Dissolved gas in a liquid - units problem

[tesser]

This may be very obvious, but not to me I'm afraid.
I have a pump data sheet in front of me. It gives the product as crude oil with an SG of 0.752 @ 22ºC and a Vapour Pressure of 8.5 Bara at 22ºC. Under H2S concentration it gives '92250ppm v'.

My question is what does '92250 ppm v' really mean? I assume that it means ppm by volume. I have no problem with ppm w/w for liquids (or gases) or ppm v/v for gas mixtures but what does 'ppm v' for a gas in a liquid mean. Is there some sort of convention that when expressed this way it means volume of gas under standard conditions? Or does it mean the volume of gas at the temperature and pressure at the pumping condition? If standard conditions which? (NTP or MSC or Imperial).

The question arises because I was trying to work out where it fits in the NACE MR0175/ISO 15156-2 standards for materials on fluids containing H2S.

 

[25362]

Try to look for Ostwald coefficients (ASTM D2779).
The Ostwald coefficient is defined as the ratio of the volume of dissolved gas to the volume of solvent liquid at the test temperature and pressure.

 

[mbeychok]

tesser:

What crude oil is it? Is it a petroleum crude oil? I find it virtually impossible to believe that any petroleum crude oil has a vapor pressure as high as 8.5 bara at 22 °C. I also find it even harder to believe that a petrolem crude oil has a specific gravity as low as 0.752 at 22 °C ...at 15.6 °C (60 °F), a crude oil with that specific gravity would be about 57 °API and there simply are very, very few petroleum crude oils (if any at all) that have gravities as light as that.

Milton Beychok
(Visit me at www.air-dispersion.com)
.

 

[epoisses]

or 9 vol% H2S! I'd get back to the person who produced the datasheet for clarification.

 

[SJones]

Get the process engineers involved and look at the note under Annex C, C.2 a) of ISO 15156-2

Steve Jones
Materials & Corrosion Engineer

http://www.pdo.co.om/pdo/

 

[25362]

Dissolved gas (as in gas driven recovery) may affect API gravities and vapor pressures. Broadly speaking, the lower the density (higher API) of the crude oil the greater the amount of gas it can dissolve at a given pressure and temperature.

Assuming H₂S (MW:34) is an ideal gas, its mass %, at NTP, could be roughly estimated as follows:

100[×](92,250[×]34[÷]22.4)[÷](1,000,000[×]765)=0.18% mass​

Both oil and gas taken at the same temperature of 0^oC. 765 g/L taken as density of the crude.

 

[tesser]

Thanks guys,

at least I wasn't missing something totally obvious. I should explain that I work for a mechanical seal manufacturer and as data comes down to us from process people, to rotating equipment people, to pump manufacturer to seal maker, information tends to get filtered (left out) not always helpfully. My initial reaction when somebody showed me the pump data sheet was the same as mbeychok's - though less politely expressed - then I wondered if it was the H2S elevating the bubble point pressure. Hence my question.

Regarding Steve Jones's comment. I do have a very cheap (free off the web) and nearly cheerful Vapour Liquid Equilibrium program which I can modify for several components, and could, given the information (composition, pseudocomponet properties etc), go through the procedure outlined in NACE MR0175/ISO 15156 unfortunately unless we are very lucky we don't get the information to do it.

Thanks again for the feedback.