Calculate Drying Time of Nylon Parachutes

[smbarrett3]

Hello,

I'm designing the HVAC system for a parachute drying tower. What equation can I use to estimate drying time? I know the amount of water in the parachute, the temperature and CFM across each parachute, and I can approximate the exposed surface area. I've looked at various equations for water evaporation but none appear to take into account the convective air flow across the surface. One promising way would be to calculate the BTUs required to evaporate the water but I don't see how that could include the air flow. In my mind, higher air flow would result in quicker drying times. One formula I found involved the difference in vapor pressures between the water and the vapor pressure of the air but again, this seemed to be applicable to air at a standstill. Still another was a formula for evaporation of a pool of water which did include water temperature, air temperature, surface area of pool, and CFM. However, these resulted in unrealistically short drying times. They also assume the water is freely on the surface versus entrained in a fabric, so there may be some capillary action involved and there will be numerous folds in the parachutes since they are hung from the top of their canopies.

Anyways, I have virtually all of the data since I am designing the system but I do not know of a formula or empirical data regarding nylon (or other fabric) drying rates.

If it matters, our system is 100% outside air which will be heated to 120 F, ducted into the space at the bottom of the tower, and exhausted out of the top of the tower. I do know outside air conditions if humidity of the air plays a role.

I appreciate any ideas or advice.

Thanks!

 

[racookpe1978]

Got to think about this one a bit.

 

[MikeHalloran]

Is there no way to spread the canopies a little, so you are not dealing with random multilayers of fabric adhered by the water?


Mike Halloran
Pembroke Pines, FL, USA

 

[racookpe1978]

Weigh 5x parachutes (when dry) and when wet. That will give you a measure of "how much water is in the average parachute".
That will give you a starting point for the theoretical energy to evaporate the theoretical mass of water.
Time to remove that much water remains very tightly linked to how the chute is spread out to expose the maximum surface to the dry air while NOT allowing it to deploy (billow out like a sail, or frankly, a parachute. Try a loose mesh bag with the parachute bags in a loose pile. Is tumbling the bags practical?

 

[smbarrett3]

I know the mass of water. A company that manufactures hoists has submerged and weighed the wet/dry parachutes and given me this data.

These parachutes are very large so tumbling is not practical. The design of suspending individual parachutes from hooks on hoists by their canopies and allowing them to hang down to the floor is industry standard. Water run-off and evaporation takes place during the drying time. I agree that from a practical perspective, increasing surface area is ideal and controlling airflow so that the parachutes do not billow is important. Typically, the parachutes are shook out about every hour to remove loose water and to introduce more of the chute to open air. Since these are suspended from the centerpoint of their canopies, there are numerous folds in the fabric while suspended.

I am designing the system, so I can control the surface area and airflow. Assume I can control the process as if I were in a laboratory setting. What reliable equation can I use to estimate evaporation (i.e., drying) time? I understand that theoretical and practical may vary but I am trying to get a basis.

 

[MikeHalloran]

If all else fails...

A Google search on
equations for drying fabric
produced a bunch of stuff that looked relevant to me.

Mike Halloran
Pembroke Pines, FL, USA

 

[LittleInch]

I never realised that this was such an industry. However looking at the types of structures used there is clearly a balance being made between space to spread the fabric out and cost of the building to house 20, 30 or more in one place.

Also don't forget that there are lines attached which are thicker and can take a lot longer to dry out than the canopy. As an ex skydiver I have personal knowledge that is the lines which take the time to dry, not the rather thin canopy.

In terms of your question I think there is simply too many variables on how the fabric is hung and how air flow travels past them to come up with any sort of formula. I think your only way is to do some testing of humidity levels of air entering the building versus exit and plot the decline to show when drying has been completed. You might then be able to experiment with different ways of hanging or shaking to see which is best.

Start overall with your amount of heat to evaporate the weight of water then add in an efficiency factor of say 20 to 40% and work from there. Given the number of these apparently built there must be a ROT on how long it takes to dry 30 canopies. Just need to find the grizzled old guy that's been doing it for 30 years and can still talk sense. ....

