Lecture Notes

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Lecture Number 3

Date: 9/23/98

We start with the basic compartmental equation:

q /V(c_A,i - c_A) + R_A + N_A,DA_D/V = dc_A/dt (molar form)

where q is the convective flowrate, V is the compartment volume, c_A,I is the inlet concentration, c_A is the concentration of A in the compartment, N_A,D is the inward flux of A, and A_D is the area of the compartment through which this flux occurs. Notice some assumptions: the influent concentration is spatially invariant, as it would be coming from another compartment. If it isn't you need to use the flow-average (see exercises) so that the q-c product equals the rate of inflow of A. Similarly the influx N_A,D is taken to be spatially invariant so that the total diffusional inflow is the N-A_D product.

You should be able to write the mass equivalent of this equation easily.

You should know how to derive this equation by applying the divergence theorem to the more general conservation equation that allows spatial gradients. Then you should know how to divide the flux into the purely convective part that goes through part of the surface of the compartment (leading to the q-c term) and the purely diffusive part, leading to the N-A term. In actual problems there must be a way of determining, or assigning a value to, A_D. It could be the whole compartment area or some part of it. The equation is set up for the usual situation: a time invariant volume compartment. If we don't divide the original equation through by V, as we did, we can treat V as a variable, but this will give rise to either a linear differential equation with variable coefficients or a non-linear differential equation, complications that we want to avoid for the time being. Likewise, unless specified otherwise, q will be treated as time invariant.

Some simple applications (all steady state in this group):

1. No reaction, no diffusion: q(c_i - c) equals zero. The trivial result is that c = c_i.
2. Reaction, no diffusion. Balance the difference between inflow and outflow against the rate of production (positive or negative) of A:

q /V(c_A,i - c_A) + R_A = 0

c_A will be higher than c_A,I if A is being made. Notice that the actual rise in concentration will depend on the q/V ratio, not on q, alone, which reduces the concentration rise when q is large, nor on V alone, which determines the overall reaction rate in the compartment.

3. You can formulate other expressions that balance reaction against diffusion, no inflow or outflow. Or flow against diffusion, no reaction. Each of these involves selectively setting one of the steady-state phenomena in the basic equation to zero as a pedagogic way of seeing how variables interdepend. Real problems can involve everything at once, but usually one of these interactions will be the most important, i.e. one of the steady-state cwill be less important than the other two.

Some simple applications that do not depend on the steady-state assumption:

4. Expanding a volume: Write the equation above for the total molar (or mass) density with no diffusion and no reaction and the inflow not equal to the outflow:

c(q_i.- q) =c dV/dt (constant molar density, and the c can be cancelled)

The compartment expands and the volume is the integral of the difference between the inflow and the outflow. Sometimes will call such a compartment an "integrator".

5. Step change in inflow concentration, at t = 0, to a compartment with time-invariant flow and volume, no diffusion, no reaction:

q /V(c_A,i - c_A) = dc_A/dt

We want to find c_A(t). We need an in initial condition for c_A. If the system had been fed steadily at an initial concentration c_A,i = c₀, the initial value of c will be c₀. If that concentration now rises to, let's say, c*, it is easy to show that

(c* - c(t))/(c* - c₀) = exp(-[q /V] t)

This expression can be solved explicitly for c(t):

c(t) = c* - (c* - c₀) exp(-[q /V] t) = c₀ + (c*-c₀)[1 - exp(-[q /V] t]

We can think of this result as the sum of two terms, c₀, a constant, and the time-variant second term, the "swing" from c₀ to c*. The system is linear. The result is the sum of the results that we would obtain by feeding the system at the steady concentration c₀ and that which we would obtain by feeding it at c = 0 until t = 0 and then at a concentration c = (c* - c₀).

