I see that you look at the equations available on internet, this is good!

Continue and you will find exactly ALL what you ask for and have the great pleasure to have solved your problem by yourself.

You should not think "energy" but power since at -30°C the plate will loose via convection a loot of "energy" per time unit. To this you add the falling snow so that you should estimate the rate and how much heat is required to melt it. This heat is used for heating from its original temperature to the melting one and after it has been reached to supply the energy for phase changing from solid to liquid and for heating a bit more over the critical temperature so that it can flow of the plate.

Although I have the strong feeling it is a homework I gave you the hints how you can solve the problem. Do not expect more before you bring results and ask for their estimation.

Agree with nick name. What you should be concentrating on is the power (i.e. rate of energy consumption) required to raise the temperature of the snow and melt it rather than the energy required to warm up the bit of aluminium.

Specific heat capacity

Latent heat

Am I on the right track here?

Given: a 2' by 2' surface with 1" of snow: V = 2304in³ or 0.0378m³.

Density of snow is 330 kgm­³ therefore Mass = 12.474Kg (D=M/V)

I estimated that by increasing snow temperature by 10ºC, it should melt (subject to atmosphere conditions)

Q=mc▲T

12.474kg * 2090 J/kg K * 10°K

Q = 260706J = 0.0724kWh = power required to melt that volume of snow.

Questions:

IS this correct?

If yes, how do I determine how much power must be given to the aluminium to have it give off this energy?

No

you must take into consideration the snow mass FLOW, if the start temperature is -30°C (given by you) will snow melt at -20°C?(-30+10=-20)

As I mentioned the heat flow you should bring has to fulfil several requirements:

1- heat snow to its melting temperature

2- supply the latent heat for change of phase

3- heat the resulting water a couple of degrees to let it come from the plate far enough before it becomes again ice!

If you consider the snow flow in kg/time unit (up to you which one you want to work with) automatically the result will be in W.

If the snow starts at -10°C you will need

12.474kg * 2050 J/kg K * 10°K = 255,717 Joules to get it to be ice at 0°C

BUT you will then need another

12.474kg * 334000 J/kg = 4,166,316 Joules to get it from ice at 0°C to water at °C

As long as the aluminium starts at 0°C the sum of these two numbers is exactly the amount of energy you will need to melt the snow.

Your piece of aluminium is only 189 cm³ ;so it will only take

189 cm³ * 30°C * 2.422 J/cm³.K = 13,733 Joules to get it from -30°C to 0°C

(I have accepted without questioning your estimate of snow being about 1/3 the density of water, I'd have guessed it was lower.)

fantastic,

so when calculating

step1 : get the aluminium plate to 0°C using X amount energy

step2: heat the snow/ice to 0°C using Y amount energy (depending on volume or rate of snowfall)

step3: convert the snow/ice to water using z amount energy

(you write: mass of snow * 334000 J.kg etc. where does this number come from?)

step 4: take the sum of x y and z PLUS nicknames point of the water needing to flow off the tile before re-freezing.

All of this of course is subject to environmental losses. I was told to add about 20% to account for this. Too little? the bottom and sides should be very well insulated.

You've got a classic-style thermodynamic problem. The temperature of the aluminum must be above 0^oC to melt snow or ice, but the power necessary to melt it at rates that snow falls or water freezes is where the problem gets tough.

Of course, ice will form at a rate that's a function of temperature difference, and snow fall rates are a function of mother nature. So, the problem becomes one which one must place practical limits upon, such as converting 10 lb of ice at -10^oC to water at 2^oC per hour.

1 BTU is required to raise the temperature of 1 pound of water by 1^oF. 1 BTU = 0.3WH, so it takes 216 BTU's to raise the temperature of -10oC ice to 2oC, which is 65 watt-hours that must be dissipated by the aluminum plate.

That's theoretical, however. The heat transfer will not be perfect, as the side of the aluminum plate will lose energy to cold air without melting ice, so assume you'll have about 50% efficiency at best, unless the underside is well insulated.