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[smbarrett3]

Thanks Littleinch. Given your skydiving background, do you know grizzled old guys that have been doing this? I know some rigging and packing technicians that have been doing this for years but they aren't engineers, so they won't know nothing about enthalpy of vaporization. Plus they are our client so I'm supposed to be smart engineer with the equations.

Regarding the heat required to evaporate the water, allow me to ask a dumb question. Isn't the heat output from AHU energy measured in BTUs? So the air flow rate, whether stagnant or 10,000 CFM isn't accounted for because the only energy is the heating element in the AHU. I agree there are numerous complexities and hence why you suggested a generous efficiency factor. But still I'm trying to differentiate these parachutes from being in an oven versus convection oven, if that makes any sense. The air flow is going to have a dramatic impact on drying time. I feel like I would need to use a convective heat transfer equation (accounting for air speed) and set it equal to the heat of vaporization of the mass of water. This still however wouldn't account for humidity since the vapor pressure difference between the water in the chute and the relatively low humidity air is the energy which dries materials. While this could account for the heat (since hotter air can better absorb moisture, this greater vapor pressure differential) this still wouldn't account for air flow rate.

 

[EdStainless]

The airflow only matters when it come to 'fluffing' the material and keeping the air from saturating.
The incoming air has some properties, temp, RH, and corresponding enthalpy. the outlet air will be cooler and wetter.
You need to move enough air so that the outlet air does not reach 100%RH.
I would start with some Google searches, and then trial calcs based on say 16hr dry time (overnight) and see if it makes sense to remove that much water in that time period.
The lower the RH of the inlet air the more driving force there is to overcome the water stuck in the fabric.

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P.E. Metallurgy, Plymouth Tube

 

[LittleInch]

smbattett

Sadly my skydiving days were a long time ago, but even then we tried to keep our canopies dry as it was a complete pain to dry them, especially the lines, which was my point. I only ever deliberately jumped into water twice and it took days of drying in my living room to dry the bloody thing out.

The type, size and number of canopies you're looking at looking at would appear to me be wholly military application. I'm surprised you don't seem to have any real specifications or design basis to work from.

I get that you're trying to do this scientifically, but the air volume is so high and velocity you're looking at in the drying shed will be so low that I can't see it (velocity) making a big difference. The last thing you want is for any of the canopies to do anything other than gently drift a bit in your gentle wafting air flow through the shed.

In essence all you're doing is blowing warm air into a building and presumably letting it out at the top. I would have thought de-humidifying it would have helped, but clearly adds to the cost and if they are happy to leave them there for 24 hours and it's all dry then what's the issue?

Getting something "dry" is such a vague term and the density of the items is quite variable (canopy to lines) that you're on a hiding to nothing without doing some tests on an existing facility to make sure you're not over or undersizing your heater.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[racookpe1978]

Can you hang them upside down (lines to the top), with a modest weight in the parachute to reduce swaying and billowing of the fabric? Then the air at 1-2 meters/sec will flow around the set of parachutes, but each parachute will be somewhat restrained in its motion.

 

[MintJulep]

Likely empirical and experience based.

Contract someone who does this for a living to design the process and you can select the equipment.

Like these guys maybe:

http://www.paradrysystems.com/

 

[smbarrett3]

We are working with ParaDry on this project. They design the hoist systems. I spoke with their principal and he said he's seen lots of different system designs for the HVAC systems in drying towers. It doesn't sound like there is a standard design. He wasn't able to given me any names for references. As far as he knew, general HVAC engineers design the systems. There isn't a firm or manufacturer that specializes in this.

 

[LittleInch]

What sort of time are you getting / aiming for / required by the spec to obtain?

I would have thought you were looking at 12 to 18 hours to do this for a bunch of canopies all hanging next to each other getting shaken out at regular intervals.

