TEMPERATURE INCREASER - Page 1
Forum section: electronics / projects

author: Matias wet
author_rank: Newbie
timestamp: April 10, 2025, 08:02:07 pm
content: Hi everyone, I'm working on a project that requires raising the temperature of a liquid to 85-90 degrees Celsius in 3 minutes using lithium batteries. The current model uses four 4000 mAh batteries and reaches that temperature in 17 minutes, with a 60% charge consumption. Any recommendations on how to improve efficiency to achieve this in 3 minutes and with lower consumption? I'm open to suggestions on battery types, circuit design, or any other components that might help. Thanks in advance!

author: RoGeorge
author_rank: Super Contributor
timestamp: April 10, 2025, 08:07:24 pm
content: This is an English forum. Please post in English only.

author: BennoG
author_rank: Frequent Contributor
timestamp: April 10, 2025, 08:18:23 pm
content: main question how much liquid and what kind of liquid.10ml no problem 10l big problem.Benno

author: themadhippy
author_rank: Super Contributor
timestamp: April 10, 2025, 08:21:28 pm
content: first thing would be to calculate how much energy is actually needed to raise the temperature of your liquid in the given time,for that you'll need the volume of liquid and its specific heat capacity

author: Matias wet
author_rank: Newbie
timestamp: April 10, 2025, 08:25:54 pm
content: The liquid is water, I want to raise the temperature like for example an electric kettle that is plugged into the mains but without plugging it in, with lithium batteries, the system I have now does it in 17 minutes, half a liter but my question is if I add double the lithium batteries, that is, 8000 mAh, could I, I also want to do it for one liter

author: TimFox
author_rank: Super Contributor
timestamp: April 10, 2025, 08:30:02 pm
content: Start by calculating the totalenergyrequired to raise your (unspecified) volume of water, then divide by the maximum time allowed to calculate thepowerduring the heating.The specific heat of water is 1, meaning that raising the temperature of 1 L by 1 K requires 1 kcal of heat.In terms of energy, 1 kcal = 4184 J, where 1 J = 1 W-s.A resistor submerged in water will transfer all the electrical energy to heat the water.Match the resistor to your battery and see what happens.

author: Matias wet
author_rank: Newbie
timestamp: April 10, 2025, 08:32:19 pm
content: The thermos is half a liter. In 4 minutes, starting with a full battery, the temperature can be raised by 45 degrees, leaving the battery at 84%. After a second heating cycle, it drops to 62%, then 56% after the third, and 40% after the fourth. To fully heat half a liter of water to 85 degrees, it currently takes 17 minutes, consuming 60% of the battery. I need to reduce this time to 3 minutes per cycle, allowing for two cycles per day at 85 degrees.”

author: TimFox
author_rank: Super Contributor
timestamp: April 10, 2025, 08:35:08 pm
content: So what power is required? This is a quantitative problem.

author: themadhippy
author_rank: Super Contributor
timestamp: April 10, 2025, 08:35:36 pm
content: raising half a liter of water by 90 degrees needs around 188000 joules,thats about 1Kw if its done in 3 minutes

author: Picuino
author_rank: Super Contributor
timestamp: April 11, 2025, 07:48:37 am
content: 1kW is 71 Amperes at 14 Volts (4s battery)188000 joules = 52 Ah at 14 Volts

author: bitwelder
author_rank: Super Contributor
timestamp: April 11, 2025, 07:57:02 am
content: In addition to the amount of water you want to heat, and the starting temperature of the liquid (ambient?)do you have a constraint also on the shape of your container, i.e. are you able to place the heating element in the best position?You can probably study a common houshold eletric kettle, as you have probably the best geometries and you know their power and you can see how fast they are.From there you get the required power (using mains), and you only need to check: 1) if you can sustain a current that gives you that power at a low voltage (as you use batteries) and 2) how big have to be the batteries to have enough energy (= power x heating time) for doing that cycle X times

author: SteveThackery
author_rank: Super Contributor
timestamp: April 11, 2025, 08:34:37 am
content: Confirmed: you'll need to generate just over 1kW to raise half a litre by 90 degrees in 3 minutes.You asked how to "increase efficiency and lower consumption". The problem is that heating water with a submerged heating element is pretty close to 100% efficient anyway, so no, you won't be able to increase efficiencyordecrease consumption by any meaningful amount.The only way to do that by avery smallamount is to insulate the container so less heat is lost during the warm-up period.@Matias wet: it looks like it isn't going to be feasible to achieve what you want without using some serious battery power - far more than you currently have. The laws of physics dictate this, and there's no way round them.Except - you could think outside the box. Perhaps if you could give us the bigger picture of what you want to achieve, we might be able to suggest a different approach. So - please tell us the full story and perhaps we can assist.

