
This method applied and used in very few cases.
The main difficulty lies in the fact that every particle has to be traced to its original state.
Leonard Euler (1707–1783) suggested an alternative approach.
In Euler's approach the focus is on a defined point or a defined volume.
This methods is referred as Eulerian method.
Fig. 5.1 Control volume and system before and after motion.
The Eulerian method focuses on a defined area or location to find the needed information.
The use of the Eulerian methods leads to a set differentiation equations that is referred to
as Navier–Stokes equations which are commonly used.
These differential equations will be used in the later part of this book.
Additionally, the Eulerian system leads to integral equations which are the focus
of this part of the book.
The Eulerian method plays well with the physical intuition of most people.
This methods has its limitations and for some cases the Lagrangian is preferred
(and sometimes the only possibility).
Therefore a limited discussion on the Lagrangian system will be presented (later version).
Lagrangian equations are associated with the system while the Eulerian equation are associated
with the control volume.
The difference between the system and the control volume is shown in Figure 5.1.
The green lines in Figure 5.1 represent the system.
The red dotted lines are the control volume.
At certain time the system and the control volume are identical location.
After a certain time, some of the mass in the system exited the control volume
which are marked `` a '' in Figure .
The material that remained in the control volume is marked as `` b ''.
At the same time, the control gains some material which is marked as `` c ''.
5.2 Control Volume
The Eulerian method requires to define a control volume (some time more than one).
The control volume is a defined volume that was discussed earlier.
The control volume is differentiated into two categories of control
volumes, non–deformable and deformable.
Non–deformable control volume is a control volume which is fixed in space
relatively to an one coordinate system. This coordinate system may be in a relative
motion to another (almost absolute) coordinate system.

Deformable control volume is a volume having part of all of its
boundaries in motion during the process at hand.

Fig. 5.2 Control volume of a moving piston with in and out flow.
In the case where no mass crosses the boundaries, the control volume is a system.
Every control volume is the focus of the certain interest and will be dealt with the basic
equations, mass, momentum, energy, entropy etc.
Two examples of control volume are presented to illustrate difference between a deformable
control volume and non–deformable control volume.
Flow in conduits can be analyzed by looking in a control volume between two locations.
The coordinate system could be fixed to the conduit. The control volume chosen is nondeformable
control volume.
The control volume should be chosen so that the analysis should be simple and dealt with
as less as possible issues which are not in question.
When a piston pushing gases a good choice of control volume is a
deformable control volume that is a head the piston inside the cylinder
as shown in Figure 5.2.
5.3 Continuity Equation
In this chapter and the next three chapters, the conservation equations will be
applied to the control volume.
In this chapter, the mass conservation will be discussed.
The system mass change is
\begin{align}
\dfrac{D\,m_{sys}}{Dt} = \dfrac{D}{Dt} \int_{V_{sys}} \rho dV = 0
\label{mass:eq:Dmdt}
\end{align}
The system mass after some time, according Figure 5.1, is made of
\begin{align}
m_{sys} = m_{c.v.} + m_a  m_{c}
\label{mass:eq:mSysAfter}
\end{align}
The change of the system mass is by definition is zero.
The change with time (time derivative of equation qref{mass:eq:mSysAfter}) results in
\begin{align}
0 = \dfrac{D\,m_{sys}}{Dt} = \dfrac{d\,m_{c.v.}}{dt} + \dfrac{d\, m_a}{dt}  \dfrac{d\,m_{c}}{dt}
\label{mass:eq:mSysAfterdt}
\end{align}
The first term in equation qref{mass:eq:mSysAfterdt} is
the derivative of the mass in the control volume and at any given time is
\begin{align}
\label{mass:eq:masst}
\dfrac{d\,m_{c.v.} (t)}{dt} = \dfrac{d}{dt} \int_{V_{c.v.}}\rho\,dV
\end{align}
and is a function of the time.
Fig. 5.3 Schematics of velocities at the interface.
The interface of the control volume can move.
The actual velocity of the fluid leaving the control volume is the relative velocity
(see Figure 5.3).
The relative velocity is
\begin{align}
\overrightarrow{U_r} = \overrightarrow{U_f}  \overrightarrow{U_{b}}
\label{mass:eq:relativeU}
\end{align}
Where $U_f$ is the liquid velocity and $U_b$ is the boundary velocity
(see Figure 5.3).
The velocity component that is perpendicular to the surface is
\begin{align}
\label{mass:eq:reltiveA}
U_{rn} = \hat{n}\cdot \overrightarrow{U_r} = U_r \, \cos \theta
\end{align}
Where $\hat{n}$ is an unit vector perpendicular to the surface.
The convention of direction is taken positive if flow out the control volume and negative if the
flow is into the control volume.
The mass flow out of the control volume is the system mass that is not included
in the control volume.
Thus, the flow out is
\begin{align}
\label{mass:eq:madta}
\dfrac{d\, m_a}{dt} = \int_{S_{cv}} \rho_s \,U_{rn} dA
\end{align}
It has to be emphasized that the density is taken at the surface thus the subscript $s$.
In the same manner, the flow rate in is
\begin{align}
\label{mass:eq:madtb}
\dfrac{d\, m_b}{dt} = \int_{S_{c.v.}} \rho_s\,U_{rn} dA
\end{align}
It can be noticed that the two equations qref{mass:eq:madtb} and qref{mass:eq:madta} are similar
and can be combined, taking the positive or negative value of $U_{rn}$ with integration of the entire
system as
\begin{align}
\label{mass:eq:mdtCombine}
\dfrac{d\, m_a}{dt}  \dfrac{d\, m_b}{dt} = \int_{S_{cv}} \rho_s\,U_{rn} \, dA
\end{align}
applying negative value to keep the convention.
Substituting equation qref{mass:eq:mdtCombine} into equation qref{mass:eq:mSysAfterdt} results in
Continuity
\begin{align}
\label{mass:eq:intS}
\dfrac{d}{dt} \int_{c.v.} \rho_s dV =
\int_{S_{cv}} \rho\,U_{rn} \, dA
\end{align}
Equation qref{mass:eq:intS} is essentially accounting of the mass.
Again notice the negative sign in surface integral.
The negative sign is because flow out marked positive which
reduces of the mass (negative derivative) in the control volume.
The change of mass change inside the control volume is net flow in or out of the control system.
Fig. 5.4 Schematics of flow in in pipe with varying density as a function time for example
5.1.}
The next example is provided to illustrate this concept.
The density changes in a pipe, due to temperature variation and other reasons, can be approximated as
\begin{align}
\nonumber
\dfrac{\rho(x,t)}{\rho_0} = \left( 1  \dfrac{x}{L\dfrac{}{}}\right)^2 \cos {\dfrac{t}{t_0}}.
\end{align}
The conduit shown in Figure 5.4 length is $L$ and its area is $A$.
Express the mass flow in and/or out, and the mass in the conduit as function of time.
Write the expression for the mass change in the pipe.
Solution
Here it is very convenient to choose a nondeformable control volume that is inside the conduit $dV$ is chosen
as $\pi\,R^2\,dx$.
