# 3.3: Mass Flow Rates at Entrances and Exits

There are many problems for which the control volume, $$\mathscr{V}$$, contains only a single entrance and a single exit and the flux at the interfacial area is zero. This is the type of problem considered in Example $$3.2.1$$ in terms of the word statement given by Equation $$(3.2.4)$$, and we need to be certain that no confusion exists concerning Equation $$(3.2.4)$$ and the rigorous form given by Equation $$(3.2.25)$$. For systems that contain only a single entrance and a single exit, it is convenient to express Equation $$(3.2.25)$$ as

$\frac{d}{dt} \int_{\mathscr{V}}\rho dV = \int_{A_{ entrance} }\rho \left|\mathbf{v}\cdot\mathbf{n}\right| dA - \int_{A_{ exit} }\rho \left|\mathbf{v}\cdot\mathbf{n}\right| dA \label{30}$

in which we have made use of the relations

$\mathbf{v}\cdot\mathbf{n} = -\left|\mathbf{v}\cdot\mathbf{n}\right| ,\text{ at the entrance} , \quad \mathbf{v}\cdot\mathbf{n} = \left|\mathbf{v}\cdot\mathbf{n}\right| ,\text{ at the exit} \label{31}$

The form of the macroscopic mass balance given by Equation \ref{30} is analogous to the word statement given by Equation $$(3.2.4)$$ which we repeat here as

$\left\{\begin{array}{c} \text{time rate of change } \\ \text{ of the mass contained } \\ \text{ in the control volume}\end{array}\right\} = \left\{\begin{array}{c} \text{rate at which} \\ \text{ mass }enters\text{ the} \\ \text{ control volume}\end{array}\right\}-\left\{\begin{array}{c} \text{rate at which} \\ \text{ mass }leaves\text{ the} \\ \text{ control volume}\end{array}\right\} \label{32}$

There are several convenient representations for the terms that appear in Eqs. \ref{30} and \ref{32}, and we present these forms in the following paragraph.

## Convenient forms

The mass in a control volume can be expressed in terms of the volume $$\mathscr{V}$$ and the volume average density, $$\langle \rho \rangle$$, which is defined by

$\langle \rho \rangle = \frac{1}{\mathscr{V}} \int_{\mathscr{V}}\rho dV \label{33}$

In terms of $$\langle \rho \rangle$$, the accumulation in a fixed control volume takes the form

$\frac{d}{dt} \int_{\mathscr{V}}\rho dV = \mathscr{V}\frac{d\langle \rho \rangle }{ dt} \label{34}$

To develop a convenient form for the mass flux, we make use of Equation $$(3.2.19)$$ to express the volumetric flow rate at an exit as

$\int_{A_{exit} }\left|\mathbf{v}\cdot\mathbf{n}\right| dA = Q_{ exit} \label{35}$

Given this result, we can define an average density at an exit, $$\langle \rho \rangle_{b, exit}$$, according to

$\langle \rho \rangle_{b, exit} = \frac{1}{ Q_{ exit} } \int_{A_{ exit} }\rho \left|\mathbf{v}\cdot\mathbf{n}\right| dA = \frac{\int_{A_{ exit} }\rho \left|\mathbf{v}\cdot\mathbf{n}\right| dA }{ \int_{A_{ exit} }\left|\mathbf{v}\cdot\mathbf{n}\right| dA } \label{36}$

This average density is sometimes referred to as the “bulk density”, and that is the origin of the subscript $$b$$ in this definition. In addition, $$\langle \rho \rangle_{b, exit}$$ is sometimes referred to as the “cup mixed density” since it is the density that one would measure by collecting a cup of fluid at the outlet of a tube and dividing the mass of fluid by the volume of fluid. Equations \ref{35} and \ref{36} also apply to an entrance, thus we can use these results to express Equation \ref{30} in the form

$\mathscr{V} \frac{d\langle \rho \rangle }{ dt} = \langle \rho \rangle_{b, entrance} Q_{entrance} - \langle \rho \rangle_{b, exit} Q_{exit} \label{37}$

If the process is steady, the volume average density will be independent of time and this result simplifies to

$\langle \rho \rangle_{b, entrance} Q_{entrance} = \langle \rho \rangle_{b, exit} Q_{exit} \label{38}$

If the density is constant over the surfaces of the entrance and exit, the average values in this result can be replaced with the constant values and one recovers the mass balance given by Equation $$(5)$$ in Example $$3.2.1$$.

