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Energy Balance for closed rigid container

  1. Oct 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Water, initially saturated vapor at 4 bar, fills a closed, rigid container, The water is heated until its temperature is 400C. For the water, determine the heat transfer in KJ per kg of water .

    2. Relevant equations


    3. The attempt at a solution
    Assume potential and kinetic energy=0, W=0,Steady state

    I got the energy balance right: Q/m=(u2-u1)

    Found v1 and u1 from steam tables:
    specific volume: v1=v2=0.4625 m3/kg
    u1 = 2553.6 KJ/kg

    I know that state two is super heated, but how do I know what the pressure is?? What values do I interpolate with to get u2??
     
  2. jcsd
  3. Oct 19, 2015 #2

    SteamKing

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    Well, you have a closed, rigid container. What does that tell you about the kind of heating process which is going on when the vapor is heated to 400° C?
     
  4. Oct 19, 2015 #3
    I'm not really sure, the change in internal energy is what causes the increase in heat? Or is there an increase in temperature because the pressure increases?
     
  5. Oct 19, 2015 #4

    SteamKing

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    Think about the container.

    It's rigid, so does its volume change with increasing pressure inside? Does the amount of water inside change while it is being heated?

    So what property must be the same at the start of heating and when it is finished?
     
  6. Oct 19, 2015 #5
    I know the specific volume is constant, but I'm still confused about how to do the interpolating. I guess I'm supposed to use specific volume and internal energy to interpolate? I'm really lost at this part.
     
  7. Oct 19, 2015 #6

    SteamKing

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    How can you use the internal energy to interpolate? Isn't that what you're looking for, the internal energy of the vapor after it has been superheated?

    If you know the specific volume is constant, what does this tell you about finding the properties of the superheated vapor?
     
  8. Oct 19, 2015 #7
    I really don't know :( In the solution I have they used the table p=5bar and p=10 bar to interpolate. I don't understand how they knew to use these tables.
     
  9. Oct 19, 2015 #8
    But then another solution said to use p=5bar and p=7bar instead...
     
  10. Oct 19, 2015 #9

    SteamKing

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    Different steam tables are tabulated differently. Some may tabulate superheated properties for pressure increments of 5 bar, some for closer increments of pressure.

    You've answered your own question, and apparently you don't realize it. If the vapor has a constant specific volume, that's the property you search for when interpolating the superheated vapor tables at T = 400° C.
     
  11. Oct 19, 2015 #10
    I think both solutions are using the same tables. What you said makes sense, so I just look for the pressure table that has a value close to my constant specific volume at my given temperature.

    What formula are they using here though: (0.4397-0.4625)/(0.4397-0.6173)=(2960.9-u2)/(2960.9-2963.2)
    I feel like this is different than the formula I was given to use.
     
  12. Oct 19, 2015 #11

    SteamKing

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    It looks like someone is using linear interpolation to find the value of u for T = 400 °C and v = 0.4625 m3 / kg

    https://en.wikipedia.org/wiki/Linear_interpolation

    Check the property values in the superheated vapor table.
     
  13. Oct 20, 2015 #12
    Thanks so much!
     
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