Energy requirements to get rid of ice

[phoenix221]

Hello everyone,

I have an interesting problem which I have been grappling with for a while. Thermodynamics is not my specialty, so this problem proves exceedingly difficult. I need a sanity check on my computations and even some help if it proves that I am out to lunch

The Problem:

I have a small surface exposed to 120mph wind, high moisture in an ambient temperature no lower than -4F. Ice forms of varrying thickness on this exposed surface, this ice needs to be removed in cycles. Each ice melt cycle will unfortunately be very short between 3-5 seconds. The idea I am considering is to use a kapton flexible heater if possible. To determine the viability of this idea, I need to determine the watts/square inch required to be outputted by the heater.

Assumptions:

1. There is no need to melt the whole ice. It is sufficient to liquify a thin layer, say 1/16th, the ice then gets literally blown off, i.e. mechnically removed, by the wind, or slides of due to its own weight!

2. Energy requirements need to be calculated for

a) the heat needed to warm the ice to 32F
b) the heat needed to melt the ice, i.e. heat fusion
c) the heat needed to warm the water to 33F

3. Need to determine the wind-chill equivalent temperature

The Reasoning (faulty as it may be ):

1. Wind-chill equivalent temperature according to the new windchill index: -47.33F

2. The temperature change is Dt=-47.33-32=-79.33 (I'll ommit the (-) sign for calculations).

3. The weight of the ice being warmed is determined by using an average density of ice of 57.67 pounds/qubic foot. The weight of 1 qubic inch of ice is 57.67/1730=0.0333 pounds. Now as per my assumption I am really melting 1/16th of an inch of the total ice only so my weight for computational purposes is M = 0.0333/16 = 0.00208 pounds

4. I will use the following formula to determine the watts required for warming the ice to 32F:

watts=(M x Csp x Dt)/(3.42btu/watt hr x Th)

where
M - weight of material (lb)
Csp - specific heat of ice (Btu/lb F) = 0.5
Dt - temperature rise (F)
Th - heatup time (h)

Watts = (0.00208 x 0.5 x 79.33) / (3.42 x 1) = 0.02412 watts


5. I will use the following formula to determine the watts required for converting the ice to water at 32F without a change in temperature:

watts=(M x Csp)/(3.42btu/watt hr)

where
M - weight of material (lb)
Csp - specific heat for heat fusion (Btu/lb F) = 144
Dt - temperature rise (F)
Th - heatup time (h)

Watts = (0.00208 x 144) / (3.42) = 0.08757 watts

6. I will use the same formula as in step 4. to determine the watts required for warming the ice to 33F except we'll work with Csp=1 and Dt=1 so...

Watts = (0.00208 x 1 x 1) / (3.42 x 1) = 0.00081 watts

7. the total:
So this gives me a total of 0.02412 + 0.08757 + 0.00081 = 0.112508 watts/hour/square inch.
For a 5 second performance I will compute (0.112508*3600)/5 = 81.005 watts/square inch
For a 3 second performance I will compute (0.112508*3600)/3 = 135.009 watts/square inch


However I have the feeling that something is wrong with this approach! In addition to whatever errors I may have comitted, I suspect one also would have to account for the wattage for heat loss between the layer being flash melted and the rest of the ice... and perhaps there are some other things I am missing as well...

I am also seeking advice on what a reasonable padding of these numbers would be to account for unforseen issues...

Any help would be appreciated...

Thank you in advance

 

[IRstuff]

SAE-AIR-1168 vol4 is the current manual for aircraft deicing.

Not sure about your premise, e.g., where do you account for conductive heat removal from the heated surface by the unmelted ice? Also, where are you accounting for the heat loss from convection/radiation on the interior surface of the heater strip?

From the aforementioned manual, you only need a STEADY-STATE value of about 3.5W/in^2 to keep a surface ice-free at 150 kt, so I wonder why the need for such an aggressive approach, particularly since such high power densities come with extremely high temperatures, thereby inducing all sorts of secondary effects such as heat damage to the strip heater, debonding of the bondline, warpage of the deiced surface, etc.

The steady-state solution can be readily adapted to a temperature controlled algorithm to simply ensure that the surface temperature is always above the maximum dewpoint.


TTFN

 

[phoenix221]

Thank you IRStuff...

>SAE-AIR-1168 vol4 is the current manual for aircraft
>deicing

Someone else also mentioned this before. I tried to find this on the internet, no such luck... I don't know where to order it from... would be good to have

>Not sure about your premise, e.g., where do you account for
>conductive heat removal from the heated surface by the
>unmelted ice?

I am not at this point, this is just one of the things I mentioned I was aware of that I am missing from my calculations at this point. Not sure how to compute that though...

>Also, where are you accounting for the heat loss from
>convection/radiation on the interior surface of the heater
>strip?

