# Percentage change in volume of a fluid parcel

1. Jan 24, 2010

### ColdFusion85

1. The problem statement, all variables and given/known data

The velocity components in a plane (i.e., 2-D) flow are measured at four points as indicated in the sketch at the bottom of this page (see attached doc. I wrote in the coords). The velocity components at the respective points are, in units of cm/sec:

$$u_a=5, v_a=2, u_b=7, v_b=3, u_c=6.5, v_c=3.5, u_d=5.5, v_d=1.5$$

a) Estimate the percentage change in volume of a fluid parcel at the central point o.
b), c) no questions thus far on my part

2. Relevant equations

divergence of V = 0 for incompressible fluid (assumption made in problem since we haven't done compressible flow yet)

3. The attempt at a solution

So, I figured that since the divergence of V = partial(u)/partial(x) + partial (v)/partial(y), the partial(v)/partial(x) and partial(u)/partial(y) components don't need to be calculated (i.e., v_a, v_b, u_c, u_d are not relevant). Is this a correct assumption? If so, I took

(u_b-u_a)/2(dx)
= (7-5)/2(3) = (1/3)

and (v_d-v_c)/2(dy)
= (1.5-3.5)/2(2) = (1/2)

so div(V) = (1/3)-(1/2)
= -(1/6)

Thus, the total change in volume at the central point is (-1/6).

Does this make sense, or am I approaching the problem incorrectly?

Part b) asks whether this flow satisfies conservation of mass for an incompressible fluid. I would think that since there is a net change in volume, that it wouldn't satisfy it. Is this correct?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Jan 25, 2010

### ColdFusion85

Can anyone offer some help?

3. Jan 26, 2010

### ColdFusion85

One more plea for help?

4. Jan 28, 2010

### minger

Does this help. It's copied from notes of mine written in tex, so pardon the inline math type.

The divergence of velocity is written as:
$$\label{div01} \vec{\nabla}\cdot\vec{V}\equiv \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}$$
This term shows up often in the governing equations. It essentially tells us the rate of change of the volume. See Figure ???? using the following notation:
$$\begin{split} &V = \mbox{Control Volume} \\ &S = \mbox{Control Surface} \\ &dS = \mbox{Infinitesimal Control Surface Element} \\ &\vec{V} = \mbox{Flow Velocity at }ds \\ &\vec{n} = \mbox{Unit Normal Vector pointing \emph{out} of }V \end{split} \nonumber$$
We consider a control volume $V$ moving with the fluid (Lagrangian approach). Since it's moving with the fluid, no mass crosses the boundary $S$ as the fluid moves. Because of this, the total mass in the control volume is fixed, and invariant with time. However, the volume $V$ and its surface $S$ change in time, as the volume moves with the fluid. Consequently, the density $\rho$ \emph{can} change with time.

We will now further examine the infinitesimal control surface element $ds$. The change in the volume of the control volume, $\Delta V$, due to the movement of $dS$ over a time $\Delta t$ is equal to the long thin element with a base area $dS$ and altitude $(\vec{V}\Delta t\cdot\vec{n})$ as shown in Figure ????. We can write:
$$\label{div002} \Delta V = \left[ (\vec{V}\Delta t)\cdot\vec{n}\right] dS$$
We write $d\vec{S} = dS\vec{n}$
$$\label{div003} (\vec{V}\Delta t)\cdot d\vec{S}$$
Over the time increment $\Delta t$, the total change in volume $V$, of the entire element is the summation of all the changes from all of the infinitesimal elements, $dS$. We take the limit as $dS$ approaches zero and write the total change in volume:
$$\label{div003} \Delta V = \iint_S (\vec{V}\Delta t)\cdot d\vec{S}$$
We divide by $\Delta t$ to obtain the time rate of change of the control volume $V$ or how fast its changing. Do note that because the element is moving (recall Lagrangian approach), we \emph{must} use the total derivative.
$$\label{div004} \frac{DV}{Dt} = \frac{1}{\Delta t} \iint_S (\vec{V}\Delta t)\cdot d\vec{S} = \iint_S \vec{V}\cdot s\vec{S}$$
We are allowed to pull the $\Delta t$ out because it doesn't change with a change in control surface $S$.

At this point we need to recall Divergence Theorem (Gauss's Theorem) from vector calculus. This theorem states that the outward flux of a vector field through a surface is equal to the triple integral of the divergence on the region inside the surface. Simply put, it states that the sum of all sources minus the sum of all sinks gives the ne flow out of a region. In equation form:
$$\label{divergencetheorem} \iiint_V (\Delta \cdot \vec{F})dV = \iint_{\delta V} \vec{F}\cdot \vec{n}dS$$

If we now apply the Divergence Theorem, Equation \ref{divergencetheorem}, we get:
$$\label{div005} \frac{DV}{Dt} = \iint_S \vec{V}\cdot d\vec{S} \equiv \iiint_V (\vec{\nabla}\cdot\vec{V})dV$$
We then shrink the control volume $V$ to an infinitesimal control volume $\delta V$.
$$\label{div006} \frac{D(\delta V)}{Dt} = \iiint_{\delta V} (\vec{\nabla}\cdot\vec{V})dV$$
If we can then assume that $\delta V$ is small enough that the term $(\vec{\nabla}\cdot\vec{V})$ is essentially the same value throughout $\delta V$, then we can write:
$$\label{div007} \frac{D(\delta V)}{Dt} = \iiint_{\delta V} { (\vec{\nabla}\cdot\vec{V}) } dV = (\vec{\nabla}\cdot\vec{V})\delta V$$
Or:
$$\label{div008} \vec{\nabla}\cdot\vec{V} = \frac{1}{\delta V}\frac{D(\delta V)}{Dt}$$
The divergence of the velocity is \emph{physically} the time rate of change of the volume of a \emph{moving} fluid element, per unit volume.

5. Jan 28, 2010

### Redbelly98

Staff Emeritus
You are on the right track, but there is a problem in your interpretation of the answer, -1/6.

What are the units associated with this number? If it's not obvious, review how you calculated it, and the units of the numbers that went into the calculation.

Also, recall that the question asks for a percentage change.