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

 

[phoenix221]

>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

I don't know enough to do the transient analysis you mentioned myself :-(. Is there a software program that facilitates/automates this analysis?

 

[JKEngineer]

I don't know enough to do the transient analysis you mentioned myself :-(. Is there a software program that facilitates/automates this analysis?

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

http://www.KleinfeldTechnical.com

 

[sailoday28]

PHEONIX221(computer) STATES
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.

 

[sailoday28]

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

 

[crysta1c1ear]

>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...

===================

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?

==

Sorry to be a pain rather than a positive help!