Vapor Pressure of Water Below Freezing: How Does It Affect Grain Storage?

[Gopher72]

MOD Note: Moved from Member intros to here hence no homework template.

While its not homework it fits the spirit of homework.

If ambient air is 10 deg C with relative humidity of 75%. And air exiting a grain bin is 18 deg C with relative humidity of 85% at a rate of 3000 cubic feet per second (cfm). How much water is being removed from the bin every hour?

 

[jedishrfu]

Have you attempted to solve this problem yourself? If so can you show us your work?

 

[Gopher72]

Have you attempted to solve this problem yourself? If so can you show us your work?

No, I have not found the right formula yet.

 

[Gopher72]

This is not homework, this is a farmer trying to dry grain in his bin properly.

 

[jedishrfu]

Yes, I know but the problem has elements of homework and so we placed it here.

There's not some magic formula, rather someone familiar with the problem might work out a solution or more likely will provide some hints or links to a solution.

In the past, we have had some crafty students posing as interested parties ie "we're not students" hoping we would do their homework. The only thing we have to go by is the question they ask and if it looks like homework, smells like homework it becomes homework.

 

[jedishrfu]

SmarterEveryday did a grain bin video:



You might get something from it.

Maybe you could fire him a email about it? and he could suggest a resource to pursue.

 

[jedishrfu]

Also I found this reference:

https://www.ag.ndsu.edu/graindrying/documents/eb35.pdf

 

[haruspex]

MOD Note: Moved from Member intros to here hence no homework template.

While its not homework it fits the spirit of homework.

If ambient air is 10 deg C with relative humidity of 75%. And air exiting a grain bin is 18 deg C with relative humidity of 85% at a rate of 3000 cubic feet per second (cfm). How much water is being removed from the bin every hour?

First thing you need is some table to tell you what 100% relative humidity translates to as absolute humidity at each temperature.
https://www.engineeringtoolbox.com/maximum-moisture-content-air-d_1403.html

Multiplying those by the relative humidities gives you the absolute humidities. From that you can find the quantity of water in each cu ft of air. Taking the difference tells you how much water is being removed in ea cu ft:
water per unit volume out = (water per unit volume at 100% RH at temperature out)*(actual RH out)
water per unit volume in = (water per unit volume at 100% RH at temperature in)*(actual RH in)
Water lost per unit time = (volume of air per unit time)*((water per unit volume out) - (water per unit volume in))

 

[gmax137]

I like what @haruspex came up with; I might add a question about the flow rate: is it 3000 cfm at the inlet, at the outlet, or ??

I imagine you have some kind of fan and you know the flow thru the fan, but is it blowing the air into the bin, or is it sucking it out? And the number you have, is it "actual" CFM or is it "SCFM"

I haven't really worked through this, maybe these finer points don't change the answer by much.

 

[Gopher72]

Thx haruspex, I’ve been continuing my research on my own and it seems to always come back to plotting points on a psychometric table. But I am trying to design a computer model that can take the units expressed and convert it to lbs of water/hr removed which can then be calculated against the bin moisture approximate to come up with the time to run my fans. Atmospheric conditions vary very greatly day to day, and by monitoring these changes over time with a computer my accuracy can be greatly improved. So is there actual formulas that the psychometric charts are based upon?

also cfm is measured at output which I realize will not be the same as input but based upon the values given input volume should also be able to be calculated but might not be statistically significant.

 

[haruspex]

Thx haruspex, I’ve been continuing my research on my own and it seems to always come back to plotting points on a psychometric table. But I am trying to design a computer model that can take the units expressed and convert it to lbs of water/hr removed which can then be calculated against the bin moisture approximate to come up with the time to run my fans. Atmospheric conditions vary very greatly day to day, and by monitoring these changes over time with a computer my accuracy can be greatly improved. So is there actual formulas that the psychometric charts are based upon?

also cfm is measured at output which I realize will not be the same as input but based upon the values given input volume should also be able to be calculated but might not be statistically significant.