In addition to expressing the mass flux in terms of a density and a volumetric flow rate, there are many problems in which it is convenient to work directly with the mass flow rate. We designate the mass flow rate by $$\dot{m}$$ and represent the terms on the right hand side of Equation \ref{30} according to

$\int_{A_{ entrance} }\rho \left|\mathbf{v}\cdot\mathbf{n}\right| dA = \dot{m}_{ entrance} , \quad \int_{A_{ exit} }\rho \left|\mathbf{v}\cdot\mathbf{n}\right| dA = \dot{m}_{exit} \label{39}$

This leads to the following form of the macroscopic mass balance for one entrance and one exit

$\mathscr{V} \frac{d\langle \rho \rangle }{ dt} = \dot{m}_{ entrance} - \dot{m}_{ exit} \label{40}$

and when the process is steady this reduces to

$\dot{m}_{ entrance} = \dot{m}_{ exit} \label{41}$

When systems have more than one entrance and more than one exit, the mass flow rates can be expressed in precisely the manner that we have indicated by Eqs. \ref{37} through \ref{39}; however, one must be careful to include the flows at all entrances and exits.

Example $$\PageIndex{1}$$: Polymer coating

As an example of the application of Equation $$(3.2.25)$$, we consider the process of coating a polymer optical fiber with a polymer film. The process is similar to that discussed in Example $$3.2.1$$; however, in this case, we wish to determine the coating thickness rather than the fiber diameter. To determine the thickness of the polymer film, we make use of the control volume illustrated in Figure $$\PageIndex{1}$$. In constructing this control volume, we have followed the rules given earlier which are listed here along with the commentary that is appropriate for this particular problem.

Rule I. Construct a cut (a portion of the surface area $$\mathscr{A}$$) where information is given.

• In this case we have constructed a cut at a position downstream from the take-up wheel where the thickness of the polymer coating will normally be specified and the velocity of the polymer film will be equal to the velocity of the optical fiber. Thus we have created a cut where information is normally given.

Rule II. Construct a cut where information is required.

• In this case we have created a cut at the entrance where the coating polymer comes in contact with the optical polymer since we need to know the volumetric flow rate of the polymer that will produce the desired thickness of the coating. Thus we have created a cut where information is required.

Rule III. Join these cuts with a surface located where $$\mathbf{v}\cdot\mathbf{n}$$ is known.

• In this case we have joined the two cuts along the polymer-polymer interface on the basis of the assumption that $$\mathbf{v}\cdot\mathbf{n} = 0$$ at this interface. Thus we have joined these cuts with a surface located where $$\mathbf{v}\cdot\mathbf{n}$$ is known.

Rule IV. Be sure that the surface specified by Rule III encloses regions in which volumetric information is either given or required.

• For this particular problem, there is no volumetric information either given or required, and this rule is automatically satisfied.

We begin our analysis of the coating operation by assuming that the process operates at steady-state so that Equation $$(3.2.25)$$ reduces to

$\int_{\mathscr{A}}\rho \mathbf{v}\cdot\mathbf{n} dA = 0 , \quad \text{ steady state} \tag{1}\label{1a}$

When no mass transfer occurs at the portion of $$\mathscr{A}$$ located at the glass-polymer interface, we need only evaluate the area integral at the entrance and exit and Equation \ref{1a} reduces to

$\int_{A_{ entrance} }\rho \mathbf{v}\cdot\mathbf{n} dA + \int_{A_{ exit} }\rho \mathbf{v}\cdot\mathbf{n} dA = 0 \tag{2}\label{1b}$

At this point we can follow Example $$3.2.1$$ to obtain

$\left(\rho_{ o} Q_{ o} \right)_{ polymer} = \left(\rho_{1} Q_{ 1} \right)_{ polymer} \tag{3}\label{1c}$

At the exit, the velocity of the polymer coating is determined by the take-up wheel and is identical to the velocity of the optical fiber. This leads to

$\left(Q_{ 1} \right)_{ polymer} = (A_{1} )_{ polymer} \left. \langle { v}\rangle_{1} \right|_{\text{ take-up wheel}} \tag{4}\label{1d}$

in which $$(A_{ 1} )_{ polymer}$$ represents the area of the annular region occupied by the coating polymer.

Use of Equation \ref{1d} in Equation \ref{1c} allows one to determine the cross-sectional area of the polymer coating to be

$\left(A_{ 1} \right)_{ polymer} = \frac{\left(\rho_{ o} Q_{ o} \right)_{ polymer} }{ \left. \left(\rho_{1} \right)_{ polymer} \langle { v}\rangle_{1} \right|_{\text{take-up wheel} }} \tag{5}\label{1e}$

This expression can be used to determine the coating thickness in terms of the operating variables, and this will be left as an exercise for the student.

In both Examples $$3.2.1$$ and $$\PageIndex{1}$$, we illustrated applications of the macroscopic mass balance in order to determine the characteristics of a steady fiber spinning and coating process. Many practical problems are transient in nature and can be analyzed using the complete form of Equation $$(3.2.25)$$. The consumption of propane gas in rural areas represents and important transient problem, and we analyze this problem in the next example.

