Physics Help

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1. The total enthalpy of 15.000kg of steam at 2250.000kPa is 34191.510kJ. Determine the dryness of the steam.
ANS = %


2. Determine the quantity of heat required, to raise 11.100kg of water at 80.000 degrees Celsius, to saturated steam at 2250.000kPa and 71.400% dry.
ANS = kJ

Not sure where to start. I need a little help please. I do have a chart with the properties of saturated steam. Thank you.

  • Physics Help -

    Since you've posted this question about eight times in the last few days, I think we can assume that none of our tutors can help you.

    Sorry.

  • Physics Help -

    (A)Ht=Hs•ζ+(1-ζ)Hw,
    where
    Ht = total (actual) enthalpy (kJ/kg),
    Hs = enthalpy of steam (kJ/kg),
    Hw = enthalpy of saturated water or condensate (kJ/kg)
    Using steam-pressure table
    http://enpub.fulton.asu.edu/ece340/pdf/steam_tables.PDF
    we can find the magnitudes for p=20 bar
    (the given data in the problem is p=2250 000 Pa = 22.5 bar)
    Hs =2800 kJ/kg
    Hw = 908.8 kJ/kg and.
    Ht=34191. 510/15 =2279.434 kJ/kg

    Ht=Hs•ζ+ Hw –Hw•ζ
    ζ(Hs-Hw)=Ht-Hw
    ζ= (Ht-Hw)/ (Hs-Hw)=
    =(2279.434-908.8)/(2800-908.8) =
    =1370.634/1891.2=0.725
    Answer: 72.5% (the answer may differ due to using of 20 bar instead of 22.5 bar)
    (B)
    Q=Q₁+Q₂+Q₃
    Q₁=cmΔt =4186•11.1•20 = 929292 J = 929.292 kJ
    Q₂= = mL =11.1•2260000 =25086000 J= 25086 kJ
    Q₃= Ht ≈34191.510 kJ (I believe that we can use the result of part (a)
    Q = 929.292+25086+ 34191.510 ≈ 60206.802 kJ

  • Physics Help -

    Thank you, Elena. I was wrong.

  • Physics Help -

    Another table http://www.chem.mtu.edu/~tbco/cm3230/steamtables.pdf
    gives more precise values
    Hs =2801.7 kJ/kg
    Hw = 936.48 kJ/kg
    As a result
    ζ= (Ht-Hw)/ (Hs-Hw)=
    =(2279.434-936.48)/(2801.7-936.48) =
    =13.42.954/1865.22=0.7199 => ≈72%

  • Physics Help -

    Using Problem 5 page 234 from
    http://books.google.com.ua/books?id=PQRo3XuHHvYC&pg=PA262&lpg=PA262&dq#v=onepage&q&f=false

    From steam table
    at 22.5 bar
    Hw= 936.48 kJ/kg
    He = 1865.2 kJ/kg - enthalpy of evaporation (Latent heat)
    Enthalpy of 1 kg of wet steam is
    H=Hw+ζ •He=
    =936.48 +0.714•1865.2 = 2268.23 kG/kg
    Heat already at water (water is at 80℃) = 4.186•80= 334.9 kJ/kg
    Heat required per 1 kg of steam = {Enthalpy (or) heat required to raise 1 kg of steam from water at 0℃} – {Heat already present in water} = 2268.23 -334.9 = 1933.33 kJ/kg
    Heat supplied per kg = 1933.33 kJ/kg.
    Heat required for 11. 1 kg =11.1• 1933.33= 21459.963 kJ

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