Four of these, and you make a cup of tea while you wait. You have to keep them in the water.

 

http://www.ebay.co.uk/itm/Hot-Water-12v-volt-heater-kettle-element-tea-coffee-NEW-/130462652665?pt=UK_Car_Accessories_Touring_Travel&hash=item1e602e18f9

I think you have just solved my problem of cold cups of tea when working in the garage !!

Four of these, and you make a cup of tea while you wait. You have to keep them in the water.

 

http://www.ebay.co.uk/itm/Hot-Water-12v-volt-heater-kettle-element-tea-coffee-NEW-/130462652665?pt=UK_Car_Accessories_Touring_Travel&hash=item1e602e18f9

 

Thanks D8veh, it does seem you solved all my problems at once. I found this:

http://www.ebay.co.uk/itm/Travel-Heater-Element-Mini-Boiler-Hot-Water-Coffee-Immersion-500W-220V-Adapter/181489761029?_trksid=p2047675.c100005.m1851&_trkparms=aid%3D222007%26algo%3DSIC.MBE%26ao%3D1%26asc%3D20140122125356%26meid%3D771ce095ef1a47df94e5340b658d4724%26pid%3D100005%26rk%3D2%26rkt%3D6%26mehot%3Dpp%26sd%3D261522527171&rt=nc

 

So it's using 120W at 12V, so 480W at 48V. So if I buy 8 of these I'd have:

4P * 2S = 480*4p / 2s = 960W high discharge.

Or 8S = 480/8s = 60W low discharge.

 

I'd put all of these into a big bucket of water while discharging, make my own hot shower lol

 

Is the calculation correct?

the calculation is incorrect, but the principle of using these heaters is sound. The first step is working out the resistance of the heating element.

P = V * I or 500W =220V * I

therefore I = 2.2A

Next, use the equation V= R * I or 220V = R * 2.2A, therefore R=100 Ohms

If you connect one heating element to your 54V (48V nominal) battery, you discharge V * V/R = 54V * 54V / 100 Ohms = 29.16W

To discharge at 50W, connect two of those in parallel.

To discharge at 900W, connect 30 of those in parallel.

the calculation is incorrect, but the principle of using these heaters is sound. The first step is working out the resistance of the heating element.

P = V * I or 500W =220V * I

therefore I = 2.2A

Next, use the equation V= R * I or 220V = R * 2.2A, therefore R=100 Ohms

If you connect one heating element to your 54V (48V nominal) battery, you discharge V * V/R = 54V * 54V / 100 Ohms = 29.16W

To discharge at 50W, connect two of those in parallel.

To discharge at 900W, connect 30 of those in parallel.

 

Sorry here is another one:

http://www.ebay.co.uk/itm/Portable-12V-Car-Immersion-Heater-Tea-Coffee-Water-Auto-Electric-Heater-New-Q-/261522527171?pt=LH_DefaultDomain_3&hash=item3ce3f513c3

 

12V for 120W. So I = 10A.

 

R = 12V/10A = 1.2 ohm.

 

And on a discharge battery, 54*54/1.2 = 2430W.

 

I think I did something wrong somewhere because I can't discharge at 2430W on a 12V heater?

No, you would burn it. You'll need to put 5 of them in series (5S), the resistance will then be 5 * 1.2 Ohms = 6 Ohms, the total dissipative power is 5 *120W = 600W, you will send into this array 54V * 54V /6 Ohms = 486W, well within its dissipative capability.

To make a bigger system, put 10 of them into 5S2P array for 1kW.

edit

for slow 60W discharge, use two of the 220V heating elements in the previous example.

Edited February 8, 201511 yr by trex

You got it correct, Cwah. Trex used the wrong voltage.

 

As they're water cooled, you should be able to over-volt them a bit, so you probably only need to buy 4. R = 1.2 ohm, so four make 5.8 ohms. At 72v, you get approximately 500w (72 ×72/5.8).

72 ^2/5.8=893.79

anyway, 4 * 1.2 = 4.8

I've got it wrong in the previous post, can't do mental arithmetics with this sort of numbers anymore, had to use the calculator this time.

I'm an engineer, not a mathemaician. I always round numbers to the nearest convenient one to suit my needs.

Purchased 6 of them.

 

So with R= 1.2 I have 6*R=7.2ohm

 

So on my 54V battery I have 54*54/7.2=400W discharge

72V battery I have 72*72/7.2=720W discharge.

 

If I need higher discharge, I do 2s3p = 54*54/2.4 = 1215W or 72*72/2.4 = 2160W (ouch)

 

Thank you guys, hopefully when I'll received these I'll share my tests

They're rated at 200W. You can probably go as high as 300W, but I wouldn't go any higher, so that's 1800W max.. 300W at 12v is 25 amps, so use good wires and connectors, like from your battery to the controller.

I have two 36v 9Ah bottle batteries and I want to be able to discharge them effectively so I can 'condition' the batteries again according to the manual.

 

Using d8veh's 12v heaters how many would I need?

How do I know when it's discharged, will the water stop boiling?

Anything else I need to watch out for if I use this method?

 

I've got this arriving soon, how would I wire this into the circuit since it's got two wires each side? - http://www.ebay.co.uk/itm/390932345418?_trksid=p2059210.m2749.l2649&ssPageName=STRK:MEBIDX:IT

 

Thanks

I have two 36v 9Ah bottle batteries and I want to be able to discharge them effectively so I can 'condition' the batteries again according to the manual.

 

Using d8veh's 12v heaters how many would I need?

How do I know when it's discharged, will the water stop boiling?

Anything else I need to watch out for if I use this method?

 

I've got this arriving soon, how would I wire this into the circuit since it's got two wires each side? - http://www.ebay.co.uk/itm/390932345418?_trksid=p2059210.m2749.l2649&ssPageName=STRK:MEBIDX:IT

 

Thanks

Don't waste your time. Your battery doesn't require any conditioning. Who's manual is telling you to do it and why do you think they need conditioning?

Don't waste your time. Your battery doesn't require any conditioning. Who's manual is telling you to do it and why do you think they need conditioning?

 

From the Cyclotricity manual -

 

Before you use the battery for the first time it is best to fully condition your battery. This is achieved by charging your battery for 12 hours and then using the bike until the battery is completely drained.Repeat this process at least three times. After this “conditioning” process, you can leave the battery charging as and when you require.

 

Under the Troubleshooting section it says that you can repeat the process if you think the battery is not performing correctly.

I wouldn't take any notice of that. It's a lithium battery, the same as any other one. What other manufacturers tell you to do that. On an old battery, it'll probably do more hatm than good.

 

If you've lost capacity, you need to open it up and check the cell voltages. Those batteries are not waterproof. It's easy to get water in that drains down the cells at the botyom and knocks it out of balance more than what the BMS can cope with.