The tables are based on actual measurements. There are formulas that give a reasonable approximation over some range. See e.g. https://www.conservationphysics.org/atmcalc/atmoclc1.html
You could just hold one of the tables in your program and interpolate as necessary.

 

[Tom.G]

The link in the previous post has, in its last paragraph, a link to a downloadable calculator that runs in your browser. The formulas can be found in the calculator after you download and save it to a file.

 

[jedishrfu]

Did our responses help you figure it out?

 

[Gopher72]

Ok i got it.

First you must use this formula to find the saturation pressure at the given temperature.
Saturation vapour pressure, ps, in pascals:
ps = 610.78 *exp( t / ( t + 238.3 ) *17.2694 )
where t is the temperature in degrees Celsius

At 18 deg C it = 2054 Pa

Relative Humidity (RH) = vapor pressure/saturation vapor pressure (ps) therefore
vapor pressure (p) = ps * RH

p=2054 * .85 (85 percent) = 1746 Pa

kg/m3 = 0.002166 *p / ( t + 273.16 ) where p is the actual vapour pressure...temp is converter to Kelvin

kg/m3 = 0.002166*1746/(18+273.16)
kg/m3 = 0.012989 for air exiting the bin.

Repeat for air entering the bin to get 0.007025 kg/m3

1 ft = 0.3048 m or 1 m = 3.28084 ft
3.28084 cubed = 35.31467 ft cu
3000 (cfm)/35.31467 = 84.95053 m3

If we assume air entering and exiting is the same volume (which I am not sure is a valid assumption because i know that air roughly doubles in volume for every 10 deg C it is raised in temperature).
Then (0.012989-0.007025)*84.95053 = 0.50664496 kg/min or 1.117 lb/min or 67 lbs/hr.

Does anyone know the actual air expansion ration? and is it linear or quadratic?

 

[Tom.G]

air roughly doubles in volume for every 10 deg C it is raised in temperature).

According to this site:
https://en.wikipedia.org/wiki/Density_of_air
10°C results in 3.3% density change, because the temperature used is Kelvin.

Cheers,
Tom

 

[jbriggs444]

know that air roughly doubles in volume for every 10 deg C it is raised in temperature

The rule of thumb you may be thinking of is that reaction rates double for every 10 degree Celsius rise in temperature. For instance, see the description here.

For ideal gasses (and air at room temperature is a decent approximation), volume is directly proportional to absolute temperature -- temperature measured on either the Kelvin or Rankine scales.

For room temperatures (around 300 Kelvin), 10 degrees is about 3 percent.

 

[Gopher72]

Ok, so air only changes 3.3% for every 10 deg C. That’s great. I would call that statistically insignificant so I can ignore it.

the rule of thumb that I guess I was referring to was that humidity drops roughly in half (assuming no additions or removal of water) for ever 10 deg C increase in air temp. I guess I just assumed the reason that the humidity dropped in half was because the volume doubled. But apparently that is incorrect.

thx for all the help

 

[haruspex]

I guess I just assumed the reason that the humidity dropped in half was because the volume doubled

The relative humidity drops in half if the saturation vapour pressure doubles (and no liquid water is present). That is roughly each 10C at room temperature but falls off quite a lot as you go up. From 90 to 100 it goes up by less than 50%.
SVP is the gas pressure that would exist in a sealed reservoir containing only water and water vapour at the given temperature. If you let air in, the total gas pressure is just the sum of the air pressure and the SVP.

 

[Gopher72]

Oh, so it’s not a linear relationship. Good to know. What about as air drops below freezing to -10 or -20?

 

[jbriggs444]

Oh, so it’s not a linear relationship. Good to know. What about as air drops below freezing to -10 or -20?

You can Google up a chart. For instance. That's for the vapor from ice.

You can also think about super-cooled water. Like this one. That's for the vapor from super-cooled water.

[I Googled "vapor pressure of water below freezing" to get both hits]