Summary of Week 2

1. Some Examples of Process Modeling

- Mass Balance:
Rate of Change of Mass in System =
    Mass Flow Rate In - Mass Flow Rate Out
- Component Balance:
Rate of Change of Component j in System =
    Flow of Component j In - Flow of Component j Out
    + Rate of Production of Component j
    -  Rate of Depletion of Component j
- Energy Balance:
Rate of Change in Enthalpy in System =
    Bulk Flow of Enthalpy In - Bulk Flow of Enthalpy Out
    + Rate of Heat Generated
    + Rate of Heat Transferred to Surroundings
    ( - Rate of Shaft Work on surroundings )


2. Solution of Linear ODEs with Constant Coefficients ( a quick review )

General Form:
Solution:
x = xc+x p
where xc is the complementary solution and xp is the particular solution.
In solving for the complementary solution, the procedure involves obtaining the characteristic equation:
whose n roots are the eigenvalues of the system.
3. Mulitiple Linear ODE System
 
Main Idea: First reduce the system to a high order differential equation involving only one state variable, say x1. Then, solve this equation for x1(t).  Use this result to reduce the number of state variables to be solved for and repeat the process for x2(t), etc.

For the special case of two simultaneous system,

we have
4. Analysis of Process using Models
 
a) Steady States: fixed values of process variables (states) as t approaches infinity
- Given x(t):  then xss=lim t ® ¥  x(t)
- Altenatively, if given the ODEs,
i. Set all time derivatives to zero
ii. Replace the state variables, say x, by the steady state, xss
iii. Solve xss from the resulting equation and take limit as t approaches infinity
Note: For some nonlinear systems, there can be multiple steady states !
b) Dynamic behavior of linear systems with constant coefficients
5. Some special parametric forms of first and second order systems:
 
First order systems:   t dx/dt + x = K
    where t is the time constant.

    [ x( t=t ) - x(0) ] / [ xss - x(0) ] = 1 - exp(-t/t ) = 0.632
 

Second order systems:  tn2 d2x/dt2 + 2 z tn dx/dt + x = K
where  1/tn = natural frequency
            z = damping coefficient