Using equation qref{mass:eq:intS}, the flow out (or in) is
\begin{align*}
\dfrac{d}{dt} \int_{c.v.} \rho dV =
\dfrac{d}{dt} \int_{c.v.}
\overbrace{\rho_0 \left(1  \dfrac{x}{L} \right)^2 \cos\left( \dfrac{t}{t_0} \right)} ^{\rho(t)}
\overbrace{\pi\,R^2\, dx}^{dV}
\end{align*}
The density is not a function of radius, $r$ and angle, $\theta$ and they can be taken out the integral as
\begin{align*}
\dfrac{d}{dt} \int_{c.v.} \rho dV =
\pi\,R^2 \dfrac{d}{dt} \int_{c.v.}\rho_0 \left(1\dfrac{x}{L\dfrac{}{}} \right)^2 \cos \left(
\dfrac{t}{t_0 \dfrac{}{}}\right) dx
\end{align*}
which results in
\begin{align*}
\text{Flow Out} = \overbrace{ \pi\,R^2}^{A}
\dfrac{d}{dt} \int_{0}^{L} \rho_0 \left(1\dfrac{x}{L \dfrac{}{}}\right)^2 \cos {\dfrac{t}{t_0}} dx =
 \dfrac{\pi\,R^2\,L\, \rho_0}{3\,t_0} \, \sin\left( \dfrac{t}{t_0} \right)
\end{align*}
The flow out is a function of length, $L$, and time, $t$, and is the change of the mass in the control volume.
5.3.1 Non Deformable Control Volume
When the control volume is fixed with time, the derivative in equation qref{mass:eq:intS} can enter
the integral since the boundaries are fixed in time and hence,
Continuity with Fixed b.c.
\begin{align}
\label{mass:eq:intSrho}
\int_{V_{c.v.}} \dfrac{d\,\rho}{dt} dV = \int_{S_{c.v.}} \rho\,U_{rn} \, dA
\end{align}
Equation qref{mass:eq:intSrho} is simpler than equation qref{mass:eq:intS}.
5.3.2 Constant Density Fluids
Further simplifications of equations qref{mass:eq:intS} can be obtained by assuming constant density and the
equation qref{mass:eq:intS} become conservation of the volume.
5.3.2.1 Non Deformable Control Volume
For this case the volume is constant therefore the mass is constant, and hence the mass change of the
control volume is zero.
Hence, the net flow (in and out) is zero.
This condition can be written mathematically as
\begin{align}
\label{mass:eq:cvCmCV}
\overbrace{\dfrac{d\,\int}{dt}}^{ = 0} \longrightarrow
\int_{S_{c.v.}} V_{rn} dA = 0
\end{align}
or in a more explicit form as
Steady State Continuity
\begin{align}
\label{mass:eq:cvCmCV1}
\int_{S_{in}} V_{rn}\, dA = \int_{S_{out}} V_{rn}\,dA = 0
\end{align}
Notice that the density does not play a role in this equation since it is canceled out.
Physically, the meaning is that volume flow rate in and the volume flow rate out have to equal.
5.3.2.2 Deformable Control Volume
The left hand side of question qref{mass:eq:intS} can be examined further to develop a simpler
equation by using the extend Leibniz integral rule for a constant density and result in
\begin{align}
\dfrac{d}{dt} \int_{c.v.}\rho \,dV =
\overbrace{\int_{c.v.} \overbrace{\dfrac{d\,\rho}{dt}}^{=0} \,dV }^{\mbox{thus, =0}} +
\rho\,\int_{S_{c.v.}} \hat{n} \cdot U_{b}\, dA =
\rho\,\int_{S_{c.v.}} U_{bn}\, dA
\label{mass:eq:dmasst}
\end{align}
where $U_b$ is the boundary velocity and $U_{bn}$ is the normal component of the boundary velocity.
Steady State Continuity Deformable
\begin{align}
\label{mass:eq:cvmV}
\int_{S_{c.v.}} U_{bn}\, dA = \int_{S_{c.v.}} U_{rn}\, dA
\end{align}
The meaning of the equation qref{mass:eq:cvmV} is the net growth (or decrease) of the Control volume is
by net volume flow into it. Example 5.2 illustrates this point.
Fig. 5.5 Filling of the bucket and choices of the deformable control volumes for example .
Liquid fills a bucket as shown in Figure 5.5.
The average velocity of the liquid at the exit of the filling pipe is $U_p$ and cross section of
the pipe is $A_p$.
The liquid fills a bucket with cross section area of $\mathbf{A}$ and instantaneous height is $h$.
Find the height as a function of the other parameters.
Assume that the density is constant and at the boundary interface $A_{j} = 0.7\,A_{p}$.
And where $A_j$ is the area of jet when touching the liquid boundary in bucket.
The last assumption is result of the energy equation (with some influence of momentum equation).
The relationship is function of the distance of the pipe from the boundary of the liquid.
However, this effect can be neglected for this range which this problem.
In reality, the ratio is determined by height of the pipe from the liquid surface in the bucket.
Calculate the bucket liquid interface velocity.
Solution
This problem requires two deformable control volumes.
The first control is around the jet and second is around the liquid in the bucket.
In this analysis, several assumptions must be made.
First, no liquid leaves the jet and enters the air.
Second, the liquid in the bucket has a straight surface.
This assumption is a strong assumption for certain conditions but it will be not discussed here
since it is advance topic.
Third, there are no evaporation or condensation processes.
Fourth, the air effects are negligible.
The control volume around the jet is deformable because the length of the jet shrinks with the time.
The mass conservation of the liquid in the bucket is
\begin{align*}
\overbrace{\int_{c.v.} U_{bn} \, dA}^{\mbox{boundary change}} =
\overbrace{\int_{c.v.} U_{rn}\, dA}^{\mbox{flow in}}
\end{align*}
where $U_{bn}$ is the perpendicular component of velocity of the boundary.
Substituting the known values for $U_{rn}$ results in
\begin{align*}
\int_{c.v.} U_{b}\, dA =
\int_{c.v.} \overbrace{ \left( U_{j} + U_{b}\right) }^{U_{rn}} dA
\end{align*}
The integration can be carried when the area of jet is assumed to be known as
\begin{align}
\label{bucket:a}
U_{b}\, A = A_{j}\, \left(U_{j} + U_{b} \right)
\end{align}
To find the jet velocity, $U_{j}$, the second control volume around the jet is used as the following
\begin{align}
\label{bucket:b}
\overbrace{U_p \, A_p}^{\text{flow in }}
 \overbrace{ A_{j} \left( U_{b} + U_{j} \right) }^{\text{flow out }} =
\overbrace{ A_{j}\, U_{b} } ^{ \text{ boundary change} }
\end{align}
The above two equations qref{bucket:a} and qref{bucket:b} are enough to solve for the two unknowns.
Substituting the first equation, qref{bucket:a} into qref{bucket:b} and using the ratio of
$A_{j} = 0.7\,A_{p}$ results
\begin{align}
\label{bucket:c}
U_p \, A_p  U_b\,A = 0.7\,A_p\,U_b
\end{align}
The solution of equation qref{bucket:c} is
\begin{align*}
U_b = \dfrac{A_p}{A 0.7\,A_p}
\end{align*}
It is interesting that many individuals intuitively will suggest that the solution is $U_b A_p/A$.
When examining solution there are two limits.
The first limit is when $A_p = A/0.7$ which is
\begin{align*}
U_b = \dfrac{ A_p}{ 0} = \infty
\end{align*}
The physical meaning is that surface is filled instantly.
The other limit is that and $A_p/A \longrightarrow 0$ then
\begin{align*}
U_b = \dfrac{ A_p}{ A}
\end{align*}
which is the result for the ``intuitive'' solution.
It also interesting to point out that if the filling was from other surface (not the top surface), e.g.
the side, the velocity will be $U_b = U_p$ in the limiting case and not infinity.