It's one of those peoblems that may be most easily solved by experimentation in the environment you plan to use it.

Great advice. Obviously testing is the best and easiest way to do this. The problem I am facing right now is that I must order heaters in order to test, and heaters cost money so I cannot order a whole bunch for testing. I'm trying to get an approximation of the power required to order the heater for testing. So now do I order a heater that can supply the plate with 25W per ft², 100W per ft², 1500W per ft² ?? If the result is above 30W per ft² then I have failed to better the competition and have no purpose in pursuing this.

And so I am faced with quite a dilemma. Obviously the theoretical calculations for such an application are ambiguous at best and, like you say, the only practical way is through testing in the field.

Nick name: I thank you very much for your guidance, unfortunately, I think that your path is beyond my level of understanding!! As you and others have pointed out, Thermodynamics are never as "simple" as they seem. I think that I have found my solution: If my boss wants engineering calculations of this magnitude, he ought to hire an engineer!!

Thanks a lot guys

Can you buy one of the least expensive heaters and test it? It's performance should scale up without a lot of error. Do you have an environmental chamber to test it in, or do you have to wait for the weather to provide various conditions?

lol an environmental chamber. that would be awesome!

If you saw my work-place you'd chuckle with me!

Unfortunately must wait for winter conditions. I can buy the least expensive ones of course and work my way up. My goal was to avoid doing this by getting a rough idea of what I need first. If I cant, well, it is what it is!

"Thermodynamics are never as "simple" as they seem."

Your calculations are actually a subject called heat transfer. This is a very hard subject and it's usually a "make or break" course for mechanical engineers.

An equation you can try using is Q = (kAΔT)/L. The Q is actually transfer not heat energy but I don't know how to show the difference.

k is the thermal conductivity of the plate (a physical property of the material), A is cross sectional area of the plate, ΔT is temperature on 1 side of the plate minus the temperature on the other, and L is the thickness of the plate.

I'm not sure but it sounds like you know the temperature on one side of the plate and you can select the temp on the other (let's say 4°C to make sure the ice melts and flows). This will give you the power required to heat the plate.

A word of warning though, it's been many years since I've done these calculations so you will want to look this up before you take my word. Also, this makes assumptions that might not apply to what you are doing.

If your boss wants you to do this then start by looking up conduction, convection, radiation, and thermal resistance concept. But in general, it's best to get someone familiar with the subject to solve it.

Agree.

But it's still easy for Noideabob to understand.

Dear Noideabob, you see, your question is about "select a exchanger", a common therodynamic engineering project.

To satisfy your boss's requirments, there are two necessary factors: (1) Generate the heat you need to meltdown and heat the paticular mass of snow (it's defferrent if it is snow rather than water or else); (2) Thansfer the heat to the snow.

So the question you need to solve can be concluded as: A)Select a heat source; B)Calculate the Capacity (or call it "Load"); C) According to the capacity and work condtion to select a exchanger;

Details:

Follow the point (1), it's easy to find a steady heat source something like hot water from a Heat-pump or just vapour from the boiler. Different heat source can supply you different "Temperature", Heat-pump usually can afford hot water of 55-65 ℃, the Boiler can afford hot gas of 150℃ or higher. Then calculate how much heat you need to generate, specially for the worst condtion you may meet in future. It also helps to reduce the unncessary cost (each boss cares the cost). As mentioned, when the snow melts down, it absorbs extra heat, we call it Latent heat, during this course, the temperature remains 0 ℃. And this Capacity meanwhile is also the basic Capacity of the Heat Exchanger.

To calculate the Sensible Heat: Q = c M △t，Q = Capacity，c = coefficiency，M = mass of the substance your need to heat, or the mass flow if the substance is liquid, △t=t1 - t2 , the temperature you need to raise.

To calulate the Latent Heat: Snow turns into gas, it needs 335 kJ / kg, (1 a.t.m). You can look up any conducts, books, website or manuals to get the parameter. Attention: You have to turn the ice into the gas !! Or it surely will fail becase the water may turn to the ice, I'll explain it later.

