Thermodynamics - 7-12 Isentropic Efficiency example 1

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  • Опубліковано 13 гру 2024

КОМЕНТАРІ • 18

  • @grandlong5462
    @grandlong5462 3 роки тому +3

    I finally know where the h2s came from. Thank you very much!

  • @dottosebastianophilipo2693
    @dottosebastianophilipo2693 3 роки тому

    Congratulation for solving well this problem
    Much respect from TANZANIA 🇹🇿🇹🇿

  • @duakhan5882
    @duakhan5882 4 роки тому +6

    Why are the actual and isentropic pressures the same?

    • @yolo3004
      @yolo3004 10 місяців тому

      same question

  • @sunmichoi6888
    @sunmichoi6888 2 роки тому +3

    Thank you so much!

  • @Jack-nv3to
    @Jack-nv3to 3 роки тому +1

    Thank you for this wonderful course. Quick question: I think we should not use the actual Work (2000KW) to get actual flow rate. Should use Isentropic Work.

  • @deyardeyar8163
    @deyardeyar8163 4 роки тому +3

    Dear Doctor,
    why S1 was equal to S2s and not S2a?

    • @engineeringdeciphered
      @engineeringdeciphered  4 роки тому +11

      Because S2s is the entropy if the process was isentropic. So it makes sense that if it was isentropic S2s would be the same as at the beginning, S1. The actual entropy is S2a and since it’s not really isentropic S2a is not the same as S1.

  • @yolo3004
    @yolo3004 10 місяців тому

    in problem one, why do we keep the pressures the same for the isentropic process? Couldnt we keep the tempratures the same instead?

  • @johng3792
    @johng3792 3 роки тому

    back here helping me review for thermo 2

  • @kylecatman7738
    @kylecatman7738 4 місяці тому

    If it's reversible ONLY does that mean S2 - S1 = Q/T ?? Because Sgen would be zero at that point.

    • @engineeringdeciphered
      @engineeringdeciphered  3 місяці тому

      yes, correct. Sgen is a measure of irreversibility. So a reversible heat transfer would be S2-S1 = Q/T

  • @saravmohan9852
    @saravmohan9852 11 місяців тому

    why turbine work output is negative it should be positive since work done by the system is positive while work done on the system for compressor is negative right???

    • @engineeringdeciphered
      @engineeringdeciphered  11 місяців тому

      I like to teach Q+W=deltaH and say everything (Q, W, H) in is positive and everything out is negative. So if using Q+W=dH, work done by the system (out) is negative and work done on the system (in) is positive. If you use Q-W=deltaH then work out is positive and work in is negative.

    • @saravmohan9852
      @saravmohan9852 11 місяців тому

      @@engineeringdeciphered now I'm super clear thanks for the instant reply

  • @donsupra
    @donsupra Рік тому

    Same problem but 2.5 Mpa and 450°C at inlet, and 60kpa and 100°C at outlet pls 🥲 I can't calculate the value of h2 in the tables