Solution steady-state

It will be assumed that a sufficiently long time has passed since the stress was first applied to allow steady-state conditions to be established. We can use techniques similar to those described in Sect. 3.11, except that the present case is easier in some respects. Consider first the case where the crack is closed. Let this be the first cycle of closure after t = 0, partly occupying the time interval [n/co,27t/co]. We expect 6 to be somewhat larger than n/co. Let us denote it by 0. Also, 62 will be in the vicinity of / = 0. We denote it by Og. For steady-state conditions. 

We now give more explicit expressions for these quantities in the case of a standard linear solid. Equation (3.11.12) gives that, for odd /, 

The conditions that q(dc) and q(0s) vanish take the form Aroe sin (co6c)= - - 

The condition determining the next time of opening, namely ds+A, has the form 

In a continuous wave (CW) magnetic resonance experiment, the radiation field B is continuous and BQ is changed only slowly compared with the relaxation rates (so-called slow passage conditions). Thus a steady-state solution to eqns 

Notice that as B approaches zero, u and v go to zero and Mz approaches M0, as expected. That is, it is the transverse oscillating field that causes the magnetization to have a non-equilibrium value. On the other hand, as B increases, Mz decreases (moves away from equilibrium) u and v at first increase with increasing B, but eventually they decrease as the third term in the denominator begins to dominate. 

When the microwave or radiofrequency power, proportional to B-f, is small so that y1B T T2 1, eqn (5.13b) becomes  

When the absorption is detected via small amplitude field modulation, the signal is proportional to the first derivative of absorption  

In first-derivative spectra, it is most convenient to describe the line width as the separation between derivative extrema. This width may be computed by taking the second derivative and finding the zeros, obtaining  

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There are tliree steps in the calculation first, solve the frill nonlinear set of hydrodynamic equations in the steady state, where the time derivatives of all quantities are zero second, linearize about the steady-state solutions third, postulate a non-equilibrium ensemble through a generalized fluctuation dissipation relation. 

A] = b/a (equation (A3.4.145)) is stationary and not [A ] itself This suggests d[A ]/dt < d[A]/dt as a more appropriate fomuilation of quasi-stationarity. Furthemiore, the general stationary state solution (equation (A3.4.144)) for the Lindemaim mechanism contams cases that are not usually retained in the Bodenstein quasi-steady-state solution. 

In analogy to equation Bl.5.3. we can write the steady-state solution to equation B 1.5.10 for tlie SFIG process as... 

The amplitude of the response, x 2ca), is given by the steady-state solution of equation B1.5.10 as... 

The steady-state solution without saturation to this equation is obtained by setting the time derivatives to zero and taking the tenns linear in as in equation (B2.4.11). 

Reactions in porous catalyst pellets are Invariably accompanied by thermal effects associated with the heat of reaction. Particularly In the case of exothermic reactions these may have a marked influence on the solutions, and hence on the effectiveness factor, leading to effectiveness factors greater than unity and, In certain circumstances, multiple steady state solutions with given boundary conditions [78]. These phenomena have attracted a great deal of interest and attention in recent years, and an excellent account of our present state of knowledge has been given by Arls [45]. 

To pursue this question we shall examine the stability of certain steady state solutions of Che above equaclons by the well known technique of linearized stability analysis, which gives a necessary (but noc sufficient) condition for the stability of Che steady state. 

Using different types of time-stepping techniques Zienkiewicz and Wu (1991) showed that equation set (3.5) generates naturally stable schemes for incompressible flows. This resolves the problem of mixed interpolation in the U-V-P formulations and schemes that utilise equal order shape functions for pressure and velocity components can be developed. Steady-state solutions are also obtainable from this scheme using iteration cycles. This may, however, increase computational cost of the solutions in comparison to direct simulation of steady-state problems. 

Adding time-dependent terms to the equations the simulation is treated as an initial value problem in which at a given reference time all stresses are zero the steady-state solution can be found iteratively. 