6. Same problem but a step change in inflow concentration at t = 0 that is sustained only for T time units.

The solution to the problem just discussed can be generalized a bit if we ask how we would calculate the concentration vs. time profile if the step change in concentration persisted only between t = 0 and t = T. For this case we can imagine the response to be the sum of the individual responses to inputs of c₀ over all time, plus that to c = (c* - c₀) from t = 0 for all positive time, plus that to c = - (c* - c₀) for all time greater than T. The latter solution is:

c(t) = - (c* - c₀) [1 - exp(-[q /V] [t - T])], t > T, c(t) = 0, t < T

Thus the overall solution is:

C(t) = c₀, t < 0,

c(t) = c₀ + (c*-c₀) [1 - exp(-[q /V] t], 0<t<T, and

c(t) = c₀ + (c*-c₀) [1 - exp(-[q /V] t] - (c* - c₀) [1 - exp(-[q /V] [t - T])], t > t

We have implicitly used the multiplicity property of linear systems: the solution for a step of size +(c* - c₀) if multiplied by (here) -1 is the solution for a step of size - (c* - c₀). In general the solution for a step n times that of another step will be n times greater, and n can be either positive or negative. We have also used the fact that in a stationary linear system (one whose properties or parameters do not change with time), if one time-shifts (here, by T time units), the response will be the same response except that it is shifted by the same number of time units.

7. Same problem but only one input, c_i = C sin (2 pi t/T) over all time. (Sinusoidal steady-state response.)

In this problem we look at the system behavior to a steady, periodic variation in c_i. T is the period of the sinusoidal variation. It is convenient to substitute the angular frequency, w for 2 pi / T. We assume a solution, c = C' sin (w t + phi). It is easy to show that C'/C (the amplitude ratio of the responding sinusoid to that of the forcing sinusoid) is: 1/sqrt ( 1 + [wV/q] ²). The phase angle, phi, is - wV/q. What does this mean? The quantity V/q is the time constant or characteristic time, of the flow system. (Look at the solutions above, the speed of response of the system is inversely proportional to V/q.) When the system is disturbed by a slow moving sinusoid (wV/q << 1), the amplitude ratio is nearly unity and the system "tracks" closely follows the waveform of the excitation. The phase angle is a small negative number which implies that the system lags only a small distance behind the excitation waveform, another sign that the system closely tracks the excitation waveform. This type of response is often called quasistatic. At each moment the responding variable, c, has a value close to the value it would have if the value of c_i that exists then had existed for all previous time. The changes are so slow (small w, large T) relative to the response time of the system (V/q), that the system cannot "remember" previous conditions, only the condition existing at the moment.

Linear systems have three properties of importance for us: additivity, multiplicity, and frequency orthogonality. Additivity means that if the system response to an input c_i¹ (t) is c¹ (t) and the system response to another input c_i²(t) is c² (t), then the response to c_i¹ (t) + c_i²(t) will be c¹ (t) + c² (t). Multiplicity means that if the system response to an input c_i¹ (t) is c¹ (t) then the system response to an input n*[ c_i¹ (t) ] is n*[ c¹ (t) ]. Frequency orthogonality means that the system response to an arbitrary, pure sinusoid C sin (wt + phi ) will be a pure sinusoid of the same frequency, C' sin (wt + phi' ). Only the amplitude and phase angle will be different between the forcing and responding variables. There will be no harmonics or subharmonics.

Same problem but c is affected by changes in both c_Ai and R_A. In this case, the concentration in the compartment can change because of two independent inputs. A step change in R_A will cause an exponential change from some initial concentration c₀ with the same characteristic time (V/q) as a step change in c_A,i. The only unknown about the transient is its amplitude and that is a steady state calculation to find out what the ultimate new concentration will be. If we call that c_A**, then the steady-state concentration gives: c_A** = c₀ + V R_A / q. the concentration transient following a change in R_A at some time T will be, simply:

c₀ + (c** - c₀) [1-exp(-[q/V] [t - T])]

and if there is a concentration transient due to a change in c_A,i it is simply added to the R_A - driven change, just given.

Not much biology, granted, but that's about what you need to know about linearity in compartments.

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