Example $$\PageIndex{2}$$: Delivery schedule for propane

In rural areas where natural gas is not available by pipeline, compressed propane gas is used for heating purposes. During the winter months, one must pay attention to the pressure gauge on the tank in order to avoid running out of fuel. Deliveries, for which there is a charge, are made periodically and must be scheduled in advance. The system under consideration is illustrated in Figure $$\PageIndex{2a}$$. The volume of the tank is 250 gallons, and when full it contains 750 kg of propane at a pressure of 95 psi (gauge). If the gas is consumed at a rate of 400 scf/day, we need to know whether a monthly or a bi-monthly delivery is required. Here the abbreviation, scf, represents standard cubic feet, and the word standard means that the pressure is equivalent to 760 mm of

mercury and the temperature is 273.16 K. A standard cubic foot represents a convenience unit 6 for the number of moles and this is easily demonstrated for the case of an ideal gas. Given the pressure ($$p =$$ one atmosphere), the temperature ($$T = 273.16$$ K), and the volume ($$V =$$ one cubic foot), one can use the ideal gas law to determine the number of moles according to

$n = pV / RT \nonumber$

If the gas under consideration is not an ideal gas, one must use the appropriate equation of state7 to determine what is meant by a standard cubic foot.

The propane inside the tank illustrated in Figure $$\PageIndex{2a}$$ is an equilibrium mixture of liquid and vapor, and we have illustrated this situation in Figure $$\PageIndex{2b}$$. There we have shown a control volume that has been constructed on the following basis:

1. A cut has been made at the exit of the tank where information is given.
2. There is no required information that would generate another cut.
3. The surface of the control volume is located where $$\mathbf{v}\cdot\mathbf{n}$$ is known.
4. The surface specified by #3 encloses the region for which volumetric information is required.

The proportions of liquid and vapor inside the tank are not known; however, we do know the initial mass of propane in the tank and we are given (indirectly) the average mass flow rate leaving the tank. Our analysis of the transient behavior of this system is based on the macroscopic mass balance given by

$\frac{d}{dt} \int_{\mathscr{V}}\rho dV + \int_{\mathscr{A}}\rho \mathbf{v}\cdot\mathbf{n}dA = 0 \tag{1}\label{2a}$

for which the fixed control volume, $$\mathscr{V}$$, is illustrated in Figure $$\PageIndex{2b}$$. Regardless of how the propane is distributed between the liquid and vapor phase, the mass of propane in the tank is given by

$\left\{\begin{array}{c} \text{mass of} \\ \text{ propane}\\ \text{ in the tank}\end{array}\right\} = \int_{\mathscr{V}}\rho dV = m \tag{2}\label{2b}$

and Equation \ref{2a} takes the form

$\frac{dm}{dt} + \int_{A_{ exit} }\rho \mathbf{v}\cdot\mathbf{n} dA = 0 \tag{3}\label{2c}$

On the basis of the analysis in Sec. $$3.3$$, we can express this result as

$\frac{dm}{dt} + \rho_{exit} Q_{exit} = 0 \tag{4}\label{2d}$

We are given that the volumetric flow rate is 400 standard cubic feet per day and this can be

converted to SI units to obtain

$Q_{exit} = 400 \frac{ft^{3} }{ day} \times \left(2.83 \times 10^{-2} \right) \frac{ m^{3} }{ { ft}^{3} } = 11.32 \frac{ m^{3} }{ day} \tag{5}\label{2e}$

In order to determine the density of the propane vapor leaving the tank, we first recall that one mole of gas at standard conditions occupies a volume of 22.42 liters. This leads to the molar concentration of propane given by

$c_{propane} = \frac{ mol}{ 22.42 \ L} \times \frac{L}{ 10^{ 3} \ cm^{ 3} } \times \left(\frac{100 \ cm}{m} \right)^{3} = 44.60 \frac{mol}{m^{ 3} } \tag{6}\label{2f}$

The molecular mass8 of propane is given in Table $$A2$$ of Appendix $$A$$, and use of that molecular mass allows us to determine the mass density of the propane as

$\rho_{propane} = c_{propane} MW_{propane} = 44.60 \frac{ mol}{ m^{ 3} } \times 44.097 \frac{ g}{ mol} = 1967 \frac{ g}{ m^{ 3} } \tag{7}\label{2g}$

The mass flow rate at the exit of the control volume illustrated in Figure $$\PageIndex{2b}$$ can now be calculated as

$\dot{m}_{exit} = \rho_{exit} Q_{exit} = 22,266 \ g/day = 22.27 \ kg/day \tag{8}\label{2h}$

Use of this result in Equation \ref{2d} leads to the governing differential equation for the mass in the tank

$\frac{dm}{dt} = - 22.27 \ kg/day \tag{9}\label{2i}$

This equation is easily integrated, and the initial condition imposed to obtain,

$m (t) = m(t = 0) - \left(22.27 \ kg/day \right) t \tag{10}\label{2j}$

For an initial mass of 750 kg we find that the tank will be empty ($$m = 0$$) when $$t = 33.7$$ days. For these circumstances, we require one delivery per month to insure that the propane tank will never be empty.