The reason for this difference is that the liquid already fill the bucket and has not to move into
bucket.
Balloon is attached to a rigid supply in which is supplied by a constant the mass rate, $m_i$.
Calculate the velocity of the balloon boundaries assuming constant density.
Solution
The applicable equation is
\begin{align*}
\int_{c.v.} U_{bn} \, dA =
\int_{c.v.} U_{rn}\, dA
\end{align*}
The entrance is fixed, thus the relative velocity, $U_{rn}$ is
\begin{align*}
U_{rn} =
\left\{
\begin{array}{ccc}
U_p &@ \;\mbox{the valve} & \
0 & \mbox{every else} &
\end{array}
\right.
\end{align*}
Assume equal distribution of the velocity in balloon surface and that the
center of the balloon is moving, thus the velocity has the following form
\begin{align*}
U_b = U_x\,\hat{x} + U_{br}\,\hat{r}
\end{align*}
Where $\hat{x}$ is unit coordinate in $x$ direction and $U_x$ is the velocity of the center and
where $\hat{r}$ is unit coordinate in radius from the center of the balloon and
$U_{br}$ is the velocity in that direction.
The right side of equation qref{mass:eq:cvmV} is the net change due to the boundary is
\begin{align*}
\int_{S_{c.v.}} \left( U_x\,\hat{x} + U_{br}\,\hat{r} \right) \cdot \hat{n}\, dA
= \overbrace{\int_{S_{c.v.}} \left( U_x\,\hat{x} \right) \cdot \hat{n}\, dA}^{\mbox{center movement}} +
\overbrace{\int_{S_{c.v.}} \left( U_{br}\,\hat{r} \right) \cdot \hat{n}\, dA}^{\mbox{net boundary change}}
\end{align*}
The first integral is zero because it is like movement of solid body
and also yield this value mathematically (excises for mathematical oriented student).
The second integral (notice $\hat{n} = \hat{r}$) yields
\begin{align*}
\int_{S_{c.v.}} \left( U_{br}\,\hat{r} \right) \cdot \hat{n}\, dA =
{4\,\pi\, r^2}\,{U_{br}}
\end{align*}
Substituting into the general equation yields
\begin{align*}
\rho\, \overbrace{4\,\pi\, r^2}^{A}{U_{br}} = \rho\,U_p\,A_p = m_i
\end{align*}
Hence,
\begin{align*}
U_{br} = \dfrac{m_i}{\rho\,4\,\pi\, r^2}
\end{align*}
The center velocity is (also) exactly $U_{br}$.
The total velocity of boundary is
\begin{align*}
U_t = \dfrac{m_i}{\rho\,4\,\pi\, r^2} \left( \hat{x} + \hat{r}\right)
\end{align*}
It can be noticed that the velocity at the opposite to the connection to the rigid pipe which
is double of the center velocity.
5.3.2.3 One–Dimensional Control Volume
Additional simplification of the continuity equation is of one dimensional flow.
This simplification provides very useful description for many fluid flow phenomena.
The main assumption made in this model is that the proprieties
in the across section are only function of $x$ coordinate .
This assumptions leads
\begin{align}
\label{mass:eq:1Dim}
\int_{A_2} \rho_2\,U_2\, dA 
\int_{A_1} \rho_1\,U_1\, dA = \dfrac{d}{dt} \int_{V(x)} \rho(x)\,\overbrace{A(x) \, dx}^{dV}
\end{align}
When the density can be considered constant equation qref{mass:eq:1Dim} is reduced to
\begin{align}
\label{mass:eq:1DimMay}
\int_{A_2} U_2\, dA 
\int_{A_1} U_1\, dA = \dfrac{d}{dt} \int A(x) dx
\end{align}
For steady state but with variations of the velocity and variation of the density reduces
equation qref{mass:eq:1Dim} to become
\begin{align}
\label{mass:eq:1DimSteadyState}
\int_{A_2} \rho_2\,U_2\,dA = \int_{A_1} \rho_1\,U_1\,dA
\end{align}
For steady state and uniform density and velocity equation qref{mass:eq:1DimSteadyState} reduces further to
\begin{align}
\label{mass:eq:1DconstantRho}
\rho_1\,A_1\,U_1 = \rho_2\,A_2\,U_2
\end{align}
For incompressible flow (constant density), continuity equation is at its minimum form of
\begin{align}
\label{mass:eq:1DconstantRhoIncompressible}
U_1\,A_1 = A_2\,U_2
\end{align}
The next example is of semi one–dimensional example to illustrate equation qref{mass:eq:1Dim}.
Fig. 5.6 Height of the liquid for example .
Liquid flows into tank in a constant mass flow rate of $a$.
The mass flow rate out is function of the height.
First assume that $q_{out} = b \,h$ second
Assume as $q_{out} = b \,\sqrt{h}$.
For the first case, determine the height, $h$ as function of the time.
Is there a critical value and then if exist find the critical value of the system parameters.
Assume that the height at time zero is $h_0$.
What happen if the $h_0 =0$?
Solution
The control volume for both cases is the same and it is around the liquid in the tank.
It can be noticed that control volume satisfy the demand of one dimensional since the flow is only function of
$x$ coordinate.
For case one
the right hand side term in equation qref{mass:eq:1Dim} is
\begin{align*}
\rho \dfrac{d}{dt} \int_0^L h\,dx = \rho \, L \dfrac{d\,h}{dt}
\end{align*}
Substituting into equation equation qref{mass:eq:1Dim} is
\begin{align*}
\rho\,L\,\dfrac{d\,h}{d\,t} = \overbrace{b_1\,h}^{\mbox{flow out}}  \overbrace{m_i}^{\mbox{flow in}}
\end{align*}
solution is
\begin{align*}
h= \overbrace{\dfrac{m_i}{b_1}\,\text{ e}^{\dfrac{b_1\,t}{\rho\,L} }}^{\text{homogeneous solution}} +
\overbrace{c_1\,\text{e}^{\dfrac{b_1\,t}{\rho\,L}}}^{\text{ private solution}}
\end{align*}
The solution has the homogeneous solution (solution without the $m_i$) and the solution of the $m_i$ part.
The solution can rearranged to a new form (a discussion why this form is preferred will be provided
in dimensional chapter).
\begin{align*}
\dfrac{ h\, b_1}{m_1} = \text{e}^{\dfrac{b_1\,t}{\rho\,L} } +
c\,\text{ e}^{\dfrac{b_1\,t}{\rho\,L}}
\end{align*}
With the initial condition that at $h(t=0) = h_0$ the constant coefficient can be found as
\begin{align*}
\dfrac{h_0\,b_1}{m_1} = 1  c \Longrightarrow c = 1  \dfrac{h_0\,b_1}{m_i}
\end{align*}
which the solution is
\begin{align*}
\dfrac{ h\, b_1}{m_1} = \text{ e}^{\dfrac{b_1\,t}{\rho\,L} } +
\left[ 1  \dfrac{h_0\,b_1}{m_i\dfrac{}{}}\right]\,\text{ e}^{\dfrac{b_1\,t}{\rho\,L}}
\end{align*}
It can be observed that if $ 1 = \dfrac{h_0\,b_1}{m_i}$ is the critical point of this solution.
If the term $\dfrac{h_0\,b_1}{m_i}$ is larger than one then the solution reduced to a negative number.
However, negative number for height is not possible and the height solution approach zero.
If the reverse case appeared, the height will increase.
