Net growth rate

The kinetic of cell growth for prediction of growth rate is projected by the net growth rate, which is ... 

The top panel of Fig. 17.2 (Ts = 800 K) reveals that there is very little decomposition of the silane in the gas phase, which is a result of the relatively low temperature. As a result the net growth rates should be expected to be quite low, since the silane sticking coefficient is so low. At a surface temperature of Ts = 1300 K, however, the decomposition of silane to silylene in the gas-phase boundary layer is nearly complete. The relatively high silylene concentrations should lead to high growth-rates. The peak in the silylene profile at about 1.5 mm above the surface results from the competition between production by the homogeneous decomposition reaction and consumption at the surface by heterogeneous reaction. 

The growth of layer i — 1, on the other hand, will require an analogous decomposition of layer i by solid-state reaction at the interface xt = 0. This will lead to a negative contribution to the growth kinetics of layer i. Thus, from the time-dependence of the cation interstitial balance at the interfaces we can write the net growth rate of any inner layer i as... 

This result, evaluated for layer i as well as layer i 4- 1, can in turn be used in eqns. (216)—(218) to evaluate the net growth rate of any inner layer i. 

Next, we must develop growth equations for the two oxide layers Lx and Ln. A decomposition of layer N is required in forming layer N — 1, as required by eqn. (261), and so in this respect layer N does not differ from the inner layers. On the other hand, there is no decomposition of a layer N + 1 required to obtain the requisite oxygen, since layer N is already in contact with the gaseous oxygen phase. Thus, there is no equivalent cation vacancy current to be considered, in contrast to the other layers. Neither is there an actual cation vacancy current J l, to be considered. However, there may be oxide evaporation which must be compensated for by the net growth rate of layer N. Thus we obtain... 

This result for the decomposition of layer i — 1, together with its equivalent for layer i obtained by replacing i by i + 1, can be substituted into eqns. (295)—(297) to obtain the net growth rate of any inner layer i, viz., T... 

The rate (dL /df)evap is considered to be positive in sign if evaporation is indeed occurring. Utilizing eqn. (302) in eqn. (301), with i = N, and then substituting into the net growth rate given by eqn. (308) leads to the result... 

Consider, also, that layer i will likewise undergo solid-state decomposition at the interface xt = L, in the formation of layer i + 1, yielding a negative contribution (cLL,/df) to the growth of layer i. The net growth rate of layer j will be given by the difference in the formation rate of layer i at xt = 0 and the decomposition rate of layer i at xt = L,... 

To obtain the decomposition rate expressions (dL,-/df) and (dL(, /dt) required in this equation for the net growth rate of any interior layer i, we equate the two expressions (333) and (337) which were obtained for (dL,/dt)+ by means of the cation and the anion balances at the phase boundary xt = 0. This gives... 

We note in passing that this result holds for one of the extreme layers (namely, oxide layer 1 in contact with the parent metal) in addition to the inner layers this will be needed in developing the expression for the net growth rate of layer 1. 

The net growth rate for the inner layers now follows by substituting eqn. (343), together with its equivalent obtained by replacing i by i + 1, into eqn. (338). This gives... 

The growth rate for layer 1 at x =0 requires an anion vacancy current J(jav) through layer 1. There are no growing oxide layers i for i < 1, so J)av) contributes entirely to the growth of layer 1. In addition, there is no oxide decomposition reaction at the phase boundary xt = 0 since this represents the metal interface thus, there is no equivalent current jJ°xv> to be considered. Therefore, we can write (dLj/d )+ = R av)J av), so the net growth rate of layer 1 is given by... 

This ignores any evaporation of oxide at the gas interface (xN = LN). If such occurs, the evaporation rate (dLN /dt)evap, considered to have a positive sign if evaporation is indeed occurring, will decrease the net growth rate of layer N, so that (dLN /dt) in eqn. (359) should then be replaced by [(cLLN /dt) + (dLN /dt)evaP ] Equivalently, we can write... 

We have estimated nitrate-specific net growth rates (/iN) of Phaeocystis for each experimental treatment between days 25 and 31, using the mean values of nitrate + nitrite drawdown, relative to the starting seawater, at these timepoints. This calculation assumes that (1) growth was exponential in all bottles between days 25 and 31, and (2) decreases in nitrate + nitrite were directly... 

Equation (7-93) may have to be modified by subtraction of a death-rate term idCx. may well increase during the batch fermentation in which case the net growth rate of (viable) cells eventually becomes negative, and the concentration of (viable) cells will start to decrease. 

A complete analysis of the properties of these equations is quite difficult since the coefficients of P and Z are time variables and also functions of P and Z. However, the behavior of the solution becomes more accessible if the variation of these coefficients is studied as a function of time. The expressions GP — (DP + Q/V) and Gz — (Dz + Q/V) can be considered the net growth rates for phytoplankton and zooplankton. The advective or flushing rate, Q/V, is included in these expressions since it acts as a death rate in one segment system. 

The sign and magnitude of the net growth rate controls the behavior of the solution. For a linear equation, for which the net growth rate is not a function of the dependent variable (i.e., P or Z), the type of solution obtained depends on the sign and magnitude of the net growth rate. That is, for the equation... 

Competition among two species means that the increase in one of the populations decreases the net growth rate of the second one, and vice versa. This happens when they feed on the same resources, or if they produce substances (toxins) that are toxic for the other species. A classical competition model was also introduced in Volterra (1926), and considered in a more general parameter range by Lotka (1932). It is known as the competitive Lotka-Volterra system ... 

S > 1 and a constant net growth rate of 7, Nj. At saturation (S = 1) all 7i+1 /2 = 0, whereas at steady-state nucleation conditions all 7,+i/2 = 7. There is a third distribution that we will not explicitly introduce until the next section. It is the hypothetical, equilibrium distribution of clusters corresponding to S > 1. Thus it corresponds to all 7i+1/2 = 0, but 5 > 1. Because of the constraint of zero flux, this third distribution is called the constrained equilibrium distribution, Nf. We will distinguish this distribution by a superscript e. 

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