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Specific enthalpy at dryness fraction

  1. Apr 27, 2011 #1
    1. The problem statement, all variables and given/known data
    Hi there,
    I am trying to complete a thermodynamics assignment. I am doing fine until I get to the following question.

    "Hx = Specific enthalpy at dryness fraction = hf + x . hfg" - I am not quite sure where Hf came from, I guess they want a value for Saturated liquid from the Enthalpy tables but which one?

    2. Relevant equations

    I have two Hf's I have already calculated.

    Hf @ Ti 63 kJ/kg
    Hf @ Tb 419.2 kJ/kg

    These are just the values for saturated water at 15°C and 100°C.

    Other relevant equations:

    X = Dryness fraction = Mev / Mw = 0.013kg / 1.805kg = 0.007
    Hfg at boiling temperature = 2256.4
    Hev = Hx - Hf @ Ti
    Qev = Mw * Hx

    3. The attempt at a solution

    I would just like to add that I have calculated all of the above, they were not given to me.

    When I put the numbers I assume is supposed to be there I get.

    Hx = Hf + x * Hfg = 419.2 - 0.007 * 2256.4 = 435
    Hev = Hx - Hf @ Ti = 435 - 63 = 372
    Qev = Mw * Hx = 1805grams * 435 = 785175J

    Energy delivered by kettle
    Qk = Pk * t = 2100watt * 310.4 = 651840J

    Efficiency of my kettle
    Nev = Qev / Qk = 785175 / 651840 = 1.20 or 120% lol

    That can't be 120% so I have gone wrong somewhere and I think it is to do with "Hx = Specific enthalpy at dryness fraction = hf + x . hfg"

    If anyone could shed some light that would be great.

    Many Thanks.

    Sorry for the long post.
  2. jcsd
  3. Apr 28, 2011 #2


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    Gold Member

    Just to make sure i understand the experiment so that i can help -- So you had ~1.8kg of water in a kettle at 15°C which you evaporated fully? And you measured how long it took, along with the kettles power rating, to find the energy required which you then compare with the theoretical energy required? Is that the experiment you did?
  4. Apr 28, 2011 #3
    Hi there,

    You are nearly right. It's my fault for not explaining the situation properly.

    I had 1.8kg of water which was at 15°C, I boiled that in a kettle and timed how long it took to boil and switch off (310.4s). Then I did some calculations using specific heats to work out my kettles efficiency. Then I did calculations using Steam Tables. I got 98.4% for the former and 98.6 for the latter methods respectively.

    Then I had to do the calculations using steam tables and taking evaporation into account; of which there was 13g (1792g total after kettle had boiled) of water lost to evaporation.

    This is where I am getting stuck and I have calculated 120% efficiency which must be wrong surely?

    Many thanks.
  5. Apr 28, 2011 #4


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    Gold Member

    So you had water at 15 degrees, it was heated to 100 degrees and then 13g was evaporated?

    Have i interpreted that correctly? I did the calculation and got an efficiency of more than 100% which as you have noted is not possible.

    I also noticed that you have used 2256.4 kJ/kg for the vapour enthalpy, however my textbook (Engineering and Chemical Thermodynamics, Koretsky) says that it should be 2676.0 kJ/kg which is significantly different. Did you maybe accidentally read the enthalpy of vaporisation instead of the vapour enthalpy (my textbook quotes the enthalpy of vaporisation as 2257.0 kJ/kg, so this is a possibility)?
  6. Apr 28, 2011 #5
    13g of water evaporated during the heating of the water to 100°C yes

    You are right I have used enthalpy of vaporisation, is this not correct? Is this also known as Hfg?

    For your information, I have just tried the calculations with 2676 kJ/kg and it has raised the efficiency from 120% to 121%

    This really has me stumped. I keep going over everything and I'm starting to mix things up because I've been staring at it for too long.
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