Can a 2500 Gallon Water Tank Sufficiently Cool a Condenser Without a Chiller?

[LarrySputnik]

Hello! Hopefully someone here can help with this problem:

I have a condenser that holds 35 gallons of water to cool the copper coil on my still. I need to know if I can create a closed loop system of cooling water that allows me to avoid running a chiller. I know that there is probably an equation of Input and Output versus coil temperature and water temperature, but I have unfortunately never come across such an equation throughout my experience.

I can add the following information to help get things started:

-The condenser holds 35 gallons of cooling water
-I will be pumping water from a 2500 gallon tank to the condenser at 5GPM
-The water will be traveling through 3/4" hoses to and from the condenser
-I want to have the water pass through the condenser and return back to the top of the tank without using a chiller.

I simply need to know if 2500 gallons of water stored at, say, 72 degrees F can circulate through a condenser to cool a coil at 5GPM that may reach temperatures of 180 degrees F without heating up too much. If there is a way to calculate that I would appreciate your help.

Thanks!

 

[CWatters]

You would need to know the flow rate of the liquid to be cooled (eg the alcohol). I'm assuming the alcohol it enters the condenser at around 180F but what temperature does it leave at? Alcohol boils at 172F so obviously you want it lower than that but how much lower? This info would allow you to calculate how much energy has to be removed from the alcohol per min.

Then you can calculate how much the cooling water would warm up when pumped through at 5GPM and how fast the 2500Gal tank would warm up.

PS: This info is provided for educational purposes only!

 

[CWatters]

If this is a continuous process (eg like a factory) then the 2500Gal tank will eventually warm up. The temperature it reaches will depend on how much heat the tank can dissipate to it's surroundings, local air temperature etc.

 

[LarrySputnik]

You would need to know the flow rate of the liquid to be cooled (eg the alcohol). I'm assuming the alcohol it enters the condenser at around 180F but what temperature does it leave at? Alcohol boils at 172F so obviously you want it lower than that but how much lower? This info would allow you to calculate how much energy has to be removed from the alcohol per min.

Then you can calculate how much the cooling water would warm up when pumped through at 5GPM and how fast the 2500Gal tank would warm up.

PS: This info is provided for educational purposes only!

Thank you CWatters!

This is very helpful. So the alcohol will move along into the condenser around 172+F and then be cooled down by the water. If the cooling water is always kept at room temperature then the cooling water is always variable from day to day, season to season. Nevertheless, I would like the cooling water to remove about 100F from the alcohol from the top of the condenser to the output spout. This would give me product coming off the still about 70 to 80 degrees F.

So how do I then calculate how much energy has to be removed per min?

Thank you for getting the ball rolling on this!

 

[LarrySputnik]

If this is a continuous process (eg like a factory) then the 2500Gal tank will eventually warm up. The temperature it reaches will depend on how much heat the tank can dissipate to it's surroundings, local air temperature etc.

The process is only over the course of 4 to 8 hours. I was hoping that a large tank, such as 2500 gal, would allow me at least 8 hours. If I have to go larger though, I just need to know how large I need to go, which is where these calculations will really come in handy.

Thanks!

 

[CWatters]

I would like the cooling water to remove about 100F from the alcohol from the top of the condenser to the output spout. This would give me product coming off the still about 70 to 80 degrees F.

OK so it goes in at 172-180F and you want it to come out at 70-80F.

We still need the flow rate though.

 

[LarrySputnik]

OK so it goes in at 172-180F and you want it to come out at 70-80F.

We still need the flow rate though.

I'm not sure how to accurately arrive at the flowrate. I am using a copper alembic pot still and all of the calculators I have found are for reflux stills. If I run those calculators it gives me a flow rate of 1.13 g/s, which is highly unlikely.

Perhaps we can suppose that the flow rate of the distillate coming off the condenser is 0.5gpm. It would seem reasonable to collect 1 gallon every 2 minutes for the sake of running this scenario.

