# 7.06: Work and Power Revisited

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In this chapter we have considered several different types of work or power. Specifically, we have discussed mechanical work, shaft work, \(\text{PdV}\) work, and electrical work. Each of these work transfers of energy satisfies the definition of thermodynamic work. In this section, we want to introduce another characteristic of work and power, the concept of a *quasiequilibrium *work transfer of energy.

To explore this concept in depth, we will consider electrical work. Specifically, we will examine electrical work for two situations: charging a battery and energizing a resistor. First, we will consider the electrical work required to charge a 12-volt battery. From electrochemistry, we know that the 12-volt designation is an indication of what the electrochemist might call the cell potential, \(\Delta V_{\text {cell}}\). This is an intensive property of the battery and depends solely on the chemical makeup of the battery. To recharge the battery, we must run a current backwards through the battery to reverse the chemical reaction. The differential amount of electrical work to recharge the battery is \[\delta W_{\text {electric, in}} = i \cdot\left(V^{+}-V^{-}\right) \cdot dt = i \cdot \Delta V_{\text {terminals}} \cdot dt \nonumber \] where \(\left(V^{+}-V^{-}\right)=\Delta V_{\text {terminals}}\) is the voltage difference between the battery terminals and \(i\) is the current. (Note that \(i \cdot dt\) is \(dq\), the differential amount of charge entering the system.) When the battery is being recharged, the voltage difference between the battery terminals must be at least equal to the cell potential, i.e. \(\Delta V_{\text {terminals}} \geq \Delta V_{\text {cell. }}\).

If the battery is recharged in such a fashion that \(\Delta V_{\text {terminals}} = \Delta V_{\text {cell}} + \varepsilon\), where \(\varepsilon\) is very, very small, we would say that this is quasiequilibrium electrical work and \[ \left. \delta W_{\text {electric, in}} = \underbrace{\Delta V_{\text {cell}}}_{\begin{array}{c} \text {An intensive} \\ \text { property } \end{array}} \cdot \underbrace{i \cdot d t}_{\begin{array}{c} =dq \\ \text {Change in an} \\ \text { extensive } \\ \text { property } \end{array}} \right| \begin{array}{c} \text { Quasiequilibrium } \\ \text { electric work } \end{array} \nonumber \] The word *quasiequilibrium* implies that the transfer of energy occurs in such a fashion that the system, as far as this work is concerned, is passing through a series of equilibrium states. For quasiequilibrium electric work, this implies that the electric current flows through the battery in such a fashion that the terminal voltage just equals the cell potential. Note that quasiequilibrium electric work is the product of an intensive property (cell potential) of the system and the change in an extensive property (charge) of the system. If \(\varepsilon\) is still very, very small but negative, the direction of the current would reverse as would the direction of the work transfer of energy. The ability to reverse the direction of the energy transfer with only an infinitesimal change in the terminal voltage difference is also a characteristic of quasiequilibrium work.

To see the significance of this, consider the case of the electric work done on a resistor: \[\delta W_{\text {electric resistor}} = i \cdot \Delta V_{\text {terminals}} \cdot dt = i \cdot (iR) \cdot dt = i^{2} R \cdot dt \geq 0 \nonumber \] where \(R\) is the electrical resistance of the resistor. Unlike the case for Eq. \(\PageIndex{1}\), making an infinitesimal change in the terminal voltage does not change the direction of the current or the direction of the energy transfer. In fact, electrical work transfer of energy to a resistor is always positive regardless of the direction of the current. This is an example of a *non-quasiequilibrium* work transfer of energy.

Similarly, if we examine the work done to compress a substance in a piston-cylinder device, we could write the differential amount of work as follows: \[\delta W_{\text {piston, in}}=F_{\text {piston}} \ dx \nonumber \] where \(F_{\text{piston}}\) is the force that the piston exerts on the system and \(dx\) is the differential displacement of the boundary in the direction the external force acts. Under conditions where the piston moves slowly enough so that the force exerted by the piston just equals the uniform pressure in the system times the piston area, \(F_{\text {piston}} = P A_{\text {piston}}\), we have quasiequilibrium compression-expansion work:

\[ \left. \delta W_{PdV} = - \underbrace{P}_{\begin{array}{c} \text{Intensive} \\ \text{property} \end{array}} \cdot \underbrace{d V\kern-0.8em\raise0.3ex-}_{\begin{array}{c} \text{Change in an} \\ \text{extensive property} \end{array}} \quad \right| \begin{array}{c} \text{Quasiequilibrium} \\ \text{compression-expansion work} \end{array} \nonumber \] where \(P\) is the uniform pressure in the system and \(d V\kern-0.8em\raise0.3ex-\) is the differential change in the system volume. For this to be accurate, the system pressure (an intensive property) must be spatially uniform during the process. Again, a change in the direction of the volume change causes the sign of the work transfer of energy to change.

For comparison, consider the case where work is done on a system by a frictional force, say a block sliding on a plane. \[\delta W_{\text {friction, in}} = -F_{\text {friction}} \cdot dx = -\mu F_{\text {normal}} dx \nonumber \] where \(F_{\text {normal}\) is the normal force at the surface and \(\mu\) is the appropriate coefficient of friction. Unlike \(\text{PdV}\) work, the transfer of energy into a system by frictional work is always the same sign and cannot be reversed by changing the direction of motion on the boundary.

So what conclusions can we draw from this discussion? First, there is a distinction between quasiequilibrium and non-quasiequilibrium work transfers of energy. Second, there are several distinctive characteristics that distinguish between these two types of work transfers of energy. These are summarized in the table below.

Quasiequilibrium (Reversible) Work | Non-quasiequilibrium (Irreversible) Work | |
---|---|---|

External force | Related to an intensive property of the system. | Related to the transport rate of an extensive property measured on boundary of the system. |

Displacement | Related to a change of an extensive property of the system. | Related to a change of an extensive property of the system. |

Constraints on system | Pertinent intensive property is spatially uniform. | No constraint on intensive properties of the system. |

Generalized form | Can be written in terms of a generalized external force \(F_{\mathrm{K}}\) (an intensive property) and a generalized displacement \(X_{\mathrm{K}} (an extensive property) \[\delta F_{\mathrm{K}} = \pm \underbrace{F_{\mathrm{K}}}_{\begin{array}{c} \text{Intensive} \\ \text{Property} \end{array}} \cdot \underbrace{d X_{\mathrm{K}}}_{\begin{array}{c} \text{Extensive} \\ \text{Property} \end{array}} \nonumber \] | Frequently, most easily written as an energy transfer rate, i.e. power. |

Direction of work transfer | Can change directions (reversible). | Cannot change directions (irreversible). |

Third, as we will discuss in the next chapter these reversible work transfers of energy have much to do with the limits on the performance of real systems. In a later course, (ES202—Fluid & Thermal Systems), you will learn how these processes play a key role in our study of the thermophysical properties of substances. In this course, our main goal is to introduce the differences between quasiequilibrium and non-quasiequilibrium work transfers so that you better understand the consequences of modeling a work transfer using these assumptions.

In the discussion above, you have seen how a work transfer of energy may or not be quasiequilibrium depending on how the work transfer of energy is related to the properties of the system. Consider shaft work under two situations:

- Case I - Shaft work drives a paddlewheel inside a tank of water.
- Case II - Shaft work is used to twist a thin wire within its elastic range. (Assume the wire acts as a torsional spring.)

Which of these cases involves a quasiequilibrium work transfer of energy? Why?