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Time taken to freeze ice vertically from top exposed surface

Time taken to freeze ice vertically from top exposed surface

Time taken to freeze ice vertically from top exposed surface

(OP)
Hi guys,
New to the forum wanted a little help in designing an insulated ice mold. Basically the objective is to force the water in the mold to freeze axially or in one direction in order to make clear ice. Ideally, the time taken for the ice cube to freeze from the top surface exposed to the ambient air should be less than the time taken for the inside surfaces of the mold to reach zero.

Assuming:
Air:-
Ta = -10°C
P = 1ATM
g = 9.8

Water:-
Ti = 20°C
Tf = 0°C
Cp = 4.18
h(fg) = 355 KJ/kg (Latent Heat of Fusion)
rho = 1000 kg/m^3

Cube Dimension:-
L = 0.05m
m = 0.05^2*1000 = 0.125 kg

Initially, I've calculated the energy required to freeze the cube as:

Q = m[Cp*∆T - m*h(fg)]
= 0.125[4.18(0-20) - 355]
= -52.14375 KJ

Then I get stuck on determining the convective coefficients. I've tried calculating the Nusselt number for free convection on a hot horizontal plate top surface (top exposed surface) but Rayleigh number is less than 10^4. And for the sides (assuming the bottom is sitting on the inside surface which is practically adiabatic) I get:

Tf = (Ti-Ta) / 2 = 278°K
Beta = 1/Tf = 3.597*10^-3 °K^-1

From Linear Interpolation of Data for air @ Tf (Fundamentals of Heat and Mass Transfer, Frank P. Incropera: Table A.4)
Nu = 13.932*10^6 m^2 / s
Alpha = 19.596*10^6 m^2 / s
Pr = 0.71272
k = 24.14 *10^-3 W / m.K

Ra = g*Beta*(Ti-Ta)*L^3 / (Alpha*Nu)
= 4.842*10^5
*as Ra~<10^9 use laminar equation. C=0.59, n=1/4

Nu = 0.68 + 0.67*Ra^(n) / [1+(0.492/Pr)^(9/16)]^(4/9)
= 14.251

Nu = h*L/k
h = 6.8804 W / m^2.K (Convection Coefficient)

Then developing a thermal circuit for the side walls which is 3mm (tpe) LDPE sandwiched between two 2mm (tsi) layers of silicone rubber.

Ksi = 0.2 W / m.K
Kpe = 0.034 W / m.K

The source would be the water and the sink would be the ambient freezer temp.

1/Rt = 2*tsi/Ksi +tpe/Kpe +1/h
Rt = 3.94 W / m^2.K

q = Rt*A*∆T
= 3.94*(0.05)^2*((-10)-20)
= 0.2958 W

I know if you divide that through Q/q = 176298.49 sec which works out to ~48 hours. That seems a little excessive to me especially considering that q is an instantaneous rate when the temperature difference is greatest and will only decrease meaning that it would take ~>50 hours for the inside surface temperature to reach 0°C (unless it is an average rate, I forget). Knowing my luck it's possible I'm going in the wrong direction for determining this altogether. Am I doing the right thing and how do I work out the time taken for the whole ice cube to freeze from the top surface?

Regards,

Josh

RE: Time taken to freeze ice vertically from top exposed surface

How did you address the thermal conduction through the Ice as it gets progressively thicker? Ice can be a good insulator compared to water. Also the heat loss through the ice will have to be greater than the heat coming from the walls or you will reach an equilibrium state.

RE: Time taken to freeze ice vertically from top exposed surface

(OP)
Dougt115 I hadn't addressed that yet. I'm just trying to scratch out a basic model on paper so I can decide how much insulation would be needed for a prototype and in the future I can refine it further. I've just assumed that if the time taken to freeze ice 50mm thick is a few (maybe 4 considering the thermal conductivity is about 4 times greater) times less than the time taken for the inside walls to reach 0°C that will more than account for the change in thermal conductivity (I could be wrong). I'm just not sure how to approach the change in temp axially. Can I treat it as an infinite plane freezing to 50mm depth? Or does the small geometry dominate infinite plane assumptions? What is a reasonably simple method for determining the time taken to freeze axially?

EnergyMix. That is the exact website that I had used in the first place. Having run a few cocktail bars in the past I used those methods to make large clear ice cubes for drinks. The problem is that the process is very time and space intensive for small bars that need a lot of them and this is an issue shared by a lot of others I have spoken to. Hence, I wanted to take that concept and apply it to the mould itself in order to simplify the process and allow higher production. A lot of people have asked and it seems like a worthwhile venture.

RE: Time taken to freeze ice vertically from top exposed surface

(OP)
I also haven't taken into account the effects of radiation but again I figure that can be refined later.

RE: Time taken to freeze ice vertically from top exposed surface

I agree with IRstuff and that is to make clear ice, you'll need to freeze the water from the bottom of the container; as the ice thickens, the trapped air will be forced upward.

RE: Time taken to freeze ice vertically from top exposed surface

I think that it's quite fortunate for the evolution of life on Earth that ice is less dense than liquid water.

you must get smarter than the software you're using.

RE: Time taken to freeze ice vertically from top exposed surface

(OP)
Sorry I've been a little busy this last week.
IRstuff. You're right. Conventional ice machines freeze ice much in the same way an icicle forms by running water over a mold plate and letting it freeze layer by layer. Most machines will have a vertical plate that is timed so that the refrigeration cycle running through the plate will reverse after a certain time period to melt the surface of the ice against the mold allowing the cubes to slide out via gravity. The machine we use is a Hoshizaki which are internationally renowned for making the largest and most dense ice cubes due to an upside down horizontal plate that require upward facing jets to fill the molds rather than allowing water to trickle down a vertical plate.

These machines are however too expensive for most bars and the cubes are only 20mm in width. I'm looking to create a cheap ($10-$15) solution for small bars that want larger clear ice (generally for cocktails). It is a bit of a niche market but it also has potential for peoples home bars.

Back to the calculations. It's been a while since I've done any thermo calculations and I realised the other day that the second half of my approach is wrong and I forgot to treat it as a constant surface convection transient conduction problem (which yields ~13.5 hoursfor the inside surface to reach zero). The top surface has too low a Rayleigh number for determining free convection constants and treating it as a constant surface temperature case yields ~15 minutes to reach Zero degrees at the bottom with k values for water, and ice is even quicker. Can it be treated as an infinite plane conduction problem for the top surface?

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