Actual rankine cycle - Work done by pump

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SUMMARY

The work done by the pump in an actual Rankine cycle can typically be neglected unless a high degree of accuracy is required. This is particularly true when comparing the pump work (h5-h4) to the heat added by the boiler and rejected by the condenser, as well as the work output from the turbine. The pump's work is minimal because it operates on liquid condensate rather than gas, which would necessitate a compressor and significantly higher energy input.

PREREQUISITES
  • Understanding of Rankine cycle thermodynamics
  • Familiarity with thermal efficiency calculations
  • Knowledge of heat transfer principles
  • Basic concepts of fluid mechanics
NEXT STEPS
  • Calculate the work done by the pump in various Rankine cycle scenarios
  • Explore the differences between ideal and actual Rankine cycles
  • Study the impact of pump work on overall thermal efficiency
  • Investigate the role of condensate properties in pump performance
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Students and professionals in mechanical engineering, thermodynamics researchers, and anyone involved in the design or analysis of power generation systems using the Rankine cycle.

Jameseyboy
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Hi,

In what cases do I neglect the work done by the pump in an actual rankine cycle?

I am reading that h5-h4 is negligible but in what circumstances do I accept this.

Is this only for ideal rankine cycles?

As I understand thermal efficiency is (work out + work in)/Heat in

Thanks
 
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Unless you are doing a very specific analysis which requires a good degree of accuracy, you can almost always neglect the work done by the pump. If you are unsure why, I'd recommend that you calculate the work term and compare it to the heat added and rejected by the boiler and condenser, respectively, and the work out from the turbine. You will see notice that the work done by the pump is very insignificant compared to the other processes. The reason the work done by the pump is so small is because it is pumping liquid condensate as opposed to a gas (or vapour) which would require a compressor and a much larger energy input.
 

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