A perfectly stirred, constant-volume tank has two input streams, both consisting of the same liquid. The temperature and flow rate of each of the streams can vary with time. Tl Stream 1 ------,1 wl '+' T2 Stream 2 w2 T3 Stream 3 w3 Figure E2.1 (a) Derive a dynamic model that will describe transient operation. Make a degrees of freedom analysis assuming that both Streams 1 and 2 come from upstream units (i.e., their flow rates and temperatures are known functions of time). (b) Simplify your model, if possible, to one or more differential equations by eliminating any algebraic equations. Also, simplify any derivatives of products of variables. Notes: w; denotes mass flow rate for stream i. Liquid properties are constant (not functions of temperature).
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Textbook Solutions for Process Dynamics and Control
Question
A perfectly stirred, constant-volume tank has two input streams, both consisting of the same liquid. The temperature and flow rate of each of the streams can vary with time.
(a) Derive a dynamic model that will describe transient operation. Make a degrees of freedom analysis assuming that both Streams 1 and 2 come from upstream units (i.e., their flow rates and temperatures are known functions of time).
(b) Simplify your model, if possible, to one or more differential equations by eliminating any algebraic equations. Also, simplify any derivatives of products of variables.
Notes:
\(w_i\); denotes mass flow rate for stream i. Liquid properties are constant (not functions of temperature).
Solution
Step 1 of 3
The number of the consider variables are \(w_{1}, w_{2}, T_{1}, T_{2}, T_{3}\) and density \(\rho\) and constant volume V.
The number of the process variables, \(N_{v}=5\)
The number of the independent equation, \(N_{E}=1\)
The degree of the freedom of the system can be calculated as,
\(\begin{array}{l}
N_{F}=N_{V}-N_{E} \\
N_{F}=5-1 \\
N_{F}=4
\end{array}\)
Write the mass balance equation for overall tank,
\(\begin{aligned}
\frac{d(m)}{d t} & =w_{1}+w_{2}-w_{3} \\
\frac{d(\rho V)}{d t} & =w_{1}+w_{2}-w_{3} \\
0 & =w_{1}+w_{2}-w_{3} \\
w_{3} & =w_{1}+w_{2}
\end{aligned}\)
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