# 5.3.2.2: Deformable Control Volume

The left hand side of question (??) can be examined further to develop a simpler equation by using the extend Leibniz integral rule for a constant density and result in

$\dfrac{d}{dt} \int_{c.v.}\rho \,dV = \overbrace{\int_{c.v.} \overbrace{\dfrac{d\,\rho}{dt}}^{=0} \,dV }^{\mbox{thus, =0}} + \rho\,\int_{S_{c.v.}} \hat{n} \cdot U_{b}\, dA = \rho\,\int_{S_{c.v.}} U_{bn}\, dA \label{mass:eq:dmasst}$

where $$U_b$$ is the boundary velocity and $$U_{bn}$$ is the normal component of the boundary velocity.

$\label{mass:eq:cvmV} \int_{S_{c.v.}} U_{bn}\, dA = \int_{S_{c.v.}} U_{rn}\, dA$

The meaning of the equation qref{mass:eq:cvmV} is the net growth (or decrease) of the Control volume is by net volume flow into it. Example 5.2 illustrates this point.

Fig. 5.5 Filling of the bucket and choices of the deformable control volumes for example .

Example 5.2

Liquid fills a bucket as shown in Figure 5.5. The average velocity of the liquid at the exit of the filling pipe is $$U_p$$ and cross section of the pipe is $$A_p$$. The liquid fills a bucket with cross section area of $$A$$ and instantaneous height is $$h$$. Find the height as a function of the other parameters. Assume that the density is constant and at the boundary interface $$A_{j} = 0.7\,A_{p}$$. And where $$A_j$$ is the area of jet when touching the liquid boundary in bucket. The last assumption is result of the energy equation (with some influence of momentum equation). The relationship is function of the distance of the pipe from the boundary of the liquid. However, this effect can be neglected for this range which this problem. In reality, the ratio is determined by height of the pipe from the liquid surface in the bucket. Calculate the bucket liquid interface velocity.

Solution 5.2

This problem requires two deformable control volumes. The first control is around the jet and second is around the liquid in the bucket. In this analysis, several assumptions must be made. First, no liquid leaves the jet and enters the air. Second, the liquid in the bucket has a straight surface. This assumption is a strong assumption for certain conditions but it will be not discussed here since it is advance topic. Third, there are no evaporation or condensation processes. Fourth, the air effects are negligible. The control volume around the jet is deformable because the length of the jet shrinks with the time. The mass conservation of the liquid in the bucket is
\begin{align*}
\overbrace{\int_{c.v.} U_{bn} \, dA}^{\mbox{boundary change}} =
\overbrace{\int_{c.v.} U_{rn}\, dA}^{\mbox{flow in}}
\end{align*}
where $$U_{bn}$$ is the perpendicular component of velocity of the boundary. Substituting the known values for $$U_{rn}$$ results in
\begin{align*}
\int_{c.v.} U_{b}\, dA =
\int_{c.v.} \overbrace{ \left( U_{j} + U_{b}\right) }^{U_{rn}} dA
\end{align*}
The integration can be carried when the area of jet is assumed to be known as

$\label{bucket:a} U_{b}\, A = A_{j}\, \left(U_{j} + U_{b} \right)$

To find the jet velocity, $$U_j$$, the second control volume around the jet is used as the following

$\label{bucket:b} \overbrace{U_p \, A_p}^{\text{flow in }} - \overbrace{ A_{j} \left( U_{b} + U_{j} \right) }^{\text{flow out }} = \overbrace{- A_{j}\, U_{b} } ^{ \text{ boundary change} }$ The above two equations (??) and (??) are enough to solve for the two unknowns. Substituting the first equation, (??) into (??) and using the ratio of $$A_{j} = 0.7\,A_{p}$$ results

$\label{bucket:c} U_p \, A_p - U_b\,A = -0.7\,A_p\,U_b$ The solution of equation (??) is
\begin{align*}
U_b = \dfrac{A_p}{A- 0.7\,A_p}
\end{align*}
It is interesting that many individuals intuitively will suggest that the solution is $$U_b A_p/A$$. When examining solution there are two limits. The first limit is when $$A_p = A/0.7$$ which is
\begin{align*}
U_b = \dfrac{ A_p}{ 0} = \infty
\end{align*}
The physical meaning is that surface is filled instantly. The other limit is that and $$A_p/A \longrightarrow 0$$ then
\begin{align*}
U_b = \dfrac{ A_p}{ 0}
\end{align*}
which is the result for the "intuitive'' solution. It also interesting to point out that if the filling was from other surface (not the top surface), e.g. the side, the velocity will be $$U_{b}=U_{p}$$ in the limiting case and not infinity. The reason for this difference is that the liquid already fill the bucket and has not to move into bucket.

Example 5.3

Balloon is attached to a rigid supply in which is supplied by a constant the mass rate, $$m_i$$. Calculate the velocity of the balloon boundaries assuming constant density.

Solution 5.3

The applicable equation is
\begin{align*}
\int_{c.v.} U_{bn} \, dA =
\int_{c.v.} U_{rn}\, dA
\end{align*}
The entrance is fixed, thus the relative velocity, $$U_{rn}$$ is
\begin{align*}
U_{rn} =
\left\{
\begin{array}{ccc}
-U_p &@ \;\mbox{the valve} & \
0 & \mbox{every else} &
\end{array}
\right.
\end{align*}
Assume equal distribution of the velocity in balloon surface and that the center of the balloon is moving, thus the velocity has the following form
\begin{align*}
U_b = U_x\,\hat{x} + U_{br}\,\hat{r}
\end{align*}
Where $$\hat{x}$$ is unit coordinate in $$x$$ direction and $$U_x$$ is the velocity of the center and where $$\hat{r}$$ is unit coordinate in radius from the center of the balloon and $$U_{br}$$ is the velocity in that direction. The right side of equation (??) is the net change due to the boundary is
\begin{align*}
\int_{S_{c.v.}} \left( U_x\,\hat{x} + U_{br}\,\hat{r} \right) \cdot \hat{n}\, dA
= \overbrace{\int_{S_{c.v.}} \left( U_x\,\hat{x} \right) \cdot \hat{n}\, dA}^{\mbox{center movement}} +
\overbrace{\int_{S_{c.v.}} \left( U_{br}\,\hat{r} \right) \cdot \hat{n}\, dA}^{\mbox{net boundary change}}
\end{align*}
The first integral is zero because it is like movement of solid body and also yield this value mathematically (excises for mathematical oriented student). The second integral (notice $$\hat{n} = \hat{r}$$) yields
\begin{align*}
\int_{S_{c.v.}} \left( U_{br}\,\hat{r} \right) \cdot \hat{n}\, dA =
{4\,\pi\, r^2}\,{U_{br}}
\end{align*}
Substituting into the general equation yields
\begin{align*}
\rho\, \overbrace{4\,\pi\, r^2}^{A}{U_{br}} = \rho\,U_p\,A_p = m_i
\end{align*}
Hence,
\begin{align*}
U_{br} = \dfrac{m_i}{\rho\,4\,\pi\, r^2}
\end{align*}
The center velocity is (also) exactly $$U_{br}$$. The total velocity of boundary is
\begin{align*}
U_t = \dfrac{m_i}{\rho\,4\,\pi\, r^2} \left( \hat{x} + \hat{r}\right)
\end{align*}
It can be noticed that the velocity at the opposite to the connection to the rigid pipe which is double of the center velocity.