Heat transfer from a container of water in a freezer

[Schott11]

I’m trying to understand the difference between how two styrofoam boxes of water cool when placed in a freezer.

see attachment for picture


Both boxes are 6’’ wide, 10’’ long and 6’’ high.
Both boxes have an internal cavity of 4x8x4 (1 inch of insulation on all 6 sides)
Box 2 has an extra piece of wall that extends into the center of the cavity, it is 1 inch wide and 2 inches long.

The boxes are made out of styrofoam that has an R value of 5 / inch.

Assume the top fits air tight and each box is filled with water at 70°F, and placed in a freezer that is a constant 10°F.

The internal surface area (area in contact with water) of box 1 is 160 in2 or 1.111 ft2
The internal surface area of box 2 is 172 in2 or 1.1944 ft2

Using
Q / t = Area * (T-hot - T-cold) / thermal resistance (Rfactor)

I show that
Box 1 transfers 13.332 BTU/Hour
&
Box 2 transfers 14.333 BTU/Hour

My questions are:
How is it that Box 2 is transferring more heat, despite more insulation? Am I missing something in the setup?
If not, and Box 2 is cooling faster, what is happening at the extra wall that extends into the cavity of water in box 2? Is it conducting more heat because of the increased surface area?

Thanks

 

[bigfooted]

Heat is conducted from the water to the styrofoam through the contact area. The heat is then conducted from the styrofoam to the outside air. When you increase the contact area with water, the heat transfer into the styrofoam will increase. This is stated by Fourier's law of conduction: the change of thermal energy is equal to the heat transfer coefficient times the surface area times the temperature gradient between the water and the styrofoam.
The conduction from the styrofoam to the outside air will also increase because of the increase in temperature difference between the styrofoam and the air.

 

[russ_watters]

Q / t = Area * (T-hot - T-cold) / thermal resistance (Rfactor)

Which area did you use for box 2?

 

[russ_watters]

Heat is conducted from the water to the styrofoam through the contact area. The heat is then conducted from the styrofoam to the outside air. When you increase the contact area with water, the heat transfer into the styrofoam will increase.

It isn't that simple. The heat that enters the "fin" has further to go to get out of the box than the heat that just passes through the wall, so you can't apply the wall thickness in the same way.

Consider the "fin" as a separate element, at the same temperature as the water. Does heat transfer as fast through the Styrofoam as does from the water to the Styrofoam? If not, then the "fin" will reduce the heat transfer rate.

 

[Schott11]

Thanks Russ, I suspected that the equation I am using doesn't properly account for how the fin is channeling heat.
For box 2 I used 172 in2 or 1.1944 ft2 as the area. I used 160in2 (1.111 ft2) for box 1. The fin adds 16 in2, but also reduces 4 in2 where the fin touches the top and bottom of the cavity, hence the extra 12 inches2.

I assume to calculate the heat transfer for box two I can use a similar setup for the 158 in2 that are the same as box 1 , then double the R value for the 2inches2 that that face the middle of the box, but not sure how to calculate the 2 walls of the fin that are 8 in2 (4in high and 2 in long).

Or do I need an entirely new approach?

 

[Schott11]

The heat will transfer faster through the water, than through the Styrofoam fin, so if I treat it as a separate element, my approach would be as follows: (See attachment for reference to Face A & B)

Q / t = Area * (T-hot - T-cold) / thermal resistance (R-factor)

For Box 1: (160 in2) 1.111 ft2 * (70 – 10) / 5 = 13.333 BTU/hour

For Box 2:
154 in2 is the same in Box 2 as Box1, so we can calculate that:
*Non-fin: (154 in2) 1.0694 ft2 * (70 – 10) / 5 = 12.8333 BTU/hour
*Face A: (2 in2) .013889 ft2 * (70-10) / 15 = .08333 BTU/hour
I used 15 as the R for face A since heat must travel 3 inches, compared to 1inch.

*Face B & C have a surface area of 16 in2 (8 each), but I’m not sure what R value to use as the path the heat travels isn't at a right angle to the gradient.

Does that mean Fourier’s law can’t be used to solve this problem?

Any suggestions? Thanks