Example 15 2 for Pinch Analysis

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  • เผยแพร่เมื่อ 22 มี.ค. 2020

ความคิดเห็น • 26

  • @hairenjeger6342
    @hairenjeger6342 6 หลายเดือนก่อน +1

    thank you for the very good and brief explanation. It would've been great if you did a lecture video about this subject

  • @dbro3ify
    @dbro3ify 3 ปีที่แล้ว +1

    Hello thank you for the video. I was wondering if the most negative value occurs at the bottom of the cascade does that mean that cold utility is not needed. Thank you

  • @Morozov97
    @Morozov97 2 ปีที่แล้ว +4

    hi. why aren't the cold streams adjusted for temperature difference (delta T)? Thank you.

  • @user-hg4up8vh5t
    @user-hg4up8vh5t 4 ปีที่แล้ว

    Could you please share the book with this example?

  • @justinkennedy9930
    @justinkennedy9930 3 ปีที่แล้ว +11

    Hello professor. I had a question on the Cp inequality for the pinch design. Should it read designing on the cold side of the pinch that the CH >= CC and vice versa on the design of the hot side of the pinch? Also great video very helpful.

    • @meet7655
      @meet7655 ปีที่แล้ว

      Same question

  • @shreyanshpagaria8201
    @shreyanshpagaria8201 3 ปีที่แล้ว

    Nice explanation

  • @somoprovahalder1819
    @somoprovahalder1819 2 ปีที่แล้ว

    Thank you so much ma'am

  • @fidanguliyeva519
    @fidanguliyeva519 4 ปีที่แล้ว +4

    I loved the lesson, very clear and informative! One question: which book the example is from?

    • @arslanmunawar7608
      @arslanmunawar7608 2 ปีที่แล้ว

      I think culson rechardson 4 edition ..🤔.

  • @Ndebelevf
    @Ndebelevf 10 หลายเดือนก่อน

    Shouldn't we put mCp as negative for hot streams as they are the ones that losing heat instead of the cold streams?

  • @sagarkacha246
    @sagarkacha246 11 หลายเดือนก่อน

    How to select approach temperature.??

  • @drew48845
    @drew48845 3 ปีที่แล้ว

    The audio is clear, but very quiet. Thank you nevertheless.

  • @ayeshaaftab6227
    @ayeshaaftab6227 3 ปีที่แล้ว

    I didn't get the calculation of mcpdeltaT

  • @gmrgmresh9665
    @gmrgmresh9665 2 ปีที่แล้ว

    At 14:10 i think the heating utility that needed to heat the cold stream is on the other side from 102 to 190. Not from 90 to 102 right ?

    • @nylepentik2696
      @nylepentik2696 6 หลายเดือนก่อน

      u mean it should be:
      Q = 700 = 8(T - 90) ?
      so it gets to T = 177.5C using the remaining kW of H1
      (if so then this is also wut i was wondering)

  • @Alexander-qi7pc
    @Alexander-qi7pc 3 หลายเดือนก่อน

    how is ΔT = 50 4:20

  • @muhammadaidilasri2314
    @muhammadaidilasri2314 3 ปีที่แล้ว +3

    Is it Cc which should be less than Ch in the cold side (below the pinch)? And Cc should be greater than Ch in the hot side (above the pinch). Refer to FCp out should be greater than FCp in.
    th-cam.com/video/xZO2aSiakuw/w-d-xo.html

    • @c.o2307
      @c.o2307 ปีที่แล้ว

      I agree

  • @od4009
    @od4009 ปีที่แล้ว

    What if I have phase change do I use enthalpy instead?

    • @pattonluks
      @pattonluks  ปีที่แล้ว

      Yes, you would need to do the actual enthalpy change. That is a significant complication, so I would definitely recommend double-checking your work!

  • @asdasdasdasd4528
    @asdasdasdasd4528 3 ปีที่แล้ว +1

    Clear explanation but didin't quite get where does number of HEX's, heaters and coolers are coming from. Thanks.

  • @javadjeddizahed1967
    @javadjeddizahed1967 2 ปีที่แล้ว

    There is a problem in the allocation of heat exchangers on the cold side 10:50 . The minimum temperature approach is 10C but you got a tempreture of 75 for the hot side which is even less than the temperature of 77.5 C for the cold side. Simply, you do not meet the requirement of the minimum temperature approach and even cross the basics of designing which means the cold side is hotter than the hot sides

  • @kamalnecefli9735
    @kamalnecefli9735 2 หลายเดือนก่อน +1

    completely misleading video. Thanks for your effort but you've made several mistakes. Firsty, temperature is not 102 celcius, it's 177.5. Secondly and must importantly, on the left side(above) Cc>Ch, not (Ch>Cc). On the right side (below) it's Ch>Cc. (not Cc>Ch) That changes the whole solution.