# 3.4: Moving Control Volumes

In Figure $$\PageIndex{12}$$ we have illustrated the transient process of a liquid draining from a cylindrical tank, and we want to be able to predict the depth of the fluid in the tank as a function of time. When gravitational and inertial effects dominate the flow process, the volumetric flow rate from the tank can be expressed as

$Q = C_{D} A_{o} \sqrt{2gh} \label{42}$

Here $$Q$$ represents the volumetric flow rate, $$A_{o}$$ is the cross-sectional area of the orifice through which the water is flowing, and $$C_{D}$$ is the discharge coefficient that must be determined experimentally or by using the concepts presented in a subsequent course on fluid mechanics. Equation \ref{42} is sometimes referred to as Torricelli’s efflux principle9 in honor of the Italian scientist who discovered this result in the seventeenth century. We would like to use Torricelli’s law, along with the macroscopic mass balance to determine the height of the liquid as a function of time. The proper control volume to be used in this analysis is illustrated in Figure $$\PageIndex{1}$$ where we see that the top portion of the control surface is moving with the fluid, and the remainder of the control surface is fixed relative to the tank.

In order to develop a general method of attacking problems of this type, we need to explore the form of Equation $$(3.2.4)$$ for an arbitrary moving control volume. In Sec. $$3.2$$ we illustrated how Equation $$(3.2.4)$$ could be applied to the special case of a moving, deforming fluid body, and our analysis was quite simple since no fluid crossed the boundary of the control volume as indicated by Equation $$(3.2.5)$$.

To develop a general mathematical representation for Equation $$(3.2.4)$$, we consider the arbitrary moving control volume shown in Figure $$\PageIndex{2}$$. The surface of this control volume moves at a velocity $$\mathbf{w}$$ which need not be a constant, i.e., our control volume may be moving, deforming, accelerating or decelerating.

An observer moving with the control volume determines the rate at which fluid crosses the boundary of the moving control volume, $$\mathscr{V}_{a} (t)$$, and thus observes the relative velocity.

Previously we used a word statement of the principle of conservation of mass to develop a precise representation of the macroscopic mass balance for a fixed control volume. However, the statement given by Equation $$(3.2.4)$$ is not limited to fixed control volumes and we can apply it to the arbitrary moving control volume shown in Figure $$\PageIndex{2}$$. The mass within the moving control volume is given by

$\left\{\begin{array}{c} \text{ mass contained in} \\ any\text{ control volume}\end{array}\right\} = \int_{\mathscr{V}_{a} (t)} \rho dV \label{43}$

and the time rate of change of the mass in the control volume is expressed as

$\left\{\begin{array}{c} \text{ time rate of change } \\ \text{ of the mass contained } \\ \text{ in }any\text{ control volume}\end{array}\right\} = \frac{d}{dt} \int_{\mathscr{V}_{a} (t)}\rho dV \label{44}$

In order to determine the net mass flow leaving the moving control volume, we simply repeat the development given by Eqs. $$(3.2.14)$$ through $$(3.2.25)$$ noting that the velocity of the fluid determined by an observer on the surface of the control volume illustrated in Figure $$\PageIndex{2}$$ is the relative velocity $$\mathbf{v}_{r}$$. This concept is illustrated in Figure $$\PageIndex{3}$$ where we have shown the volume of fluid leaving the surface element

$$dA$$ during a time $$\Delta t$$. On the basis of that representation, we express the macroscopic mass balance for an arbitrary moving control volume as

$\frac{d}{dt} \int_{\mathscr{V}_{a} (t)}\rho dV + \int_{\mathscr{A}_{a} (t)}\rho v_{r} \cdot \mathbf{n} dA = 0 \label{45}$

The relative velocity is given explicitly by

$\mathbf{v}_{r} = \mathbf{v} - \mathbf{w} \label{46}$

thus the macroscopic mass balance for an arbitrary moving control volume takes the form

Axiom:

$\frac{d}{dt} \int_{\mathscr{V}_{a} (t)}\rho dV + \int_{\mathscr{A}_{a} (t)}\rho (\mathbf{v} - \mathbf{w})\cdot \mathbf{n} dA = 0 \label{47}$

Here it is helpful to think of $$\mathbf{w}$$ as the velocity of an observer moving with the surface of the control volume illustrated in Figure $$\PageIndex{2}$$ and it is important to recognize that this result contains our previous results for a fluid body and for a fixed control volume. This fact is illustrated in Figure $$\PageIndex{4}$$ where we see that the arbitrary velocity can be set equal to zero, $$\mathbf{w} = 0$$ in order to obtain Equation $$(3.2.25)$$, and we see that the arbitrary velocity can be set equal to the fluid velocity, $$\mathbf{w} = \mathbf{v}$$ in order to obtain Equation $$(3.2.3)$$.