There would be reflective surface under the element that should reflect 95% of the radiated heat. There will be some heat loss even like that. I was thinking, rather than complicate my calculations, I could use some empirical rule of thumb and pad the number accordingly. I welcome any suggestions ...

>From the aforementioned manual, you only need a STEADY-STATE
>value of about 3.5W/in^2 to keep a surface ice-free at 150
>kt

Would this value not assume continuous running? What if you want to activate the element in a periodic intermittent fashion? ... say for x seconds on y seconds off... a heating interval/cycle being x+y

>...so I wonder why the need for such an aggressive
>approach, particularly since such high power densities come
>with extremely high temperatures, thereby inducing all sorts
>of secondary effects such as heat damage to the strip
>heater, debonding of the bondline, warpage of the deiced
>surface, etc.

In general I was also wondering about my numbers :-(. It did seem quite a lot. I am wondering how critical it is that the ice would be "flash melted" within 5 seconds? Perhaps this is not a realistic heating time considering the secondary effects?

In my mind also there really are two cases to deal with

1. Preventing ice formation
2. Removing ice already formed

I was focusing on the 2nd scenario for now. I am assuming that preventing ice formation would require less energy?!? I would like to calculate the energy requirements for scenario 1. as well ... I am open to suggestions.

Thanks

 

[JKEngineer]

I agree with some of the points raised by IRStuff.

I add the following:
The wind chill factor is not appropriate here. It is meant to reflect the equivalent rate of heat loss from skin by reporting the temperature for still conditions equivalent to the rate of loss under actual conditions. (I think.) In any rate, it is not a thermodynamic temperature. That puts your starting T at -4F. Not -47F.

However (you knew that was coming) you will lose heat from the target 1/16" thickness to the body supporting the ice and beyond it to its interior, if any, and to the bulk of the ice and beyond it to the air. Both conduction and convection will play a role. Convection will be influenced by the wind.


Jack M. Kleinfeld, P.E.
Kleinfeld Technical Services, Inc.
Infrared Thermography, Finite Element Analysis, Process Engineering

http://www.KleinfeldTechnical.com

 

[imok2]

http://www.icao.int/icao/en/anb/opsair/groundice/

 

[imok2]

This may be of some help for you

http://www.google.com/search?hl=en&...aft+deicing+manual+SAE+AIR+vol+4+&btnG=Search

 

[IRstuff]

SAE- Society of Automotive Engineers
don't ask, I don't know why they're involved in airplane stuff.

The windchill is not necessarily irrelevant, since the wind evaporation of water from the exposed ice will potentially reduce the thermodynamic temperature of the ice.

the 3.5 W/in^2 value ensures no ice formation up to 150 kt under all atmospheric temperatures of flight interest, e.g., down to -40ºC as well as under various humidity and ice content in the air. The assumption is that you couple the heater control to the ambient temperature and drive the window temperature to just above the highest dewpoint possible at that ambient temperature. If you can measure the wind speed, you can also reduce the power drive, e.g., 100 kt wind only requires a maximum of 2.5 E/in^2


TTFN

 

[phoenix221]

>the 3.5 W/in^2 value ensures no ice formation up to 150 kt
>under all atmospheric temperatures of flight interest, e.g.,
>down to -40ºC as well as under various humidity and ice
>content in the air

Let me say that I like your numbers better than mine . Having said that I would like to understand what I have done wrong.

Q. Can someone point out what errors I made in my approach to arrive at 85-131 watts/in^2?

Q. Or is it possible that melting ice that is already formed requires much more energy?

>you will lose heat from the target 1/16" thickness ... to
>the bulk of the ice and beyond it to the air. Both
>conduction and convection will play a role. Convection will
>be influenced by the wind

Q. So how do I calculate the wattage requirements for compensating for the loss from the 1/16" boundary layer?

Thanks,

 

[IRstuff]

They're not necessarily inconsistent. If you consider how long it might take to accumulate 1/16" of rime ice under typical flight conditions and the divide that time by 3 seconds multiplied by 3.5 W/in^2, I'm not sure that you'd get that different an answer.

We're only talking about the difference between running a 3.5 W/in^2 heater continuously for about 100 seconds to get the equivalent energy from 100 W/in^2 for 3 seconds.

The difference comes in whether you need to generate 126 W continuously for a 6"x6" area or 3.6 kW intermittently.


TTFN

 

[25362]

Chemicals are used in conjuction with heating to provide for de-icing/anti-icing and to avoid re-icing on aircraft exposed surfaces.

As regard to phoenix221 calculations, one can say they are not wrong in principle. However, why give results to six significant digits while the basic data are rounded figures ? And why not use tabulated enthalpy values ?
Why should the ice be at temperatures lower than 32^oF ?

