So does $W$ in the efficiency really equal $W_{turbine} - W_{pump}$
here, or is most of the work lost to inefficiencies in the
implementation?
Yes it is. But often the pump work is ignored since it is typically much less than the turbine work. In terms of the enthalpies at the points 1-4 of your diagram, the efficiency is then
$$e=\frac{W_{T}-W_{p}}{Q_{in}}$$
$$e=\frac{(h_{3}-h_{4})-(h_{2}-h_{1})}{h_{3}-h_{2}}$$
And if the $PV$ work is somehow all captured by the turbine, how do I
rectify the fact that I've been taught to view $PV$ work as being
stored as the area enclosed by the cycle, but work is only extracted
on one "edge" of the cycle?
There are two basic types of work, shaft work for open systems (like the Ranking cycle) and boundary work for closed systems (e.g piston cylinder expansion/contraction work ). Boundary work is the area enclosed by the PV diagram for a closed system. No boundary work is done in the Rankine cycle, which is a two phase cycle.
For example, for the boiler process 2-3 no work is being done since it only involves adding heat to convert water into water vapor. There is no expansion or contraction of the boiler walls (boundary work). Nor is there any shaft work. So the area under the process 2-3 does not represent either boundary work or shaft work. The same applies to the condensation process 4-1.
The only type of work done in the Rankine cycle is shaft work, the work of the pump and the turbine. The turbine work is the work done by the system (working fluid). The pump work is work done on the system.
Maybe a basic question, but why is it valid to compute work as the
differences of enthalpies like ℎ2−ℎ1 above? I thought enthalpies were
only meaningful at constant pressure, which isn't the case between,
say, points 1 and 2.
Enthalpy is not only meaningful for a constant pressure process. It happens to equal the heat transferred for a closed system constant pressure process. But it is not restricted to that.
In fact, enthalpy is most meaningful for open systems like the Rankine cycle that involve both constant pressure processes (processes 2-3 and 4-1) where it equals the heat transfer (latent heat), and non-constant pressure processes (3-4 and 1-2) where it takes into account both internal energy, $u$, and flow work, $pv$, as the specific enthalpy is defined as
$$h=u+pv$$
UPDATE:
In response to your response to @Chet Miller comment,
I am certainly familiar with the first law of thermodynamics; I'm not
sure I know about its expression for an open, continuous flow system,
though.
The expression of the first law of thermodynamics is shown in the equation directly below Fig 1 below. As indicated, for steady flow, the entering and exiting mass flow rate is the same, or $\dot m_{e}=\dot m_{i}$. Fig 1 is a generic control volume that can be simplified to represent the control volumes for the turbine, pump, condenser, and boiler of the Rankine cycle.
In effect, it says that the energy into the control volume consisting of the entering enthalpy, kinetic energy, and potential energy plus any heat crossing the boundary equals the exiting enthalpy, kinetic energy, and potential energy plus the work output of the control volume.
Fig 2 simplifies Fig 1 for the case of reversible adiabatic turbines, compressors and pumps.
Hope this helps.
However, this doesn't seem to exploit the $PV$ work that occurs when the water vapor expands (i.e., during the isobaric portions of the cycle) in any way. So does $W$ in the efficiency really equal $W_{turbine} - W_{pump}$ here, or is most of the work lost to inefficiencies in the implementation?