Clearly Equation \ref{47} is the most general form of the principle of conservation of mass since it can be used to obtain directly the result for a fixed control volume and for a body.

Example $$\PageIndex{1}$$: Water level in a storage tank

A cylindrical API open storage tank, 2 m in diameter and 3 m in height, is used as a water storage tank. The water is used as a cooling fluid in a batch distillation unit and then sent to waste. The distillation unit runs for six hours and consumes 5 gal/min of water. At the beginning of the process the tank is full of water, and since we do not want to run out of water while the distillation is going on, we want to know the level of the tank at the end of the process. The system under consideration is illustrated in Figure $$\PageIndex{5}$$ and we have constructed the control volume 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 $$(\mathbf{v}-\mathbf{w})\cdot \mathbf{n}$$ is known.
4. The control volume encloses the region for which information is required.

For the moving control volume illustrate in Figure $$\PageIndex{5}$$, the macroscopic mass balance takes the form

$\frac{d}{ dt} \int_{\mathscr{V}(t)}\rho dV + \int_{\mathscr{A}(t)}\rho (\mathbf{v} - \mathbf{w})\cdot \mathbf{n} dA = 0 \tag{1}\label{1}$

in which $$\mathscr{V}_{a} (t)$$ has been replaced with $$\mathscr{V}(t)$$ since the control volume is no longer arbitrary.

A little thought will indicate that $$\rho (\mathbf{v} - \mathbf{w})\cdot \mathbf{n}$$ is zero everywhere on the surface of the control volume except at the exit of the tank. This leads to the representation given by

$\int_{\mathscr{A}(t)}\rho (\mathbf{v} - \mathbf{w})\cdot \mathbf{n} dA = \int_{A_{ exit} }\rho \mathbf{v} \cdot \mathbf{n} dA = \rho Q_{exit} \tag{2}\label{2}$

in which we have assumed that the density can be treated as a constant. Use of Equation \ref{2} in Equation \ref{1} provides

$\frac{d}{dt} \int_{\mathscr{V}(t)}\rho dV + \rho Q_{exit} = 0 \tag{3}\label{3}$

and we again impose the condition of a constant density to obtain

$\rho \frac{d\mathscr{V}(t)}{ dt} + \rho Q_{exit} = 0 \tag{4}\label{4}$

The control volume is computed as the product of the cross sectional area of the tank multiplied by the level of water in the tank, i.e., $$V(t) = A_{ T} h(t)$$. Use of this relation in Equation \ref{4} and canceling the density leads to

$A_{ T} \frac{dh}{dt} = - Q_{exit} \tag{5}\label{5}$

and this equation is easily integrated to give

$h (t) = h \left(t = 0\right) - \frac{Q_{exit} t}{ A_{ T} } \tag{6}\label{6}$

For consistency, all the variables are computed in SI units.

$A_{ T} = \frac{\pi D^{2} }{ 4} = 3.14 \ m^{2} \nonumber$

$Q_{exit} = \frac{(5 \ gal/min ) (3.78\times 10^{-3} \ m^{3} / gal)}{ 60 \ s/min } = 3.15\times 10^{-4} \ m^{3} /s \tag{7}\label{7}$

At the end of the process, $$t = 6 \ hr = { 2.16}\times { 10}^{ 4} \ s$$, the water level in the tank will be

$h (t = 6 \ hr) = 3 \ m - 2.167 \ m = 0.833 \ m \nonumber$

Example $$\PageIndex{2}$$: Water level in a storage tank with an inlet and outlet

In Figure $$\PageIndex{6a}$$ we have shown a tank into which water enters at a volumetric rate $$Q_{1}$$ and leaves at a rate $$Q_{ o}$$ that is given by

$Q_{ o} = 0.6 A_{o} \sqrt{2gh} \tag{1}\label{1a}$

Here $$A_{ o}$$ is the area of the orifice in the bottom of the tank. The tank is initially empty when water begins to flow into the tank, thus we have an initial condition of the form

I.C. $h = 0 , \quad t = 0 \tag{2}\label{2a}$

The parameters associated with this process are given by

$Q_{ 1} = 10^{-4} \ m^{ 3} /s , \quad A_{o} = 0.354 \ cm^{ 2} , \quad D = 1.5 \ m , \quad h_{ o} = 2.78 \ m \tag{3}\label{3a}$

and in this example we want to derive an equation that can be used to predict the height of the water at any time. That equation requires a numerical solution and we illustrate one of the methods that can be used to obtain numerical results.

