Probably the most basic of fundamentals

[shoot_j]

I was wondering if anyone would be willing to breakdown a "Thermal Properties of Liquids for Idiots" summary through example using One Pint of Water as the liquid and any temperature readient. Please be as detailed as possible considering I only understand the basics of terms.
thanks

[schartruse]

The thermal properties of each and every liquid is different but basically they all have the same properties. Being from Canada I'll try to convert as much as I can. There's the standard boiling point and freezing point which is basically self explanatory so I won't go into it. There is Density which is basically how close the molicules of the water are together, the density of water at 39.2oF is 8.33lbs/gal. There is the heat of melting which is how much energy must be absorbed before your water will change from solid to liquid - 334kj/kg. Heat of vaporization to change from liquid to gas - 2270kj/kg. The temperature of your water and the energy absorbed by the water are different, you can hold water at 212oF but it may not boil until it has absorbed enough energy to change state. This is based upon barometric pressure but we'll leave that out for now. You have specific heat which is basically how much energy must be absorbed to raise the temperature of water 1oF in BTU. 1 BTU is required to raise the temperature of 1lb of water 1oF.
I'm not sure how far into this you want to get there's so much more that can be included. And it's 2am and i've been working all night so I could be completely full of it and calculated something wrong. Someone please correct me if I did. Sorry it's so dry but I'm used to writing technical documents.

[Abby Normal]

A pint is roughly a pound of water. Water pretty much defines all the heat transfer terms in the US.

Water has a specific heat of about 1 Btu per lb per degree F. Take a pint and raise or lower its temperature by a degree F and you have added or removed a Btu.

Water has a latent heat of fusion of about 144 Btu/lb. This is the heat removed at 32 F to change a pound of liquid to a pound of ice.

Take a pound of water at 33F, remove one Btu, and it will be liquid at 32F. Remove any more heat, the temperature stays at 32F, and the liquid begins to freeze.

Imagine a 2000 pound or one ton block of ice at 32F. To melt each pound of ice to a pound of water at 32F, means the ice has to absorb 144 Btu, same energy needed to freeze it.

To melt the entire ton of ice would mean the ice would have to absorb, 2000 x 144 = 288,000 Btus.

If you had that ton of ice sitting in a room and it melted at a constant rate over the period of a day, it would have absorbed heat at a rate of 288,000 Btu/24 hrs or on an hourly basis 12,000 Btu/hr.

Air conditioners are sold by 'the ton' and one ton of cooling is equal to 12,000 Btu/hr.

Water has a latent heat of vapourization as well, the heat required to evaporate a pound. At room tremperatures it will be about 1060 Btu/lb, but up at the boiling boint it is somewhere around 970 to 980 Btu/lb. Quick math is use 1000 Btu/lb and you will not be that far off.

Take a pound of water at 71 F and heat it to 211 F and you have to add 211-71-140 Btu of heat to that pound. Add one more Btu, and the water is up to 212 the boiling point, add any more heat and it will boil. Take close to another 1000 Btu to turn that water into vapour.

When steam condenses, the same amount of heat is released.

[shoot_j]

Thanks so much for the detailed responces.
So let me check to see if I am following this correctly.
If I had 1 lb of water that was 90F and I wanted to cool it to 38F it would require 52 Btus / 38-90-52Btu and to cool it from 38F to 31F it would take 7+144=151 Btus?

[Abby Normal]

conductivity is for heat flowing through it, specific heat is for raisng the temperature, not going to learn natural convection calcs on line for free but search newtons law of cooling

[shoot_j]

Understanding that I am "not going to learn natural convection calcs on line for free" I have decided to seek assistance from a private forum comprised of individuals with knowledge on the subject I am studing to which I am glad to pay with my gratitude as well as such they will recieve from future members/guests that may benifit from this thread. The exact figures have little relivance. I am simply attempting to expand my limited knowledge. Although I am somewhat familiar with Newton's Law of Cooling, it would make more sense if through example.
Conductivity being "for heat flowing through it" would refer to the effect that ambient temperature has on the contents of the container being cooled?
The suggestion to search for Newtons law of cooling did lead me to this link which shows an example of a situation in which heat is being removed from an object by a cooler exterior and solves for time to reach the desired temperature in best case senario.


ugrad.math.ubc.ca/coursedoc/math100/notes/diffeqs/cool
.html


I would like to discuss a senario involving the effect of a warmer exterior environment and the maintaining of a cooler interior environment. Lets use a refigerator for example. For simplification of math and comprehension I would like to use the 1lb of water as the liquid being refrigerated. I would also like the area inside the refrigerator to be full of only that 1lb of water so there are not other factors to worry about.
Furthermore I would like to use a high quality of insulation in this discussion, to continue the best case sinario idea. The best of which I have found to have a thermal conductivity of 0.003 W/(m·K). I am however willing to look at any other as an example if it adds to the simplification of this topic.
Perhaps I should save the energy questions until later in the conversation.

Can we instead discuss the amount of time it will take for 1 lb of water to gain 1F when it is insulated with the 0.003 W/(m·K) conductive attribute in the average ambient temperature of 70F?
Air density @ 70F 0ft elevation ~ .0745lb/ft3 = 0.0012g/cm3
Water density @ 38F ~ 62.43lb/ft3 ~1g/cm3

[Abby Normal]

search transient heat transfer and the "Biott number"

[US Iceman]

Since you are working in research and development you should also undertsand the need to list your assumptions. Don't just jump in with an immediate conclusion.

You need to break down the problem before you attempt to solve it.

The easiest way is to first look at it as a simple conduction problem.

If you know you have one pound of water and you want to change the temperature from 90F to 38F then you already know the amount of heat you need to remove is (90F - 38F) X 1 pound X 1 BTU/pound-degree F = 52 BTU.

The one thing you have not mentioned is the time rate, i.e., how fast do you want the heat to be removed?

If you can allow 1 hour for this process, then the heat removal rate (cooling rate) is 52 BTU per hour.

If you want the heat to be removed in 1 minute then the cooling rate is 52 BTU per minute. This is considerably higher than 52 BTU per hour.

After you decide what time rate you want to use, you also have to factor in the heat gain through the container (the insulated shell where the water is contained).

Know you need to know the insulation R-factor and temperature difference across the given insulation thickness to find the heat gain through the insulated shell.

This heat gain + the cooling load of the water would be the simplest model to develop. Once you have the total heat load and time rate allowed for the cooling process, then you can try to find a refrigeration system with sufficient capacity to meet the requirements.

If you want to try to get really in-depth then start to look at the problem as a transient conduction problem.

I tend to think if you approach this as a simple steady state conduction problem you will have a reasonable conclusion.

[shoot_j]

continue in new thread under "transient heat transfer"
Thank you very much for your guidance so far Abby and Iceman, you have both been more help then you know!