How Long Until a Perfectly Sealed Box Reaches 32°F?

[timbdgr]

O.D. Box 14" x 14" x 6"
I.D. Box 12" x 12" x 6"

Thermal conductivity of walls .194 Btu.in/(ft2.h.°F)

Internal temp. -109 F
block of ice (water) -109 F 2" x 12" x12"

Outside temp 80 F

No humidity,wind, leaks, perfect seal, in perfect conditions.

How long till internal temp is 32 F (ice included).

Please just explain in simple terms how to find the answer.

Thank you

 

[Mapes]

Hi timbdgr, welcome to PF. This model may be a good candidate for the "lumped capacitance" technique in heat transfer. This approach would take the ice to be a point mass; that is; all parts of the ice are assumed to be at the same temperature. This simplifies the math considerably because you only need to calculate the necessary change in energy based on the specific heat, and you don't worry about heat transfer through the ice.

One would also assume that heat is diffusing through the box walls evenly. Is there a typo in the box description? It looks like the top of the box has zero thickness.

The result of the lumped capacitance model is a simply exponential formula for temperature, which starts off rising quickly but would asymptotically approach 80°F because the driving force for heat flow drops as the contents reach the ambient temperature. Of course, in this case the ice doesn't make it past 32°F.

On board so far?

 

[timbdgr]

O.D. Box 14" x 14" x 6"
I.D. Box 12" x 12" x 6"

O.D. should be 8" sry.
I understand so far

 

[Mapes]

OK, the equation I propose applying is

\frac{T-T_\infty}{T_i-T_\infty}=\exp\left(-\frac{kA}{\rho V L c}t\right)

where T is the ice temperature, T_\infty is the ambient temperature, T_i is the initial ice temperature, k is the wall thermal conductivity, A is the effective surface area, \rho is the ice density, V is the ice volume, L is the wall thickness, c is the ice specific heat, and t is time.

There are a couple possible choices for A: surface area of the ice, inner surface area of the box, outer surface area of the box, or a weighted combination of these. I'd use the inner surface area of the box as a first pass.

Take a second to confirm that increasing or decreasing the controlling variables changes the temperature in the correct way (e.g., increasing the volume of the ice should slow the process, or predict a higher ice temperature at any given time). Also check that the units all work. Hopefully this equation will provide a good approximation for the time it takes for the ice to warm to 32°F.

There a description of the technique http://www.nd.edu/~msen/Teaching/IntHT/Slides/05A-Chapter,%20Sec%205.1%20-%205.3%20Black-1D.ppt" and at numerous places on the web. The illustrations in that presentation were taken from Incropera and DeWitt's Fundamentals of Heat and Mass Transfer, and that's the first place I'd suggest looking for more details; it's an outstanding and very readable textbook.

 

[timbdgr]

I'm sorry but, I got a couple of questions. T is the final temp? exp is for()^exp. Finally can you explain what this means ".194 Btu.in/(ft2.h.°F)" isn't k usually expressed as btu/Ft2.h.F

thanks for you help

 

[Mapes]

Yes, T is the final temperature (32°F in this case), \exp is the exponential function. Your original units were correct, if a little unusual; thermal conductivity is measured in power per unit length per unit temperature.

 

[timbdgr]

T=32 F
Tinf=80 F
T1=-109 F
k=.194 Btu.in/(ft2.h.°F)
A=864 in2
p=.0331178 lb/in3
V= 288 in3
L= 1 in
c=.485 BTU/lb

32-80
-109-80 =exp

.194 Btu.in/(ft2.h.F)*864 in2
.0331178 lb/in3 * 288 in3 * 1in * .485 btu/lb


does this look right?

or do i have to convert all the units to the same?

 

[Mapes]

The units have to all cancel in the end. It may be easier to work in SI units, where thermal conductivity is measured simply in W m^-1 °C ^-1. There are online converters to convert your BTU-based properties. I get a thermal conductivity of about 0.03 W m^-1 °C ^-1, which I recognize as quite low. Is your box made of polystyrene foam or something similar? I'm only asking because I had to ship a frozen vial of cells to England recently and it was quite an effort. We ended up using 20 lbs dry ice along with a polystyrene container with 3" walls, which kept the cells frozen for a day or two.

 

[timbdgr]

I realized that when my 1 st answer was rediculous. so the closest i have gotten i still have WJ/K is there a way to cancel them? It comes from c = 1719 J/KG C and k= W/mK or should i convert them to something else.