 

[JBA]

Rather than worry about the flow, etc I think it is easier to work with the alcohol vs water bulk volume heat capacity ratio. Either way, the heat transfer from the tank to atmosphere has to taken into account; and, it will increase the allowable running time in the below analysis.

Ethanol: Heat of vaporization = 364 BTU/lb; Specific heat of liquid = 0.65 BTU/lb °F; Density = 0.01 lb/gal
Water: Heat of vaporization = 970.4 BTU/lb; Specific heat of liquid = 1.00 BTU/lb °F; Density = 8.35 lb/gal

Mass of stored water = 2500 x 8.35 = 20,875 lb
Heat capacity for stored water with 100° temperature change = 20.875 lb x 100 x 1.00 = 20875 BTU

Ethanol condensation and 100° temperature change per gal = (0.01 x 364) + (.01 x 100 x 0.65) = 3.64 + .65 = 68.65 BTU/gal

Ethanol that can be cooled = 20875 BTU / 68.65 BTU/gal = 304 gal

Allowable running time at .5 gal/min = 304/.5 = 608 min / 60 = 10 hrs.

Verification of my calculation will be appreciated.

Edit: While the bulk allowable amount of 304 gal is accurate the actual running time required for this production is less accurate because the increasing temperature of the condenser input water from the heating tank will require increasing rates of water flow to maintain the required 100 degree temperature drop in the condenser.

 

[CWatters]

Edit: Didn't see JBA's reply before writing my effort..

Ok I think we can run some numbers. I'm afraid I work in metric so for my sanity I'll have to do a conversion and then back again..

The temperature reduction required (ΔT) is from 180F (82C) down to 80F (27C) a drop of 82-27 = 55C.

Alcohol (drinking) is Ethyl Alcohol/Ethanol which has the following properties..

Latent Heat L_A = 846 kJ/kg
Specific Heat S_A = 2.46 kJ/kg·°C
Density = 789 kg/m³

Perhaps we can suppose that the flow rate of the distillate coming off the condenser is 0.5gpm.

Convert 0.5 gal to kg...
0.5 gpm = 1.89L = 1.89*10^-3 m³

mass = volume * density
1.89*10^-3 m³ * 789 kg/m = 1.49kg

Mass flow rate (M_fr) = mass/time = 1.49/60 = 0.025kg per second.

So the energy per second (aka power) that has to be removed from the alcohol is...

P (kW) = (L_A * M_fr) + (S_A * M_fr * ΔT])
= (846 * 0.025) + (2.46 * 0.025 * 55)
= 21.15 + 3.38
= 24.5kW

So now we can look at what happens if we put that much power directly into 1200 gallons of water...

Data for the water..
Mass of Water M_w = 1200 gallons = 4524L = 4524Kg
Specific heat S_W = 4.184 kJ/kg·°C

So the temperature rise per second would be

dT/dt = P/(M_w*S_W)

= 24.5 * 10³/(4524 * 4.184 * 10³)
= 1.3 * 10^-3 degrees C per second
or
4.6C per hour
So after 8 hours the 1200 gallon tank will have risen about 37C. If it started at say 21C (70F) then it will end up at around 58C (136F).

We need to do one more check because the heat isn't dumped straight into the 1200 gal tank there is a 5gal/min transfer pump between the condenser and the 1200Gal tank. The temperature of water coming out of the condenser will be hotter than that going in...

5gal/min = 0.32L/S = 0.32Kg/S

If you put 24.5kW into 0.32Kg of water the temperature uplift will be..

ΔT = 24.5*10³ / (0.32 * 4.184 * 10³)
= 18C

So the water coming out of the condenser will be about 18C warmer than it went in.

The issue will be towards the end of an 8 hour day.. The 1200L tank may have risen to around 58C (136F) so that will be the temperature of the water going into the condenser. The water coming out of the condenser will be hotter around 58C + 18C = 76C (168F). The average temperature of the water in the condenser would be (168+136)/2 = 150F. So the alcohol wouldn't be cooled below that.

