3.2: Closed and Open Systems

While Equation 3.1 represents an attractive statement for conservation of mass when we are dealing with distinct bodies such as the cannon ball illustrated in Figure 3.1, it is not particularly useful when we are dealing with a continuum such as the water jet shown in Figure $$\PageIndex{1}$$. In considering Equation 3.1 and the water jet, we are naturally led to ask the question, “Where is the body?” Here we can identify a body, illustrated in Figure $$\PageIndex{2}$$, in terms of the famous Euler cut principle which we state as:

Not only do the laws of continuum physics apply to distinct bodies, but they also apply to any arbitrary body that one might imagine as being cut out of a distinct body.

The idea that the laws of physics, laboriously deduced by the observation of distinct bodies, can also be applied to bodies imagined as being cut out of distinct bodies is central to the engineering analysis of continuous systems. The validity of the Euler cut principle for bodies rests on the fact that the governing equations developed on the basis of this principle are consistent with experimental observation. Because of this, we can apply this principle to the liquid jet and imagine the moving, deforming fluid body illustrated in Figure $$\PageIndex{2}$$.

There we have shown a fluid body at $$t=0$$ that moves and deforms to a new configuration, $$V_{m}(t)$$, at a later, time $$t$$. The nomenclature here deserves some attention, and we begin by noting that we have used the symbol $$V$$ to represent a volume. This particular volume always contains the same material, thus we have added the subscript $$m$$. In addition the position of the volume changes with time, and we have indicated this by adding ($$t$$) as a descriptor. It is important to understand that the fluid body, which we have constructed on the basis of the Euler cut principle, always contains the same fluid since no fluid crosses the surface of the body.

In order to apply Equation 3.1 to the moving, deforming fluid body shown in Figure $$\PageIndex{2}$$, we first illustrate the fluid body in more detail in Figure $$\PageIndex{3}$$. There we have shown the volume occupied by the body, $$V_m(t)$$, along with a differential volume, $$dV$$. The mass, $$dm$$, contained in this differential volume is given by

$d m=\rho d V \label{3.5}$

in which $$\rho$$ is the density of the water leaving the faucet. The total mass contained in $$V_m(t)$$ is determined by summing over all the differential elemental that make up the body to obtain

$\left\{\begin{array}{l}{\text { mass of }} \\ {\text { the body }}\end{array}\right\}=\int_{V_{m}(t)} \rho d V \label{3.6}$

Use of this representation for the mass of a body in Equation 3.1 leads to:

Axiom $$\PageIndex{1}$$

$\frac{d}{d t} \int_{V_{m}(t)} \rho d V=0 \label{3.7}$

It should be clear that Equation \ref{3.7} provides no information concerning the velocity and diameter of the jet of water leaving the faucet. For such systems, the knowledge that the mass of a fluid body is constant is not very useful, and instead of Equation \ref{3.7} we would like an axiom that tells us something about the mass contained within some specified region in space. We base this more general axiomatic statement on an extension of the Euler cut principle that we express as

Not only do the laws of continuum physics apply to distinct bodies, but they also apply to any arbitrary region that one might imagine as being cut out of Euclidean 3-D space.

We refer to this arbitrary region in space as a control volume, and a more general alternative to Equation 3.1 is given by

Axiom $$\PageIndex{2}$$

$\left\{\begin{array}{c}{\text { time rate of change }} \\ {\text { of the mass contained }} \\ {\text { in any control volume }}\end{array}\right\}= \left\{\begin{array}{ll}{\text { rate at which }} \\ {\text { mass enters the }} \\ {\text { control volume }}\end{array}\right\}-\left\{\begin{array}{l}{\text { rate at which }} \\ {\text { mass leaves the }} \\ {\text { control volume }}\end{array}\right\} \label{3.8}$

To illustrate how this more general axiom for the mass of single component systems is related to Equation 3.1, we consider the control volume to be the space occupied by the fluid body illustrated in Figure $$\PageIndex{3}$$ . This control volume moves with the body, thus no mass enters or leaves the control volume and the two terms on the right hand side of Equation \ref{3.8} are zero as indicated by

