# 3.2: Closed and Open Systems

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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

distinctbodies, but they also apply to anyarbitrary bodythat one might imagine as beingcut outof 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:

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

distinctbodies, but they also apply to anyarbitrary regionthat one might imagine as beingcut outof 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

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.

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

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