# What can I put a kilowatt into? (Or: help me save my hairdryers)

TL;DR: I need something to discharge roughly 160A at 14V into, or 2.24 kilowatts. Any comments, or answers, with either a) something I can dump a kilowatt into, b) some way I can modify a common item to take 2kW DC at 160A, or c) another way to measure battery max continuous discharge current would be greatly appreciated.

Unfortunately, a large number of other people on the Internet that have this problem are dealing with far fewer amps (160A is pretty crazy.) Thus, any comments to "just google it" or that it is similar to previously asked questions aren't appreciated.

I recently bought a large battery, a Hobbyking Multistar 16000mAh 4 cell LiPo battery. Unfortunately, HobbyKing is notorious for inflating its products' specifications. The max continuous output is variously listed as 15C (which would be 15C*16000 mAh = 15C*16Ah =240 amps) and 10C (which would be 160A). The battery voltage should range from 4.0V to 3.2V per cell during use, so 16V to 12.8V.

I'm hoping that the continuous output is at least 10C, or 160A, but I have no idea what is is. People variously report the actual outputs of Multistar batteries as ranging from 10C to 3C, and there's a lack of actual test data and far too much anecdotal data. I'm hoping to test this myself by dumping 2kW into something and measuring the current the whole time.

Basically, I need something to discharge roughly 160A at 14V into, or 2.24 kilowatts. I've looked up things that take power in the range of a kilowatt, and found that microwaves (~1kW), ovens (~1.5kW), power tools (~500W-2kW), projectors (400W-4kW), and hairdryers (~1-2kW) are my best bets. I'm not exactly sure how to plug my battery into any of these though. Obviously the battery puts out ~2.2kW DC at 160-ish amps. I have no idea what my hairdryer wants, or how to get it to take DC, without a large amount of work. I also understand that this would be far into the range of the mad scientist, and would probably result in a cool explosion.

Is there an easier way to test my battery's capacity? Within reach I have a LiPo battery charger (max discharge rate 1A unfortunately), a decent Fluke, a lot of household equipment, a number of power supplies, a 400W projector, and a workshop with a decent number of power tools/electrical equipment.

Any way to test my battery would be greatly appreciated, including ways to get my hairdryers to take DC, ways to discharge two kilowatts into something that's not an appliance, and any other ways to generally test battery discharge characteristics.

 I know that putting a kilowatt into household appliances is pretty impractical and dangerous if you're stupid. I also now know that it's also damn hard. I've now switched to wanting to make, or buy, a large resistor. For the safety police, I know how dangerous 2kW can be. I have always intended that any test -- be it on a proven resistor that should work fine or a household appliance -- would be conducted outside, on nonflammable ground, with fire extinguishers, where if something blows up I can make a pretty video and share it with the Internet as opposed to dying from electrocution and burning down my house. I also know how 2kW can melt things and have handled energy on this scale before. I'm not an electrician and I know my limits, but I do know how to handle 2kW to the point where the worst that could go wrong is a few hundred bucks of stuff down the drain and a pretty explosion video on Youtube. I am acutely aware that there is a very high chance that the battery, or whatever I'm sticking 2kW into, might explode, and I will share the video with you all when (if) it does.

• Danger Will Robinson! 16 Ah is lots of energy, and 160 A is a huge current. You need to know exactly what you are doing in order to a) get meaningful results, and b) not hurt yourself. Your question suggests that is not the case. Jun 30, 2016 at 23:22
• The problem is that the battery can probably put out quite a bit more than either of it's ratings, but it might get really hot doing so. It's not like you hit their C rating and the power stops, a good example is the Turnigy Nanotech cells, I picked up one of the 3.7V 6600mAh cells, the short circuit current hit ~750A, so to get a better measure of whether the battery can do 10C or 15C, you might want to put a temperature probe on it and see how hot it gets under load, if it stays cool at 10C then you might be able to push it harder.
– Sam
Jul 1, 2016 at 6:07
• Note that the way you'll know that you have exceeded the maximum continuous discharge current is that the battery will catch fire, or the overpressure safety structure will permanently disable the battery. So consider this as destructive testing.
– jpa
Jul 1, 2016 at 9:19
• @OwenVersteeg "I have a decent understanding of electricity in general, and know how to be safe with the amount of power I'm handling." Both of those claims seem to be directly contradicted by your suggested plan of putting 160 amps into household electrical equipment. Jul 1, 2016 at 12:38
• Possible duplicate: electronics.stackexchange.com/q/198745/50922 Jul 2, 2016 at 12:18