$\left\{\begin{array}{l}{\text { rate at which }} \\ {\text { mass enters the }} \\ {\text { control volume }}\end{array}\right\}=\left\{\begin{array}{l}{\text { rate at which }} \\ {\text { mass leaves the }} \\ {\text { control volume }}\end{array}\right\}=0 \label{3.9}$

Under these circumstances, the axiomatic statement given by Equation \ref{3.8} takes the special form

$\left\{\begin{array}{l}{\text { time rate of change }} \\ {\text { of the mass contained }} \\ {\text { in the control volume }}\end{array}\right\}=\frac{d}{d t} \int_{V_{m}(t)} \rho d V=0 \label{3.10}$

This indicates that Equation \ref{3.8} contains Equation \ref{3.7} as a special case. Another special case of Equation \ref{3.8} that is especially useful is the fixed control volume illustrated in Figure $$\PageIndex{4}$$.

There we have identified the control volume by $$V$$ to clearly indicate that it represents a volume fixed in space. This fixed control volume can be used to provide useful information about the velocity of the fluid in the jet and the cross sectional area of the jet. As an application of the control volume formulation of the principle of conservation of mass, we consider the production of a polymer fiber in the following example.

Example $$\PageIndex{1}$$: Optical Fiber Production

The use of fiber optics is essential to high speed Internet communication, thus the production of optical fibers is extremely important. As an example of the application of Equation \ref{3.8}, we consider the production of an optical fiber from molten glass. In Figure 3.1a we have illustrated a stream of glass extruded through a small tube, or spinneret.

The surrounding air is at a temperature below the solidification temperature of the glass which is a solid when the fiber reaches the take-up wheel or capstan. A key quantity of interest in the fiber spinning operation is the draw ratio (Denn, 1980) which is the ratio of the area of the of the spinneret hole (indicated by $$A_o$$) to the area of the optical fiber leaving the take‐up wheel (indicated by $$A_1$$). Application of the macroscopic mass balance given by Equation \ref{3.8} will show us how the draw ratio is related to the parameters of the spinning operation.

We begin our analysis by assuming that the process operates at steady-state so that Equation \ref{3.8} reduces to

$\left\{\begin{array}{ll}{\text { rate at which }} \\ {\text { mass enters the }} \\ {\text { control volume }}\end{array}\right\}=\left\{\begin{array}{l}{\text { rate at which }} \\ {\text { mass leaves the }} \\ {\text { control volume }}\end{array}\right\} \label{1}$

In order to apply this result, we need to carefully specify the control volume, and this requires some judgment concerning the particular process under consideration. In this case, it seems clear that the surface of the control volume should cut the glass at both the entrance and exit, and that these two cuts should be joined by a surface that is coincident with the interface between the glass and the surrounding air. This leads to the control volume that is illustrated in Figure 3.1b where we have shown portions of the control surface at the entrance, at the exit, and at the glass-air interface where we neglect any mass transfer that may occur by diffusion. Because of this, we are concerned only with the rate at which mass enters and leaves the control volume at the entrance and exit.

The rate at which mass enters the control volume is given by

$\left\{\begin{array}{l}{\text { rate at which }} \\ {\text { mass enters the }} \\ {\text { control volume }}\end{array}\right\}=\left\{\begin{array}{l}{\text { density of }} \\ {\text { the glass }}\end{array}\right\}\left\{\begin{array}{l}{\text { volumetric flow }} \\ {\text { rate of the glass }}\end{array}\right\}=\rho_{\mathrm{o}} Q_{\mathrm{o}} \label{2}$

Here we have used $$Q_{0}$$ to represent the volumetric flow rate of the glass at the entrance, and $$\rho_{0}$$ to represent the density of the glass at the entrance.