Essentially, the critical ratio state if the flow in is larger or lower than the flow out determine
the condition of the height.
For second case, the governing equation qref{mass:eq:1Dim} is
\begin{align*}
\rho\,L\,\dfrac{d\,h}{d\,t} = \overbrace{b\,\sqrt{h}}^{\text{flow out}}  \overbrace{m_i}^{\text{flow in}}
\end{align*}
with the general solution of
\begin{align*}
\ln\left[ \left( \dfrac{\sqrt{h}\,b}{m_i\dfrac{}{}}  1 \right) \dfrac{m_i}{\rho\,L} \right]
+ \dfrac{\sqrt{h}\,b}{m_i}  1 = \left(t + c \right) \dfrac{\sqrt{h}\,b}{2\,\rho\,L}
\end{align*}
The constant is obtained when the initial condition that at $h(t=0) = h_0$
and it left as exercise for the reader.
5.4 Reynolds Transport Theorem
It can be noticed that the same derivations carried for the density can be carried for other intensive
properties such as specific entropy, specific enthalpy.
Suppose that $g$ is intensive property (which can be a scalar or a vector) undergoes change with time.
The change of accumulative property will be then
\begin{align}
\label{mass:eq:intensiveProperty}
\dfrac{D}{Dt} \int_{sys} f\,\rho dV =
\dfrac{d}{dt} \int_{c.v.} f\,\rho dV + \int_{c.v} f\,\rho\,U_{rn} dA
\end{align}
This theorem named after Reynolds, Osborne, (18421912) which is actually a three dimensional generalization of
Leibniz integral rule
.
To make the previous derivation clearer, the Reynolds Transport Theorem will be reproofed and discussed.
The ideas are the similar but extended some what.
Leibniz integral rule
is an one dimensional and it is defined as
\begin{align}
\label{mass:eq:Leibniz}
\dfrac{d}{dy} \,\int_{x_1(y)}^{x_2(y)}f(x,y) \, dx =
\int_{x_1(y)}^{x_2(y)}\dfrac {\partial f}{\partial y}\,dx
+ f(x_2,y)\, \dfrac{dx_2}{dy}  f(x_1,y)\, \dfrac{dx_1}{dy}
\end{align}
Initially, a proof will be provided and the physical meaning will be explained.
Assume that there is a function that satisfy the following
\begin{align}
\label{mass:eq:aF}
G(x,y)= \int^x f\left ( \alpha,\,y \right) \,d\alpha
\end{align}
Notice that lower boundary of the integral is missing and is only the upper limit of the
function is present
.
For its derivative of equation qref{mass:eq:aF} is
\begin{align}
\label{mass:eq:daF}
f(x,y) = \dfrac{\partial G}{\partial x}
\end{align}
differentiating (chain rule $d\,uv = u\,dv+v\,du$) by part of left hand side of the Leibniz integral rule
(it can be shown which are identical) is
\begin{align}
\label{mass:eq:daFm}
\dfrac{d\, \left[G(x_2,y)G(x_1,y) \right] }{dy}
= \overbrace{\dfrac{\partial G}{\partial x_2} \dfrac{dx_2}{dy}}^{1} +
\overbrace{\dfrac{\partial G}{\partial y}(x_2,y)}^{2} 
\overbrace{\dfrac{\partial G}{\partial x_1}\dfrac{dx_1}{dy}}^{3} 
\overbrace{\dfrac{\partial G}{\partial y}(x_1,y)}^{4}
\end{align}
The terms 2 and 4 in equation qref{mass:eq:daFm} are actually (the $x_2$
is treated as a different variable)
\begin{align}
\label{mass:eq:2and4}
\int_{x_1(y)}^{x_2(y)}\dfrac{\partial\,f(x,y) }{\partial y}\,dx
\end{align}
The first term (1) in equation qref{mass:eq:daFm} is
\begin{align}
\label{mass:eq:term1}
\dfrac{\partial G}{\partial x_2} \dfrac{dx_2}{dy} =
f(x_2,y)\, \dfrac{dx_2}{dy}
\end{align}
The same can be said for the third term (3).
Thus this explanation is a proof the Leibniz rule.
The above ``proof'' is mathematical in nature and physical explanation is also provided.
Suppose that a fluid is flowing in a conduit.
The intensive property, $f$ is investigated or the accumulative property, $F$.
The interesting information that commonly needed is
the change of the accumulative property, $F$, with time.
The change with time is
\begin{align}
\label{mass:eq:F}
\dfrac{DF}{Dt} = \dfrac{D}{Dt} \int_{sys} \rho\, f\, dV
\end{align}
For one dimensional situation the change with time is
\begin{align}
\label{mass:eq:F1D}
\dfrac{DF}{Dt} = \dfrac{D}{Dt} \int_{sys} \rho\, f\, A(x) dx
\end{align}
If two limiting points (for the one dimensional) are moving with a different coordinate system,
the mass will be different and it will not be a system.
This limiting condition is the control volume for which some of the mass will leave or enter.
Since the change is very short (differential), the flow in (or out) will be the velocity of fluid minus
the boundary at $x_1$, $U_{rn} = U_1  U_b$.
The same can be said for the other side.
The accumulative flow of the property in, $F$, is then
\begin{align}
\label{mass:eq:flowOut1D1}
F_{in} = \overbrace{f_1\,\rho}^{F_1}\,\overbrace{ U_{rn}}^{\dfrac{dx_1}{dt}}
\end{align}
The accumulative flow of the property out, $F$, is then
\begin{align}
\label{mass:eq:flowOut1D}
F_{out} = \overbrace{f_2\,\rho}^{F_2}\,\overbrace{ U_{rn}}^{\dfrac{dx_2}{dt}}
\end{align}
The change with time of the accumulative property, $F$, between the boundaries is
\begin{align}
\label{mass:eq:cvD}
\dfrac{d}{dt} \int_{c.v.} \rho(x) \,f\,A(x)\,dA
\end{align}
When put together it brings back the Leibniz integral rule.
Since the time variable, $t$, is arbitrary and it can be replaced by any letter.
The above discussion is one of the physical meaning the Leibniz rule.
Reynolds Transport theorem is a generalization of the Leibniz rule and thus the same arguments are used.
The only difference is that the velocity has three components and only the perpendicular component enters into
the calculations.
Reynolds Transport
\begin{align}
\label{mass:eq:RT}
\dfrac{D}{DT} \int _{sys} f\, \rho dV =
\dfrac{d}{dt} \int_{c.v} f\,\rho \, dV + \int_{S_{c.v.}} f\,\rho \, U_{rn}\, dA
\end{align}
5.5 Examples For Mass Conservation
Several examples are provided to illustrate the topic.
Liquid enters a circular pipe with a linear velocity profile as a function of the radius
with maximum velocity of $U_{max}$.
After magical mixing, the velocity became uniform.
Write the equation which describes the velocity at the entrance.
What is the magical averaged velocity at the exit?
Assume no–slip condition.
Solution
The velocity profile is linear with radius.
Additionally, later a discussion on relationship between velocity at interface to solid
also referred as the (no) slip condition will be provided.
This assumption is good for most cases with very few exceptions.
It will be assumed that the velocity at the interface is zero.
Thus, the boundary condition is $U(r=R) = 0$ and $U(r=0) = U_{max}$
Therefore the velocity profile is
\begin{align*}
U (r) = U_{max} \left( 1  \dfrac{r}{R} \right)
\end{align*}
Where $R$ is radius and $r$ is the working radius (for the integration).