The time it takes for a layer of ice to melt can be estimated if one knows the temperature of the "heated" surface, and one doesn't need to know the area.

As an example,

Heated surface temperature, t: 40^oC
Latent heat of melting, [λ]: 335 kJ/kg
Water ave. thermal conductivity, [κ]: 0.6 W/(m.K)
Density of water, [ρ]: 1000 kg/m³


Heat to be removed, Q: 335 kJ/kg = [λ]

The time for melting an ice film, e=1.6 mm thick under a thicker ice layer would be estimated on basis of transient heat transfer considerations, with heat transferred through a progressively increasing thickness of a water layer to be:

time = (Q)([ρ])(e²/2) / [([κ])(t)]​

time = (335)(1000)(1000)(0.0016²/2)/[(0.6)(40)]= 17.9 seconds​

Obviously the higher the temperature t, of the warm surface, the shorter the time needed for melting.

 

[JKEngineer]

I am restating the problem facing as I understand it. Everyone has assumed that airplane situations apply, but phoenix221 has not said that.

As stated:
A structure has an accumulation of ice. Heaters are to be put on the surface structure to melt, for example, a 1/16" layer so the ice slides off. Ambient conditions can be as low as -4F with winds to 120mph. The surface is small. There is a 3-5 sec window for the removal cycle.

Summarizing and annotating some of the comments:
For airplanes, it can be sufficient to keep the surface above dewpoint so that no ice forms. [Sounds reasonable, and it might apply here, but not as the problem is stated.]

The heating rate will be much higher as the time to process the layer to be melted is decreased. The faster it is set to happen, the higher the wattage, but also the total heat required actuallly goes down - fewer losses.

In heating the layer to be melted there will also be heat loss into the bulk of the ice and then through it to the air at its free surface. There will also be loss to the structure, whatever it is.

In my comments earlier I assumed that the entire system was in equilibrium with the ambient conditions when the heat is turned on. It would therefore be no colder than -4F. IRStuff points out that evaporation from the free surface will depress the temperature of the ice. I am not sure if that will be significant: at equilibrium conditions the free surface may sublime, but will not have free water as a liquid to evaporate. I am not sure how significant sublimation is in this situation. I was ignoring it. Perhaps I should not. During the melting operation, as I undertand the problem, there is also no free water at the free surface. This is because of the problem statement -- thick ice and melting the first inner layer only. So, the wind is a factor in terms of convective heat transfer from the free surface, but I do not think it is a factor in terms of evaporation or sublimation. In any event, the numerical impact of the wind is NOT the wind-chill factor, which is configured to represent a different environment.

I think phoenix221's overall approach is OK for a first degree estimate. The calculation of the heat needed to process the layer to be melted. I disagree with the choice of the starting temperature, as I have stated. The approach will be optimistic in its heat estimate because it ignores the transfer of heat from the layer to the rest of the system and on out to the world.


Jack

Jack M. Kleinfeld, P.E. Kleinfeld Technical Services, Inc.
Infrared Thermography, Finite Element Analysis, Process Engineering

http://www.KleinfeldTechnical.com

 

[aviat]

Many A/C deice systems do not actually melt the ice. They break it up and either the air stream or centrifugal force removes it. A system commonly used employed uses low a pressure pneumatic system to blow up tubes in sequence, which breaks the ice with no heat involves at all. For both thermal and pneumatic to work properly the surface should be curved and it has to cycle from the inner to outer, in other words, the inner tube inflates first and after it deflates the outer tubes inflate. A timer controls the time of cycle. Melting of the ice would cause more problems, because it would blow back over the wing are freeze, building up on top and bottom of wing.

A thermo deice system similar to that phoenix221 describes is used on small props It has a boot approx 3 feet long with an inner and two outer heating elements, which is glued to the leading edge. A manual of one A/C shows a 20A breaker, which also runs a timer, so using 17A being applied to three blades on a 24V system it comes to less than 150 Watts for each of the inner elements. The Watts on the outer elements would be half of that.

 

[phoenix221]

Based on the feedback I reworked my numbers:

1. The temperature change is Dt=-4-32=-36

2. The weight of the ice being warmed is determined by using an average density of ice of 57.67 pounds/qubic foot. The weight of 1 qubic inch of ice is 57.67/1730=0.033 pounds. Now as per my assumption I am really melting 1/16th of an inch of the total ice only so my weight for computational purposes is M = 0.033/16 = 0.002 pounds

3. the watts required for warming the ice to 32F Watts = (0.002 x 0.5 x 36) / (3.42 x 1) = 0.010 watts

4. the watts required for converting the ice to water at 32F without a change in temperature is the same, i.e. Watts = (0.002 x 144) / (3.42) = 0.084 watts

