Workfree process

Here we deduce bounds on the directions of irreversible transfers across the interface in Figure 7.4. We consider six processes workfree constant-mass heat transfer, adiabatic constant-mass work, isobaric constant-mass heat transfer, isothermal constant-mass work, isothermal-isobaric diffusion, and adiabatic-workfree diffusion. 

Heat and work are not properties of either the system or the surroundings they exist only during the interaction that carries them across the boundary. However, for certain special processes Q and W are separately related to changes in state functions. We have already seen that if no thermal interaction exists, then the adiabatic work equals AH and it can be calculated assuming a reversible change. Likewise, if only a thermal interaction connects the system to the surroundings (the process is workfree), then the heat transferred equals AH,... 

This special form, rather than (2.3.10), is more often presented in textbooks. However, (2.3.10) makes clear that even in a workfree process we expect entropy to be generated because of irreversible heat transfer. In fact, when the heat transfer term is not zero, it is usually larger that the lost work term. 

For steady-state workfree processes, (2.4.20) shows that the heat transferred can be computed from the enthalpy change between inlets and outlets common applications include steady-state heat exchangers. For steady-state adiabatic processes, (2.4.20) shows that the shaft work can be obtained from the enthalpy change these situations arise in adiabatic pumps, turbines, and compressors. 

For every workfree process between the same initial and final states, show that the heat effect is the same, regardless of how the heat is transferred. 

For any workfree isothermal process on a closed system, show that q = TAs. 

A typical set of the required five variables would be the temperatures and pressures of the inlet and outlet water streams, T, Tg, P, and Pg, plus the inlet water flow rate N. With values for these five variables, we can solve the steady-state material balance for the outlet water flow rate (the inlet and outlet mass flow rates are equal here) and we can solve the steady-state energy balance for Q. In this example the value computed for the heat duty is the actual value for the real process, regardless of reversibility, because the process is workfree. However, in the general case, when heat and work both cross a system boundary, the energy balance gives only their sum. Variations on this problem are also possible for example, if we knew values for the five variables T, Tg, P, Pg and Q, then we could solve the energy balance for the required water flow rate. Or, if we knew T, P, P , Q, and N, then we could solve for the outlet water temperature Tg. 

In the special case of workfree processes with negligible kinetic and potential energy changes, the heat can be obtained from the overall energy balance (3.6.7),... 

In workfree processes, the heat given by (3.7.3) is the actual heat 6Qgj.(, regardless of the reversibility of the process. 

How does the entropy of an isolated system respond when two system parts, initially at different temperatures, are brought into contact during a workfree process ... 

Workfree, constant-mass heat transfer. Let the interface in Figure 7.4 be impermeable, thermal conducting, and fixed in position. When the interface is impermeable, then each phase is closed when the interface is fixed in position, then the volumes of the two phases are constant = constant and VP = constant. We initiate a process by... 

This constraint applies to heat crossing the interface in either direction, but to have a particular example, say the process transfers heat from phase a to phase 3. Then 8QP > 0, and the inequality in (7.2.14) requires T > pP that is, the temperature difference (T TP) drives workfree, constant-mass heat transfer. For such processes, heat always flows from regions of high temperature to regions of low temperature. 

Coupled diffusion. But in addition, the diffusion of components may be coupled, as alluded to by Prigogine and Defay [1]. In these situations, (7.2.31) need not be satisfied term-by-term the only thermodynamic constraint is that the entire sum satisfy (7.2.31). As an example, consider binary diffusion in an isothermal workfree process then (7.2.31) reduces to two terms. 

Let a binary mixture of components 1 and 2 form each of the phases a and P in Figure 7.4. The interface between the phases is thermally nonconducting, fixed in position, and permeable to both components. The position of the interface is such that each phase has the same volume. Further, we load the same number of molecules of each component into each phase. We start an adiabatic, workfree process by adjusting the temperatures so that T > T. ... 

The positive value means that this amoxmt of heat must be added to the flash unit to maintain the temperature at 330 K. Since the process is workfree, this is the actual heat effect, regardless of any irreversibilities. 

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