# 5.5: Examples For Mass Conservation

Several examples are provided to illustrate the topic.

Example 5.5

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 5.5

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 (??). For which
$\label{mixingPipe:a} \int_{0}^R U_{max} \left( 1 - \dfrac{r}{R} \right) \, 2\,\pi\,r\,dr = U_{ave} \, \pi\,R^2 \tag{38}$
The integration of the equation (??) is
$U_{max} \,\pi\ \dfrac ParseError: EOF expected (click for details) Callstack: at (Core/Chemical_Engineering/Fluid_Mechanics_(Bar-Meir)/05:_The_Control_Volume_and_Mass_Conservation/5.5:_Examples_For_Mass_Conservation), /content/body/div[2]/div/p[2]/span, line 1, column 2  {6} = U_{ave} \, \pi\,R^2 \tag{39}$
The solution of equation (b) results in average velocity as
$U_{ave} = \dfrac{U_{max}}{6} \tag{40}$

Fig. 5.7 Boundary Layer control mass.

Example 5.6

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 5.6

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 (??) 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*}

Example 5.7

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 5.7

The mass conservation equation (??) is used. Thus, the flow out is ( 5 + 0.1 ) $$5.1kg/sec$$ 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.

Example 5.8

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 5.8

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.

Example 5.9

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 5.9

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_{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*}
substituting in the governing equation obtained the form of
\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*}
The boundary velocity is then
\begin{align*}
U_b =  \dfrac{1}{8} \dfrac{m_i\,R\,T}{  f_v \,\pi^2\, {R_b}^5}
\end{align*}

Example 5.10

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 5.10

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.

Example 5.11

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?

### Contributors

• Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.