Analyzing open steady flow systems with SFEE vs. Rankine cycle analysis

[physea]

Hello
I have a point that I don't understand.

In open steady flow systems, we use SFEE, where there's a term W in addition to a term PV, which are different things.

When we analyse a cycle like Rankine, we always consider either W or PV, but not both. I also think that W=PdV always.

Why that difference? Do we consider cycles to be close systems?

 

[SirCurmudgeon]

Generally the term "steady flow system" is applied to some single constraining structure like a compressor, a nozzle and so.
If there is work with this system it is by shaft, impellor and so. There would be pdV work only if and as the system enlarged (rare system that is).

The term Rankine applies to a cyclic forward (power-producing) device. [Heat to the system, conversion of some to work then heat out].
Cyclic reverse engine is heat at low temperature to cycle, work to the cycle then heat out at higher temperature (refrigerator or heat pump). Also Rankine addresses the entire cycle which has components linked in a cycle. Cycles as closed systems? Overall yes, but in calculations, to establish the overall one must go component to component.

 

[physea]

Generally the term "steady flow system" is applied to some single constraining structure like a compressor, a nozzle and so.
If there is work with this system it is by shaft, impellor and so. There would be pdV work only if and as the system enlarged (rare system that is).

The term Rankine applies to a cyclic forward (power-producing) device. [Heat to the system, conversion of some to work then heat out].
Cyclic reverse engine is heat at low temperature to cycle, work to the cycle then heat out at higher temperature (refrigerator or heat pump). Also Rankine addresses the entire cycle which has components linked in a cycle. Cycles as closed systems? Overall yes, but in calculations, to establish the overall one must go component to component.

Not sure if you address my question, you include general information and your answer is not clear.

 

[Chestermiller]

Hello
I have a point that I don't understand.

In open steady flow systems, we use SFEE, where there's a term W in addition to a term PV, which are different things.

When we analyse a cycle like Rankine, we always consider either W or PV, but not both. I also think that W=PdV always.

Why that difference? Do we consider cycles to be close systems?

When you analyze Rankine (or other power cycle) using SFEE, each individual piece of equipment has its own shaft work and its own $\Delta (PV)$. The $\Delta (PV)$ is included in the enthalpy term. Over the entire cycle (including all pieces of equipment), the sum of the $\Delta (PV)$s is equal to zero, and you are left only with the shaft work. You are aware that the sum of the shaft work plus the $\Delta (PV)$ for a particular piece of equipment add up to the rate of doing PdV work for that piece of equipment, right? If not, go back and study the derivation of the SFEE.

 

[physea]

In SFEE, I know that d(PV) is the flow energy, in other words the pressure energy that is responsible for the flow of the fluid.

In Rankine cycle analysis (or other cycles) do we consider that flow energy? Or we only consider the Q, W, U?

In SFEE, we know dH = Q - W, in Rankine cycle, it seems dU=Q-W. Is that right?

 

[Chestermiller]

In SFEE, I know that d(PV) is the flow energy, in other words the pressure energy that is responsible for the flow of the fluid.

In Rankine cycle analysis (or other cycles) do we consider that flow energy?

Yes. For each individual piece of equipment.

Or we only consider the Q, W, U?

No. For each individual piece of equipment, we consider Q, Ws, and H, where Ws is the shaft work. For the overall cycle, $\Delta H=\Delta U=0$

In SFEE, we know dH = Q - W, in Rankine cycle, it seems dU=Q-W. Is that right?

No. For each individual piece of equipment in the Rankine cycle, $\Delta H=Q-W_s$

 

[physea]

Yes. For each individual piece of equipment.

No. For each individual piece of equipment, we consider Q, Ws, and H, where Ws is the shaft work. For the overall cycle, $\Delta H=\Delta U=0$
No. For each individual piece of equipment in the Rankine cycle, $\Delta H=Q-W_s$

But what you say contradicts with the equations here:
https://en.m.wikibooks.org/wiki/Physical_Chemistry/Thermodynamic_Processes_for_an_Ideal_Gas
In that table it says dU=Q-W and not dH=Q-W.

 

[Chestermiller]

But what you say contradicts with the equations here:
https://en.m.wikibooks.org/wiki/Physical_Chemistry/Thermodynamic_Processes_for_an_Ideal_Gas
In that table it says dU=Q-W and not dH=Q-W.

Those equations are for a closed system, not an open system (like the equipment used in the Rankine cycle, where mass is flowing in and out). Are you not familiar with the open system (control volume) version of the first law of thermodynamics?

 

[physea]

Those equations are for a closed system, not an open system (like the equipment used in the Rankine cycle, where mass is flowing in and out). Are you not familiar with the open system (control volume) version of the first law of thermodynamics?

Yes I know that closed systems have no mass flow. But I don't know which equations we should use when analysing cycles.

 

[Chestermiller]

Yes I know that closed systems have no mass flow. But I don't know which equations we should use when analysing cycles.

For cycles, the version of the first law you use depends on whether each piece of equipment is an open system or a closed system.