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Gain-loss equation

In this form the meaning becomes particularly clear the master equation is a gain-loss equation for the probabilities of the separate states n. The first term is the gain of state n due to transitions from other states n and the second term is the loss due to transitions from n into other states. Remember that Wnn> 0 when n n and that the term with n = n does not contribute to the sum. [Pg.97]

In this form, the master equation is a gain-loss equation for the probability to find the system at time t in a given state the first sum (or integral) on the right-hand side represents the transitions into the state n (or y) from all other states, the second sum represents transitions from the state n (or y) to all other states. [Pg.89]

Using the definition of the gain (equation 9.31) and loss (equation 9.30) terms, we find... [Pg.479]

Again if this power gain is equated to the loss of energy of electrons in elastic collisions with the gas molecules, as would occur in the steady state, then... [Pg.14]

Kinetic studies such as these use the master equation to follow the flow of probability between the states of the model. This equation is a basic loss-gain equation that describes the time evolution of the probability pi(t) for finding the system in state i [24]. The basic form of this equation is... [Pg.375]

The rate of a reaetion is the number of units of mass of some partieipating reaetants that is transformed into a produet per unit time and per unit volume of the system. The rate of a elosed homogeneous reaetion (that is, no gain or loss of material during the reaetion) is determined by the eomposition of the reaetion mixture, the temperature, and pressure. The pressure from an equation of state ean be determined together with the temperature and eomposition. [Pg.110]

It must be kept in mind that the reaction term will not occur in the overall mass balance equations of reactive systems because S(R- = 0, i.e., there is no net mass gain or loss as a result of chemical reactions. [Pg.334]

The second law of thermodynamics states that energy exists at various levels and is available for use only if it can move from a higher to a lower level. For example, it is impossible for any device to operate in a cycle and produce work while exchanging heat only with bodies at a single fixed temperature. In thermodynamics, a measure of the unavailability of energy has been devised and is known as entropy. As a measure of unavailability, entropy increases as a system loses heat, but remains constant when there is no gain or loss of heat as in an adiabatic process. It is defined by the following differential equation ... [Pg.557]

Our final form for the Boltzman equation is then obtained by substituting these last two expressions for the gain and loss terms into equation 9.29 and adding the term Vijf x, v, t) to the LHS for the case where there is an external force F Vjj is the gradient operator with respect to v and m is the hard-sphere mass) ... [Pg.477]

Before you can balance an overall redox equation, you have to be able to balance two halfequations, one for oxidation (electron loss) and one for reduction (electron gain). Sometimes that s easy. Given the oxidation half-equation... [Pg.88]

Now the gain and loss of electrons must be equal. One permanganate ion utilises 5 electrons, and one iron(II) ion liberates 1 electron hence the two partial equations must apply in the ratio of 1 5. [Pg.850]

The key to writing and balancing equations for redox reactions is to think of the reduction and oxidation processes individually. We saw in Section K that oxidation is the loss of electrons and reduction the gain of electrons. [Pg.604]

When balancing redox equations, we consider the gain of electrons (reduction) separately from the loss of electrons (oxidation), express each of these processes as a halfreaction, and then balance both atoms and charge in each of the two half-reactions. When we combine the halfreactions, the number of electrons released in the oxidation must equal the number used in the reduction. [Pg.605]

An overall mass balance is written for the system as a whole. Interphase mass transfer does not appear in the system mass balance since gains in one phase exactly equal losses in the other. The net result is conceptually identical to Equation (1.3), but there are now two inlets and two outlets and the total inventory is summed over both phases. The result is... [Pg.396]

From the steady-state heat balance the heat losses can be equated to the heat gained by reaction, giving... [Pg.151]

The above equation then represents the balanced conditions for steady-state reactor operation. The rate of heat loss, Hl, and the rate of heat gain, Hq, terms may be calculated as functions of the reactor temperature. The rate of heat loss, Hl, plots as a linear function of temperature and the rate of heat gain, Hq, owing to the exponential dependence of the rate coefficient on temperature, plots as a sigmoidal curve, as shown in Fig. 3.14. The points of intersection of the rate of heat lost and the rate of heat gain curves thus represent potential steady-state operating conditions that satisfy the above steady-state heat balance criterion. [Pg.152]

Please pass the sodium chloride It is amazing that food is seasoned with an ionic compound that is composed of two deadly elements— sodium and chlorine. The gain or loss of electrons can make a big difference in properties. Reacting sodium hydrogen carbonate, which is baking soda, with hydrochloric acid (HCI), the acid found in your stomach, produces salt, carbon dioxide, and water, according to the following equation ... [Pg.61]

Note that the equation includes rates of gain and loss rather than total amounts. As a result, the mathematical form will be a differential equation rather than an algebraic one. The differential form is preferred in almost all mass transfer problems, because variations in the rates with position and time can be incorporated accurately. Each term in the equation will take on a specific functional form depending on the parameters and mass transfer characteristics of the problem of interest. [Pg.21]

Assuming no chemical reaction, what will happen if we place a piece of metal at one temperature into water at another temperature The metal will gain energy from or lose energy to the water until their temperatures are the same. Whichever was at the higher temperature initially will lose energy to the other. In the absence of any energy loss to any other body, the quantity of heat removed from one will be added to the other. We can use this fact to measure specific heats of substances, or to predict the final temperatures to which such combinations will arrive. The principle involved may be summarized in the equation ... [Pg.273]

An analogous loss-gain equation holds for the concentration cag(t)... [Pg.271]

Rips and Silbey (1991) have reexamined the thermalization of photoelectrons (of a few eV in energy) with a master equation approach for the time rate of energy loss. Their method is quite general, and it includes both direct (energy loss) and inverse (energy gain) collisions according to the principle of detailed balance. As in the Frohlich-Platzman method, they first calculate the time rate... [Pg.272]


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See also in sourсe #XX -- [ Pg.11 ]




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