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
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
RE: Time taken to freeze ice vertically from top exposed surface
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RE: Time taken to freeze ice vertically from top exposed surface
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
RE: Time taken to freeze ice vertically from top exposed surface
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RE: Time taken to freeze ice vertically from top exposed surface
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RE: Time taken to freeze ice vertically from top exposed surface
RE: Time taken to freeze ice vertically from top exposed surface
you must get smarter than the software you're using.
RE: Time taken to freeze ice vertically from top exposed surface
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?
RE: Time taken to freeze ice vertically from top exposed surface
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