These equations possess steady state solutions... 

Langmuir [18] first proposed the axial dispersion model and obtained steady state solutions from die following boundary eonditions ... 

Danckwerts [2] also obtained steady state solutions based on the same boundary conditions and various studies have since been performed by Taylor [19], Arts [20], and Levenspiel and Smith [21],... 

One way to examine the validity of the steady-state approximation is to compare concentration—time curves calculated with exact solutions and with steady-state solutions. Figure 3-10 shows such a comparison for Scheme XIV and the parameters, ki = 0.01 s , k i = 1 s , 2 = 2 s . The period during which the concentration of the intermediate builds up from its initial value of zero to the quasi-steady-state when dcfjdt is vei small is called the pre-steady-state or transient stage in Fig. 3-10 this lasts for about 2 s. For the remainder of the reaction (over 500 s) the steady-state and exact solutions are in excellent agreement. Because the concen-... 

Figure 3-10. Comparison of steady-state and exact solutions for the concentration of intermediate B (relative to Ca) for Scheme XIV, where ki = O.OI s , k t = I s 2 = 2 s . The exact. solution was obtained with Eq. (3-87), the. steady-.state solution with Eq. (3-142), where Ca was calculated with Eq. (3-90).

If the rate of sweep through the resonance frequeney is small (so-called slow passage), a steady-state solution, in which the derivatives are set to zero, is ob-tained. The result expresses M,., and as funetions of cu. These magnetization components are not actually observed, however, and it is more useful to express the solutions in terms of the susceptibility, a complex quantity related to the magnetization. The solutions for the real (x ) and imaginary (x") components then are... 

The condition p2 a] < 1 has to be satisfied to assure nucleation growth, and if y is larger, the steady-state solution can be obtained ... 

The dependence of the in-phase and quadrature lock-in detected signals on the modulation frequency is considerably more complicated than for the case of monomolecular recombination. The steady state solution to this equation is straightforward, dN/dt = 0 Nss — fG/R, but there is not a general solution N(l) to the inhomogeneous differential equation. Furthermore, the generation rate will vary throughout the sample due to the Gaussian distribution of the pump intensity and absorption by the sample... 

Exercise Obtain the steady state solution by writing down analogous equations to (5-3) for c channels, setting the derivatives equal to zero, suppressing time as the argument and solving the resulting difference equations. 

We have given an analytical method of deriving a time-dependent solution to our problem that is complicated but illustrates an important method. Frequently, steady state solutions are all that is needed. 

In the Krylov-Bogoliubov version, this steady state solution is assumed in the form... 

In general Eq. (3.69) cannot be solved to give the time dependence of nv. However, a characteristic of this equation is that at very long times (t - oo) the solution becomes time-independent, that is to say a steady-state solution exists. Therefore at long times ... 

It is now useful to show how steady-state solutions can be obtained in several other cases. We shall also present two shortcuts to simplify the algebraic steps in the derivations for complicated situations. The first example is the S>j 1 mechanism for the substitution reaction of an organic halide RX by a nucleophile Y ... 

More expediently, one may use shortcuts. One shortcut allows the writing of the steady-state solution for [Int] according to the recipe... 

Before we discuss the merits of kss versus kpe, we shall consider one other approximate solution. Some authors refer to it as the improved steady-state solution. 2 We shall present two derivations. In the first, the approximation for d[ ]/dt is refined by differentiating the steady-state expression for [I]ss, which gives... 

This expression is called the improved steady-state solution, the first-order rate constant being given as... 

Although the reaction scheme of Eq. (5-56) has a closed-form steady-state solution, the rate law cannot be integrated under conditions like these, where the concentrations of A, B, and P vary during the experiment. In this case, numerical integration has proved its value. 