This all ignores heat loss from the big tank, pipework etc and that might improve the situation. If you need additional cooling perhaps instead of using a proper compressor based chiller you might get away with some car radiators and a fan on the hot side of the condenser (eg to cool the water before it goes into the big tank). You might need something like that anyway to cool the 1200 gal tank back down to 70F overnight for the next run?

Well that's my best stab at the figures. Your mileage may vary!

 

[LarrySputnik]

Edit: Didn't see JBA's reply before writing my effort..

Ok I think we can run some numbers. I'm afraid I work in metric so for my sanity I'll have to do a conversion and then back again..

The temperature reduction required (ΔT) is from 180F (82C) down to 80F (27C) a drop of 82-27 = 55C.

Alcohol (drinking) is Ethyl Alcohol/Ethanol which has the following properties..

Latent Heat L_A = 846 kJ/kg
Specific Heat S_A = 2.46 kJ/kg·°C
Density = 789 kg/m³
Convert 0.5 gal to kg...
0.5 gpm = 1.89L = 1.89*10^-3 m³

mass = volume * density
1.89*10^-3 m³ * 789 kg/m = 1.49kg

Mass flow rate (M_fr) = mass/time = 1.49/60 = 0.025kg per second.

So the energy per second (aka power) that has to be removed from the alcohol is...

P (kW) = (L_A * M_fr) + (S_A * M_fr * ΔT])
= (846 * 0.025) + (2.46 * 0.025 * 55)
= 21.15 + 3.38
= 24.5kW

So now we can look at what happens if we put that much power directly into 1200 gallons of water...

Data for the water..
Mass of Water M_w = 1200 gallons = 4524L = 4524Kg
Specific heat S_W = 4.184 kJ/kg·°C

So the temperature rise per second would be

dT/dt = P/(M_w*S_W)

= 24.5 * 10³/(4524 * 4.184 * 10³)
= 1.3 * 10^-3 degrees C per second
or
4.6C per hour
So after 8 hours the 1200 gallon tank will have risen about 37C. If it started at say 21C (70F) then it will end up at around 58C (136F).

We need to do one more check because the heat isn't dumped straight into the 1200 gal tank there is a 5gal/min transfer pump between the condenser and the 1200Gal tank. The temperature of water coming out of the condenser will be hotter than that going in...

5gal/min = 0.32L/S = 0.32Kg/S

If you put 24.5kW into 0.32Kg of water the temperature uplift will be..

ΔT = 24.5*10³ / (0.32 * 4.184 * 10³)
= 18C

So the water coming out of the condenser will be about 18C warmer than it went in.

The issue will be towards the end of an 8 hour day.. The 1200L tank may have risen to around 58C (136F) so that will be the temperature of the water going into the condenser. The water coming out of the condenser will be hotter around 58C + 18C = 76C (168F). The average temperature of the water in the condenser would be (168+136)/2 = 150F. So the alcohol wouldn't be cooled below that.

This all ignores heat loss from the big tank, pipework etc and that might improve the situation. If you need additional cooling perhaps instead of using a proper compressor based chiller you might get away with some car radiators and a fan on the hot side of the condenser (eg to cool the water before it goes into the big tank). You might need something like that anyway to cool the 1200 gal tank back down to 70F overnight for the next run?

Well that's my best stab at the figures. Your mileage may vary!

Thank you CWatters for helping out with this. That is a ton of math to digest and I appreciate the work you put into that. It looks like I'll only barely be able to maintain a temperature that is lower than alcohol boiling point, and for my purposes I need to be much more efficient than that. So a cooler would be necessary.

But, even though this arithmetic is admittedly beyond my understanding, I still have a couple of questions about the numbers that might change the outcome.

First, I see that you are working with 1200 gallons of water in the beginning, but at the end you referenced it as 1200L. I'm sure it is just a typo, but I need to ask anyway just in case it might change the overall answer.