To dissipate $1\mathrm{kW}$ at $14\mathrm{V}$ you need a resistor with a resistance of $R = \frac{\mathrm{V}^2}{\mathrm{W}} = \frac{\left(14\mathrm{V}\right)^2}{1000\mathrm{W}} = 0.196\mathrm{\Omega}$. You can buy a $0.25\mathrm{\Omega}$ $1 \mathrm{kW}$ resistor on Digikey for $54.95 (Part no. FSE100022ER250KE). Using two or 3 of them in parallel would dissipate $2.35 \mathrm{kW}$ which is within $5\%$ of your goal of $2.24\mathrm{kW}$. If you use $0.25\mathrm{\Omega}$ resistors then the current will be $\frac{ 14\mathrm{V} }{ 0.25\mathrm{\Omega} } = 56\mathrm{A}$. So you will need a 8 AWG or larger wire going to each resistor. Alternatively you could wrap some Nichrome wire around a high temperature core (such as a cinder block) to make your own power resistor. This PDF gives some information on NiChrome wire. 14 AWG NiCr A wire has a resistance of $0.1587\mathrm{\Omega}$ per foot. NiChrome-A wire has a melting point of about $1800\mathrm{°F}$. If we run about $29\mathrm{A}$ through the wire the wire will heat to about $1400\mathrm{°F}$ which leaves $400\mathrm{°F}$ of margin. If you run 5 strands at $32\mathrm{A}$ each you will have $160\mathrm{A}$ and be somewhere in the $1400\mathrm{°F}$ range. To make $32\mathrm{A}$ we need the resistance of the wire to be $\frac{14\mathrm{V}}{32\mathrm{A}} = 0.4375\mathrm{\Omega}$. To make $0.4375\mathrm{\Omega}$ we need the wire length to be $\frac{0.4375\mathrm{\Omega}}{0.1587 \mathrm{\Omega}/{\mathrm{ft}}} = 2.76\mathrm{ft}$ (2ft 9in). $2.76\mathrm{ft} \cdot 5\ \text{parallel strands} = 13.8\mathrm{ft}$. Wrap each of the 5 strands around the cinder block so they ar not touching or alternatively use 5 separate cinder blocks. Wire each strand to the battery in parallel using at least 12 AWG wire for each connection. Dont make the connection with something that could melt such as jumper cables with plastic handles. Also, the copper wire must be run physically separate in the area near the NiChrome because it is likely that some of the insulation will melt. You can purchase a 21 ft spool of 14 AWG NiChrome wire from McMaster for$19.13. (Part no. 8880K11) Alternatively you can purchase at 20 ft spool from Jacobs Online for $15.00. • For high currents it's usually more practical to use steel wire. It's much cheaper. Jul 1, 2016 at 6:40 • @ Dmitry Grigoryev I agree that using steel wire would work also if you can find adequate temperature rating and resistance info from a specific supplier. Jul 2, 2016 at 1:28 • Worth mentioning that nichrome has a relatively low thermal coefficient of resistivity - 0.0004/C . This means that if you heat the wire by 500 C, its resistance will increase by approximately 20%. Which is small compared to other materials, but does affect the calculation. Jul 2, 2016 at 13:33 The problem is two-fold: you need a device that can provide the appropriate load, and you need to manage the heat. For my money, I would do the following (but see important warning about short circuiting batteries below!!): Take a length of thin wire (for example, the "magnet winding" wire you can buy at Radio Shack for about 9 dollars - https://www.radioshack.com/products/magnet-wire-set?variant=5717684613). This set contains about 100 feet each of 22, 26 and 30 gage copper. The resistance of these wires is 53, 134, and 339 Ohm / km, respectively. To get a current of 160 A from a 14 V source requires a total load resistance of 14/160 = 88 mOhm. That means that a little more than 1 meter of the thickest of these wires would provide the right load - but there's no way you would be able to get the heat out. You need a sufficient surface area - so I would recommend that you use the thinnest gage, double up the wires, so you end up with a number of wires in parallel providing the load. You could then solder the ends together (you have to scrape off a bit of the enamel to be able to solder to these wires) and put a piece of adhesive-lined heat shrink around the junction. Use a really thick wire (multiple strands of 6 AWG) to provide the connection to your battery, or you will get massive losses where you did not want them. Now immerse the whole thing in a large bath of water. Water is cheap, and has a remarkably high heat capacity. The insulation on the wire will ensure all the current flows through the copper, and now you have a sufficient area to dissipate the heat to the surrounding water. If you have a 16000 mAh battery, it should be able to provide 160 A for 0.1 hour or 6 minutes. In that time, you would in principle dissipate a total of 160 * 14 * 360 = 806 kJ, or roughly 200 kCal. If you immerse this contraption in 5 liters of water (a bucket), it will heat by about 40 C; that is manageable. Note that short circuiting batteries is extremely dangerous - these things have fragile chemistry and can explode. Make sure you have appropriate fire fighting equipment and personal protection. Finally - how many