Mixing Tank Example
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## 1 Introduction to the Mixing Tank Example

Figure 1: Process Scheme

This article was contributed by Asim Vodencarevic (see the forum topic). I've

There is one mixing tank and two inlet valves. Control valves can control input flow of hot and cold water. There is one outlet pump. The objective is to control both temperature and level.

## 2 Model of the Mixing Tank

We need to write differential equations which describe the dynamics of the system.

### 2.1 Water Level

The differential equation for level is a matter of mass balance.

 $LaTeX: input-output=accumulation$

 $LaTeX: \frac{dV}{dt}=F_{in}-F_{out}$

 $LaTeX: A\frac{dL}{dt}=F_{1}+F_{2}-F_{3}$

where

$LaTeX: A$ is the tank cross-sectional area and
$LaTeX: L$ is the tank level.

 $LaTeX: A\frac{dL}{dt}=\Chi_{hot}\nu_{hot}\left(t\right)+\Chi_{cold}\nu_{cold}\left(t\right)-\Delta_{out}\left(t\right)$

where

$LaTeX: \Chi$ is the Pump Capacity,
$LaTeX: \nu$ is the fraction of full capacity (can be getween [0 1]), and
$LaTeX: \Delta_{out}$ is tank output.

Finally this can be reduced to

 $LaTeX: \frac{dL}{dt}=\frac{1}{A}\left(F_{1}+F_{2}-F_{3}\right)$ Eqn. 1

### 2.2 Water Temperature

 $LaTeX: \frac{d\left(VT\right)}{dt}=F_{1}T_{1}+F_{2}T_{2}-F_{3}T$

 $LaTeX: A\frac{dL}{dt}T+AL\frac{dT}{dt}=F_{1}T_{1}+F_{2}T_{2}-F_{3}T$

There is only one fluid (water), so density and specific heat are the same in all terms. Now if we combine the last equation with the level equation (Eqn. 1)

 $LaTeX: \left(F_{1}+F_{2}-F_{3}\right)T+AL\frac{dT}{dt}=F_{1}T_{1}+F_{2}T_{2}-F_{3}T$

This yields

 $LaTeX: AL\frac{dT}{dt}=F_{1}\left(T_{1}-T\right)+F_{2}\left(T_{2}-T\right)$ Eqn. 2

 $LaTeX: \frac{dL}{dt}=\frac{1}{A}\left(F_{1}+F_{2}-F_{3}\right)$ Eqn. 1

 $LaTeX: \frac{dT}{dt}=\frac{F_{1}}{AL}\left(T_{1}-T\right)+\frac{F_{2}}{AL}\left(T_{2}-T\right)$ Eqn. 3

Table 1: Mixing Tank, Nominal Operational Parameters
Parameters Value Units

$LaTeX: \Chi_{hot}=\Chi_{cold}$

0.0042

$LaTeX: \frac{m^3}{s}$

$LaTeX: \nu_{hot}\left(0\right)$

0.4

N/A

$LaTeX: \nu_{cold}\left(0\right)$

0.5

N/A

$LaTeX: L_{final}$

1

m

$LaTeX: T_{final}$

40

deg C

$LaTeX: \Delta_{out}$

0.00378

$LaTeX: \frac{m^3}{s}$

$LaTeX: A$

1

$LaTeX: m^2$

 $LaTeX: \frac{dL}{dt}=\frac{1}{A}\left(F_{1}+F_{2}-F_{3}\right)$ Eqn. 1
 $LaTeX: \frac{dT}{dt}=\frac{F_{1}}{AL}\left(T_{1}-T\right)+\frac{F_{2}}{AL}\left(T_{2}-T\right)$ Eqn. 3