Thus the units associated with Equation \ref{2} are represented by

$\left\{\begin{array}{c}{\text { units of density }} \\ {\text { times volumetric }} \\ {\text { flow rate }}\end{array}\right\}=\left\{\frac{k g}{m^{3}}\right\}\left\{\frac{m^{3}}{s}\right\}=\frac{k g}{s} \label{3}$

and $$\rho_{0} R_{0}$$ has units of mass per unit time. At the exit of the control volume, we designate the volumetric flow rate by $$Q_{1}$$ and the density of the solidified glass as $$\rho_{1}$$. This leads to an expression for the rate at which mass leaves the control volume given by

$\left\{\begin{array}{l}{\text { rate at which }} \\ {\text { mass leaves the }} \\ {\text { control volume }}\end{array}\right\}=\left\{\begin{array}{l}{\text { density of }} \\ {\text { the glass }}\end{array}\right\}\left\{\begin{array}{l}{\text { volumetric flow }} \\ {\text { rate of the glass }}\end{array}\right\}=\rho_{1} Q_{1} \label{4}$

Use of Equations \ref{3} and \ref{4} in the macroscopic mass balance given by Equation \ref{1} leads to

$\rho_{\mathrm{o}} Q_{\mathrm{o}}=\rho_{1} Q_{1} \label{5}$

This result tells us how the volumetric flow rates at the entrance and exit are related to the densities at the entrance and exit; however, it does not tell us what we want to know, i.e., the draw ratio, $$A_o / A_1$$. In order to extract this ratio from Equation \ref{5}, we need to express the volumetric flow rate in terms of the average velocity and the cross sectional area. At the entrance, this expression is given by

$\left\{\begin{array}{l}{\text { volumetric flow }} \\ {\text { rate of the glass }}\end{array}\right\}=\left\{\begin{array}{l}{\text { average }} \\ {\text { velocity }}\end{array}\right\}\left\{\begin{array}{c}{\text { cross }} \\ {\text { sectional }} \\ {\text { area }}\end{array}\right\}=\langle\mathrm{v}\rangle_{\mathrm{o}} A_{\mathrm{o}} \label{6}$

Here we have used $$\langle\mathbf{V}\rangle_{\mathrm{O}}$$ to represent the average velocity of the glass at the entrance, and $$A_o$$ to represent the cross sectional area, thus the units associated with Equation \ref{6} are given by

$\left\{\begin{array}{c}{\text { units of velocity }} \\ {\text { times cross }} \\ {\text { sectional area }}\end{array}\right\}=\left\{\frac{m}{s}\right\}\left\{m^{2}\right\}=\frac{m^{3}}{s}$

Use of an analogous representation for the volumetric flow rate at the exit allows us to express Equation \ref{5} in the form

$\rho_{\mathrm{o}}\langle\mathrm{v}\rangle_{\mathrm{o}} A_{\mathrm{o}}=\rho_{1}\langle\mathrm{v}\rangle_{1} A_{1} \label{7}$

and from this form of the macroscopic mass balance we find the draw ratio to be given by

$\left\{\begin{array}{ll}{\operatorname{draw}} \\ {\text { ratio }}\end{array}\right\} = \frac{A_{o}}{A_1} = \frac{ \rho_1 \langle\mathrm{v}\rangle_{1} |_{\text {take-up wheel}} }{\rho_o \langle\mathrm{v}\rangle_0} \label{8}$

Here we have clearly indicated that the average velocity at the exit of the control volume is specified by the speed of the take‐up wheel. Normally, this result would be arranged as

$\left\{\begin{array}{l}{\operatorname{draw}} \\ {\text { ratio }}\end{array}\right\}=\frac{A_{\mathrm{o}}}{A_{1}}= \left(\frac{\rho_{1} A_{\mathrm{o}}}{\rho_{\mathrm{o}} Q_{\mathrm{o}}}\right) \langle\mathrm{v}\rangle_{1} |_{\text {take-up wheel}} \label{9}$

with the thought that the velocity at the take‐up wheel is the most convenient parameter used to control the draw ratio.

In the previous example, we have shown how Equation \ref{3.8} can be used to determine the velocity at the take‐up wheel in order to produce a glass fiber of a specified cross‐sectional area. In order to prepare for more complex problems, we need to translate the word statement given by Equation \ref{3.8} into a precise mathematical statement. We begin by considering the fixed control volume illustrated in Figure 3.6. This volume is identified by V , the surface area of the volume by A , and the outwardly directly unit normal vector by n. The surface A may contain entrances and exits where fluid flows into and out of the control volume, and it may contain interfacial areas where mass transfer may or may not take place.