author: Picuino
author_rank: Super Contributor
timestamp: April 11, 2025, 08:44:27 am
content: I usually use butane/propane cartridges when I go camping.EDIT:Peltier cells might give you more efficiency, but not much more.

author: tszaboo
author_rank: Super Contributor
timestamp: April 11, 2025, 08:46:58 am
content: To answer OP's question:No we don't have some sort of magic circuit that breaks the laws of the conversation of energy. Or thermodynamics.

author: 2N3055
author_rank: Super Contributor
timestamp: April 11, 2025, 09:24:51 am
content: Quote from: Picuino on April 11, 2025, 07:48:37 amFor a real life comparison, that is an average size car battery, or cca 45x21700 LiIon cells.That is a lot.Just look at LiIon battery powered welders, what currents and energies are available.1kW for 3 minutes is 50 W x hours. 4x21700 in series (16 V, 4000mAh, ) will have approx 201600 Joules stored, fully discharged.That is enough for one boil per spec, raising the temp 90°C in process.Same energy density is common in laptop batteries.Problem here is high currents and fast discharge. So cells don't self disintegrate and to last more than one use.Way to optimize this would be to run more cells in series. Less current per cell, less problems and losses.12x cells in series will work at 48V and have cca 20A current draw. That is perfectly in spec of many cells for short durations loading.It would have more than one discharge capacity too.Batteries alone would weight more than 1kg, without structure to hold them and protect them. Estimate at 1,5-2 kg total (cabling, BMS, control electronics, structural loading elements) for battery pack.Look at E-scooters and E-bikes, that have similar power. They run 72V/96V packs, 18-20 cells in series.There is a reason we all use chemical energy for mobile energy. Energy density. Simple gas bottle, white spirit, gasoline. Or fuel tablet that is having twice amount of energy in a small 30x20x10 mm tablet. 30 g or so. And a small foldable burner 100g more.Battery solution makes sense only if you are not going to carry it on foot and/or there is an easy way to recharge, rendering fuel sustainment logistics better.

author: Zero999
author_rank: Super Contributor
timestamp: April 11, 2025, 09:41:28 am
content: Quote from: Picuino on April 11, 2025, 08:44:27 amI was going to suggest a heat pump. A traditional vapor compression cycle would be more efficient, but there won't be an off the shelf solution for a 90oC hot side and room temperature, say 20oC cold side, so the OP is limited to Peltier, which isn't very efficient, unless they want to invest a lot of time any money.Definitely go for water, rather than air source. I've no idea of what sort of COP is possible.EDIT:Running the numbers through the formula gives a theoretical maximum COP of just under 5.2, but I doubt it's possible to get anywhere near that in real life.https://en.wikipedia.org/wiki/Heat_pump_and_refrigeration_cycle#Coefficient_of_performanceNote: temperatures are with respect to absolute zero, so convertoC to K by adding 273.15.

author: themadhippy
author_rank: Super Contributor
timestamp: April 11, 2025, 10:22:35 am
content: you could try getting higher,do it on the top of everest and youll only need to reach around 70c.