The control volume used to analyze this process is illustrated in Figure $$\PageIndex{6}$$ where we have shown a moving control volume with “cuts” at the entrance and exit which are joined by a surface at which $$(\mathbf{v} - \mathbf{w})\cdot \mathbf{n}$$ is zero. If the water overflows, a second exit will be created; however, we are only concerned with the process that occurs prior to overflow.

Our analysis is moved on the macroscopic mass balance for a moving control volume

$\frac{d}{dt} \int_{\mathscr{V}(t)}\rho dV + \int_{\mathscr{A}(t)}\rho (\mathbf{v} - \mathbf{w})\cdot \mathbf{n} dA = 0 \tag{4}\label{4a}$

and the assumption that the density is constant leads to

$\frac{d\mathscr{V}(t)}{ dt} + \int_{\mathscr{A}(t)}(\mathbf{v} - \mathbf{w})\cdot \mathbf{n} dA = 0 \tag{5}\label{5a}$

The normal component of the relative velocity, $$(\mathbf{v} - \mathbf{w})\cdot \mathbf{n}$$, is zero everywhere except at the entrance and exit, thus Equation \ref{5a} simplifies to

$\frac{d\mathscr{V}(t)}{ dt} + Q_{ o} - Q_{1} = 0 \tag{6}\label{6a}$

The volume of the control volume is given by

$\mathscr{V}(t) = \frac{\pi D^{2} }{ 4} h(t) + V_{jet} (t) \tag{7}\label{7a}$

and use of this expression in Equation \ref{6a} provides

$\frac{\pi D^{2} }{ 4} \frac{d h}{ dt} + \frac{dV_{jet} }{ dt} + Q_{ o} - Q_{1} = 0 \tag{8}\label{8}$

If the area of the jet is much, much smaller than the cross sectional area of the tank, we can impose the following constraint on Equation \ref{8}

$\frac{dV_{jet} }{ dt} <<\frac{\pi D^{2} }{ 4} \frac{d h}{ dt} \tag{9}\label{9}$

however, one must keep in mind that there are problems for which this inequality might not be valid. When Equation \ref{9} is valid, we can simplify Equation \ref{8} and arrange it in the compact form

$\frac{d h}{ dt} = \alpha - \beta \sqrt{h} \tag{10}\label{10}$

where $$\alpha$$ and $$\beta$$ are constants given by

$\alpha = \frac{4Q_{1} }{ \pi D^{2} } , \quad \beta = \frac{0.6\times 4\times A_{ o} \sqrt{2g} }{ \pi D^{2} } \tag{11}\label{11}$

The initial condition for this process is given by

IC. $h = 0 , \quad t = 0 \tag{12}\label{12}$

and we need to integrate Equation \ref{10} and impose this initial condition in order to determine the fluid depth as a function of time. Separating variables in Equation \ref{10} leads to

$\frac{d h}{ \alpha - \beta \sqrt{h} } = dt \tag{13}\label{13}$

and the integral of this result is given by

$\frac{2}{ \beta^{2} } \left[\left(\alpha -\beta \sqrt{h} \right) - \alpha \ln \left(\alpha -\beta \sqrt{h} \right)\right] = t + C \tag{14}\label{14}$

The constant of integration, $$C$$, can be evaluated by means of the initial condition given by Equation \ref{12} and the result is

$C = \frac{2}{ \beta^{2} } \left(\alpha -\alpha \ln \alpha \right) \tag{15}\label{15}$

This allows us to express Equation \ref{14} in the form

$\frac{2\alpha }{ \beta^{2} } \left[\left(-\beta \sqrt{h} / \alpha \right) - \ln \left(1- \beta \sqrt{h} /\alpha\right)\right] = t \tag{16}\label{16}$

and this is an equation that can be used to predict the height of the water at any time. Since Equation \ref{16} represents an implicit expression for $$h(t)$$, some computation is necessary in order to produce a curve of the fluid depth versus time.

Before we consider a numerical method that can be used to solve Equation \ref{16} for $$h(t)$$, we want determine if the tank will overflow. To explore this question, we note that Equation \ref{13} provides the result

$\left(1 - {\beta \sqrt{h_{\infty } } / \alpha } \right) = 0 , \quad t = \infty \tag{17}\label{17}$

and this can be used to determine whether $$h(t)$$ is greater than $$H$$ as time tends to infinity. One can also obtain Equation \ref{17} directly from Equation \ref{10} by imposing the steady-state condition that $$dh/dt = 0$$. We can arrange Equation \ref{17} in the form

$h_{\infty } = \left({\alpha / \beta } \right)^{2} , \quad t = \infty \tag{18}\label{18}$

and in terms of Eqs. \ref{11} we obtain

$h_{\infty } = \left({Q_{1} / 0.6\times A_{ o} \sqrt{2g} } \right)^{2} , \quad t = \infty \tag{19}\label{19}$

from the values given in the problem statement, we find that $$h_{\infty } = 2.26$$ $$m$$ at steady state, thus the tank will not overflow.