The second thing I wanted to ask, which probably ought to precede the first question, is that the water tank I want to use is 2500 gallons, which is more than twice the number we used for water in this series of equations. If we adjust that number to 2500 gallons of water, does our answer change significantly? I will try to compute this for myself, but your input is paramount here since I am not qualified to make these kinds of computations.

Thank you!

 

[LarrySputnik]

Rather than worry about the flow, etc I think it is easier to work with the alcohol vs water bulk volume heat capacity ratio. Either way, the heat transfer from the tank to atmosphere has to taken into account; and, it will increase the allowable running time in the below analysis.

Ethanol: Heat of vaporization = 364 BTU/lb; Specific heat of liquid = 0.65 BTU/lb °F; Density = 0.01 lb/gal
Water: Heat of vaporization = 970.4 BTU/lb; Specific heat of liquid = 1.00 BTU/lb °F; Density = 8.35 lb/gal

Mass of stored water = 2500 x 8.35 = 20,875 lb
Heat capacity for stored water with 100° temperature change = 20.875 lb x 100 x 1.00 = 20875 BTU

Ethanol condensation and 100° temperature change per gal = (0.01 x 364) + (.01 x 100 x 0.65) = 3.64 + .65 = 68.65 BTU/gal

Ethanol that can be cooled = 20875 BTU / 68.65 BTU/gal = 304 gal

Allowable running time at .5 gal/min = 304/.5 = 608 min / 60 = 10 hrs.

Verification of my calculation will be appreciated.

Edit: While the bulk allowable amount of 304 gal is accurate the actual running time required for this production is less accurate because the increasing temperature of the condenser input water from the heating tank will require increasing rates of water flow to maintain the required 100 degree temperature drop in the condenser.

Thank you for chiming in JBA! This assumes that the water flow rate is constant, but you mention that in actuality I would need to increase the water flow beyond 5gpm? If the water flow must remain constant (due to the pump's limitations) would a higher water flow rate of, say 7gpm, make a significant difference?

Thank you for your time and input. It is much appreciated.

 

[JBA]

At this point , I can now see a serious error of my first calculation because I failed to take into account, by oversight, that by reducing the condenser alcohol discharge to the 80°F level with water starting at 72°F would only give a tank heat capacity temperature differential of 8°F, not 100°F; and, as a result, the total heat capacity of the tank is reduced to 20,875 BTU x 8°F / 100°F = 1670 BTU.

As a result, you can only produce your 0.5 gpm of alcohol for: 1670 BTU / 34.325 BTU/min = 48 min = 0.81 Hrs. and after that, as pointed out above, the tank water and heat exchanger alcohol discharge temperatures will continue to rise. For continuous operation at a condenser output temperature of 80°F you will clearly require continuous auxiliary water cooling of 100°F.

As stated above, convection cooling, and evaporation cooling, if you have an open tank, will reduce this requirement to some extent; but, at this point, we do not have enough information to estimate that contribution and its contribution would be variable and dependent upon the ambient weather conditions.

Increasing the water flow rate would be of no value in extending your tank heat capacity limited allowable production time.

I apologize for my ridiculous oversight in my original post calculation.

 

[CWatters]

Yes my mistake. No idea why I used 1200gal. Bit busy at the moment. Will try and comment on JBA's figures and correct mine later today.

 

[CWatters]

So now we can look at what happens if we put that much power directly into 1200 gallons of water...

Data for the water..
Mass of Water Mw = 1200 gallons = 4524L = 4524Kg
Specific heat SW = 4.184 kJ/kg·°C

So the temperature rise per second would be

dT/dt = P/(Mw*SW)

= 24.5 * 103/(4524 * 4.184 * 103)
= 1.3 * 10-3 degrees C per second
or
4.6C per hour

Ok correcting this for 2500 Gal...

So now we can look at what happens if we put that much power directly into 2500 gallons of water...

