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Home › Forums › Asimptote knowledge forum › Warming / Error

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    • October 10, 2022 at 6:25 pm #2302
      henrique.fragoso da silva
      Participant

      There is one problem yet in my cycle. Someone could help me? I didn’t understand this problem.

      Thanks!

    • October 10, 2022 at 6:30 pm #2303
      henrique.fragoso da silva
      Participant

      The error is:

      ” IN EFFICIENCY CALCULATION OF APPARATUS NUMBER 1:
      – INGOING PIPE = 20, H = 1180. P = 0.65000
      – OUTGOING PIPE = 1, H = 1560. P = 15.000
      CHECK THE INLET AND THE OUTLET FOR THERMODYNAMIC PROPERTIES OF STATE! “

      • October 13, 2022 at 6:22 pm #2315
        sreekanth.manavalla
        Participant

        Is this apparatus 1 a compressor or pump? Probably the isentropic efficiency is more than 100%. In such case, this warning would be obtained. Please take a look at the inlet and outlet parameters.

    • October 14, 2022 at 6:54 pm #2320
      henrique.fragoso da silva
      Participant

      Is a compressor. I didn’t put any value to isentropic efficiency. I’m sending a picture.

    • October 15, 2022 at 5:06 am #2321
      sreekanth.manavalla
      Participant

      Isentropic efficiency is computed using the specified inlet and outlet conditions. Dont have to specify. Check whether the entropy at outlet is lower than that at inlet?
      what is the fluid in this case?
      I did not find any image attached.

    • October 25, 2022 at 11:03 pm #2379
      henrique.fragoso da silva
      Participant

      Hi! The entropy at the outlet is lower than the inlet. The fluid is Helium. I didn’t get to attach the file.

    • October 26, 2022 at 5:20 pm #2381
      sreekanth.manavalla
      Participant

      For the compressor handling Helium between pressures of 0.65 and 15 bar and the inlet temperature of 252.23 C (to give h=1180 at inlet), the outlet temperature should be 1571 C to give an isentropic efficiency of 100%. If the exit temperature is more than that, the isentropic efficiency will be less than 100%.
      At isentropic efficiency of 100%, the exit enthalpy would be 8027 kJ/kg.
      So, in your case, since the exit enthalpy is less than 8027, the isentropic efficiency is greater than 100% implying that entropy is destroyed which is against 2nd law.

    • October 26, 2022 at 5:53 pm #2382
      henrique.fragoso da silva
      Participant

      So, I have the data of the properties of helium.
      PIN = 0,65 BAR
      POUT = 15 BAR
      TIN = -46,95
      TOUT = 26,85

    • October 26, 2022 at 6:29 pm #2383
      henrique.fragoso da silva
      Participant

      I will try to attach the file.

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