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10 October, 12:08

Why in a stream containing water, the mole fraction of a given

component as calculated on a wet basis will always be less

than the mole fraction of that same component as

calculated on dry basis.

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Answers (1)
  1. 10 October, 15:24
    0
    A liquid, at any temperature, is in equilibrium with its own steam. This means that on the surface of the liquid or solid substance, there are gaseous molecules of this substance. These molecules exert a pressure on the liquid phase, a pressure known as vapor pressure.

    In chemistry, when we talk about dry basis, we talk about a state in which the presence of water in a gaseous state is denied for the calculation. So vapor pressure equals zero.

    When we talk about the wet basis, the presence of water in the steam is considered for the calculation, which normally is expressed as a percentage or moisture.

    In summary, for a gas mixture steam:

    For dry basis, we just have component A, component B ... For wet basis, we have water vapor, component A, component B ...

    So, in wet basis we have an extra component (water).

    Assuming we only have 2 components in our steam, and being X the molar fraction of eact component:

    For dry basis: Xa + Xb = 1 ... Xa = 1 - Xb For wet basis: Xa + Xb + Xwater = 1 ... Xa = 1 - Xwater - Xb

    For dry basis the mole fraction of A it is obtained by subtracting the molar fraction of B from one. And for wet basis, we have to substract the molar fraction of B AND the molar fraction of water vapor. So, logically, the mole fraction Xa will be less for wet basis.
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