How to convert reading of amps into amp hours

[Zil]

I see you John. I think butane first then propane second, gas stoves are much more efficient at cooking food than any electric device, except a microwave. I do like my microwave and will never live without one.

[O32]

Originally Posted by Sailmonkey

There is a HUGE error in measurement. Clamp meters don’t work unless on a single conductor. The OP is clamping both hot/neutral/ground at the same time.

This is correct and Bleemus mentioned same thing earlier. But thread seemed to get would up in unit conversions! All calculations were based on incorrect amp measurement.



I can't see if that meter has a DC amps setting, but for anyone with a boat, that is what you need. I think mine cost less than $50.00.


This video shows how A/C current can be measured using splitter. If cable chas two separate conductors, then clamp meter around just one of the wires. https://www.youtube.com/watch?reload=9&v=m3dAFInO7i0



For DC, same thing. Set meter to DC Amps and clamp around one conductor. And as mentioned, be sure to read manual and zero meter first. In buying a DC clamp, be sure to esyimate the current you wish to measure and compare that with the meter's capability. Some are not good for low currents.

[OldBlueBear]

Your calculation is 60 times high, you are calculating amper-minutes not Ah. amper-hours.
If you run your hot plate for say three hours your energy use would be 3 x 1.7 or 5.1 Ah. If you know the voltage (say 48 volts) you get another unit of power measurement in WattHours or more commonly shore side Kilowatt hours this is VoltsxCurrentxtime so 48V*1.7A* 3 hours = 244.8 acording to my (never wrong) calculator.
( Remember A watt equals 4.2 Joules per second (energy to raise 1 g of water 1degree C at 0 deg C)
With AC you have a further factor coming in to play Power factor, The current and voltage may not be in phase so the amount of apparent power delivered to a load expressed as VoltAmps or KiloVoltAmps may be more than the power in watts (true power) disipated by the load. The ratio TruePOWER/VOLT AMPS is the powerfactor (1 or less than one). PF is the cosine of the angle between the voltage and current. The limiting cases where the Current leads or lags the voltage occure when you have a pure capacitor or inductance (no resistance) . Ideal load is pure resistance where the KVA and Kw are the same ph = zero gegrees hence pf = 1. In real AC circuits it is common to have a pf of 0.8 or 0.9, anything below 0.7 means there is a lot more current flowing for the given Kw load, needs investigating. Your bog standard DVM or multi-meter cannot measure pf.
Hope that helps. happy Landings and Regards Capt Dave

[eemahoney]

Alan, you are not too far off, but you are looking at the wrong thing if you are concerned about the energy you need to boil water or cook.


The energy consumed is measured in Watt-Hours, which is Amps x Volts x Time.


In Boats most people look at only Amp Hours because they are looking at 12 volt devices. However if you start converting the voltage, then you need to work in Watt Hours.

In your example, if you have 120 Volt apliances, then the 1.7 amps converts to 120 Volts x 1.7 Amps x 1 hour or 204 watt hours. If you have a 1000 watt solar panel, you would need 12.24 minutes of full exposure to replenish your energy usage.


Most induction cooking apliances come in 1200 or 1800 Watt I think more investigation on the actual current consumed is in order, I use a NuWave PIC, and on high it draws about 9 amps. When I run it on my Inverter, the current draw on the battery bank is about 95 Amps.


Good Luck, you are on the correct path by measuring, just do a reality check on your numbers before making a decision.

[dcruz99801]

Alan, I have a tri in the Guaymas yard and would like to help (background in radar, TV and computers) but I am currently in SoCal for the summer, returning Oct or Dec. Are you currently in San Carlos or did you also run away from summer heat?

It sounds like the issues of amp hours -- normally used only in discussing 12v battery drain as AC tends to be limited only by max current/watts, at least until the generator fuel runs out -- has been clarified, and you've been enlightened as to the difficulties of providing enough battery juice to run any 110v appliance for very long on battery/inverter.

I see you have a Condor 30. Don't forget that any large-scale additions for livability will add weight that hurts performance; my 45' tri now sails like a mono.