wires do you need if you have a total length of 100 feet? If we assume that you cut N wires of length $\ell$ such that the final resistance is $R$, then for a resistance per unit length $\rho$ we write $$\frac{\ell\rho}{N} = R$$ We also know that the total length is $N\ell$, which is given as 100 feet ($L$). We can now solve for $\ell$: $$\frac{\ell\rho}{\frac{L}{\ell}} = R\\ \ell = \sqrt{\frac{RL}{\rho}}$$ With the numbers above, you would want to cut the 30 gage wire into 11 pieces with a length of 92 cm each; these 11 wires in parallel would give you a resistance of 84 mOhm, very close to the value you need. And I'm sure you will have a few more mOhm of losses elsewhere. Finally - you charge the battery, determine the amount of water in the bucket, connect the whole thing up and stand clear. When the current stops flowing you will be able to measure the temperature rise in the bucket and you will know how much energy you were able to transfer from the battery to the water. If the weight of the empty bucket is E, and of the full bucket is F, then the mass of water is F - E, and if the temperature rise (in °C) is $\Delta T$, then the total energy is $$\frac{F-E}{\Delta T} \cdot 4200~J$$ Where the weight is in kg and temperature difference in Celsius. Divide the energy by the time in seconds and you will have the average power. I don't have a good suggestion for measuring such large currents directly unless you have the appropriate tool (see for example this article for some pointers). A regular Fluke won't do it... You don't want to put anything directly in the path of the large current. UPDATE The question - "can such a thin wire dissipate this heat" can be answered analytically. According to this paper, a thin wire in water (where that water is allowed to boil) can dissipate $2\cdot 10^5~\rm{W/m^2/C}$. If we assume your magnet wire is rated up to 180°C and the water is at 30°C, you have a thermal gradient of 150°C. To dissipate 2 kW , the area we need is $$A = \frac{P}{h\Delta T} = 7.3 \cdot 10^{-5}~\rm{ m^2}$$ The 30 AWG wire has a diameter of 0.254 mm, so a surface area of $8\cdot 10^{-4}~\rm{m^2}$ per meter length. The total length of 30 m gives it an area of $2.4\cdot 10^{-2}~\rm{m^2}$; this is much more than we needed. So even if the coefficient of thermal conductivity is much lower (say, the "non-boiling" value of $8\cdot 10^3~\rm{W/m^2/C}$ from the same article) it is still sufficient to get the heat out. Note that two wires carrying current in the same direction will attract: this may result in a reduction in available area for heat dissipation. You might want to experiment a bit with this (perhaps string little beads on the wires to separate them). • He could use a shunt & some math to determine the current. Jul 3, 2016 at 15:33 • The likelihood of this not working is considerable. Tiny magnet wire and huge currents are a recipe for a very fast failure, even sandwiched in heatsinks and dunked underwater. I haven't calculated this, but I would expect the energy density to be simply too high - that you could not conduct the heat away fast enough through the tiny surface area that such thin wire affords. – J... Jul 3, 2016 at 15:50 • @J... - Valid concern. I have updated my answer with a calculation of heat transfer. Your inputs are welcome. Jul 3, 2016 at 18:59 • This is the same solution Mikeselectricstuff uses. He's made a video about a similar setup here: youtube.com/watch?v=WECW88rJYrE Jul 8, 2016 at 9:23 • @grahamparks that is indeed almost exactly what I had in mind - thank you for finding that link, it validates that this approach works in practice! Jul 8, 2016 at 12:09 I am choosing to answer this part of your question which has been neglected by other answers "Is there an easier way to test my battery's capacity?" Yes, you already have the means to test your battery capacity with the discharge function on your LiPo battery charger at 1A rate(follow mfr directions). Or just discharge at 1A rate and time it with a stopwatch. It should be something near 16000mAh at low rates of discharge and significantly less at higher rates. Please do measure capacity first, at the low rate, to make sure that you do indeed have a 16000mAh pack. The max discharge rate 10C, 15C, etc is specified that way for a reason. It is not a fixed Amp value, it depends on the capacity and condition of the particular pack at that moment. It is a 'fuzzy' spec that is chosen for safety and reliability, not measured. That is why you never see an 11.2C maximum discharge rate. Just because you can discharge at a certain rate, does not mean that you should. It is entirely possible to discharge at a very high rate one time without anything apparently terrible happening. Yet the heat and stress could have created a weak point that will cause a violent fire the next time you try the same test. All loads are not equivalent. A real automotive carbon pile load