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

 

[sailoday28]

I belive transient effects must be taken into account.
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.

 

[phoenix221]

>To just start melting dt=36F 32--4
>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...

 

[IRstuff]

The transient analysis theory derives a "thermal diffusivity" that's independent of external parameters; effectively, it's RC time constant of the material.


TTFN

 

[sailoday28]

Time is about 41 seconds to just start melting if the heat flux and conditions are as I stated. The internals of the heater is another question.
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.

 

[25362]

To my grasping, phoenix221, both you and sailoday28 are right. The time to melt the ice depends on the temperature of the heating medium. Assuming the ice is at a constant temperature all through, and being an unsteady state heat transfer problem, I gave an example of time estimation on Aug 24. No comments yet received.

 

[IRstuff]

I'm failing to grasp why you need to make it so complicated. 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.


TTFN

 

[sailoday28]

25363(chemical)
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.

 

[phoenix221]

>I'm failing to grasp why you need to make it so complicated.
>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

 

[TangoCleveland]

It's possible that someone at NASAs Icing Research group (Cleveland, NASA Glenn Research Center) could help. e-mail: info@icebox.grc.nasa.gov might get you some information.

Website is

http:\\icebox.grc.nasa.gov

Larry

 

[IRstuff]

As you say, the key is the time. 2.5 W/in^2 through a 1" 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


TTFN

 

[25362]

Just for clarity, 1 W equals 1 J/s.
Since the estimation for heat conduction was done over a 1 in² area, the heat flux is then 1 J/(s.in²).

 

[sailoday28]

I may have missed it in all these posts, but what is the geometry of the exposed surface? ie, sphere, cylider, plane wall, parabolic shape, etc? AND what shape should we be considering for the ice growth?

 

[phoenix221]

>what is the geometry of the exposed surface? ie, sphere,
>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...

 

[phoenix221]

>It's possible that someone at NASAs Icing Research group
>(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

 

[phoenix221]

>As you say, the key is the time. 2.5 W/in^2 through a 1"
>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?

 

[IRstuff]

That's YOUR presumption that it's a requirement.


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

 

[phoenix221]

>That's YOUR presumption that it's a requirement.
>
>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 ...

 

[IRstuff]

The point is, if your ambient temperature really is at -4ºC, you have excess heat available for the heat of fusion. If the outer surface temperature is only -4ºC, with the inner surface at 0ºC, only 0.25W/in^2 is conducted away, leaving 2.25 W/in^2 going into melting the 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

 

[phoenix221]

> The point is, if your ambient temperature really is at
> -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?

 

[JKEngineer]

A few comments for phoenix221:

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

http://www.KleinfeldTechnical.com