# 5.3.2.3: One–Dimensional Control Volume

Additional simplification of the continuity equation is of one dimensional flow. This simplification provides very useful description for many fluid flow phenomena. The main assumption made in this model is that the proprieties in the across section are only function of $$x$$ coordinate. This assumptions leads
$\label{mass:eq:1Dim} \int_{A_2} \rho_2\,U_2\, dA - \int_{A_1} \rho_1\,U_1\, dA = \dfrac{d}{dt} \int_{V(x)} \rho(x)\,\overbrace{A(x) \, dx}^{dV} \tag{20}$
When the density can be considered constant equation (??) is reduced to
$\label{mass:eq:1DimMay} \int_{A_2} U_2\, dA - \int_{A_1} U_1\, dA = \dfrac{d}{dt} \int A(x) dx \tag{21}$
For steady state but with variations of the velocity and variation of the density reduces equation (??) to become
$\label{mass:eq:1DimSteadyState} \int_{A_2} \rho_2\,U_2\,dA = \int_{A_1} \rho_1\,U_1\,dA \tag{22}$
For steady state and uniform density and velocity equation (??) reduces further to
$\label{mass:eq:1DconstantRho} \rho_1\,A_1\,U_1 = \rho_2\,A_2\,U_2 \tag{23}$
For incompressible flow (constant density), continuity equation is at its minimum form of
$\label{mass:eq:1DconstantRhoIncompressible} U_1\,A_1 = A_2\,U_2 \tag{24}$
The next example is of semi one–dimensional example to illustrate equation (??).

Fig. 5.6 Height of the liquid for example.

Example 5.4

Liquid flows into tank in a constant mass flow rate of $$a$$. The mass flow rate out is function of the height. First assume that $$q_{out}=b \,h$$ second Assume as $$q_{out} = b \,\sqrt{h}$$. For the first case, determine the height, $$h$$ as function of the time. Is there a critical value and then if exist find the critical value of the system parameters. Assume that the height at time zero is $$h_0$$. What happen if the $$h_{0}=0$$?

Solution 5.4

The control volume for both cases is the same and it is around the liquid in the tank. It can be noticed that control volume satisfy the demand of one dimensional since the flow is only function of $$x$$ coordinate. For case one the right hand side term in equation (??) is
\begin{align*}
\rho \dfrac{d}{dt} \int_0^L h\,dx = \rho \, L \dfrac{d\,h}{dt}
\end{align*}
Substituting into equation equation (??) is
\begin{align*}
\rho\,L\,\dfrac{d\,h}{d\,t} = \overbrace{b_1\,h}^{\mbox{flow out}} -  \overbrace{m_i}^{\mbox{flow in}}
\end{align*}
solution is
\begin{align*}
h= \overbrace{\dfrac{m_i}{b_1}\,\text{ e}^{-\dfrac{b_1\,t}{\rho\,L} }}^{\text{homogeneous solution}} +
\overbrace{c_1\,\text{e}^{\dfrac{b_1\,t}{\rho\,L}}}^{\text{ private solution}}
\end{align*}
The solution has the homogeneous solution (solution without the $$m_i$$) and the solution of the $$m_i$$ part. The solution can rearranged to a new form (a discussion why this form is preferred will be provided in dimensional chapter).
\begin{align*}
\dfrac{ h\, b_1}{m_1} = \text{e}^{-\dfrac{b_1\,t}{\rho\,L} }    +
c\,\text{ e}^{\dfrac{b_1\,t}{\rho\,L}}
\end{align*}
With the initial condition that at $$h(t=0) = h_0$$the constant coefficient can be found as
\begin{align*}
\dfrac{h_0\,b_1}{m_1} = 1 - c \Longrightarrow c =  1 - \dfrac{h_0\,b_1}{m_i}
\end{align*}
which the solution is
\begin{align*}
\dfrac{ h\, b_1}{m_1} = \text{ e}^{-\dfrac{b_1\,t}{\rho\,L} }  +
\left[ 1 - \dfrac{h_0\,b_1}{m_i\dfrac{}{}}\right]\,\text{ e}^{\dfrac{b_1\,t}{\rho\,L}}
\end{align*}
It can be observed that if $$1 = \dfrac{h_0\,b_1}{m_i}$$ is the critical point of this solution. If the term $$\dfrac{h_0\,b_1}{m_i}$$ is larger than one then the solution reduced to a negative number. However, negative number for height is not possible and the height solution approach zero. If the reverse case appeared, the height will increase. Essentially, the critical ratio state if the flow in is larger or lower than the flow out determine the condition of the height. For second case, the governing equation (??) is
\begin{align*}
\rho\,L\,\dfrac{d\,h}{d\,t} = \overbrace{b\,\sqrt{h}}^{\text{flow out}} -  \overbrace{m_i}^{\text{flow in}}
\end{align*}
with the general solution of
\begin{align*}
\ln\left[ \left( \dfrac{\sqrt{h}\,b}{m_i\dfrac{}{}} - 1 \right) \dfrac{m_i}{\rho\,L} \right]
+  \dfrac{\sqrt{h}\,b}{m_i}  - 1 = \left(t + c \right) \dfrac{\sqrt{h}\,b}{2\,\rho\,L}
\end{align*}
The constant is obtained when the initial condition that at $$h_{0}=0$$ and it left as exercise for the reader.

### Contributors

• Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.