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Ideal gas law and its use

  1. Feb 13, 2012 #1
    I am trying to solve a problem but I am not sure if I am going about it the right way. I am using the ideal gas law to find out the amount of steam it will take to open a relief valve. Using the law I have calculated the amount of steam that will be in a volume but I am not sure if that is also the point the relief valve will open or is there a formula to account for the steam compression in the volume.

    Data
    volume of container = 213.75m3
    area of container = 0.21375m2
    steam temperature = 573 kelvin
    thermal conductivity of five walls of container = 0.003 W/m/k
    heat loss through one wall = 10 000 watts
    relief valve = 86 bars
     
  2. jcsd
  3. Feb 14, 2012 #2

    jim hardy

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    saturated steam is not ideal gas.

    use steam tables to find specific volume of steam at your T & P.
     
  4. Feb 14, 2012 #3
    thank you for your answer i thought that i should use steam tables. this is the way i thought of using the table. say the specific volume is 0.123m3 at a unamed Temp and Pressure and my container has a volume of 0.006m3. 0.123/0.006=? does the answer mean that the container will be filled up at ?. can i use the real gas law that takes into account the compresablity factor
     
  5. Feb 15, 2012 #4

    jim hardy

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    specific volume is 1/density.

    Sorry , i usually work in English units, cubic feet and BTU's and Fahrenheit
    but i'll try.

    at the temperature you gave in first post
    573K


    from this steam table calculator
    http://www.efunda.com/materials/water/steamtable_sat.cfm

    i get specific volume of 46.053 kg/m^3 and pressure of 85.698 bars, which i took to be your 86 bars rounded to nearest integer. Note it's absolute pressure.
    Since you mentioned cooling through the walls i assumed you meant to use saturated steam.
    An enclosure of one cubic meter would contain 46 kg of saturated steam.
    Well, 46.053 more exactly.

    Were your container 213.75 meters in volume it would contain about 213.75 X 46.053 = ~9844 kg of steam,

    were it .006 m^3 it would contain 0.276 kg

    Perhaps you'll convert those masses to moles and see how they compare to ideal gas law? Molecular weight of water is 18.015 g/mol.

    "a problem well stated is half solved"
     
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