author: SteveThackery
author_rank: Super Contributor
timestamp: April 11, 2025, 06:20:42 pm
content: So look, although I'm still struggling to properly understand what you want, you seem not to be understanding what we have told you.The laws of physics determine how much energy is needed to raise half a litre of water by a given amount. This has been explained before, but let me have one more go. Let's assume the water is starting at 20C and you want to heat it to 90C. So you are increasing the temperature by 70 degrees.A litre of water requires 4184 joules of energy to raise it by 1 degree celsius. Now, you have half a litre, so you need 2092 joules for every degree C.In total you want to raise it by 70 degrees, so the total energy needed is: 2092 x 70 =146440 joules.Now, this is fixed by the laws of physics, so no matter what you do, you will need to find that much energy from somewhere. From what I can tell, you are starting with a 3.7V 4000mAh battery. Let's find out the energy content in watt-hours:Watt hours = volts x amp-hours = 3.7 x 4 = 14.8 watt-hours. There are 3600 seconds in an hour, so:14.8 watt-hours x 3600 = 53280 watt-seconds, which is53280 joules. This is approximately1/3rd of the energyrequired to achieve your aim of raising the water by 70 degrees C.So you will need three full charges of your 3.7V, 4000mAh battery.At this point I should point out that you may get less than 4000mAh from the battery; it depends on how much current you are drawing. In general, the more current you draw, the lower the capacity of the battery. So in reality you might needfour full charges, maybe more.You mentioned substituting a 3.7V 10,000mAh battery. This contains 2.5 times as much energy, so:53280 x 2.5 =133200JThis is well on the way to the energy you need: 146440J needed, 133200J available. You will still need to recharge the battery to get all the way to 90C, but that's better than before.Here is a possible strategy: deliver half the required energy, then recharge and deliver the second half (but see the caveats later).146440 / 2 =73220J. As a proportion of the total charge in your 10,000mAh (133200J) battery:73220 / 133200 = 0.55. That means you could discharge the battery down to 45%, recharge it and discharge it again down to 45%. In theory this should get your water up to 90C (if you started at 20C).CAVEATSBatteries only give their rated amp-hours under specified conditions (typically when discharging them over a period of 10 hours). If you are hoping to heat your water in minutes, then the battery capacity will be much lower. This is just one of those things you have to live with. The manufacturer will publish charts showing the capacity at different discharge rates.Second, the water will be losing heat to the environment all the time you are heating it, and also while you are recharging the battery. I don't know how much will get lost, but in reality you will need more than 146440J of energy.Third, I don't know what is determining the 17 minute period. It could be the resistance of the heating element, or it could be the internal resistance of the battery. Or both. Not enough information. I'm assuming it's the resistance of the element for the purpose of this post.HOW FAST?You said you wanted to heat the water up faster than the 17 minutes it currently takes. The only way to achieve this is to add the energy more quickly. In your case, that means pushing more current through the heating element. For example, if you put two 3.7V batteries in series, so 7.4V to the element, you will double the current and double the voltage, so multiply the power input by four, so 17 minutes comes down to 4 minutes 15 seconds (call it 5 minutes).This suggests another possible solution.1/ Get two 3.7V 10,000mAh batteries, fully charged.2/ Connect them in series to the existing heating element.3/ With luck, you should warm up the water from 20C to 90C in five minutes.In theory. In practice those batteries will be hammered pretty hard by the high current. But give it a go and see what happens.If it doesn't work, get two more 10,000mAh batteries, so that you've got four. Connect a pair in parallel, and connect the other pair in parallel. Then connect those paired units in series. This will give you approximately 7.4V and 20,000mAh.You should be getting close.

author: TimFox
author_rank: Super Contributor
timestamp: April 11, 2025, 07:09:30 pm
content: By the way, the analyses by myself and SteveThackery above are correct for resistive heating.In theory, one could use less energy through a heat pump, where the thermodynamics are more complicated, but that's not practicable for the application.

author: Bud
author_rank: Super Contributor
timestamp: April 11, 2025, 08:32:12 pm
content: Your boiled water is going to be very expensive.Now, think about how you will be charging such battery.Also watch some electric car fires on youtube.

author: TimFox
author_rank: Super Contributor
timestamp: April 11, 2025, 08:40:25 pm
content: He only wants 90oC water.To boil the water after heating to 100o, he still needs to add the latent heat of evaporation, 2256 J/g.

author: Smokey
author_rank: Super Contributor
timestamp: April 12, 2025, 12:07:00 am
content: Quote from: Matias wet on April 10, 2025, 08:02:07 pmStep 1: Dissolve a bunch of salt in the water.Step 2: Toss the batteries directly into the water.Step 3: Stand back while they go into thermal runaway.Step 4: ...Step 5: Profit! Hot water!