Some water measurements for the Flood

The text says the firmament is heaven. No problem there.But then there is a vault of the sky that separates the waters in heaven from the waters on earth. Which means that your look at the commentaries does not resolve the issue.Then the text indicates that on day four lights separating day and night and indicating the seasons were placed in vault of the sky.

I'll check, but I think you are taking it outside the valid range. But the calculations are correct. I'll put a more complete answer together, hopefully today.{ABE} I started on the full explanation, and figured out where you are going wrong. See the first word in cell B12. The pressure under consideration is normal atmospheric pressure (assumed due solely to oxygen and nitrogen and argon and a few trace gasses) *plus* the pressure due to water vapor in the atmosphere. (Actually it could be partly liquid, clouds, above the Earth's surface without affecting the result as long as it's all vapor at the Earth's surface).The boiling point of water at 15 psi is 100 C. The boiling point of water at (15 + ε ) psi, where ε is small compared to 15 is 100 C within the precision we need here.And we are looking at pure vapor. Vapor can obviously exist at temperatures lower than 100 C and pressure 15 psi, since humidity exists, but water can also exist; it depends on other factors. But for just vapor no matter what other factors you have to be in the vapor area. A Pascal (Pa) is 6,895 psi and a Kelvin (K) (not degrees K, just K) is C + 273.15. The red dots are "critical points". The brownish area to the right of the liquid area is "supercritical fluid", both liquid-ish and vapor-ish. The point where all three phases touch is a "triple point" where the three phases are indistinguishable. The little blue X is standard temperature and pressure. If you move right from the blue X you get to the vapor/liquid line at (surprise) 373 K or 100 C.It's clearer on a pressure-enthalpy diagram with lines of constant temperature, from STEAM P-H DIAGRAM: The purple-bounded "bell" area is mixed liquid and vapor. To the left of the bell is liquid, to the right of the bell is vapor. The triple point is at the top of the bell Note that the lines of constant temperature are horizontal within the "bell". When a system is mixed water and steam, somewhere along one of those horizontal lines, the percentage of how far along that line it is is the "quality" of the steam. My spreadsheet is calculating the temperature where a line of constant pressure intersects the right edge of the "bell". That's the minimum temperature at which you can have pure vapor. Increase the temperature by the eentsiest bit or more and you must have pure vapor.Clear as mud?{ABE2}Cells A12, A13, and A14 should be set to display two figures to the right of the decimal point.{ABE3} The C to F conversions are correct, multiplication and division take precedence over addition. See the online calculator at Online Conversion - Temperature Conversion.Edited by JonF, : No reason given.Edited by JonF, : No reason given.Edited by JonF, : No reason given.Edited by JonF, : No reason given.

Sure. I understand how phase diagrams work; that is not the issue I was objecting to. What I was saying is that your spread sheet says that for there to be ANY vapor in the atmosphere, the temperature would need to be 100^oC. We know that is not the case as there is water vapor in the atmosphere at normal temperatures.So I am not overly convinced of the Temperature required to maintain that water as vapor calculations. I suppose that once water vapor exceeds the saturation point (past equilibrium point), it would require temperatures over 100^oC to maintain that high level of water vapor. Yes, they are correct if you are converting absolute temperatures from ^oC to ^oF. However , in cell A20, you are converting a change in ^oC to a change in ^oF. For this conversion, the formula is [delta]^oC * 1.8 = [delta]^oF.For example, a 10^oC change in temp converted to ^oF, say from 10^oC to 20^oC.10^oC = 50^oF
20^oC = 68^oF68^oF - 50^oF = 18^oFSo, a 10^oC change in temp is an 18^oF change, not 50^o that you would get if you converted using a standard (^oC * 9/5)+32 conversion.Anyway, I tried a slightly different approach to this problem. I calculated the total volume of the atmosphere and calculated the total amount of water that could be vapor in the atmosphere if the entire troposphere was at 4.25% saturation. At saturation levels greater than 4.25%, water begins to condense and precipitate out.The volume of the troposphere is about 3.0e20 ft³ so, 4.25% yields 1.28e19 ft³ of water vapor which converts to 1.03e16 ft³ of liquid water. This is enough to cover the surface of the earth to a depth of 2 feet! Using your spreadsheet calculation, this would raise the temperature of the earth 105^oC (189^oF).I also found this source The USGS Water Science School that says:

quote:

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One estimate of the volume of water in the atmosphere at any one time is about 3,100 cubic miles (mi3) or 12,900 cubic kilometers (km3). That may sound like a lot, but it is only about 0.001 percent of the total Earth's water volume of about 332,500,000 mi3 (1,385,000,000 km3), as shown in the table below. If all of the water in the atmosphere rained down at once, it would only cover the globe to a depth of 2.5 centimeters, about 1 inch.

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My calculations above are 23 times the actual estimated volume according to the USGS and still can’t come even close to covering anything close to a mountain.HBDEdited by herebedragons, : No reason given.

I'll go into this more, but my spreadsheet is not saying that it must be 100 C to have any water vapor in the atmosphere. It must be 100C for liquid water to not exist in the atmosphere. The initial condition was solely vapor in the atmosphere.

Yeah, that's what I gathered after figuring how much vapor could actually exist in the atmosphere under normal conditions. When you are calculating the water vapor to produce 60 - 4000 feet of rain, the water vapor that is in the atmosphere initially is insignificant. No need. It makes sense to me.HBD

OK.You are correct about the delta T conversions, I've fixed it and updated the online copy.