The magical averaged velocity is obtained using the equation qref{mass:eq:cvCmCV1}.
For which
\begin{align}
\label{mixingPipe:a}
\int_{0}^R U_{max} \left( 1  \dfrac{r}{R} \right) \, 2\,\pi\,r\,dr = U_{ave} \, \pi\,R^2
\end{align}
The integration of the equation qref{mixingPipe:a} is
\begin{align}
U_{max} \,\pi\ \dfrac{{R}^{2}}{6} = U_{ave} \, \pi\,R^2
\end{align}
The solution of equation (b) results in average velocity as
\begin{align}
U_{ave} = \dfrac{U_{max}}{6}
\end{align}
Fig. 5.7 Boundary Layer control mass.
Experiments have shown that a layer of liquid that attached itself to the surface and it is
referred to as boundary layer.
The assumption is that fluid attaches itself to surface.
The slowed liquid is slowing the layer above it.
The boundary layer is growing with $x$ because the boundary effect is penetrating further into fluid.
A common boundary layer analysis uses the Reynolds transform theorem.
In this case, calculate the relationship of the mass transfer across the control volume.
For simplicity assume slowed fluid has a linear velocity profile.
Then assume parabolic velocity profile as
\begin{align*}
U_x(y) = 2\,U_0\left[ \dfrac{y}{\delta\dfrac{}{}} + \dfrac{1}{2}\left(\dfrac{y}{\delta\dfrac{}{}} \right)^2 \right]
\end{align*}
and calculate the mass transfer across the control volume.
Compare the two different velocity profiles affecting on the mass transfer.
Solution
Assuming the velocity profile is linear thus, (to satisfy the boundary condition) it will be
\begin{align*}
U_x(y) = \dfrac{U_0 \, y}{\delta}
\end{align*}
The chosen control volume is rectangular of $L\times\delta$.
Where $\delta$ is the height of the boundary layer at exit point of the flow as shown in
Figure 5.7.
The control volume has three surfaces that mass can cross, the left, right, and upper.
No mass can cross the lower surface (solid boundary).
The situation is steady state and thus using equation qref{mass:eq:cvCmCV1} results in
\begin{align*}
\overbrace{\overbrace{\int_0^{\delta} U_0\, dy}^{in} 
\overbrace{\int_0^{\delta} \dfrac{U_0 \, y}{\delta}\,dy}^{out}}^{\mbox{x direction}} =
\overbrace{\int_0^{L} U{x} dx }^{\mbox{y direction}}
\end{align*}
It can be noticed that the convention used in this chapter of ``in'' as negative is not
``followed.'' The integral simply multiply by negative one.
The above integrals on the right hand side can be combined as
\begin{align*}
\int_0^{\delta} U_0 \left( 1  \dfrac{y}{\delta} \right) \,dy = \int_0^{L} U{x} dx
\end{align*}
the integration results in
\begin{align*}
\dfrac{U_0\, \delta}{2} = \int_0^{L} U{x} dx
\end{align*}
or for parabolic profile
\begin{align*}
{\int_0^{\delta} U_0\, dy} 
{\int_0^{\delta} U_0\left[ \dfrac{y}{\delta\dfrac{}{}} + \left(\dfrac{y}{\delta\dfrac{}{}} \right)^2 \right] } dy
= \int_0^{L} U{x} dx
\end{align*}
or
\begin{align*}
\int_0^{\delta} U_0\left[ 1 \dfrac{y}{\delta\dfrac{}{}}\left(\dfrac{y}{\delta\dfrac{}{}}\right)^2\right] dy =
U_0
\end{align*}
the integration results in
\begin{align*}
\dfrac{U_0\, \delta}{2} = \int_0^{L} U{x} dx
\end{align*}
Air flows into a jet engine at $5\,kg/sec$ while fuel flow into the jet is at $0.1\,kg/sec$.
The burned gases leaves at the exhaust which has cross area $0.1\,m^2$ with velocity of $500\, m/sec$.
What is the density of the gases at the exhaust?
Solution
The mass conservation equation qref{mass:eq:cvCmCV1} is used.
Thus, the flow out is ( 5 + 0.1 ) $5.1\, kg/sec$
The density is
\begin{align*}
\rho = \dfrac{\dot{m}}{A\,U} = \dfrac{5.1\,kg/sec}{0.01\, m^2\; 500\, m/sec}
= 1.02 kg/m^3
\end{align*}
The mass (volume) flow rate is given by direct quantity like $x\,kg/sec$.
However sometime, the mass (or the volume) is given by indirect quantity such as the effect of flow.
The next example deal with such reversed mass flow rate.
The tank is filled by two valves which one filled tank in 3 hours and the second by 6 hours.
The tank also has three emptying valves of 5 hours, 7 hours, and 8 hours.
The tank is 3/4 fulls, calculate the time for tank reach empty or full state when all the valves are open.
Is there a combination of valves that make the tank at steady state?
Solution
Easier measurement of valve flow rate can be expressed as fraction of the tank per hour.
For example valve of 3 hours can be converted to 1/3 tank per hour.
Thus, mass flow rate in is
\begin{align*}
\dot{m}_{in} = 1/3 + 1/6 = 1/2 tank/hour
\end{align*}
The mass flow rate out is
\begin{align*}
\dot{m}_{out} = 1/5 + 1/7 + 1/8 = \dfrac{131}{280}
\end{align*}
Thus, if all the valves are open the tank will be filled.
The time to completely filled the tank is
\begin{align*}
\dfrac{ \dfrac{1}{4} } { \dfrac{1}{2}  \dfrac{\strut 131}{280} } = \dfrac{70}{159} hour
\end{align*}
The rest is under construction.
Inflated cylinder is supplied in its center with constant mass flow.
Assume that the gas mass is supplied in uniformed way of $m_i\,[kg/m/sec]$.
Assume that the cylinder inflated uniformly and pressure inside the cylinder is uniform.
The gas inside the cylinder obeys the ideal gas law.
The pressure inside the cylinder is linearly proportional to the volume.
For simplicity, assume that the process is isothermal.
Calculate the cylinder boundaries velocity.
Solution
The applicable equation is
\begin{align*}
\overbrace{\int_{V_{c.v}}\dfrac{d\rho}{dt}\, dV}^{\mbox{increase pressure}} +
\overbrace{\int_{S_{c.v.}} \rho \, U_{b} dV}^{\mbox{boundary velocity}}
= \overbrace{\int_{S_{c.v.}} \rho U_{rn} \,dA}^{\mbox{in or out flow rate}}
\end{align*}
Every term in the above equation is analyzed but first the equation of state and volume to pressure
relationship have to be provided.
\begin{align*}
\rho = \dfrac{P}{R\,T}
\end{align*}
and relationship between the volume and pressure is
\begin{align*}
P = f \, \pi\, {R_c}^2
\end{align*}
Where $R_c$ is the instantaneous cylinder radius.
Combining the above two equations results in
\begin{align*}
\rho = \dfrac{f \, \pi\, {R_c}^2}{R\,T}
\end{align*}
Where $f$ is a coefficient with the right dimension.
