Energy requirements to get rid of ice
Energy requirements to get rid of ice
(OP)
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
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





RE: Energy requirements to get rid of ice
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
RE: Energy requirements to get rid of ice
>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
RE: Energy requirements to get rid of ice
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
www.KleinfeldTechnical.com
RE: Energy requirements to get rid of ice
RE: Energy requirements to get rid of ice
http://www.google.com/search?hl=en&lr=&ie=ISO-8859-1&q=federal+aircraft+deicing+manual+SAE+AIR+vol+4+&btnG=Search
RE: Energy requirements to get rid of ice
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
RE: Energy requirements to get rid of ice
>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
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,
RE: Energy requirements to get rid of ice
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
RE: Energy requirements to get rid of ice
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 32oF ?
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: 40oC
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:
Obviously the higher the temperature t, of the warm surface, the shorter the time needed for melting.
RE: Energy requirements to get rid of ice
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
www.KleinfeldTechnical.com
RE: Energy requirements to get rid of ice
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.
RE: Energy requirements to get rid of ice
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?
RE: Energy requirements to get rid of ice
RE: Energy requirements to get rid of ice
http://www.engineeringtoolbox.com/24_576.html
TTFN
RE: Energy requirements to get rid of ice
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.
Again, the remaining question is what is the external ice temperature. To assume the extreme case of -4oF 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.
RE: Energy requirements to get rid of ice
Jack
Jack M. Kleinfeld, P.E. Kleinfeld Technical Services, Inc.
Infrared Thermography, Finite Element Analysis, Process Engineering
www.KleinfeldTechnical.com
RE: Energy requirements to get rid of ice
RE: Energy requirements to get rid of ice
>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?
RE: Energy requirements to get rid of ice
>http://www.engineeringtoolbox.com/24_576.html
Very cool link IRstuff
RE: Energy requirements to get rid of ice
Consider Fouries equation in time and distance.
For a Semi infinite solid with a sudden constant heat flux at X=0, The temperature at that surface can be expressed in terms of thermal diffusivity (alpha), k(conductivity), input heat flux q, and temp diff dt as
sqrt(alph*theta/PI)=k*dt/2/q
Reference Temperature Response Charts By P.J. Schneider
John Wiley and Sons
Chart 41 page 114
Using alpha(k/rho/c) =0.049 ft^2/H
k=1.28 B/H/ft/F
q=1720 B/H/ft^2
To just start melting dt=36F 32--4
time is about 41 seconds.
RE: Energy requirements to get rid of ice
>time is about 41 seconds.
Doesn't the time required to melt depend on the temperature generated by the heater which is directly related to its output?
Since one can varry the output/heat, the time to melt can also be controlled...
RE: Energy requirements to get rid of ice
TTFN
RE: Energy requirements to get rid of ice
My calculation infers that transient effects must be taken into account.
The referenced equation was not for an on off device and perhaps that should also be taken into account.
Again, I don't believe a quasi-steady analysis is adequate.
RE: Energy requirements to get rid of ice
RE: Energy requirements to get rid of ice
TTFN
RE: Energy requirements to get rid of ice
re:your 8/24 response
Your calculations are based on quasi steady (a very slow transient)AND I feel that transient effects must be considered.
Again, something as I have calculated but to include on-off. My calc is specifically for a constant heat flux on a semi-infinite body.
RE: Energy requirements to get rid of ice
>If the air temperature is truly no lower than -4ºC, you
>probably need less than 2 W/in^2 steady state to keep the
>surface ice-free. No fuss, no muss; plus, you don't need to
>depend on the wind to push any remnants off the surface.
Belive me I would like to simplify! I do like the 2-3 W/in^2 value, however what happens if you start when ice is already formed? An ounce of prevention is worth a lot, and it is also true in this case. As I understand it the problem has two characteristic scenarios:
1. Ice prevention; 2-3 W/in^2
2. Ice buildup; 20-100 W/in^2 depending on the length of a heat cycle
Ice prevention does seem to take less energy than removing it! Nevertheless, I have to compute for worst case scenario! One may require as much as 10-50 times as much energy for removal depending on the length of the heat cycle (based on my current approach to calculate which seems to be accepted as reasonable for now). I am not sure how transient effects would affect these numbers :-(, since as indicated they are not dependent on outside factors I am not quite sure how to integrate it into my calculations :-(
In any case, if anyone can give me some practical values for "ice removal" as well, backed by research, that I can use without having to go through all this, I am all ears
RE: Energy requirements to get rid of ice
Website is
http:\\icebox.grc.nasa.gov
Larry
RE: Energy requirements to get rid of ice
2.5W/m-K ~ 2.5W/in-39.4K
TTFN
RE: Energy requirements to get rid of ice
Since the estimation for heat conduction was done over a 1 in2 area, the heat flux is then 1 J/(s.in2).