Personally I think you want to look at lowering the incoming RH by a bit of de-humidification so you don't roast the packers as they shake the canopies and it would also account for any rally humid days you get on site.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[btrueblood]

Agree with Mint, you're going to need some empircal data points, hopefully from existing facilities and specifications for same. Some anecdotal data exists in the attached pdf document, maybe enough to do some calculations. Interesting that dehumidification and fairly modest (30C) temperature was effective...then again it makes sense when you think about it, get the water out of the circulating air is the idea, right?

http://www.apricus.com/upload/userf...ial-Project-Patrick-Airforce-Base-Florida.pdf

 

[smbarrett3]

My initial thought was also to dehumidify the air (either subcooling or desiccant wheel) then heat it but that is a big waste of money. I think littleinch is on point in the drying time is in the order of 12-48 hours, so the system doesn't need to be overkill, but I can't reliably say that it will be dry in that time span. Right now I'm showing heating to 120 F with 100% OA and all of it exhausted out the top. Humidity shouldn't be an issue since the air isn't being recirculating. I think this is an effective design but I don't have much in the way of calculations.

Recirculating the air is tricky because the air would need to be blown back down, outside of the building in a very large insulated duct, then recirculated through the AHU for moisture removal. It seems like a mess and I'm not sure that much energy would be saved in the long run. If I were to do this, I would probably not heat and just ventilate/dehumidify the space. Keep in mind the entire tower (say 35 ft x 35 ft) is acting like a large duct for air flow.

 

[EdStainless]

So start with your worst case local weather (hottest and most humid) and figure out the absolute humidity, then when you heat that air to 120F what is the difference between the incoming humidity and saturation.
You know how much water you need to remove, so a little math tells you how much air you need to flow. I would assume that real amount is double or triple this because of the factors already mentioned.
Does this total volume of air over 12-18-24 hours sound reasonable? Could you manage those flows?

I wouldn't try recirculating, but if the incoming air was very humid I might look at dehumidification. It uses a lot of power but you can at least capture some it in helping warm the air a little.

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P.E. Metallurgy, Plymouth Tube

 

[smbarrett3]

Yes, I used the ASHRAE Humidification DP/MCDB for the humidity ratio of the heated air and used an equation from the Engineering Toolbox which is something like: g[sub]h[/sub] = Θ A (x[sub]s[/sub] - x)

where
g[sub]h[/sub] = amount of evaporated water per hour (kg/hr)
Θ = (25 + 19 v) = evaporation coefficient (kg/m[sup]2[/sup]h)
v = velocity of air above the water surface (m/s)
A = water surface area (m[sup]2[/sup])
x[sub]s[/sub] = humidity ratio of saturated air at the same temperature as the water surface (kg/kg) (kg H[sub]2[/sub]O in kg Dry Air)
x = humidity ratio air (kg/kg) kg H[sub]2[/sub]O in kg Dry Air

Link

So first problem is this is an empirical equation for pools (open water surfaces). Problem two, I have no idea what this equation comes from. Problem three, I am somewhat confused by the two humidity ratios. Is x[sub]s[/sub] the fully saturated air at 120 F? And x is the humidity ratio of 0.4% MCDB air heated to 120 F, as taken off the psychrometric? [0.4% MCDB 80.2 at 76.0 DP]

Based on an estimated exposed area of 43.8 m[sup]2[/sup] per chute (3 ft dia x 50 ft length), a velocity of 0.02 m/s, a humidity ratio of saturated air of 0.08 and a humidity ratio of air at 0.017, I got 70 kg/hr of evaporation. Our biggest chute holds 175 gal of water, so this equation (if I'm using it correctly), predicts the chute will be dried in just over an hour. Even before a heavy safety factor, that seems high.

If any kind souls want to let me know if I'm using this formula correctly, I sure would appreciate that.

 

[EdStainless]

That formula does not apply, so why are you trying to use it? This equation assumes a free surface and in your case the movement of the water in the cloth will limit the evaporation rate.
You need to take your worst case inlet air (hot and humid) and figure out its moisture content at 120F.
The exhaust air won't be at 120F, it will be a little cooler, so you will need to assume some evaporation rate and do some iteration.

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P.E. Metallurgy, Plymouth Tube

 

[smbarrett3]

Ed, the reason I started this thread is because I don't know what formula I should be using. I could use the required heat of vaporization of water at my design temperature 120 F (43000 J/mol). If my 175 lb of water is 4410 mol, then it requires 189626894 J or 179,731 BTU per chute, so given the output of my heater (BTU/hr) and some correction for efficiency of evaporation, I could back out a drying time.

Does anyone think this is a valid approach?