It also can be noticed that boundary velocity is related to the radius in the following form
\begin{align*}
U_b = \dfrac{dR_c}{dt}
\end{align*}
The first term requires to find the derivative of density with respect to time which is
\begin{align*}
\dfrac{d\rho}{dt} = \dfrac{d}{dt} \left( \dfrac{f \, \pi\, {R_c}^2}{R\,T} \right)
= \dfrac{2\,f\,\pi\,R_c}{R\,T} \overbrace{\dfrac{dR_c}{dt}}^{U_b}
\end{align*}
Thus the first term is
\begin{align*}
\int_{V_{c.v}}\dfrac{d\rho}{dt}\, \overbrace{dV}^{2\,\pi\,R_c} =
\int_{V_{c.v}} \dfrac{2\,f\,\pi\,R_c}{R\,T} U_b
\overbrace{dV}^{2\,\pi\,R_c\,dR_c} =
\dfrac{4\,f\,\pi^2\,{R_c}^3}{3\,R\,T} U_b
\end{align*}
The integral can be carried when $U_b$ is independent of the $R_c$
. The second term is
\begin{align*}
\int_{S_{c.v.}} \rho \, U_{b} dA = \overbrace{\dfrac{f \, \pi\, {R_c}^2}{R\,T}}^{\rho}\, U_b \,
\overbrace{2\,\pi R_c}^{A}
= \left(\dfrac{f \, \pi^3\, {R_c}^2}{R\,T}\right) U_b
\end{align*}
substituting in the governing equation obtained the form of
\begin{align*}
\dfrac{f\,\pi^2\,{R_c}^3}{R\,T} U_b + \dfrac{4\,f \, \pi^2\, {R_c}^3}{3\,R\,T}\, U_b = m_i
\end{align*}
The boundary velocity is then
\begin{align*}
U_b = \dfrac{m_i}{\dfrac{7\,f \, \pi^2\, {R_c}^3}{3\,R\,T} }
G = \dfrac{3\,m_i\,R\,T} {7\,f \, \pi^2\, {R_c}^3}
\end{align*}
A balloon is attached to a rigid supply and is supplied by a constant mass rate, $m_i$.
Assume that gas obeys the ideal gas law.
Assume that balloon volume is a linear function of the pressure inside the balloon
such as $P = f_v\, V$.
Where $f_v$ is a coefficient describing the balloon physical characters.
Calculate the velocity of the balloon boundaries under the assumption of
isothermal process.
Solution
The question is more complicated than Example 5.10.
The ideal gas law is
\begin{align*}
\rho = \dfrac{P}{R\,T}
\end{align*}
The relationship between the pressure and volume is
\begin{align*}
P = f_v \, V = \dfrac{4 \, f_v \,\pi\, {R_b}^3 }{3}
\end{align*}
The combining of the ideal gas law with the relationship between the pressure and volume results
\begin{align*}
\rho = \dfrac{4 \, f_v \,\pi\, {R_b}^3 }{3\,R\,T}
\end{align*}
The applicable equation is
\begin{align*}
\int_{V_{c.v}}\dfrac{d\rho}{dt}\, dV + \int_{S_{c.v.}} \rho \, \left(U_c\,\hat{x} + U_{b}\hat{r} \right) dA
= \int_{S_{c.v.}} \rho U_{rn} \,dA
\end{align*}
The right hand side of the above equation is
\begin{align*}
\int_{S_{c.v.}} \rho U_{rn} \,dA = m_i
\end{align*}
The density change is
\begin{align*}
\dfrac{d\rho} {dt} = \dfrac{12\,f_v\,\pi\,{R_b}^2}{R\,T}
\overbrace{\dfrac{dR_b}{dt}}^{U_b}
\end{align*}
The first term is
\begin{align*}
\int_{0}^{R_b} \overbrace{\dfrac{12\,f_v\,\pi\,{R_b}^2}{R\,T}\, {U_b}}^{\neq f(r)}
\overbrace{4\,\pi\,r^2\,dr}^{dV} =
\dfrac{16\,f_v\,\pi^2\,{R_b}^5}{3\,R\,T}\, {U_b}
\end{align*}
The second term is
\begin{align*}
\int_{A} \dfrac{4 \, f_v \,\pi\, {R_b}^3 }{3\,R\,T} \,U_b\, dA =
\dfrac{4 \, f_v \,\pi\, {R_b}^3 }{3\,R\,T} \,U_b\, \overbrace{4\,\pi\,{R_b}^2}^{A}
= \dfrac{8 \, f_v \,\pi^2\, {R_b}^5 }{3\,R\,T} \,U_b
\end{align*}
Subsisting the two equations of the applicable equation results
\begin{align*}
U_b = \dfrac{1}{8} \dfrac{m_i\,R\,T}{ f_v \,\pi^2\, {R_b}^5}
\end{align*}
Notice that first term is used to increase the pressure and second the
change of the boundary.
Open Question: Answer must be received by April 15, 2010
The best solution of the following question will win 18 U.S. dollars and your name will be associated with the
solution in this book.
Solve example 5.10 under the assumption that the process is isentropic.
Also assume that the relationship between the pressure and the volume is $P = f_v\, V^2$.
What are the units of the coefficient $f_v$ in this problem?
What are the units of the coefficient in the previous problem?
5.6 The Details Picture – Velocity Area Relationship
Fig. 5.8 Control volume usage to calculate local averaged velocity in three coordinates.
The integral approach is intended to deal with the ``big'' picture.
Indeed the method is used in this part of the book for this purpose.
However, there is very little written about the usability of this approach to
provide way to calculate the average quantities in the control system.
Sometimes it is desirable to find the averaged velocity or velocity distribution
inside a control volume.
There is no general way to provide these quantities.
Therefore an example will be provided to demonstrate the use of this approach.
Consider a container filled with liquid on which one exit opened and the liquid flows out
as shown in Figure 5.8.
The velocity has three components in each of the coordinates under the assumption that flow is uniform
and the surface is straight
.
The integral approached is used to calculate the averaged velocity of each to the components.
To relate the velocity in the $z$ direction with the flow rate out or the exit the velocity mass balance is
constructed.
A similar control volume construction to find the velocity of the boundary velocity (height) can be carried out.
The control volume is bounded by the container wall including the exit of the flow.
The upper boundary is surface parallel to upper surface but at $Z$ distance from the bottom.
The mass balance reads
\begin{align}
\label{mass:eq:containerZs}
\int_V \dfrac{d\rho}{dt} \,dV + \int_A U_{bn}\,\rho\,dA + \int_A U_{rn}\,\rho\,dA = 0
\end{align}
For constant density (conservation of volume) equation
and ($h>z$) reduces to
\begin{align}
\label{mass:eq:containerZrho}
\int_A U_{rn}\,\rho\,dA = 0
\end{align}
In the container case for uniform velocity equation
ef{mass:eq:containerZrho} becomes
\begin{align}
\label{mass:eq:Zu}
U_z\,A = U_{e}\,A_e \Longrightarrow U_z =  \dfrac{A_e}{A} U_e
\end{align}
It can be noticed that the boundary is not moving and the mass inside does not change
this control volume.
The velocity $U_z$ is the averaged velocity downward.
Fig. 5.9 Control volume and system before and after the motion.
The $x$ component of velocity is obtained by using a different control volume.
The control volume is shown in Figure 5.9.
The boundary are the container far from the flow exit with blue line projection
into page (area) shown in the Figure 5.9.
The mass conservation for constant density of this control volume is
\begin{align}
\label{mass:eq:containerXs}
 \int_A U_{bn}\,\rho\,dA + \int_A U_{rn}\,\rho\,dA = 0
\end{align}
Usage of control volume not included in the previous analysis
provides the velocity at the upper boundary which is the same as the velocity at $y$ direction.
