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How to convert Nm3/hr (normal cubic meter) of nitrogen to kg/hr?

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  • How to convert Nm3/hr (normal cubic meter) of nitrogen to kg/hr?

    Haii.. Can anyone explain to me how to convert Nm3 (normal cubic meter) of nitrogen to kg?
    I have 820 Nm3/hr of Nitrogen at 1.4 bar(g) and 76oC with the density is 2,3256 kg/m3. The molecular weight of nitrogen is 28,01 g/mole. The normal condition refers to 1 bar and 25oC. Is it correct that the answer is 1965.34 kg/hr? or it's about 952 kg/hr?

    Thank you for help me out ^_^

  • #2
    Originally posted by loveline719 View Post
    Haii.. Can anyone explain to me how to convert Nm3 (normal cubic meter) of nitrogen to kg?
    I have 820 Nm3/hr of Nitrogen at 1.4 bar(g) and 76oC with the density is 2,3256 kg/m3. The molecular weight of nitrogen is 28,01 g/mole. The normal condition refers to 1 bar and 25oC. Is it correct that the answer is 1965.34 kg/hr? or it's about 952 kg/hr?

    Thank you for help me out ^_^
    You need the density of your gas at your "normal" conditions. Then multiple by flow rate.

    The problem is slightly over-specified and there are multiple ways to approach this, both with the Ideal Gas Law and better equations of state for gases that include departure from ideality (compressibility factor):
    1. Calculate density from molecular weight, ideal gas law and gas constant at your normal conditions
    2. Cast your measured density to your "normal" conditions.
    3. Cast published density at 1.01325 bar, 0 C to your "normal" conditions

    I am not going to bother with compressibility factor, I will assume Ideal Gas Law is "good enough" and use published density of 1.2506 kg/m at 1.01325 bar (1 atm), 0 C.
    At your "normal" conditions:

    1.2506 kg/m x 1 bar/1.01325 bar x 273.15 K/298.15 K = 1.130754 kg/m (with a couple of guard digits)
    Multiple by 820 Nm/h, getting 927 kg/h.

    This published figure, cast to 2.4 bar(a) and 74 C, yields a density of 2.3308 kg/m, not in perfect agreement with your measured density, but close. I think it supports a three significant figure answer. Anything better is going to require compressibility factor and the calculation is pretty messy.

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    • #3
      Thank you so much John, I get your point ^_^

      Comment


      • #4
        Thank you so much John, I get your point ^_^

        Originally posted by JohnS View Post

        You need the density of your gas at your "normal" conditions. Then multiple by flow rate.

        The problem is slightly over-specified and there are multiple ways to approach this, both with the Ideal Gas Law and better equations of state for gases that include departure from ideality (compressibility factor):
        1. Calculate density from molecular weight, ideal gas law and gas constant at your normal conditions
        2. Cast your measured density to your "normal" conditions.
        3. Cast published density at 1.01325 bar, 0 C to your "normal" conditions

        I am not going to bother with compressibility factor, I will assume Ideal Gas Law is "good enough" and use published density of 1.2506 kg/m at 1.01325 bar (1 atm), 0 C.
        At your "normal" conditions:

        1.2506 kg/m x 1 bar/1.01325 bar x 273.15 K/298.15 K = 1.130754 kg/m (with a couple of guard digits)
        Multiple by 820 Nm/h, getting 927 kg/h.

        This published figure, cast to 2.4 bar(a) and 74 C, yields a density of 2.3308 kg/m, not in perfect agreement with your measured density, but close. I think it supports a three significant figure answer. Anything better is going to require compressibility factor and the calculation is pretty messy.

        Comment

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