tester(which I would recommend if you carry out the test) is a purely resistive load, but motors are highly inductive loads with back EMF spikes and other complex components that may or may not be carried back to the battery over the ESC depending on how well it is filtered. In conclusion, you probably do not need to run the test you were planning. Figure out how much current your application actually draws worst case. If it is less than 32A, you are good. If it is more, you may be ok, but the best test is just to try it out on the actual hardware and see how long it runs. In the neighborhood of 160A, this next warning is NOT just boilerplate. In no case should you exceed the current rating of any wiring, connector, or component. Test on a non flammable surface away from anything you cannot afford to have burned up. If you really want "c) another way to measure battery max continuous discharge current"(not safe discharge current) and are not willing or able to provide additional parameters like load impedance, then there is really only one way. A dead short into thick short cable or busbar. Measure the current with a clamp on inductive meter until it melts. Any method with load resistance, even a very small current shunt resistor will not reach the true maximum. This is almost certainly a destructive test and the value of any results are dubious. If we know more about what information you are trying get by performing this test, we can give you more useful answers. • How do you know what features his charger has? Can you add a few paragraph breaks to your wall of text? We won't think you're showing off. ;^) Jul 1, 2016 at 17:53 • @transistor thanks, I did try(unsuccesfully) to add linebreaks to the original with shift- return. You prompted me to look up the correct way(double return). OP stated "Within reach I have a LiPo battery charger (max discharge rate 1A unfortunately)" If LiPo charger has discharge mode, it very likely exists as a consequence of capacity measuring mode. Once you have discharge hardware built in, it only requires firmware to implement capacity measure. Jul 1, 2016 at 18:34 • I'm not concerned about the Ah rating but rather the max cont. discharge rating. Unfortunately, as long as the Ah rating is above half of its rated value I'll be fine. Typically no company will inflate the Ah rating by that much for the type of battery I have, and if they did I'd be shocked. The max continuous discharge rating is what I'm worried about, as that is a far more commonly inflated number. Jul 1, 2016 at 18:59 • Think about that for a minute. If a mfr inflates Ah rating(capacity), then the run time is slightly shorter and maybe a few people are disappointed. If they inflate max continuous discharge rating, and someone actually discharges at that rate, the pack catches fire, burns down a house, maybe kills someone. And mfr said it was safe, so they are liable for damages in court. Max continuous discharge rating is determined by lawyers, not physics. It is a theoretical maximum, not a guarantee. Jul 1, 2016 at 19:32 • If they claim a 15c max discharge rate for the cells and build in a pack current limiting resistor that limits actual discharge to 10c, they are within their rights to add a safety device. It sounds like your application is fine, you are just upset that you may not be getting the advertised maximum. You won't get it. and that is because the max discharge rate will be limited by your application, not the battery pack. mAh is routinely inflated, prepare to be shocked. Jul 1, 2016 at 19:41 Google for "dynamic braking resistor". They are not cheap, but they are available down to just an ohm or two and up to multi-kilowatts. They are basically large heaters, but the nice thing is that you can specify the resistance, current and power that you need. • Those can be scavenged for free if you don't mind visiting a junkyard near a railroad / tramway / subway depot. Oh, and the biggest models with forced cooling go well into megawatt range. Jul 1, 2016 at 6:33 • Well don't just take them off units you see disused. A lot of those are stored or museum pieces awaiting their turn at restoration. Talk to a scrapyard that actually scraps units (Larry's Truck & Electric comes to mind) but, do your homework on size or you'll get a component of wildly wrong value. I personally say "make it out of steelwire" is the better plan for such a small wattage resistor. Jul 1, 2016 at 23:46 • Electronic surplus stores/warehouses are a reasonable place to search for power resistors at a lower budget that the suggested new-from Digikey. Jul 2, 2016 at 1:48 You might use a load tester. These devices are designed for testing vehicle batteries and alternators and are intended to handle hundreds of amps in your voltage range. They're basically a big carbon pile resistor in a box with an ammeter and voltmeter. A 500A one can be had for about$50 (examples 1 2).

