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Throttling Calorimeter water equivalent

  1. May 9, 2017 #1
    1. The problem statement, all variables and given/known data
    Steam runs through a pipe, and the state of the steam is required to be found. To do this, some of the steam was condensed inside a container filled with water.

    The container weighs 850g, and has 8kg of liquid water inside it initially. The whole system was measured to be in 16 degrees celsius.

    The steam is inserted into the container, and after condensation the new total weight of the container+water+condensed steam is 9.04kg and the new temperature is 31.5 degrees celsius.

    The steam was originally in the pipe at 3.0 MN/m2. To safely pass the steam into the container, it was first passed through an isenthalpic throttle device to reduce the pressure to atmospheric (0.1 MN/m2).

    2. Relevant equations

    Determine the state of the steam in the pipe.

    specific heat of container = 0.384 kJ/kgK and specific heat of liquid water = 4.2 kJ/kgK

    3. The attempt at a solution

    So I did this:
    Inside the container,
    Heat transferred by steam = Heat absorbed by water

    specific enthalphy of steam after throttling x steam mass - steam mass x 4.2 x 31.5 (i.e. the steam that has condensed) = water mass * 4.2 * (31.5-16) + container mass * 0.384 * (31.5-16).

    The steam mass is, from mass conservation, equal to 9.04-8-0.85=0.19 kg

    Solving for specific enthalphy I got it to be 2899.88 kJ/kg
    Since throttling doesn't change the specific enthalphy of the steam, the specific enthalphy of steam after throttling=before throttling.

    And so now we know the pressure and the specific enthalphy at the pipe, we can go to steam tables and realise that the steam is superheated.

    Now however in textbook I saw that they do this instead to the heat transfer equation:

    specific enthalphy of steam after throttling x steam mass - steam mass x 4.2 x 31.5 (i.e. the steam that has condensed) = water mass * 4.2 * (31.5-16) + (container mass *4.2 ) * 0.384 (31.5-16)

    in which they claim that it is the water equivalent of container. Doing it their way gives specific enthalphy of 2985.188 kJ/kg.

    Where did I go wrong in my method? what is this water equivalent thing?
    I also realised I did not use the information about the pressure after throttling. Did I do something wrong?
     
  2. jcsd
  3. May 10, 2017 #2
    In my judgment, you are right and they are wrong. But I don't like the way either you or they have set up the problem.
     
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