3 replies to this topic

[jorgefherrero]

Dear All,

My question is on the background of Appendix A in API 2000 5th Edition

I would like to understand the physics behind API 2000 rules.

Looking at API 2000 5th Edition, in Appendix A, it says the calculation for outbreathing flow for liquids with flash point below 37,8ºC is 2,02 Nm3/h, where:

- 1,01 Nm3/h per cubic meter represents vapor displacement due to liquid movement
- 1,01 Nm3/h per cubic meter is due to partial evaporation of the incoming liquid.

It considers 0,5% evaporation rate, and specific gravity of the vapor phase is 1,5 times that of air. Air density in Normal condition is around 1,29 kg/m3. So the hydrocarbon considered as example has a density of 1,29 x 1,5 = 1,95 kg/m3

so we have an evaporating mass flow of: 1,95 *1,01 = 1,97 kg/h per cubic meter incoming liquid.

If this is 0,5% of total incoming flow, we have as total flow: 1,97 / 0,005 = 394 kg/h per cubic meter incoming liquid

so the liquid density of the considered hydrocarbon is: 394 / 1 = 394 kg/m3. Which hydrocarbon has been considered? In the Introduction of this Standard it refers to studies carried out on hexane. But this venting calculation can not be hexane, since hexane has 655 kg/m3 liquid density and specific gravity of vapor phase around 3 times that of air

Are my calculations right, or am I missing some point? Can anyone help?

[fallah]

jorgefherrero,

API 2000 did state that the evaporation rate of 0.5 % is selected on the basis of the gasoline being pumped into an essentially empty tank.

[kkala]

I cannot state experience on API 2000, having only once used its tables without much thinking. For want of something better, following notes could be of some help, based on data mentioned in previous posts.
One has to suppose that a volatile liquid hydrocarbon (HC, or rather a mixture of them) comes into the tank, where evaporation 0.5% w/w occurs for some reason (degasing?).

1. For 1 m3 of incoming liquid HC, evaporation produces 1.01 Nm3 of HC vapor = 1.01/22.414=0.045061 kgmol. This is 0.5% of incoming liquid, being 0.045061/0.005=9.01 kgmol. So its "density" is 9.01 kgmol/m3, indicating a liquid like C5H12 (8.69 kgmol/m3). This "density" is lower for C6H14 (7.66 kgmol/m3) or C7H16 (6.84 kgmol/m3) <http://en.wikipedia.org/wiki/Alkane>.

HC liquid can be a mixture of light components, "average" composition approaching C5H12.

2. Density of HC vapor is understood as 1.5 in respect to air, so (average) MW = 1.5×28.96=43.44; this corresponds to C3H8 (MW=44), .more volatile than pentane

3. Suppose that 1 m3 of liquid C5H12 (626 kg) comes into the tank. Evaporation will be 626×0.5%=3.13 kg, with C3H8 as an average composition. Volume of the latter 3.13/44×22.414=1.59 Nm3. Total volume for displacement  V = 1×0.995+1.59=2.585 Nm3 (versus 2.02 Nm3).

4. An interpretation for a more clear picture would be welcomed, as well as explanatory notes. No interpretation has been given so far, that is why this post is sent.

Note 1: Tank, apparently fixed roof, has to be nitrogen blanketed, since flash point is lower than 37.8 oC . This can make issue more complex  (displaced gas will entrain also N2).

Note 2: API2000 is known to be conservative, at least in some points

Edited by kkala, 09 February 2013 - 09:24 AM.

[jorgefherrero]

Thanks, kkala. I find very interesting your hint that a mixture of hydrocarbons has been considered, where the 0,5% portion evaporating is the most volatile part.

 

But I do not agree with one part of your calculation. Where you say:

 

<< For 1 m3 of incoming liquid HC, evaporation produces 1.01 Nm3 of HC vapor = 1.01/22.414=0.045061 kgmol. This is 0.5% of incoming liquid, being 0.045061/0.005=9.01 kgmol.>>

 

I understand 0,5% is meant as weight proportion, not as molar proportion. We get 0.045061 kgmol evaporating hydrocarbon, but this does not mean the incoming liquid is 9.01 kgmol. (It would be OK, if we had the same composition in the evaporating part as in the remaining liquid phase) But as you say later, a heterogeneous mixture of hydrocarbon as been considered, where the 0,5% evaporating part is the most volatile one, with different composition as the remaining liquid.