At this point we would like to use Equation \ref{16} to develop a general solution for the fluid depth in the tank as a function of time, i.e., we wish to know $$h(t)$$ for the specific set of parameters given in this example. It is convenient to represent this problem in dimensionless variables so that

$H(x,\theta ) = -\left[x + \ln (1-x)\right] - \theta \tag{20}\label{20}$

in which $$x$$ is the dimensionless dependent variable representing the depth and defined by

$x = \beta \sqrt{h}/\alpha \tag{21}\label{21}$

In Equation \ref{20} we have used $$\theta$$ to represent the dimensionless time defined by

$\theta = \frac{\beta^{2} t}{ 2\alpha } \tag{22}\label{22}$

and our objective is to find the root of the equation

$H(x,\theta ) = 0 , \quad x = x^{*} \tag{23}\label{23}$

Here we have identified the solution as $$x^{*}$$ and it is clear from Equation \ref{20} that the solution will require that $$x^{*} < 1$$. In Appendix $$B$$ we have described several methods for solving implicit equations, such as Equation \ref{23}, and in this example we will use the simplest of these methods.

## Bisection method

Here we wish to find the solution to Equation \ref{23} when the parameter $$\theta$$ is equal to 0.10. A sketch of $$H(x,\theta )$$ is shown in Figure $$\PageIndex{6b}$$ where we see that $$x^{*}$$ is located between zero and one. The bisection method begins by locating values of $$x$$ that produce positive and negative values of the function $$H(x,\theta )$$ and these values are identified as $$x_{ o}$$ and $$x_{1}$$ in Figure $$\PageIndex{6b}$$. The next step in the bisection method is to bisect the distance between $$x_{ o}$$ and $$x_{1}$$ to produce the value indicated by $$x_{2} = (x_{1} +x_{ o} )/2$$ Next one evaluates $$H(x_{2} ,\theta )$$ in order to determine whether it is positive or negative. From Figure $$\PageIndex{6b}$$ we see that $$H(x_{2} ,\theta ) > 0$$, thus the next estimate for the solution is given by $$x_{3} = (x_{ 2} +x_{ 1} )/2$$. This type of geometrical construction is not necessary to carry out the bi-section method. Instead one only needs to evaluate $$H(x_{2} ,\theta ) H(x_{1} ,\theta )$$ to determine whether it is positive or negative. If $$H(x_{2} ,\theta ) H(x_{1} ,\theta ) < 0$$ the next estimate is given by $$x_{3} = (x_{ 2} +x_{ 1} )/2$$.

However, if $$H(x_{2} ,\theta ) H(x_{1} ,\theta ) > 0$$ the next estimate is given by $$x_{3} = (x_{ 2} +x_{ o} )/2$$. This procedure is repeated to achieve a converged value as indicated in Table $$\PageIndex{1}$$.

Table $$\PageIndex{1}$$: Converging Values for $$x = \beta \sqrt{h}/\alpha$$ at $$\theta = \beta^{2} t / 2\alpha = 0.10$$

Starting Values: $$x_{ o} = 0.90, \quad H(x_{ o} ) = +1.5026 \quad \text{ & } \quad x_{1} = 0.10, \quad H(x_{1} ) = -0.0946$$

$$n$$ $$x_n$$ $$H(x_n)$$
2 0.5000 0.09315
3 0.3000 – 0.04333
4 0.4000 0.01083
5 0.3500 – 0.01922
6 0.3750 – 0.00500
7 0.3875 0.00271
8 0.3812 – 0.00120
9 0.3844 0.00074
10 0.3828 – 0.00023
11 0.3836 0.00026
12 0.3832 0.00001
13 0.3830 – 0.00011
14 0.3831 – 0.00005
15 0.3832 – 0.00002
16 0.3832 0.00000

Given the solution for the dimensionless depth defined by Equation \ref{21}, and given the specified dimensionless time defined by Equation \ref{22}, we can determine that the fluid depth will be 0.435 meters at 11 hours and 9.3 minutes after the start time. The results tabulated in Table $$\PageIndex{1}$$ can be extended to a range of dimensionless times in order to produce a curve of $$x$$ versus $$\theta$$, and these results can be transformed to produce a curve of $$h$$ versus $$t$$ for any value of $$\alpha$$ and $$\beta$$.

In Examples $$\PageIndex{1}$$ and $$\PageIndex{2}$$ we have illustrated how one can develop solutions to transient macroscopic mass balances. For the system analyzed in Example $$\PageIndex{2}$$ an iterative method of solution was required to find the root of an implicit equation for the fluid depth. More information about iterative methods is provided in Appendix $$B$$.