I wanted to use the following unit

cMax = Quantity[6.8221*^-9, "L/s/bar"];

But when calculating with it, I recognised there is some strange behaviour. So I tried to convert it to L/s/Pa

cMax = Quantity[6.8221*^-9, "L/s/bar"]
UnitConvert[%, "L/s/Pa"]

enter image description here

So why do I get the unit IndependentUnit[bars difference]?

I also tried "Liters/Bars/Seconds", "L/s/Bars", "L/s/Bar", ....


Wolfram Alpha had trouble interpreting your first unit and gave "bar" as IndependentUnit["bars difference"], i.e. it did not know it was pressure.

You can use the proper syntax and WA will not need to be called:

UnitConvert[Quantity[6.8221*^-9, "Liters"/("Bars" "Seconds")], "L/s/Pa"]
Quantity[6.8221*10^-14, "Liters"/("Pascals" "Seconds")]

I found the proper syntax by evaluating each unit separately:

FullForm /@ {Quantity[1, "L"], Quantity[1, "s"], Quantity[1, "bar"]}
{Quantity[1, "Liters"], Quantity[1, "Seconds"], Quantity[1, "Bars"]}
  • $\begingroup$ The tip with FullForm was exactly what I needed. I often had such a Problem and never knew how to get the exact form for the unit. $\endgroup$
    – Phab
    Mar 5 '15 at 8:58

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