Only problem is you'd be going much longer than their intended duty cycle. They're designed to handle that load for 30 seconds or so (peak current engine starting) rather than the 4-6 minutes you'd be measuring for, though these units are designed for 6KW, so you could run it for a minute or so, then let it cool for a while, and repeat until the battery is drained.

• Or put a high powered fan on the carbon piles to help them dissipate the heat. :) Jul 3, 2016 at 15:34
• @DIYser - Higher capacity ones do just that, but they cost about twice as much as the battery he's testing. Of course, you could always jury rig something. Jul 4, 2016 at 2:29
• I'm sure you are right! You never know though, sometimes hackers have unused turbo fans laying around. Never hurts to make a suggestion. Jul 4, 2016 at 14:06

Your mains-powered appliances (hair dryers, et.al.) aren't really suitable loads for a 14V battery. They are designed for 120V (or 240V) power and will consume < 10% of their rated power down at 14V.

When I checked Ebay a few minutes ago, there appear to be plenty of those 100W power resistors in nice aluminum cases that could be bolted to a big heat sink. You could get 25 of those things and wire them in parallel. Chose a resistance that will parallel to 0.0875 Ohms. Dunno if this is worth the expense?

Or you could try to find a place that sells appliance repair parts and get a roll of heavy Nichrome wire and make your own 0.0875 Ohms resistor.

But, as others have already said, fooling around with 160A is no place for amateurs. You could kill yourself AND burn your house down at the same time. EXTREME CAUTION IS STRONGLY URGED!