RE: Energy requirements to get rid of ice
RE: Energy requirements to get rid of ice
>cylider, plane wall, parabolic shape, etc
almost a perfect cylinder... Does this matter through?... the shape of the exposed surface will affect the shape of the ice buildup... however for the sake of the computation I idealized the ice to be uniform 1" thick...
RE: Energy requirements to get rid of ice
>(Cleveland, NASA Glenn Research Center) could help. e-mail:
>info@icebox.grc.nasa.gov might get you some information.
Would be nice... but I doubt that my insignificant problem will be of interest
RE: Energy requirements to get rid of ice
>thick ice layer will drop about 40ºC, which means that in
>steady-state conditions, that amount of power applied to that
>ice layer with a surface temperature kept at warmer than
>-35ºC, should result in melted ice at the inner surface.
>
>2.5W/m-K ~ 2.5W/in-39.4K
Could it be this simple? I don't fully realize the reasoning behind your suggestion :-( ... What about the heat fusion of ice to water... where is the energy comming for that?
Depending on the ambient temperature and the density of ice (which I also idealized/homogenized for the purposes of calculations) I suspect that there is a minimum power output that would result in melting the 1/16" layer of ice, even if it's constantly applied without interruption. I would suspect that 2.5W / in^2 is well below that treshold... Am I wrong?
RE: Energy requirements to get rid of ice
All I'm saying is that a 1" thick layer of ice will have a 40ºC temperature drop when 2.5 W/in^2 is supplied.
TTFN
RE: Energy requirements to get rid of ice
>
>All I'm saying is that a 1" thick layer of ice will have a
>40ºC temperature drop when 2.5 W/in^2 is supplied.
Are you saying that heat fusion is optional? As I understand it is an unavoidable phenomenon when state transition occurs, say from ice to water. I don't mind being wrong on this one since it takes a lot of power to achieve it
Interestingly enough, the number I got for conduction heat transfer loss to the 1" thick ice is 2.384W/in^2 and is more or less the same as your 2.5W/in^2... so. This amount of energy will be sufficient to achieve a temperature difference of 36F between the inner and outer layer of the ice... If I toss heat fusion out of the equation, this is my number right here... problem is that the ice has not melted yet... it is still ice and sticks to the surface... to turn it into water heat fusion is required :-(
Please tell me I'm wrong ...
RE: Energy requirements to get rid of ice
Your 1/16" melt layer requires 342J/in^2, which, with 2.25W/in^2 excess input, takes 152 seconds to melt the layer.
TTFN
RE: Energy requirements to get rid of ice
> -4ºC ...
All the temperatures given are in Farenheight, i.e. -4F is -20C ... how does that change your numbers?
> Your 1/16" melt layer requires 342J/in^2,
How did you get this value? Can you provide the formula?
> which, with 2.25W/in^2 excess input, takes 152 seconds to
> melt the layer.
How did you get to the 152 seconds? Could you provide the formula/rationale?
Your way of calculating seems to be much simpler! Does it account for all energy loss to be dealt with as well?
RE: Energy requirements to get rid of ice
There is a difference between power and energy. Power is energy per time. Watts is a power quantity. As the time frame you try to accomplish your melting of the 1/16" layer goes down the power will increase. However, I believe the total energy will actually go down. This is because the energy required in the process is a combination of the energy to achieve the goal, that is melt the specific volume of ice, and the energy lost to the surroundings, that is heating up the supporting structure, the rest of the ice, and the world beyond both. Your orginal calculation was structured to estimate the energy required to melt the layer only, with the qualifications that I stated in previous posts. This will be an optimistic estimate of both the energy and power required. In the limit, that is, with the job done in zero time, it approaches the actual value, but of course is both impossible and requires an infinitely large, infinite wattage heater. To do this at all accurately you need to either get empirical guidance, and perhaps some of the sources cited provide that, or do a transient analysis. There are more than one way to do the transient analysis. The approach that I would favor, since it is what I am equipped with and do for a living, would be to do a transient finite element analysis.
HTH
Jack
Jack M. Kleinfeld, P.E. Kleinfeld Technical Services, Inc.