5. the watts required for warming the ice to 33F

Watts = (0.002 x 1 x 1) / (3.42 x 1) = 0.0005 watts. Since I am only computing to 3 digit precision, I will ignore this value

6. to determine the wattage for the loss due to conduction heat transfer in the ice I'll use the formula

watts=-K x A x (Dt / Dx

Where

K - thermal conductivity (W/mk)
A - area (m^2)
Dt - temperature difference between inner and outer layer (F)
Dx - The thickness of the ice layer (m)

working with an average thermal conductivity for ice of 363 W/mk, a maximum ice thickness of 1 inch=0.025, and the area of 1 in^2=0.0006

watts = -363 x 0.0006 x (36/0.025) = 31.36 / m^2 = 0.020 / in^2

7. I consider the wattage required due to losses from convection in the 1/16" layer of water negligable

8. to determine the wattage for the loss due to conduction heat transfer in the body the formula from step 6 can also be used! However the material is glass fiber composite, a good insulator... working with an average thermal conductivity for glass fiber composite of 1 W/mk

watts= -1 x 0.0006 x (36/0.025) = 0.086 / m^2 = 0.0005 / in^2 or negligable

8. So this gives me a total of 0.010 + 0.084 + 0.020 = 0.114 watts/hour/square inch.
For a 5 second performance I will compute (0.114*3600)/5 = 82 watts/square inch
For a 3 second performance I will compute (0.114*3600)/3 = 136.8 watts/square inch

So when all is said and done... the numbers are pretty much the same, though this approach is more accurate, and perhaps reassuring

Now the wattage required is very high and will tend to heat the composite surface past 100F. This is a problem!

The easyest way to manage this with lower temperatures is by lengthening the time of a heating cycle.
For a 10 second performance the power required is (0.114*3600)/10 = 41 watts/square inch
For a 15 second performance the power required is (0.114*3600)/15 = 27.36 watts/square inch

These values are more realistic and products delivering this performance are also readily available!

Have I missed anything?

 

[25362]

I may be wrong but the thermal conductivity of ice would be about 2.3 W/(m.K), not as given. Please comment.

 

[IRstuff]

Thermal properties of ice:

http://www.engineeringtoolbox.com/24_576.html

TTFN

 

[25362]

For heat conduction to take place one has to assume that the ice layer is at a gradually droping temperature towards the exposed surface probably cooled below 32 deg F by sublimation effects due to the 120 mph cold wind.

I have one comment to your point 6. The steady state heat conduction Fourier's formula is right, but the estimate appears to be wrong. See, please:

A= 1 in² = 0.0006 m²
dT = 36^oF = 18^oC = 18 K
k = 2.4 W/(m.K) (on the average)
dx = 1 in = 0.0254 m

Q = kA dT/dx = (2.4)(0.0006)(18)/(0.0254) ~ 1 W

and this is already for 1" thickness, and 1 in² area.

Again, the remaining question is what is the external ice temperature. To assume the extreme case of -4^oF may be on the safe side for estimating the needed heat input.

I pressume, however, it may be too large a safety factor, since the wind is already moist (saturation would involve about 0.1% vol. water in the air). It all would depend on the time lapse between the melt (de-icing) cycles. Longer intervals would tend to bring down the exposed ice surface temperature. Shorter periods would make the heat "lost" by conduction almost negligible.

Your comments appreciated.

 

[JKEngineer]

Without commenting on any other content, 25362's post shows a dT of 36F=18C. In fact a dT of 36F = 20C.

Jack

Jack M. Kleinfeld, P.E. Kleinfeld Technical Services, Inc.
Infrared Thermography, Finite Element Analysis, Process Engineering

http://www.KleinfeldTechnical.com

 

[25362]

JKEngineer, you are absolutely right. What happened with me ? Believe me, in my notes I correctly wrote 20 deg C, and all of a sudden I wrote 18 in my post instead of dividing by 1.8. Why ? Probably because I'd had to re-write the message since it didn't accept it the first time. A lapsus memoriae ? God knows. Thanks a lot.

 

[phoenix221]

>I have one comment to your point 6. The steady state heat
>conduction Fourier's formula is right, but the estimate
>appears to be wrong. See, please:
>
>A= 1 in2 = 0.0006 m2
>dT = 36oF = 18oC = 18 K
>k = 2.4 W/(m.K) (on the average)
>dx = 1 in = 0.0254 m
>
>Q = kA dT/dx = (2.4)(0.0006)(18)/(0.0254) ~ 1 W
>
>and this is already for 1" thickness, and 1 in2 area.

Ok... what is the unit of time for the heat transfer calculated ? hour, minute, second?

 

[phoenix221]

>Thermal properties of ice:
>

http://www.engineeringtoolbox.com/24_576.html

Very cool link IRstuff ... you know of a decent reference site for structural engineering?