Irreversible processes steady flow

THROTTLING - An irreversible adiabatic steady flow process in which the fluid is caused to flow through an obstruction in a pipe with a resulting drop in pressure. 

Real irreversible processes can be subjected to thermodynamic analysis. The goal is to calciilate the efficiency of energy use or production and to show how energy loss is apportioned among the steps of a process. The treatment here is limited to steady-state, steady-flow processes, because of their predominance in chemical technology. 

For a real (irreversible) flow process through the control volume CV between fluid states X and Y (Fig. 2.4), with the same heat rejected at temperature T [Q x = [0rev]x)> the work output is [WcvJx. Heat [Qq x Iso be transferred from CV directly to the environment at Tq. From the steady-flow energy equation,... 

In many tubular reactors cooling or heating occurs as the process fluid flows through the reactor. This produces a major difference between an adiabatic and a nonadiabatic tubular reactor. In an adiabatic reactor, with an exothermic irreversible reaction, the maximum steady-state temperature occurs at the end of the reactor. In a cooled reactor, the maximum steady-state temperature usually occurs at some axial position part way down the reactor. Thus the temperature does not change monotonically with length. 

Calculation of the entropy produced in systems undergoing different flow processes (called irreversible processes) is key for considering steady-state systems. In order to measure the entropy produced in the system, we think of it as transported to the surroundings in a reversible manner and measure the entropy changes in the surroundings. From Eqs. (5) and (7),... 

The second law of thermodynamics not only gives us a direction for time but also gives us a macroscopic explanation for the direction for irreversible processes in steady-state systems. For heat flow and diffusion, give a microscopic reason why the flows are in the direction opposite to the gradient of temperature and concentration, respectively. 

The irreversibility accompanying a steady flow process may be computed for any zone by noting the fluxes of entropy into and out of the zone. Leaving aside the problems posed by semi-permeable membranes, which introduce ambiguities into the meanings of heat and work, (2), equation (3) provides such a balance ... 

Therefore, the total entropy produced within the system must be discharged across the boundary at stationary state. For a system at stationary state, boundary conditions do not change with time. Consequently, a nonequilibrium stationary state is not possible for an isolated system for which deS/dt = 0. Also, a steady state cannot be maintained in an adiabatic system in which irreversible processes are occurring, since the entropy produced cannot be discharged, as an adiabatic system cannot exchange heat with its surroundings. In equilibrium, all the terms in Eq. (3.48) vanish because of the absence of both entropy flow across the system boundaries and entropy production due to irreversible processes, and we have dJS/dt = d dt = dS/dt = 0. 

If we consider the change of local entropy of a system at steady state ds/dt = 0, the local entropy density must remain constant because external and internal parameters do not change with time. However, the divergence of entropy flow does not vanish div J, = . Therefore, the entropy produced at any point of a system must be removed or transferred by a flow of entropy taking place at that point. A steady state cannot be maintained in an adiabatic system, since the entropy produced by irreversible processes cannot be removed because no entropy flow is exchanged with the environment. For an adiabatic system, equilibrium state is the only time-invariant state. 

Heat conduction, as well as heat radiation and convection, is an irreversible process. Heat can only flow from a place of higher temperature to a place of lower temperature, that is, the flow of heat cannot take place in the absence of a temperature difference or a local temperature gradient. This leads to the conclusion that the entropy must increase. As can be shown by the case of steady-state heat flow through a bar, if the heat flow rate into the bar at one side (at temperature Ti)... 

We ehoose as our system a sample of n moles of gas that flows through the porous plug after a steady state is established. We assume that irreversible processes take place only within the porous plug so that the initial and final states of our system can be treated as equihbrium states. The work done on the gas on the left side is given by... 

Transport properties are studied off equilibrium, thus investigating the irreversible or the steady-state process. The flux (J) may be considered the time-dependent change of any nondifferentiated state variable (X) in the Gibbs free energy fnnction (Eqnation 8.84) divided by the cross sectional area through which the flow occurs. The flow is induced by the gradient of the conjngated differentiated state variable (T)- The flnx J is thns defined as... 

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