2 replies to this topic

[mbeychok]

Your question is very unclear. What do you mean by vapour "quality"?

If you divide an amount of vapour in kg-mol/hr by the total amount of liquid and vapour (kg-mol/hour), you simply get the mol fraction of vapor in the total liquid and vapor stream.

If the liquid and vapor each have the same molecular weight, then the mol fraction of vapor in the total stream equals the weight fraction of vapor in the total stream.

If the vapor and liquid molecular weights are not the same then:

weight fraction of vapor = (vapor flow, kg/hr) / (vapor flow, kg/hr + liquid flow, kg/hr)

where:
vapor flow, kg/hr = (vapor flow, kg-mol/hr) (vapor molecular weight)
liquid flow, kg/hr = (liquid flow, kg-mol/hr) (liquid molecular weight)

[siretb]

Yes, your approach is correct
What is called the vapor quality is the fraction of vapor divided by total, on a molar basis.
For pure water, since the molecular weight is the same for steam and liquid , it does not make any difference to calculate on a molar or mass value.
For a distillate feed that would contain multiple components, then use molar ratios, not massic ratios.

So vapor, at dewpoint has a quality of 1
Liquid, at bubble point has a quality of 0

Super heated steam can be viewed as having a quality greater than one.

My advice is to rely either on compostion T,P (or P,enthalpy) rather than quality to define the "quality" of steam.