A New Thermodynamics

Blog: Rankine cycle

By Kent W. Mayhew

www.newthermodynamics.com

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 Rankine Cycle

 

The traditional interpretations of this cycle can be found in texts or on-line, and for the most part are complications of reality, all due to the traditional entropy based considerations. Herein will be shown a much simpler interpretation based upon our new perspectives.

The Rankine cycle is often used in power plants, wherein electricity-generating steam turbines are powered by anything from a nuclear reaction to the combustion of oil, natural gas, or coal, with water commonly used as the working fluid. The Rankine cycle generally consists of four steps:

Step 1): Water is pressurized via a pump. Since water is incompressible, the energy required is considered as being minimal in comparison to power generated. The work required for the waterís isometric pressure increase is:    

         Wstep1)=VdP      eqn 1

Step 2): High-pressure water enters a constant pressure & volume boiler then is heated into a dry saturated steam. Energy is required to break the liquid waterís intermolecular bonds that being; dEbond. 

 

Step 3): The dry saturated steam is then allowed to expand thus spinning the turbine and generating power. As the steam expands it not only turns the turbine but it also displaces Earthís atmosphere, hence does work, hence both the steamís pressure and temperature decreases. We can treat this in terms of changes to the steamís ability to do work, thus write:

 

   Wstep3)=-d(PV)steam = PatmdVsteam+Wturbine    eqn 2

 

In real life applications, it may be more useful to rewrite eqn 2 in terms of molecular volumes (v), number of molecules passing through the turbine (N), as well as final (f) and initial (i) states, i.e.:

 

Wstep3)=N[(PV)i(steam)-(PV)f(steam)]=PatmN(vf-vi)steam+Wturbine eqn 3)

 

Generally the final pressure will be atmospheric pressure thus we can write:

 

Wstep3)=N[(PV)i(steam)-PatmVf(steam)]=PatmN(vf-vi)steam+Wturbine eqn 4)

 

Step 4): The vapours then enter a condensate chamber, wherein they return to the liquid state. Here energy from changes to the waterís intermolecular bonding energy could be extracted that being;  -dEbond.

 

If the Rankine process were as efficient as possible, the efficiency would still be limited because of step 3). This is due to the lost work (Wlost) required to displace our atmosphereís weight, as the expanding steam that drives the turbine creating its energy,. (see lost work) Consider that the turbine is frictionless than the work to drive the turbine would equal the energy extracted from the turbine, i.e. Wturbine=Eturbine. Accordingly, strictly in terms of work the optimal efficiency  for the Rankine cycle is:

 

Efficiency = Eturbine/(PatmdVsteam+Wturbine)         eqn 5)

 

What about the fact that only 66.67% of a gasís energy can do work? Assuming that the flux of molecules that interact with the turbineís blades is similar to that of a flat wall, then that flux is defined by eqn B.7.14: Flux=(1/4)nv (see blog Work vs Energy). In which case the actual the optimal efficiency for the Rankine cycle becomes:    

  

Efficiency = 2Eturbine/[3(PatmdVsteam+Wturbine)]       eqn 6)

 

 Note turbines may have a higher value for molecular flux dependent upon design. This would most likely occur with an overlap of the blades and  the design of housing around the turbine in which case  higher efficiencies are possible.  

 

In a real power plant there would be additional energy loses such as frictional loses in flowing fluids, heat loses to the surroundings, as well as frictional loses in machinery. 

 

Such processes are most efficient when the steam is supercritical because the work required to displace the atmosphere remains constant, so the greater the overall energy within the steam the higher the efficiency should be. Accordingly, steam turbines typically operating around 838 K (565 oC), and condensation occurs at slightly above room temperature i.e. above 300 K. Furthermore by being in the supercritical regime there are also no issues of energy loses due to water being in the vapour state, or even tensile layer formation.

 

 

Copyright Kent W. Mayhew

 

 

This website is copyright of Kent W. Mayhew who in 2018 resides in Ottawa Ontario Canada
 
   This website is full of new ideas, which are the property of Kent W. Mayhew.  
 
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