It is actually Divinycell H45 made by the DIAB Group. The website actually lists .028 W/mK. It only comes in 4 x 8' sheets for around 170.

 

[Mapes]

You can cancel °C and K because the increment size is the same. And don't forget 1 W = 1 J s^-1!

 

[timbdgr]

ok i got t= -19.46?

 

[Mapes]

ok i got t= -19.46?

Yeah... no.

We'd expect somewhere between an hour and a day.

 

[timbdgr]

T= 273.15K
Ti= 194.82K
Tinf= 299.82K
A= .371612 m2
p= 916.7 Kg/m3
c= 1719 J/Kg C
K= .194 W/mK
V= .03097 m3
L= .0254 m


-.0094389 ^.254
1239.58 =1/t

t= -19.9554

where did i go wrong?

 

[Mapes]

Hard to tell; can you format the equation in LaTeX? (Click on my equation above to see the format.)

 

[timbdgr]

i can see the format.

 

[timbdgr]

-.0094389 ^.254
1239.58 =1/t
<br /> {t^\ (-1)} = \left(-\frac{.0094389 }{ \ 1239.58}\right)^\ (.254)<br />


t=-19.95546

 

[Mapes]

Are you working from

t=-\frac{\rho V L c}{kA}\ln\left(\frac{T-T_\infty}{T_i-T_\infty}\right)\quad\mathrm{?}

I can't tell.

 

[timbdgr]

\frac{T-T_\infty}{T_i-T_\infty}=\exp\left(-\frac{kA}{\rho V L c}t\right)
this is the one you sent me I inputed the values divided by t and solved. what is ln?

 

[Mapes]

It's the same equation, solved for t. \ln is the natural logarithm (e.g., \ln(e)=1, \ln(1)=0, etc.).

 

[timbdgr]

I didnt look b4 I signed up is this sight for students only? If it is I am sorry I am 29 and haven't ben in school for 11 years. I am trying to design a bag for my job could you possibly tell me just an approximate time frame ie with in an hour or 2.

 

[timbdgr]

if not thanks for your help.

 

[Mapes]

I'm 34 and it feels like I've been (back) in school for the last 11 years! I get about 30,000 seconds, or about 8 hours. But we've made a bunch of assumptions, both in choosing the solution method and in using temperature-dependent material properties, so make sure you build in a factor of safety (e.g., assume that the time could be anything from a few hours to a day) and/or do some testing. I hope the answer calculated here isn't too far off. Good luck!

EDIT: In reading this over, I don't mean to say our discussion today felt like 11 years; I mean I've actually been in grad school for the last few years. Today was a fun exercise in practical heat transfer.

 

[timbdgr]

Again thanks for your help. just out of curosity do you receive any compensation for your time? Not trying to impose, I just think mechanically and could use some mathmatical insight. again thanks.

timbdgr@yahoo.com

 

[timbdgr]

T= 273.15K
Ti= 194.82K
Tinf= 299.82K

p= 916.7 Kg/m3
c= 1719 J/Kg C
K= .194 W/mK

L= .0254 m

my original conversions were wrong do you mind one more calc.?
what happens if

A= .030946 m2
and V= .004719 m3

 

[Mapes]

Try Google: http://www.google.com/search?hl=en&...+*+.0254+*+1719+/+0.194+/+0.0309&btnG=Search"

 

[timbdgr]

The link was awesome. thanks. my conversion ability not so much. I was thinking if we input the area of the inside of the container doesn't that tell the equation that the ice is spread out approx 1 in thick around the container? I did some calculations and determined that if i make the ice 9.3" x 9.3" x 3.15" (thats close the numbers are in my car) that i would have 1.3" space between the ice and the sides. so... if air is .025 W/kM and the insulation is .028 W/kM can i ?? ((1.3*.025)+(1*.028))/2.3" = .026 W/kM for a combined k?

 

[Mapes]

The air is definitely a complicating factor that would slow down the warming process because of the extra thermal insulation it provides. I'm not sure how best to alter the model to handle the air, but your approach looks like a good way to handle it.

Still, I wouldn't trust the predicted time (to get to 32°F) to a precision closer than an order of magnitude. Fortunately, it shouldn't be too hard to try an experiment to confirm.

 

[timbdgr]

the values i gave you were converted incorrectly so the time is definatly off. I think i got the correct values in and got about 11.3 hrs. is my assumption about the ice being spread out correct or close?

 

[Mapes]

Sure, it seems like a reasonable assumption.