How long does it take for the liquid in a 55-gallon drum to thaw?

[pennym72]

TL;DR Summary

Rate of Thaw for Ice at a Given Temperature

I am trying to calculate the rate of thaw for a 55 gallon drum of frozen juice. I know the surface area of the drum is 20,772 sq cm. I know the starting temperature of the frozen juice (-10C) and I know the ambient temperature the drum is currently in (10C). What is the equation that will allow me to calculate the time required for the entire contents of the drum to rise to 1C. Thank you!

 

[Bystander]

What kind of juice? Un/Sweetened? Air circulation? Drum material plastic/metal?....

 

[Vanadium 50]

How much heat are you putting into the room?

If you don't see why this is important, consider a thermally ioslated room with a total volume of 110 gallons, half of which is occupied by the drum. This will never thaw.

A good start is to calculate how much heat you need to thaw it and compare that to how much you are putting into the room.

 

[erobz]

If you assume a relatively thin metal drum and water, you can neglect thermal gradients across the drum.

Assuming convection is the dominant form of heat transfer:

There will be two stages of heat transfer The firs being taking the ice from -10 C , to 0 C. The time $t_1$ would be given by:

$$ M c_p \frac{dT}{dt_1} = - h A ( T - T_{\infty}) \tag{1}$$

Where
$A$ is the surface area of the drum
$h$ is the average convection coefficient
$M$ is the mass of juice (ice)
$c_p$ is the specific heat
$T_{\infty}$ is the temp of the environment (assumed large and constant temperature)

The solution to (1):

$$ t_1 = \frac{M c_p}{hA} \ln \left| \frac{ T_o - T_{\infty} }{ T_f-T_{\infty}} \right| $$

The next phase ( 0 C ice to 0 C water) would be heat transfer at a constant rate as the ice completed the phase change to water.

$$ \frac{dE}{dt_2} = hAT_{\infty} \tag{2} $$

Solving (2):

$$ \Delta E = h A T_{ \infty} t_2 $$

The change in energy would be from the mass $M$ and the latent heat of freezing $q_L$, such that:

$$ t_2 = \frac{M q_L }{ h A T_{ \infty} } $$

The total time $t$:

$$ t = t_1 + t_2 $$

Thats a ballpark figure I would start with.

Addendum: Convert all temperature to absolute temps ( kelvin scale )

 

[Baluncore]

Welcome to PF.

If you assume a relatively thin metal drum and water, you can neglect thermal gradients across the drum.

There may be an internal food-grade bag or drum lining. That is to protect the drum from the chemistry of the juice, but any special barrier will also insulate the drum to some extent.

 

[erobz]

To @Baluncore 's point. The ability to neglect thermal gradients across the drum is usually characterized by the Biot Number being less than a certain value. Its known as the criterion for validating the lumped capacitance method.

$$Bi = \frac{hL_c}{k} < 0.1$$

Where $L_c$ is the characteristic length of the solid (The ratio of the solids volume to its surface area). $h$ and $k$ are the average convection coefficient and thermal conductivity, respectively.

$L_c = \frac{V\llap{-}}{A_s}$

If that criterion isn't satisfied, don't expect a good approximation.

For $t_2$ we could improve by considering steady state conduction through a cylinder, with thermal conductivity $k$, in contact with two fluids (internal - the juice, extremally the environment). We can explore that if interested.

But the transient conduction analysis $t_1$ for that geometry would be significantly more involved, you might get away with figuring out how to solve:

$$ \frac{1}{r} \frac{ \partial}{\partial r} \left( kr \frac{ \partial T}{ \partial r}\right) = \rho c_p \frac{\partial T }{ \partial t }$$

If you want to go that route, probably wait for someone who knows what they are doing in that arena!

 

[Vanadium 50]

All these differential equations are nice, but I sure wouldn't start there,

You have 200 kg of ice. It takes 20 J per gram to warm up the ice to 0C and 334 J/g to melt it. So you need 71 MJ to do this. Where does this energy come from? It's not going to melt until you get it there somehow.

 

[russ_watters]

Considering the patriotic unit of a Ton of refrigeration, if the ice thaws in a day that's equivalent to half a typical window air conditioning unit. Yes, the room will need to be heated, but that isn't a lot of heating, and I suspect the free convection in that situation to be slower.

 

[erobz]

All these differential equations are nice, but I sure wouldn't start there,

You have 200 kg of ice. It takes 20 J per gram to warm up the ice to 0C and 334 J/g to melt it. So you need 71 MJ to do this. Where does this energy come from? It's not going to melt until you get it there somehow.

In the equations the heat is coming from surroundings via whatever is maintaining the surroundings at $T_{\infty}$.

 

[Vanadium 50]

but that isn't a lot of heating

But the question is "how fast?". You may not need a lot of heating, but if you have a lot of heating it will go faster. I would argue that this dominates - certainly more so than temperature gradients in the container.

Before modern refrigerators you had ice-boxes and ice-houses that worked on this principle: if you keep the heat from coming in, the ice lasts a long time.

 

[erobz]

But the question is "how fast?". You may not need a lot of heating, but if you have a lot of heating it will go faster. I would argue that this dominates - certainly more so than temperature gradients in the container.

Before modern refrigerators you had ice-boxes and ice-houses that worked on this principle: if you keep the heat from coming in, the ice lasts a long time.

But the OP is asking how fast it will thaw given a constant temperature surroundings left to its own devices (passively), not how fast it can thaw with a heat cannon?

At least that's how I interpret the problem.

