10
$\begingroup$

I am using a data set from Justin Sheffield at Princeton that includes specific humidity. The description is as follows:

"Reanalysis 6-hourly specific humidity interpolated to 1.0deg with account for elevation changes and scaled to maintain consistency with reanalysis relative humidity and CRU TS3.0 monthly air temperature";
The units are = "kg kg-1"

What does this mean? Does it mean kg/kg so essentially dimensionless? Because I have seen around specific humidity measured also in g/kg, maybe that was wrong? These two pages give the same definition of specific and absolute humidity, I thought they were different things: http://www.nzifst.org.nz/unitoperations/drying3.htm http://tigge.ecmwf.int/tigge/d/show_object/table=parameters/name=specific_humidity/levtype=pl/

Improve this question
$\endgroup$
2
  • $\begingroup$ Humidity is generally expressed as a ratio, which is dimensionless. E.g $kg/kg$ as in your example. $\endgroup$
    – user2821
    Commented Jun 23, 2015 at 8:32
  • $\begingroup$ $1\,{\rm g}/{\rm kg} = 0.001$. Yes, that's technically a valid (dimensionless) unit. $\endgroup$
    – Ilmari Karonen
    Commented Jun 23, 2015 at 10:13

3 Answers 3

Reset to default
12
$\begingroup$

Yes, the unit is kilogram per kilogram, but it means kilogram of water (moisture) per kilogram of air.

Improve this answer
$\endgroup$
6
$\begingroup$

Just to expand on yo's answer:

It's unit-less so, mathematically, you could omit the kg/kg. However then the reader wouldn't know if you're expressing a quantity as a percent weight or a percent volume which are different.

Improve this answer
$\endgroup$
5
$\begingroup$

The basic idea is that by mentioning $\mathrm{kg}\, \mathrm{kg}^{-1}$ you make it clear that you measure the weight. You can think of it as

$$ 0.01\, \mathrm{kg}\, \mathrm{kg}^{-1} = 1\, \mathrm{wt.}\%. $$

This is to avoid confusion with $\mathrm{vol.}\,\%$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ kg is mass, N would be weight, right? $\endgroup$
    – Thomas Weller
    Commented Jun 23, 2015 at 20:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.