Section 4.4.9: DAE Systems

The Flow Controller revisited

So far, ODEs only:

$\begin{displaymath}\frac{d{\bf x}}{dt} = {\bf f}({\bf x},t) \end{displaymath}$

E.g.

$\begin{displaymath}\frac{dM}{dt} = f(M) \end{displaymath}$

If f(M) involves non-trivial calculation of other variables we may be better to `open out' the problem and make all these variables explicitly part of the formulation.

E.g. flow controller:
f(M) = F₁ - F₂ where F₂ depends on M and k_v, the valve opening.
But k_v depends on F₂ because of the control algorithm.

DAE formulation of FC

Full set of equations in section 4.4.9.1

Instead of 1 ODE (mass balance) in 1 variable (M),
we have 12 equations in 12 variables.

1 equation is an ODE (the mass balance).
The other 11 are algebraic equations.
The 6 simple specification equations are included to show the full structure that is possible, though we might prefer to regard the corresponding variables as constants or explicit functions of time instead.

Terminology

General DAE system:

$\frac{d{\bf x}}{dt} = {\bf f}({\bf x,y},t)$ : M_D ODEs
${\bf g}({\bf x,y},t) = {\bf0}$ : M_A algebraic equations.

x: variables whose time derivatives appear in the model,
y: variables whose time derivatives do not appear in the model.

x are called differential variables,
y are called algebraic variables.

We have M_D differential and M_A algebraic variables.

For a DAE system to be easy to solve,
it must be possible to solve the AEs

$\begin{displaymath}{\bf g} = {\bf0} \end{displaymath}$

for y, if fixed values of x and t are given.

If not, `high-index' difficulties occur.
Will not be discussed in this course.
Solution of high-index problems remains an active research topic.
It's always best to make sure that a DAE system you formulate satisfies the above condition.

Flow Controller example

M is a differential variable because $\frac{dM}{dt}$ appears in the model.
The other 11 variables are algebraic variables: F₁, F₂, k, k_v, $\phi$ , k_v,max, y, K, e, F_s and F_range.

If M is given the AEs can be used to solve for the algebraic variables.

For example:

1.: Guess F₂.
2.: Calculate e, then y, then $\phi$ , then k_v; substituting the specification equations as required.
3.: Calculate the residual $F_2 - \sqrt{\frac{M}{1/k^2 + 1/k_v^2} }$
4.: Use a secant method (say) to get a new estimate for F₂.
5.: Repeat from 2. until converged.

This loop gives all the algebraic variables except F₁ which is easily calculated from its specification equation.

Explicit solution

Explicit Euler:
Discretise the ODE as

M(n+1) = M(n) + h ( F₁(n) - F₂(n) )

Needs values for all the algebraic variables at t_n.
Integration gives only the differential variable(s) at t_n+1.

Then solve the AEs for the algebraic variables at t_n+1 using the value calculated for M(n+1) by explicit Euler.

Solution is no different from treating the system as an ODE:

M(n+1) = M(n) + h f(n)

where $f(n) \equiv f(M(n))$ is calculated in exactly the same way as F₁(n) - F₂(n), using for example the iterative scheme on the previous slide.

Initial conditions

If the flow controller is treated as an ODE the initial condition is M(0) at time t₀.

This must remain true if the problem is treated as a DAE system.

General Rule for ICs:
One IC for each differential variable.

A safe choice is to give the initial value of each differential variable (as here). In other cases this may be inconvenient - see next section.

What about the algebraic variables?
Consistent initial values must be found by solving the AEs at time t₀ with specified values ${\bf x}(0)$ for the differential variables.
I.e. solve

$\begin{displaymath}{\bf x} - {\bf x}(0) = 0 \end{displaymath}$

$\begin{displaymath}{\bf g}({\bf x,y},t_0) = 0 \end{displaymath}$

This is equivalent to replacing the ODEs with IC specifications.

DO NOT GIVE INDEPENDENT
INITIAL CONDITIONS
FOR THE ALGEBRAIC VARIABLES.

However, initial guesses may be needed for the algebraic variables when solving the initial point equations above, e.g. by Newton's method.

Implicit solution

Implicit Euler:
Discretise the ODE as

M(n+1) = M(n) + h ( F₁(n+1)) - F₂(n+1) )

M(n+1), F₁(n+1) and F₂(n+1) are all unknown. So it makes sense to solve the discretised ODE simultaneously with the AEs at t_n+1.
The unknowns in the AEs are the differential and algebraic variables at time t_n+1:

$\begin{displaymath}{\bf x}(n+1), {\bf y}(n+1) \end{displaymath}$

This is different from the method used if we regard the system as just an ODE.

If the flow controller is treated as an ODE

$\begin{displaymath}\frac{dM}{dt} = f(M) \end{displaymath}$

and solved with implicit Euler, we must solve

M(n+1) = M(n) + h f(n+1)

for M(n+1), given that f is a function of M.

Solving for M(n+1) involves an iterative loop, if f is non-linear.
Each iteration involves calculation of f at an iterated value M^(k) for M(n+1).
But f = F₁ - F₂ and, in the flow control example, iterative solution for F₂ is required at a given M.
Thus we have an inner iterative loop for F₂,
and an outer iterative loop for M.

Which is better?

Nested loops each for one unknown,
or a single loop for 12 variables
(though a partitioning scheme can easily be devised which reduces the number of simultaneous unknowns in the implicit DAE solution to 2: M and F₂.)

An open question.
Probably depends on the problem.

Experimental results

The flow controller was solved for 200 time steps of 10 s, by which time (2000 s) steady state had been reached.

(Feed F₁ and initial holdup M₀ as in section 7 level control example).

Solving as a DAE needed 418 simultaneous linear equation solutions for the 12 variables, i.e. 418 Newton iterations of the full system.
(This could have been reduced to a 2 variable system with special programming.)

3 iterations per step to t=540 s, 2 iterations per step to 1610 s, 1 iteration per step thereafter.

Solving as an ODE took 464 outer loop iterations:
3 outer to 710 s, 2 outer to 1910 s, then 1 outer. All inner loops converged in 3 secant iterations.

Conclusion

A simple flow control problem was seen to be expressible either as a
single ODE or as a DAE system.
For explicit solution of the ODEs there is no practical difference in the solution for the two formulations.
For implicit solution of the ODEs
the ODE formulation tends to give nested iterative loops, but
the DAE formulation allows fully simultaneous solution.
There must be the same number of initial conditions as ODEs, and they may safely be given as initial values for the differential variables.

Next - Section 4.4.10: Towards Process Simulation
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Last modified: Wed Aug 5 16:41:48 BST