Infrared Thermography, Finite Element Analysis, Process Engineering
www.KleinfeldTechnical.com
RE: Energy requirements to get rid of ice
>equipped with and do for a living, would be to do a transient
>finite element analysis
I don't know enough to do the transient analysis you mentioned myself :-(. Is there a software program that facilitates/automates this analysis?
RE: Energy requirements to get rid of ice
There are many. Some are quite expensive. Some are free, but I am not familiar with those in any detail. In any event, you should really know what you are doing before you start trying it on a computer.
Jack
Jack M. Kleinfeld, P.E. Kleinfeld Technical Services, Inc.
Infrared Thermography, Finite Element Analysis, Process Engineering
www.KleinfeldTechnical.com
RE: Energy requirements to get rid of ice
almost a perfect cylinder... Does this matter through?... the shape of the exposed surface will affect the shape of the ice buildup... however for the sake of the computation I idealized the ice to be uniform 1" thick...
For further analysis, the basis of 1"thick should be justified.
Consider the wind direction causing the ice to form on the cylinder as a wedge type fin in the direction of the prevailing wind.
What might be the shape and length of the "fin" without heat transfer before the prevailing wind changes?
If a fin shape occurs, then the heat flux will not be uniformlly distributed over the cyliderical surface. The surface temperature of the cyliderical surface will vary AND because of the find effect, the heat conduction model will have to be modified. That analysis will also give a better handle on specifying the heat generation device.
Typically fin analysis for "steady state" problems are analyzed in one dimension, along the length of the fin. However, with high heat transfer coefficients on the fin surface, I believe that heat conduction in two dimensions should be accounted for.
If this reply makes sense, I believe other contributors will have additional thought.
RE: Energy requirements to get rid of ice
Ambient temp -4F, wind 120 mph
Refer to
Thread1-102277 in aerodynamics section
aerodynamic heating in local areas where fludid has stagnated (0 velocity), the locat temperature will increase about 2 to 2.5F -4+2=-2
There temperature in vicinity of stagnation is about -2F
RE: Energy requirements to get rid of ice
>cylider, plane wall, parabolic shape, etc
almost a perfect cylinder... Does this matter through?... the shape of the exposed surface will affect the shape of the ice buildup... however for the sake of the computation I idealized the ice to be uniform 1" thick...
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If you have a missile you want to launch and it is a cylinder shape with little stabilizing wings at the bottom, then you'll just end up with a ring of ice separated from the cylinder, sliding down it till it sits on the wings, but not disengaged from the projectile.
So the shape does matter, but since you've not mentioned it that makes it your problem - maybe already considered or taken care of - and not ours.
Taking my made up example further ...
The fact that 3W per quare inch could keep it ice free once launched doesn't mean that there should be a continuous power drain before use while its just sitting there, so the short preparation time should be considered too: the originator says its important, so it is.
I'm not sure I go for all these thermal conductivity Watts per metre per Kelvin calculations. That seems to be about temperature gradients in materials and heat being conducted away. I can hold a metal spoon in the flame of my gas cooker (I guess, I'll try it later) and the heat takes some time to get to my fingers. So if we take the 3 Watts for 100 seconds or 100 Watts for 3 seconds argument, surely with the 3 Watts for 100 seconds, there is time to build up a temperature gradient and heat being cinducted away need to be considered, whereas in the 100 Watts for 3 seconds, the heat doesn't have the time to dissipate through the ice.
I guess what I am saying, is a blitz approach would use less energy than gradual approach. Indeed, if 3W (per sq inch) was enough to keep ice at bay, then say 2.5W might allow ice to build up and so never be able to remove the ice regardless of time.
Also, the 1/16 of an inch figure seems to have been pulled out of thin air like a rabbit from a hat. While I can see a ball park figure is useful, why not use 1/8 or 1/32 and maybe halve or double the answer?
We have taken it for granted that the surface the ice attaches to is a good conductor of heat. That seems likely. I don't suppose heat would get conducted away by melting ice at one spot, while it remains strongly attached elsewhere. But there should be some calculation to check there are no sticky points resulting from heating conducting away more easily through ice and water than spreading across the heated attachment surface.
The ice attachement surface is presumably heated on one side or internally, and the ice is on the other side, so there will be a thermal gradient in its material. Don't we need to consider its specific heat capacity to know the energy needed to heat it up, and its thermal conductivity to know what sort of termperatur gradient will build up inside it, lowering the temperature at its surface? Or does it have a sort of efficiency figure or warm up time of its own?
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Sorry to be a pain rather than a positive help!