• If you use nichrome wire, you could immerse it in a large container of water (e.g., a large trash bin). At 10 or 15C, you are not going to be running it long enough for the water to heat up all that much. Then the only thing you have to worry about is the battery catching on fire. Jul 1, 2016 at 5:29
• Using 10 times lower voltage than rated will not effect in 10 times lower power consumption. It's gonna be 100 times lower. That's because Power = Voltage * Current and Current = Voltage / Resistance therefore Power = Voltage^2 / Resistance. Jul 1, 2016 at 11:11
• @jms, how fast is the process? Would the OP be able to conduct the test once without the wire completely failing? Copper is so conductive that the OP would need to use very fine wire, I suspect. Or a very long wire. Jul 1, 2016 at 17:10
• @mkeith You are right, nichrome would work fine for a few tests which only take a few minutes in total to do. I am no chemist so I cannot say how long the wire will last, however I wouldn't want to deal with the resulting electrolyzed chromium sludge. As for the thinness of the wire, 160 A at 14 V implies a resistance of just 87 mΩ, so the high conductivity of copper is actually a good thing in this context. 4.1 m of 1mm enameled copper wire would have the correct resistance, and would work fine submerged in a tub of water.
– jms
Jul 1, 2016 at 18:38
• @RichardCrowley, jms is talking about nichrome submerged in water. There would definitely be electrolysis with 12V end-to-end. I think jms has made the case that copper wire is a much better choice if the OP wants to use a wire submerged in water (as a thermal mass). I think this is actually the easiest way to do the test. Jul 1, 2016 at 21:06

Do you really have to discharge the battery at that rate? If you just want to check what Ah you really have, why not do it at a lower current, but over more time? I did something similar with some 18650 Li batteries I bought on ebay. I wanted to check what I really had, so just set up a circuit to drain them at around 500mA and measured how long that took. A lot easier (and safer) than basically shorting them out!

You could use a couple (or 3) of those 100W resistors to give you 10A-20A drain and see what happens. That would at least give you a 'ballpark' figure for the battery.

• A non destructive way of testing a battery, rather than to suck the life out of it. Takes more time but is worth it. Jul 1, 2016 at 17:52
• I'm not concerned about the Ah rating but rather the max cont. discharge rating. Unfortunately, as long as the Ah rating is above half of its rated value I'll be fine. Typically no company will inflate the Ah rating by that much for the type of battery I have, and if they did I'd be shocked. The max continuous discharge rating is what I'm worried about, as that is a far more commonly inflated number. Jul 1, 2016 at 18:57

14vdc @ 160a is in the range of a standard car starting battery. Get a 3KW 12VDC to 120VAC power inverter (google it - these exist) then use a 2KW 120V heater as load. You will have to use the shortest length of massive #0 or #00 gauge copper wire to connect it to the battery. You will also need a 100Amp-to- 1Amp shunt resistance standard (it is an electrical tool) to measure this much current accurately. If you attach your Fluke meter to the shunt and it reads 1.6Amp then 160Amps is flowing across the shunt. The only problem is if the battery is too incapable then it may not support the 3KW inverter for very long. Hopefully this is not some hobby battery, these specs are for a full sized electric vehicle segment lithium battery. These also exist. Don't forget 16.000 amp-hours is also 1amp for 16hours as well. How fast is determined by the battery cell's internal resistance.

There are electronic loads out there that can dissipate that kind of power indefinitely, like the EL series from Kepco: http://www.kepcopower.com/el.htm

They aren't cheap, but they are very good at pulling constant current, voltage, power, pretty much whatever you need. I'm pretty sure they are controllable over a serial connection as well.

• This is what I've done for something similar (especially since we had one laying around). The one I used was from B&K Precision . Jul 1, 2016 at 14:41

If you have an open coil resistor handy, but it's too high resistance and too low current capacity for your needs, you can tap it thusly:

You're dividing the resistor into n segments, which will make the whole suitable for the same wattage at 1/n voltage.

The gory details: If the pack is resistance r, then resistance of each segment will be, obviously, r/n. So with all of them in parallel, resistance is (r/n)/n. Sorry, can't find superscript in the app.

.

