# Understanding work in thermodynamic cycles and defining different systems

I am confused about defining systems and calcuating work in a thermodynamic cycle, say a typical power cycle. Such a cycle might have a pump, boiler, turbine, and condenser.

Now the typical assumption is if we define a system as a boiler or as a condenser, we say that no work is done on those systems and the only relevant terms in an energy balance would be heat transfer and the enthalpies of the masses coming in and out. In this kind of scenario, would the efficiency of the cycle be defined as Wnet / Qin where Qin is the heat added to the boiler and work net is the difference between the work done by the turbine and the work added to the pump?

But if I look at the gas as a system rather than each individual component, I would want to assert that the gas in the boiler and condenser does work because of the change in its volume (dW = P dV). Am I correct to assume this? And if so, would the gas as a system on its own in the turbine do a different amount of work than the amount of work the turbine (if the turbine is the system) might produce in the same cycle?

And if so, would asking what the efficiency of the cycle yield a different answer when defining the system as the gas in the cycle versus each block of the cycle as a system?

EDIT: In the last part, would the net work we use for defining efficiency if the system is the gas alone the net work performed by the gas in all 4 blocks of the cycle (its two expansions and compressions)? Contrast this with my previous assumption where work net is the difference in work the turbine outputs minus the work input into the compressor/pump if the systems are each block.

• Are you familiar with the open system (control volume) version of the first law of thermodynamics, and its derivation from the closed system version of the first law? If so, then you know that (1) the focus of the closed system version is on the working fluid (2) the focus of the open system version is the equipment through which the working fluid is flowing, (3) the open system version splits the work into shaft work and work to force fluid into or out of each piece of equipment, and (4) that the two versions give equivalent results. Oct 1, 2017 at 1:10
• By equivalent you mean the cycle's efficiency will be the same? And also the net work? And if so, how does the fluid technically do work if it's stuck inside a component like a turbine as its volume increases? Oct 1, 2017 at 1:27
• To your first question, yes. To your second question, the work required to push fluid into and out of each component cancels with the work into and out of the adjacent components. So the only work remaining is the shaft work in the compressor and pump, and the shaft work in the turbine. These devices function by the gas doing work on (or receiving work from) rotating shafts that have attached blades. Oct 1, 2017 at 1:40
• I see the cancellation now when leaving one component and entering the next. But where does this "extra" work come from if everything might cancel itself when looking at the system as a fluid? Does it just always work out in the end that not all of it is cancelled out when looking at the turbine and compressor? Oct 1, 2017 at 2:18
• In applying the open system version of the first law to the turbine, the rotor and shaft are regarded as external to the open system (even though they are internal to the turbine shell), and only the gas space within the shell is regarded as the open system. So the moving turbine blades are part of the surroundings, and the gas does work on them as it flows through the turbine. Thus, the gas space within the system has a moving boundary, even though the volume of the gas space is constant. The turbine blades turn the rotor which delivers mechanical energy. This is the "extra" work. Oct 1, 2017 at 12:50

For each piece of equipment in the loop, the open system version of the first law of thermodynamics tells us that: $$\dot{Q}-\dot{W_s}=\dot{m}(u_{out}-u_{in}+(Pv)_{out}-(Pv)_{in})$$where $\dot{m}$ is the mass flow rate, $\dot{Q}$ is the rate of heat addition to the piece of equipment, $\dot{W_s}$ is the rate of shaft work done by the piece of equipment on the surroundings, u is the internal energy per unit mass, P is pressure, and v is specific volume. The rate of work done to push mass into and out of the piece of equipment is $\dot{m}((Pv)_{out}-(Pv)_{in})$.
Note that, for each individual piece of equipment in the loop, the right hand side of the equation is not necessarily zero. Therefore, the change in internal energy between inlet and outlet plus the net work done required at the inlet and outlet to push material in and out of the equipment captures the combined effect of the heat added to the individual piece of equipment and the shaft work done by the equipment. But, if we add up these heat balance equations for all the pieces of equipment in the closed loop (operating at steady state), we obtain: $$\sum{\dot{Q}}-\sum{\dot{W_s}}=0$$ and $$\sum{\dot{m}(u_{out}-u_{in}+(Pv)_{out}-(Pv)_{in})}=0$$