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Standard Storm Page
- Thread starter Burning_Full_Fingers
- Start date
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- 1,314
This again? Pretty sure we had the same debate in the past, but I have no idea where it went.
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Assault was planning to make a revision about what density we should use, the really is that 1.003 is not the density of clouds if not the density of the air at the height lf the clouds; but you known, he left.
- 15,181
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- Thread starter
- #4
I don't really know if it was discussed before. I'm just asking if the link in our page should be replaced with one that shows 1.003, because the current link shows 0.0005.
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The calc about the 1.003 kg/m┬│:
In fact at 2 km the atmosphere density is of 1.007 km so the 1.003 km definitely isn't the "density of the atmosphere at this altitude".
Of course, this is, not only including water droplets, but also water vapor. We have already had two discussions on this subject, but I find that no one remembers them or they are "lost" even though they are relatively recent (I am also surprised that the link was not updated and in the previous discussion the calculation was also shown).
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| The ideal gas law states that: DENSITY = P / TR (pressure(P) divided by temperature(T) multiplied by a constant(R)) From the average values found in our atmosphere, we can use a temperature at about 2 km altitude of 275.15K (+2.15C) and a pressure at 2 km of 79.495KPa (795 mb). For dry air, the constant(R) equals 287 J/K*kg. This gives us a density of 1.007 kg/m┬│ for dry air. For pure water vapor, we would use a constant(R) of 461 J/K*kg which ends up giving us a density for pure water vapor of 0.627 kg/m┬│. The density of the cloud droplets in the cloud comes out to .0005kg/m┬│. Because the cloud is made up of a small amount of water vapor and a large amount of air, we would need to calculate the partial pressure of the water vapor, which in this case comes out to around 7 mb, or 0.9% of the total pressure (795 mb). So, [99.1 X (dry air density) + 0.9 X (moist air density)]/100 is approximately the density of the cloud itself. Calculating this gives a density in the cloud of 1.003 kg/m3, compared to 1.007 kg/m3 in the dry air surrounding the cloud. |
Of course, this is, not only including water droplets, but also water vapor. We have already had two discussions on this subject, but I find that no one remembers them or they are "lost" even though they are relatively recent (I am also surprised that the link was not updated and in the previous discussion the calculation was also shown).
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I think you means kg/m^3, not km; and to be fair 1.003 and 1.007 is practically the same, varying only 0.4%
Anyway, if you find a source for the 1.003 kg/m^3 better place it in there for now.
Anyway, if you find a source for the 1.003 kg/m^3 better place it in there for now.