• Won't the "divide by 2" actually give you by divide-by-4? (R/2 in parallel with R/2.) Same for the rest of the diagrams... Jul 2, 2016 at 16:59
• Sounds right. Divide by 2 will work out right for half the voltage same wattage. Jul 2, 2016 at 20:24
• Ah ok - if your captions mean "you can divide the voltage by 2 to achieve the same power" then yes. Just not particularly clear what the intent was. Jul 2, 2016 at 21:23

You seem to be missing some simple calculations. The test resistance required is $$R = \frac {V}{I} = \frac {14}{160} = 0.0875~\Omega$$

This will give a load of $$P = VI = 14 \times 160 = 2240~W$$

A mains device with this resistance would have a power rating of $$P = \frac {V^2}{R} = \frac {120^2}{0.0875} = 164~kW$$

At 240 V the power would be four times that (due to the square term) = 658 kW = 0.6 MW. You won't find these in your kitchen.

Since you have an application already I suggest you come up with a method to use your real load as the test.

I would use a step-by-step approach. That means using loads incrementally and measuring voltage drop and calculating the internal resistance of the battery, then extrapolating theoretically for larger loads. And ony after that, if calculations shown good safety margin, i would try a high current load approach.

Lets give some example: To calculate the battery internal resistance is very simple using an ammeter (not necessay precision) and a digital voltmeter (here is necessary to have at least four digits value display, but again, precision is not very important, only the number of digits), and two low-value resistors. Can be two 12V 55W car bulbs. The approach involve using the voltmeter in parallel with the battery, ammeter in series with the bulbs, and taking two measurements: one with only one bulb, and the second with two bulbs in parallel. From the current and voltage results we can calculate the internal resistance of the battery: Ri = dV / dI ; (dV = V1-V2; dI = I2-I1).

Now that you have calculated the battery internal resistance, you can approximate the battery internal dissipation of power (heat) at 160A using the well-known formula: P=I2R, where: P will be internal battery dissipation, in watts; I squared is 160A squared that means 25600; R is the previously calculated resistance, in ohms.

If P result is larger then 100W for a small battery (200-400 grams), i would not even try taking 160A from it. If is a quite large battery (more than a kilogram for example) is should safely absorb the 100W for a few minutes that means it will work okay. Of course, other detrimental effects may appear at high currents, but i'll give it a try.

I looked up the specs for your battery. There is one serious difference. The spec calls for a constant discharge rate of 10C, not a continuous discharge rate of 10C. What this means is that you can discharge the battery at 160A until the battery is discharged.
They also have a peak discharge rate of 20C for 10secs. Using this to estimate the length of time at 10C, I would guess 40secs.
I had a chat with a CSR of Hobbyking and he reassured me that it was safe to use a discharge rate of 160A, however, he was evasive as to how long the battery would be able to deliver it (the theoretical is 6 min max).
I would be surprised if it even last a minute. This is a "far cry" from continuous. You may not even have enough time to measure anything.

Another approach (and safer), is to determine the battery's internal resistance(Ri). Once this is done, it is very simple to calculate the max discharge rate (Is = Vo/Ri).
To find the internal resistance, measure the no load voltage (Vo). Use a 5 ohm load resistor and measure the voltage under load (Vl) and the current(I). Ri = (Vo-Vl)/I.
As an example, Vo = 16v, Vl = 14.55v, and I = 2.91A. Ri = (16 - 14.55)/2.91 = .498ohms. Using this value of Ri, one gets the peak discharge rate of (16/.498 = ) 32.18A.

An automotive starter motor (especially for a big engine) can easily draw that amount of current, and in the 14V range too. The only problem is that all that energy has to go somewhere. If you rigged up a starter motor with the shaft locked to prevent rotation, it will draw its peak (stall) current, but all that energy is going to go into heat in the windings, so it can't operate like that for more than a few seconds at a time without cooking itself. If you could rig it to drive some sort of mechanical load - maybe a big fan or something, then it could run longer because most of the power will be dissipated into the load, but then you'd need a really big motor and hefty mechanical load or it won't draw the target current.

If you're working on a drone and you already have the motors and propellers for it, maybe the solution is to build a static test rig to lock down the drone so it can't move, insert your electrical performance monitoring (current/voltage) equipment into the circuitry, and "fly" your drone in the rig.