Substituting into qref{mass:eq:containerXs} results in
\begin{align}
\label{mass:eq:containerXU}
\int_{{A_{x}}^{}} \dfrac{A_e}{A} U_e \,\rho\,dA + \int_{A_{yz}} U_{x}\,\rho\,dA = 0
\end{align}
Where ${A_{x}}^{}$ is the area shown the Figure under this label.
The area $A_{yz}$ referred to area into the page in Figure under the blow line.
Because averaged velocities and constant density are used transformed equation qref{mass:eq:containerXU} into
\begin{align}
\label{mass:eq:containerXUa}
\dfrac{A_e}{A} {A_{x}}^{} U_e + U_{x}\,\overbrace{Y(x)\,h}^{A_{yz}} = 0
\end{align}
Where $Y(x)$ is the length of the (blue) line of the boundary.
It can be notice that the velocity, $U_{x}$ is generally increasing with $x$
because ${A_{x}}^{}$ increase with $x$.
The calculations for the $y$ directions are similar to the one done for $x$ direction.
The only difference is that the velocity has two different directions.
One zone is right to the exit with flow to the left and one zone to left with averaged velocity to right.
If the volumes on the left and the right are symmetrical the averaged velocity will be zero.
Fig. 5.10 Circular cross section for finding $U_x$ and various cross sections.}
Calculate the velocity, $U_{x}$ for a cross section of circular shape (cylinder).
Solution
The relationship for this geometry needed to be expressed.
The length of the line $Y(x)$ is
\begin{align}
\label{Ccontainer:yLength}
Y(x) = 2\,r\, \sqrt{ 1  \left(1\dfrac{x}{r}\right)^2 }
\end{align}
This relationship also can be expressed in the term of $\alpha$ as
\begin{align}
\label{Ccontainer:yLengthalpha}
Y(x) = 2 \, r\,\sin\alpha
\end{align}
Since this expression is simpler it will be adapted.
When the relationship between radius angle and $x$ are
\begin{align}
\label{Ccontainer:rx}
x = r (1 \sin\alpha)
\end{align}
The area ${A_{x}}^{}$ is expressed in term of $\alpha$ as
\begin{align}
\label{Ccontainer:area}
{A_{x}}^{} = \left( \alpha  \dfrac{1}{2\dfrac{}{}},\sin(2\alpha) \right) r^2
\end{align}
Thus, the velocity, $U_{x}$ is
\begin{align}
\label{Ccontainer:Uxs}
\dfrac{A_e}{A} \left( \alpha  \dfrac{1}{2\dfrac{}{}}\,\sin(2\alpha) \right) r^2\, U_e +
U_{x}\, 2 \, r\,\sin\alpha\,h = 0
\end{align}
\begin{align}
\label{Ccontainer:Uxf}
U_x = \dfrac{A_e}{A} \dfrac{r}{h}
\dfrac{\left( \alpha  \dfrac{1}{2\dfrac{}{}}\,\sin(2\alpha) \right) }{\sin\alpha} \, U_e
\end{align}
Averaged velocity is defined as
\begin{align}
\label{Ccontainer:UxaveDef}
\overline{U_x} = \dfrac{1}{S}\int_S U dS
\end{align}
Where here $S$ represent some length.
The same way it can be represented for angle calculations.
The value $dS$ is $r\cos\alpha$.
Integrating the velocity for the entire container and dividing by the angle, $\alpha$ provides the averaged
velocity.
\begin{align}
\label{Ccontainer:Uxtotal}
\overline{U_x} = \dfrac{1}{2\,r} \int_0^{\pi}\dfrac{A_e}{A} \dfrac{r}{h}
\dfrac{\left( \alpha  \dfrac{1}{2\dfrac{}{}}\,\sin(2\alpha) \right) }{\tan\alpha} \, U_e\, r\, d\alpha
\end{align}
which results in
\begin{align}
\label{Ccontainer:Uxtotala}
\overline{U_x} = \dfrac{\left( \pi 1\right)}{4} \dfrac{A_e}{A} \dfrac{r}{h} \, U_e
\end{align}
$\,$\[15pt]
Fig. 5.11 $y$ velocity for a circular shape
What is the averaged velocity if only half section is used.
State your assumptions and how it similar to the previous example.}
Solution
The flow out in the $x$ direction is zero because symmetrical reasons.
That is the flow field is a mirror images.
Thus, every point has different velocity with the same value in the opposite direction.
The flow in half of the cylinder either the right or the left has non zero averaged velocity.
The calculations are similar to those in the previous to example 5.12.
The main concept that must be recognized is the half of the flow must have come from one side and the
other come from the other side.
Thus, equation qref{mass:eq:containerXUa} modified to be
\begin{align}
\label{ene:eq:containeryU}
\dfrac{A_e}{A} {A_{x}}^{} U_e + U_{x}\,\overbrace{Y(x)\,h}^{A_{yz}} = 0
\end{align}
The integral is the same as before but the upper limit is only to $\pi/2$
\begin{align}
\label{Ccontainer:Uytotal}
\overline{U_x} = \dfrac{1}{2\,r} \int_0^{\pi/2}\dfrac{A_e}{A} \dfrac{r}{h}
\dfrac{\left( \alpha  \dfrac{1}{2\dfrac{}{}}\,\sin(2\alpha) \right) }{\tan\alpha} \, U_e\, r\, d\alpha
\end{align}
which results in
\begin{align}
\label{Ccontainer:Uytotala}
\overline{U_x} = \dfrac{\left( \pi 2\right)}{8} \dfrac{A_e}{A} \dfrac{r}{h} \, U_e
\end{align}
5.7 More Examples for Mass Conservation
Typical question about the relative velocity that appeared in many fluid mechanics exams is the following.
$\,$
Fig. 5.12 Schematic of the boat for example .
The inboard engine uses a pump to suck
in water at the front $A_{in} = 0.2\, m^2$ and eject it through
the back of the boat with exist area of $A_{out}=0.05\,m^2$.
The water absolute (relative to the ground) velocity leaving the back is 50$m/sec$, what are the
relative velocities entering and leaving the boat and the pumping rate? }
Solution
The boat is assumed (implicitly is stated) to be steady state and the density is constant.
However, the calculations have to be made in the frame of reference moving with the boat.
The relative jet discharge velocity is
\begin{align*}
U_{r_{out}} = 50 + 10 = 60[m/sec]
\end{align*}
The volume flow rate is then
\begin{align*}
Q_{out} = A_{out} \,U_{r_{out}} = 60\times 0.05 = 3 m^3/sec
\end{align*}
The flow rate at entrance is the same as the exit thus,
\begin{align*}
U_{r_{in}} = \dfrac{A_{out}}{A_{in}} \,U_{r_{out} } = \dfrac{0.05}{0.2}{60} = 15.0 m/sec
\end{align*}
In this case (the way the question is phrased), the velocity of the river has no relavence.
The boat from Example 5.14 travels downstream with the same relative exit jet speed (60$m/s$).
Calculate the boat absolute velocity (to the ground) in this case, assume that the areas to the pump did not change.
Is the relative velocity of the boat (to river) the same as before?
If not, calculate the boat reletive velocity to the river.
Assume that the river velocity is the same as in the previous example.
Solution
The relative exit velocity of the jet is 15$m/sec$ hence the flow rate is
\begin{align*}
Q_{out} = A_{out} \,U_{r_{out}} = 60\times 0.05 = 3 m^3/sec
\end{align*}
The relative (to the boat) velocity into boat is same as before (15$m/sec$).