 

[jrmichler]

I am trying to calculate the rate of thaw for a 55 gallon drum of frozen juice. I know the surface area of the drum is 20,772 sq cm. I know the starting temperature of the frozen juice (-10C) and I know the ambient temperature the drum is currently in (10C). What is the equation that will allow me to calculate the time required for the entire contents of the drum to rise to 1C. Thank you!

Starting with the assumptions that you want to:
1) Get an estimate of the time, and
2) Not mess with differential equations,
then
Here is how an engineer tackles this problem. Sorry about the English units, but that's what I'm familiar with. Start with some assumptions:
1) Assuming that the heat transfer coefficient is roughly equal to that of a single pane glass window, the R-value equals $1.0 {hr-ft^2-deg F}/BTU$.
2) The total heat required to melt the juice is equal to the heat required to melt an equal volume of water.
3) Ice has a specific heat of 0.5.
4) The drum contains 55 gallons of juice after melting.

Then the total heat required to raise 55 gallons of water from 14 deg F to 32 deg F, and melt it at 32 deg F is $55 gallons * 8.34 lbs/{gallon} * ((18 deg F * 0.5) + 144) = 70,200 BTU$

Since the heat to raise the temperature of the ice to 32 deg F from 14 deg F is small compared to the heat to melt the ice, simplify by assuming the average temperature of the drum is 32 deg F. Then the average temperature difference between the room and the drum is 18 deg F (10 deg C).

Then the rate of heat flow into the juice is $22 ft^2 * 18 deg F / 1.0 {hr-ft^2-deg F}/BTU = 400 BTU/hr$.

And the time = $70,200 BTU / 400 BTU/hr = 180 hours.$

If you could let us know how long it did take when it finally melts, we would appreciate the feedback.

 

[Vanadium 50]

how fast it will thaw given a constant temperature surroundings left to its own device

First, it is not necessary to quote the entire message, and it just clutters the thread.

Second, the OP hasn't come back. so it looks like a post-and-run.

Third, one can calculate the minimum size of the "room" to thaw it without introducing extra heat. Tbe "room" has to be about the size of the White House. Air has almost no heat capacity - that's why you can stick your hand in the oven without burning it, so log as you don't touch anything.

Oops - forgot a point (editing) - the OP never says the room is held at constant temperature, and in fact, suggests it is not.

For this this to melt at all, he needs to be adding heat. To answer his question, we need to know how much. Physics is not math.

 

[erobz]

I get about 68 hrs, using a convection coefficient $h = 1 \rm{ \frac{W}{m^2 K} }$ ( that's bottom end of free convection coefficient according to my heat transfer text , which states they are usually between 2 and 25).

I'd say that is as fast as it could be done.

If it is a relatively thick plastic drum, I could believe a significant increase in the elapsed time.

 

[erobz]

I am noticing an issue though. With that low of a convection coefficient, if it is in this plastic drum [1] HDPE with thermal conductivity $k = 0.48 \rm{ \frac{W}{m K}}$ [2] the time until it would thaw ( just considering the steady state portion 0C ice to 0 C water) is an order of magnitude or two larger!

So depending on drum material realistically, the range of possible times is quite large!
[1] https://www.uline.com/Product/Detail/S-10757BLU/Drums/Plastic-Drum-55-Gallon-Closed-Top-Blue?pricode=WB3684&utm_source=Bing&utm_medium=pla&utm_term=S-10757BLU&utm_campaign=Drums,+Pails+&+Containers&utm_source=Bing&utm_medium=pla&utm_term=S-10757BLU&utm_campaign=Drums,+Pails+&+Containers&msclkid=5dfb7ee2e82d1de7eb921d31aba235c7

(Click additional specs for drum dimensions.)

[2]https://www.substech.com/dokuwiki/doku.php?id=thermoplastic_high_density_polyethylene_hdpe#:~:text=Thermoplastic Thermal expansion (20 ºC) , ºF 10 more rows

 

[jrmichler]

That calculation of 1115 hours is over 46 days, which fails a sanity check. Using that number and working backwards:

$70,200 BTU / 1115 hours = 63 BTUH$.

$63 BTUH / 22 ft^2 = 2.86 BTU/{hr-ft^2}$

Average temperature difference of $18 deg F / 2.86 BTU/{hr-ft^2} = 6.3 hr-ft^2-deg F/BTU$. That's the R-value.

You can get an R-value of 6.3 from 2 inches of fiberglass insulation, or 1.3 inches of foam insulation, but not 2.2 mm of solid anything. The source of the error is thermal conductivity as a material property is in $W/m-K$, not $m^2$.

 

[erobz]

The source of the error is thermal conductivity as a material property is in $W/m-K$, not $m^2$.

I just used the wrong units in the explanation, the program wouldn't allow that error. Notice the $k$ and units of it in the picture of the computation.

The "problem" is there is a very strong dependency $h$. By making $h = 5 \rm{ \frac{W}{m^2 K}}$. Which is still absolutely reasonable I get:

 

[erobz]

If I allow for the convection from the top of the drum as well (air to material to air boundary conditions) That trims it down a smidge.

I also added the convection from fluid to wall in the radial term for completeness. It's negligible, but I might as well show it.

This is just the time for 0C ice to 0C water.

 

[phinds]

Guys this sure is a lot of effort for an OP who just asked the question as a drive-by and then left the scene.

 

[erobz]

Guys this sure is a lot of effort for an OP who just asked the question as a drive-by and then left the scene.

I personally would let it go, but people keep bringing up how math isn't physics and saying don't use what I've figured out, implying that it is incorrect somehow. My methodology is clearly being attacked without justification. Furthermore, when someone says things like "this is how an engineer would do it", I resent that. I am an engineer! These were the very hoops we were made to jump through to earn our pitiful degrees...and frankly, I like jumping through them.