We begin our analysis of Equation \ref{3.8} for the control volume illustrated in Figure $$\PageIndex{6}$$ by considering the differential volume, $$dV$$, and denoting the mass contained in this differential volume by $$dm$$. In terms of the mass density $$\rho$$, we have

$d m=\rho d V \label{3.11}$

and the mass contained in the control volume can be represented as

$\left\{\begin{array}{c} {\text { mass contained }} \\ {\text { in the control }} \\ {\text { volume }} \end{array}\right\}=\int_{V} \rho d V \label{3.12}$

This allows us to express the first term in Equation \ref{3.8} as

$\left\{\begin{array}{l}{\text { time rate of change }} \\ {\text { of the mass contained }} \\ {\text { in the control volume }}\end{array}\right\}=\frac{d}{d t} \int_{V} \rho d V$

The determination of the rate at which mass leaves the control volume illustrated in Figure 3.6 requires the use of the projected area theorem (Stein and Barcellos, 1992, Sec. 17.1), and before examining the general case we consider the special case illustrated in Figure 3.7. In Figure 3.7 we have shown a control volume, V, that can be used to analyze the flow rate at the entrance and exit of a tube. In that special case, the velocity vector, v, is parallel to the unit normal vector, n, at the exit. Thus the volume of the fluid, V , leaving a differential area dA in a time

$\Delta V=\mathrm{v} \Delta t d A=\mathbf{v} \cdot \mathbf{n} \Delta t d A$

flow orthogonal to an exit

From this we determine that the volume of fluid that that flows across the surface $$dA$$ per unit time is given by

$\underbrace{\frac{\Delta V}{\Delta t}=\mathbf{v} \cdot \mathbf{n} d A}_{\text{volumetric flow rate orthogonal to an area} \, dA} \label{3.15}$

Following the same development given by Equation \ref{4} in Example 3.1, we express the mass flow rate as

$\frac{\rho \Delta V}{\Delta t}=\frac{\Delta m}{\Delta t}=\rho \mathbf{v} \cdot \mathbf{n} d A$

In order to determine the total rate at which mass leaves the exit of the faucet shown in Figure $$\PageIndex{7}$$, we simply integrate this expression for the mass flow rate over the area A exit to obtain

$\left\{\begin{array}{l} {\text { rate at which mass }} \\ {\text { flows out of the faucet }} \end{array}\right\}=\int_{A_{\text {exit }}} \rho \mathbf{v} \cdot \mathbf{n} d A \label{3.17}$

When the velocity vector and the unit normal vector are parallel, the flow field has the form illustrated in Figure $$\PageIndex{7}$$ and the mass flow rate is easily determined. When these two vectors are not parallel, we need to examine the flow more carefully and this is done in the following paragraphs.

3.1.1 General flux relation

In order to determine the rate at which mass leaves a control volume when $$\mathbf{v}$$ and $$\mathbf{n}$$ are not parallel, we return to the differential surface area element illustrated in Figure 3.6. A more detail version is shown in Figure 3.8 where we have included the unit vector that is normal to the surface, $$\mathbf{n}$$, and the unit vector that is tangent to the velocity vector, $$\mathbf{λ}$$. The unit tangent vector, $$\mathbf{λ}$$, is defined by

$\mathbf{v}=v \lambda \label{3.18}$

in which $$\mathbf{v}$$ represents the fluid velocity vector and v is the magnitude of that vector. In Figure 3.8, we have “marked” the fluid at the surface area element, and in order to determine the rate at which mass leaves the control volume through the area dA , we need to determine the volume of fluid that crosses the surface area dA per unit time.

In Figure 3.9, we have illustrated this volume which is bounded by the vectors v t that are parallel to the unit vector λ . One should imagine an observer who is fixed relative to the surface and who can determine the velocity v as the fluid crosses the surface A . The magnitude of v t is given by v t and this is the length of the cylinder that is swept out in a time t . The volume of the cylinder shown in Figure 3.9 is equal to the length, v t, times the cross sectional area, dA , and that concept is illustrated in Figure 3.10. We express the volume that is swept out of the control volume in a time \( as

$\Delta V=(\mathrm{v} \Delta t) d A_{c s} \label{3.19}$