The full capacity= sensible heat + latent heat, it also helps you to select the heat source like Heat-pump, (a kind of chiller)

And then, follow the point (2), now you need to transfer the heat to the snow effectively(adquate and continously). The exchanger also has the formula of the capacity calculating:

Q= k F △t (or an earlier expressing = k A △t, just the same), Q=capacity, F=squarmeter of the heat exchanging surface( the bigger the better), △t = coefficiency(the smaller, the better)differential of two sides of the exchanger, k = heat exchanging resistance. Each piece of your Aluminum plate has a stable "k" and "F" and the confirmed condition has a stable "△t", so it's easy to calculate the quatity of the pieces you need.

But we usually use another convenient method: Look up the catalogs or manuals or Selection softwares of other famous exchangers producer, like SWEP. Although the products are different and characteristics may be little different ( the most different is the quality), the calcalating and theory is just the same.

P.S.: Why we need to turn the snow into gas rather than water in winter? Although when we making the designing calculating we may keep a "spare space" of the capacity to meet the worst conditon, however, with running time accumulated, the coefficiency of the exchanger will decline slowly. It is caused by rust, dirty, bacterium or scale. If you use refrigerant system as the heat souce, it may be remains more problems to be maintenanced. So the effeciency of the heat source may decline. If the lack of the heat results in you fail to keep the surface of the exchanger drying. Then you will see "ice" on the exchanger, the ice is much more dense or thick than snow, it reduce the efficiency of the exchanger rapidly. that means more mass to melt down and hard to melt them down. And the system FAIL!

In facts, many of the earlier Heat-pump-chiller like YORK's AWHC or Carrier's 30AQA, they use the aluminum-fin air-refrigerant exchanger as the condenser(chilling)/evaporator(Heating) of the system. In winter, during the heating process, the system need "Defrost" from time to time, Frost = Snow, but if the system is not adjusted or maintenanced properly, the Defrost will fail to make it dry enough at the surface, then when it begins heating, the water turns into ice. Then more frequetly Defrosting... and recycling.

For you, if you meet this problem in future, the only way for you is to raise the temperature of the heat source if you could. If not (already reached the limit), it fails!

So, the cost of maintenance is also need to be considered in designing work :-) If I were the sales-engineer, I will demand you to think of this point.

So much for this, it's pleased to discuss with you. Wait for your reply and emails: Jimmy.qian@refcomp.cn

Sorry, I fogot something important. Select heat source CAREFULLY, it must match the exchanger. Because the technology and constuction of the exchanger ( specially for piece-plate exchanger), it may have a limit of the temperature and pressure. For examples, not every exchanger can be used as a brine exchanger ( CaCl2) or as a refrigerant exchanger.

Take Care of the leakage and rust!

Thank you Jimmy, your advice will certainly be followed. There is one thing I think you may have missed. In this situation, I have my exchanger already and I am trying to size my source (electric). Most of your advice is still pertinant only now I must work backwards on some things.

Thanks very much, we will be in touch soon!

Hehe.. :-P sorry for missing, I was thinking that you are preparing to assemble a heat exchanger with alleminum plate pieces by your self and making the calculating.

Now the difficulty for you, I think, is to confirm the work conditon. (It's about to confirm the parameter T, M ...)

The ambient temperature, the Temp of the snow, the rate of the snowfall and the depth of the snow (if you start the melting after snow). You may need two modes, because the melting starting after snow or in the heavy snowfall needs more energy and higher Temp to prevent from re-freezing, to keep a Temp of the piece in the snowfall or start the melting before/in snowfall may need less energy for economic operation (less dynamic mass flow).

Using your original numbers, I believe you will need at least 0.5 kW. I'd go a little higher and test150W per square foot to keep the 2'x2' plate clear of snow (at <=1" of snow per hour).

Winter will be here soon enough.