The absolute velocity of boat (relative to the ground) is
\begin{align*}
U_{boat} = 15+ 5 = 20 m/sec
\end{align*}
The relative velocity of boat to the river is
\begin{align*}
U_{\text{reltive to river} } = 205 = 15 m/sec
\end{align*}
The relative exit jet velocity to the ground
\begin{align*}
U_{in} = 60  15 = 45 m/sec
\end{align*}
The boat relative velocities are different and depends on the directions.
Liquid A enters a mixing device depicted in at 0.1 [$kg/s$].
In same time liquid B enter the mixing device with a different specific density at 0.05 [$kg/s$].
The density of liquid A is 1000[$kg/m^3$] and liquid B is 800[$kg/m^3$].
The results of the mixing is a homogeneous mixture.
Assume incompressible process.
Find the average leaving velocity and density of the mixture leaving through the
2O [$cm$] diameter pipe.
If the mixing device volume is decreasing (as a piston pushing into the chamber) at rate of
.002 [$m^3/s$], what is the exit velocity?
State your assumptions.
Solution
In the first scenario, the flow is steady state and equation qref{mass:eq:intSrho} is applicable
\begin{align}
\label{mixChamber:gov}
\dot{m}_{A} + \dot{m}_B = Q_{mix}\,\rho_{mix} \Longrightarrow = 0.1 + 0.05 = 0.15 [m]
\end{align}
Thus in this case, since the flow is incompressible flow, the total volume flow in is equal to volume flow out as
\begin{align*}
\nonumber
\dot{Q}_A + \dot{Q}_B = \dot{Q}_{mix}
\Longrightarrow = \dfrac{\dot{m}_{A}}{\rho_A} + \dfrac{\dot{m}_{A}}{\rho_A} =
\dfrac{0.10}{1000} + \dfrac{0.05}{800}
\end{align*}
Thus the mixture density is
\begin{align}
\label{mixChamber:rhoMix}
\rho_{mix} = \dfrac{\dot{m}_{A} + \dot{m}_B }{ \dfrac{\dot{m}_{A}}{\rho_A} + \dfrac{\dot{m}_{B}}{\rho_B}}
= 923.07 [kg/m^3]
\end{align}
The averaged velocity is then
\begin{align}
\label{mixChamber:Umix}
U_{mix} = \dfrac{Q_{mix}}{A_{out}} =
\dfrac{ \dfrac{\dot{m}_{A}}{\rho_A} + \dfrac{\dot{m}_{B}} {\rho_B} } {\pi\,0.01^2}
= \dfrac{1.625}{\pi } [m/s]
\end{align}
In the case that a piston is pushing the exit density could be changed and fluctuated depending
on the location of the piston.
However, if the assumption of well mixed is still holding the exit density should not affected.
The term that should be added to the governing equation the change of the volume.
So governing equation is qref{mass:eq:cvmV}.
\begin{align}
\label{mixChamber:deformable}
\overbrace{U_{bn}\,A\,\rho_{b}}^{Q_b\,\rho_{mix}}
= \overbrace{\dot{m}_A + \dot{m}_B}^{in}  \overbrace{\dot{m}_{mix}}^{out}
\end{align}
That is the mixture device is with an uniform density
\begin{align}
\label{mixChamber:defCV}
 0.002 [m^/sec]\,923.7[kg/m^3] = 0.1 + 0.05  m_{exit}
\end{align}
\begin{align*}
m_{exit} = 1.9974 [kg/s]
\end{align*}
A syringe apparatus is being use to withdrawn blood
.
If the piston is withdrawn at O.01 [$m/s$].
At that stage air leaks in around the piston at the rate 0.000001 [$m^3/s$].
What is the average velocity of blood into syringe (at the tip)?
The syringe radios is 0.005[m] and the tip radius is 0.0003 [m].
Solution
The situation is unsteady state (in the instinctive c.v. and coordinates)
since the mass in the control volume (the syringe volume is not constant).
The chose of the control volume and coordinate system determine the amount of work.
This part of the solution is art.
There are several possible control volumes that can be used to solve the problem.
The two ``instinctive control volumes'' are the blood with the air and the
the whole volume between the tip and syringe plunger (piston).
The first choice seem reasonable since it provides relationship of the total to specific
material.
In that case, control volume is the volume syringe tip to the edge of the blood.
The second part of the control volume is the air.
For this case, the equation qref{mass:eq:cvmV} is applicable and can be written
\begin{align}
\label{syringe:blood}
U_{tip}\,A_{tip} \, \cancel{\rho_b} = U_b\,A_{s}\,\cancel{\rho_b}
\end{align}
In the air side the same equation can used.
There several coordinate systems that can used, attached to plunger, attached to the blood edge, stationary.
Notice that change of the volume do not enter into the calculations because the density of the air
is assumed to be constant.
In stationary coordinates two boundaries are moving and thus
\begin{align}
\label{syringe:airS}
\overbrace{U_{plunger} \,{A_s} \,\rho_a  U_b\,{A_{s}}\,\rho_b}^
{\text{moving b.c.}} = \overbrace{\rho_a \dot{Q}_{in}}^{\text{ in/out}}
\end{align}
In the case, the choice is coordinates moving with the plunger, the relative plunger velocity is
zero while the blood edge boundary velocity is $U_{plunger}  U_b$.
The air governing equation is
\begin{align}
\label{syringe:airP}
\overbrace{\left(U_{plunger} U_b\right)}^{\text{blood b. velocity}}
\,{A_{s}}\,\rho_b
= \overbrace{\rho_a \dot{Q}_{in}}^{\text{in/out}}
\end{align}
In the case of coordinates are attached to the blood edge similar equation is obtained.
At this stage, there are two unknowns, $U_b$ and $U_{tip}$, and two equations.
Using equations qref{syringe:blood} and qref{syringe:airP} results in
\begin{align}
\begin{array}{rcl}
\label{syringe:Ub}
U_b = U_{plunger}  \dfrac{\rho_a\,Q_{in}}{A_s\,\rho_b} \
U_{tip} = \dfrac{U_b\,A_s}{A_{tip}} =
\dfrac{\left( U_{plunger}  \dfrac{\rho_a\,Q_{in}}{A_s\,\rho_b}\right) \, A_s}{A_{tip}}
\end{array}
\end{align}
Fig. 5.13 Water jet Pump.
The apparatus depicted in Figure 5.13 is referred in the literature sometime
as the water–jet pump.
In this device, the water (or another liquid) is pumped throw the inner pipe at high velocity.
The outside pipe is lower pressure which suck the water (other liquid) into device.
Later the two stream are mixed.
In this question the what is the mixed stream averaged velocity with $U_1= 4.0[m/s]$ and $U_2= 0.5[m/s]$.
The cross section inside and outside radii ratio is $r_1/r_2=0.2$.
Calculate the mixing averaged velocity.
Solution
The situation is steady state and which density of the liquid is irrelevant (because it is the same
at the inside and outside).
\begin{align}
\label{jetPumpeq:gov1}
U_1\,A_1 + U_2\,A_2 = U_3\, A_3
\end{align}
The velocity is $A_3 = A_1+A_2$ and thus
\begin{align}
\label{jetPumpeq:gov}
U_3 = \dfrac{ U_1\,A_1 + U_2\,A_2 } { A_3}
= U_1 \dfrac{A_1}{A_3} + U_2 \left( 1  \dfrac{ A_1}{A_3\dfrac{}{}} \right)
\end{align}
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