The next step should clarify why the unstable growth of the variable x occurs through a stable state at the bifurcation point. To determine the stability of the bifurcation point, it is necessary to examine the linear stability of the steady-state solution. For Eq. (1), the steady-state solution at the bifurcation point is given as jc0 = 0. So, let us examine whether the solution is stable for a small fluctuation c(/). Substituting Jt = b + Ax(f) into Eq. (1), and neglecting the higher order of smallness, it follows that... 

Figure 10. Adsorbed cation coverage as a function of electrode potential, assuming a cation interaction parameter / = 6.18 The solid line is the steady-state solution, whereas the broken line is the quasi-steady solution. Open circles indicate the unstable area. (From G. L. Griffin, J. Electrochettu Soc. 131, 18, 1984, Fig. 1. Reproduced by permission of The Electrochemical Society, Inc.)...

A steady-state solution of Eq. (3.30) exists only for < 2. At z > 2 hydrodynamic thermal explosion occurs and oscillatory flow takes place. 

The polymerization system for which experiments were performed is represented by the mathematical model consisting of Equations 1 and 7. Their steady state solutions are utilized for kinetic evaluation of rate constants. Dynamic simulations incorporate viscosity dependency. 

It should be noted that the steady-state solution of Equation (12) is not necessarily unique. This can easily be seen in the case of the four-reservoir system shown in Fig. 4-7. In the steady state all material will end up in the two accumulating reservoirs at the bottom. However, the distribution between these two reservoirs will... 

Even if satisfactory equations of state and constitutive equations can be developed for complex fluids, large-scale computation will still be required to predict flow fields and stress distributions in complex fluids in vessels with complicated geometries. A major obstacle is that even simple equations of state that have been proposed for fluids do not always converge to a solution. It is not known whether this difficulty stems from the oversimplified nature of the equatiorrs, from problems with ntrmerical mathematics, or from the absence of a lamirrar steady-state solution to the eqrratiorrs. 

This set of first-order ODEs is easier to solve than the algebraic equations where all the time derivatives are zero. The initial conditions are that a ut = no, bout = bo,... at t = 0. The long-time solution to these ODEs will satisfy Equations (4.1) provided that a steady-state solution exists and is accessible from the assumed initial conditions. There may be no steady state. Recall the chemical oscillators of Chapter 2. Stirred tank reactors can also exhibit oscillations or more complex behavior known as chaos. It is also possible that the reactor has multiple steady states, some of which are unstable. Multiple steady states are fairly common in stirred tank reactors when the reaction exotherm is large. The method of false transients will go to a steady state that is stable but may not be desirable. Stirred tank reactors sometimes have one steady state where there is no reaction and another steady state where the reaction runs away. Think of the reaction A B —> C. The stable steady states may give all A or all C, and a control system is needed to stabilize operation at a middle steady state that gives reasonable amounts of B. This situation arises mainly in nonisothermal systems and is discussed in Chapter 5. 

In this example, an initial steady-state solution with a = 0 is propagated downstream. At the fourth axial position, the concentration in one cell is increased to 16. This can represent round-off error, a numerical blunder, or the injection of a tracer. Whatever the cause, the magnitude of the upset decreases at downstream points and gradually spreads out due to diffusion in the y-direction. The total quantity of injected material (16 in this case) remains constant. This is how a real system is expected to behave. The solution technique conserves mass and is stable. 

Solution The reactions are the same as in Example 12.5. The steady-state performance of a CSTR is governed by algebraic equations, but time derivatives can be useful for finding the steady-state solution by the method of false transients. The governing equations are... 

Meiron (12) and Kessler et al. (13) have shown that numerical studies for small surface energy give indications of the loss-of-existence of the steady-state solutions. In these analyses numerical approximations to boundary integral forms of the freeboundary problem that are spliced to the parabolic shape far from the tip don t satisfy the symmetry condition at the cell tip when small values of the surface energy are introduced. The computed shapes near the tip show oscillations reminiscent of the eigensolution seen in the asymptotic analyses. Karma (14) has extended this analysis to a model for directional solidification in the absence of a temperature gradient. 

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