Data for the water..
Mass of Water M_W = 2500 gallons = 9463L = 9463Kg
Specific heat S_W = 4.184 kJ/kg·°C

So the temperature rise per second would be

dT/dt = P/(M_w*S_W)

= 24.5 * 10³/(9463 * 4.184 * 10³)
= 0.62 * 10^-3 degrees C per second
or
2.2C per hour

So if the tank starts at 21C (70F) and must not go above 27C (80F) that gives you only about 3 hours running.

 

[CWatters]

However...

If the flow rate of the pump is 5gal/min then it would take 2500/5 = 500 mins or 8 hours for all the water to be circulated through the condenser. So if you can keep the 2500gal tank stratified (eg not stirred so the hot water stays at the top) then it should work for a lot longer. Ideally you want a tall tank and must feed the returning hot water into the top of the 2500 tank and extract cold water at the bottom. Run the circulating pump as slow as possible while still achieving the required condensate temperature.

So I'm not quite as pessimistic as JBA. If you already have the tank and pump then perhaps worth giving it a go. If you need to buy one then I think you need to weigh up the risks. There is still the issue of cooling down the hot tank over night for the next days run.

 

[LarrySputnik]

However...

If the flow rate of the pump is 5gal/min then it would take 2500/5 = 500 mins or 8 hours for all the water to be circulated through the condenser. So if you can keep the 2500gal tank stratified (eg not stirred so the hot water stays at the top) then it should work for a lot longer. Ideally you want a tall tank and must feed the returning hot water into the top of the 2500 tank and extract cold water at the bottom. Run the circulating pump as slow as possible while still achieving the required condensate temperature.

So I'm not quite as pessimistic as JBA. If you already have the tank and pump then perhaps worth giving it a go. If you need to buy one then I think you need to weigh up the risks. There is still the issue of cooling down the hot tank over night for the next days run.

This sounds promising. My thoughts were right in line with yours in terms of keeping the warm water at the top. The tank would be pumped from the bottom nozzle over to the bottom of the condenser drum. The warm water at the top of the condenser drum will be fed through a hose that connects to a port at the top of the water tank. While a run will likely last 8 hours, the good news is that the condenser will not be at the temperature of 170+ degrees for more than 4 or 5 of those hours, which means I won't really need to start circulating the water until the condenser starts to get up to temp. This buys me a little more time.

I haven't purchased the water tank yet, but seeing as I'm trying to make moves in order to get my distillery as green as possible you have given me some hope. Thank you! I will post updates to let you know the progress.

 

[CWatters]

Before buying a big tank might be better to just build a big fan cooled radiator. Will need something like that anyway to cool the tank down again. Unless you have another use for hot water?

 

[NTL2009]

I just skimmed most of this, but I think I can add a few practical considerations, based on my related experience chilling 5-6 gallons of wort (raw beer) from near boiling temps to room temperature.

I am not familiar with the design of your condensor, but you will want a "counter-flow" design - the coldest cooling water enters at the exit of the product you are chilling, and the (now warmer) cooling water exits where the hot product enters. That maintains a maximum temperature differential, and allows you (depending on flow rates) to get the product very close to the temperature of your cooling water. Maybe it i already designed this way?

Also, it would be best to not return the warmed water back to the tank. You want to keep that cool, so keep hot/cold separate. And/or, start with tap water and empty that into the tank from the condensor output, your tap water may be cooler than room temp (mine is)?

You could use a two stage condensor - if the limits of the first stage only get you down to say, 85F, and you want 70F~80F, then you could use a chiller, or just a simple ice bath to get those last few degrees out in the second stage. That will take far less ice/energy than applying it to the whole temperature delta. Or three stage - tank water in first, tap water in second, ice/chiller in last/third stage.

A key is to maintain a maximum temperature delta between the two, and good thermal transfer as well of course. And don't underestimate the power of some turbulence in each, without turbulence, a boundary layer builds up, and heat transfer is reduced very significantly.

http://www.engineersedge.com/heat_transfer/parallel_counter_flow_designs.htm