Chapter 2, Problem 2.1E

Interpretation Introduction

(a)

Interpretation:

The dynamic model of the given process is to be described along with the degree of freedom analysis.

Concept introduction:

For chemical processes, dynamic models consisting of ordinary differential equations are derived through unsteady-state conservation laws. These laws generally include mass and energy balances.

The process models generally include algebraic relationships which commence from thermodynamics, transport phenomena, chemical kinetics, and physical properties of the processes.

The degree of freedom analysis of a process model ensures that its model equations are solvable. The expression to calculate the degree of freedom is:

$N_F=N_V-N_E$ ..... (1)

Here, $N_F$ is the number of degree of freedom for the process model, $N_V$ is the total number of process variables, and $N_E$ is the total number of independent equations written for the process model.

Answer to Problem 2.1E

The dynamic model for the given process is:

$\left(\rho \overline{V}\right)\frac{d\left(T_3-T_{ref}\right)}{dt}+\left(w_1+w_2\right)T_3=w_1{T}_1+w_2{T}_2$

Degree of analysis results in 4 degrees of freedom for the given process.

Explanation of Solution

Given information:

A process model containing two input streams of the same liquid is stirred perfectly in a constant-volume tank. For each of the streams, the temperature and flowrate vary with time.

Assume that both the input streams come from the upstream units and their flowrates and temperature are known as the function of time.

For the given tank, apply the degree of freedom analysis. The parameters taken for the given process are the density $\left(\rho \right)$ of the liquid and the constant volume $\left(\overline{V}\right)$ of the tank.

Variables taken for the given process model are $w_1,w_2,T_1,T_2\text{, and }{T}_3$ according to the conditions specified. Thus, $N_V=5$.

For the given process model, there will be 1 independent equation for the energy balance. Thus, $N_E=1$.

Use equation (1) to calculate the degree of freedom as:

$\begin{array}{c}{N}_F=N_V-N_E\\ =5-1\\ =4\end{array}$

Therefore, the degree of freedom analysis results in 4 degree of freedom for the given process model.

For the tank system given, write the overall mass balance equation as:

$\frac{d\left(m\right)}{dt}=w_1+w_2-w_3$ ..... (2)

Here, $w_i$ is the mass flowrate of the stream $i$ and $m$ is the mass accumulated in the tank. Mass accumulated in the tank can be written as:

$m=\rho \overline{V}$ ..... (3)

Substitute above equation in equation (2) as:

$\frac{d\left(\rho \overline{V}\right)}{dt}=w_1+w_2-w_3$

It is assumed that the tank is a constant-volume system and its density does not vary with time and temperature, thus,

$\begin{array}{c}\frac{d\left(\rho \overline{V}\right)}{dt}=w_1+w_2-w_3\\ 0=w_1+w_2-w_3\\ w_3=w_1+w_2\end{array}$

Therefore, the mass balance on the given process model gives,

$w_3=w_1+w_2$ ..... (4)

Now, apply energy balance and write its equation as the total energy of the tank is conserved.

$mC\frac{d\left(T_3-T_{ref}\right)}{dt}=C{w}_1\left(T_1-T_{ref}\right)+C{w}_2\left(T_2-T_{ref}\right)-C{w}_3\left(T_3-T_{ref}\right)$

Substitute equation (3) in above equation and simplify as:

$\begin{array}{c}\left(\rho \overline{V}\right)C\frac{d\left(T_3-T_{ref}\right)}{dt}=C{w}_1\left(T_1-T_{ref}\right)+C{w}_2\left(T_2-T_{ref}\right)-C{w}_3\left(T_3-T_{ref}\right)\\ \left(\rho \overline{V}\right)\frac{d\left(T_3-T_{ref}\right)}{dt}=w_1{T}_1-w_1{T}_{ref}+w_2{T}_2-w_2{T}_{ref}-w_3{T}_3+w_3{T}_{ref}\\ \left(\rho \overline{V}\right)\frac{d\left(T_3-T_{ref}\right)}{dt}=w_1{T}_1+w_2{T}_2-w_3{T}_3+w_3{T}_{ref}-\left(w_1+w_2\right)T_{ref}\end{array}$

Substitute equation (4) in the above equation and simplify the expression to get the dynamic model of the given process as:

$\begin{array}{c}\left(\rho \overline{V}\right)\frac{d\left(T_3-T_{ref}\right)}{dt}=w_1{T}_1+w_2{T}_2-\left(w_1+w_2\right)T_3+\left(w_1+w_2\right)T_{ref}-\left(w_1+w_2\right)T_{ref}\\ \boxed{\left(\rho \overline{V}\right)\frac{d\left(T_3-T_{ref}\right)}{dt}+\left(w_1+w_2\right)T_3=w_1{T}_1+w_2{T}_2}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The derived dynamic model of the system is to be simplified by eliminating any algebraic equations.

Concept introduction:

For chemical processes, dynamic models consisting of ordinary differential equations are derived through unsteady-state conservation laws. These laws generally include mass and energy balances.

The process models generally include algebraic relationships which commence from thermodynamics, transport phenomena, chemical kinetics, and physical properties of the processes.

Answer to Problem 2.1E

The simplified dynamic model of the given system is:

$\left(\rho \overline{V}\right)\frac{d\left(T_3\right)}{dt}+\left(w_1+w_2\right)T_3=w_1{T}_1+w_2{T}_2$

Explanation of Solution

Given information:

A process model containing two input streams of the same liquid is stirred perfectly in a constant-volume tank. For each of the streams, the temperature and flowrate vary with time.

Assume that both the input streams come from the upstream units and their flowrates and temperature are known as the function of time.

From part (a), the dynamic model of the given process is derived as:

$\left(\rho \overline{V}\right)\frac{d\left(T_3-T_{ref}\right)}{dt}+\left(w_1+w_2\right)T_3=w_1{T}_1+w_2{T}_2$

Now, let the reference temperature for the system is kept constant. So, $\frac{d{T}_{ref}}{dt}=0$. Thus,

$\begin{array}{c}\left(\rho \overline{V}\right)\frac{d\left(T_3\right)}{dt}-\left(\rho \overline{V}\right)\frac{d\left(T_{ref}\right)}{dt}+\left(w_1+w_2\right)T_3=w_1{T}_1+w_2{T}_2\\ \left(\rho \overline{V}\right)\frac{d\left(T_3\right)}{dt}-\left(\rho \overline{V}\right)\left(0\right)+\left(w_1+w_2\right)T_3=w_1{T}_1+w_2{T}_2\\ \left(\rho \overline{V}\right)\frac{d\left(T_3\right)}{dt}+\left(w_1+w_2\right)T_3=w_1{T}_1+w_2{T}_2\end{array}$