[JOULESTOO]

I have a 36' trip and I use induction cooking. Decided early on not to use propane. I will say propane is the lowest cost. I have 400 amphr useable Li house battery with 20amp@110v inverter. The boat has 700watts of solar and 2000watt generator. Electrical use items are induction plate, toaster oven, coffee pot, hot water heater, and soon AC. I can use these one at a time for total time of about 2 hours per day. Depending on solar gain this doesn't require generator support. Not a cheap system but it does work.

[O32]

Originally Posted by eemahoney

Most induction cooking apliances come in 1200 or 1800 Watt I think more investigation on the actual current consumed is in order, I use a NuWave PIC, and on high it draws about 9 amps. When I run it on my Inverter, the current draw on the battery bank is about 95 Amps.

Sounds about right - 10A or 15A depending on wattage. OP's 1.7A was simply wrong because of how he measured it. Assuming 10A, going to 12/13V, 95A seems about right.



Now that must require some serious cable! My windlass doesn't draw that much. Probably longer run, but even still. But I suppose wire is sized for inverter anyway.

[john61ct]

Yes heavy inverter loads require in many cases heavier gauge cables for the DC side than charge sources, especially since distances can be longer.

Best of course to put high-watt inverters close to the bank and do the long run via AC wiring.

[Thumbs Up]

Can we back up a minute? Those amp readings on the power cable to the induction cooktop, if measured over the non split power cord, should measure zero! And while we are at it, if you clamp it over your shore power cord, it should read zero. Is that one strand of the induction cord or is it the power cable? (I can't tell from the photo, it looks thin). Also if you clamp it over just the safety ground wire, it should read zero! If not, you have leakage. 1.7 amps A/C110V leaking into freshwater is more than enough to cause electric shock drowning. While you are at it, go down the dock and test everyone"s shore power cable. They should all read pretty close to zero.

[Thumbs Up]

I am now trying to lookup leakage current associated with induction cooktops. They do in fact seem to trip gfci and rcd devices. 1.7 amps seems like a lot though. Does the reading change when you touch the pan? This seems like a good riddle for all of the electrical engineer types here.

[Adelie]

Originally Posted by deblen

Correct. Amps x hrs = amp hrs
Also Watts x hrs = watt hrs


Attachment 171271

Your formulas are correct but the OP is not.

[Adelie]

Originally Posted by alansmith

Ok, the rabbit hole just swallowed another clueless victim. So, let me see here if I understood. On dry land at 110 voltage my meter shows me 1.7 amps. If I used that induction stove top for one hour it would be 1.7 amp hours. If I used it for 30 minutes it would be 0.85 amp hrs. Do I have that correct?

Okay, if I take one of these induction tops onto a boat with 600 amp hr lithium batteries at 12 volts what am I doing with amp hrs after I run it through an inverter? Can the inverter handle that intense of a draw?

Ok

Batteries run at 12v, induction cooktop runs at 110v or 220v. You should probably figure out which.

I would go look at the circuit panel for the house. If there is a double circuit breaker for the cooktop it is probably 220v.

Let’s assume the cooktop runs on 110v. 110Vx1.7A=187Watts

187w/12v=15.6amps

There is going to be losses in the inverter so let’s just say there is a 10:1 ratio of amps for 12v vs 110v so the 12v draw in this case would be 17amps.

If you use the stove for 12min to boils some water that’s 1/5 or 0.2hr.

17a x 0.2hr = 3.4a-hr

Inverters tend to be rated in watts not amps or amp-hr since they deal with 2 voltages and 2 amperages.

A 200w inverter would be pretty small and cheap and would adequately handle your 187w cooktop.

I see 2 problems.

1. I bet you cooktop is really 220v. That doubles your wattage and the current draw on the 12v side. Not a big deal, just get a bigger inverter. I have a 1,000w inverter under the seat of my car to run a laser printer for work that’s under the other front seat. $100 or so, big. The problem is getting a 220v inverter. Fine in the EU, doable but more expensive in the US.
2. The mid range inverters supply a modified sine wave AC. More expensive inverters ($400ish) will give you something approaching a true sine wave. I don’t know if an induction cooktop needs any better than a modified sine wave but it’s something to look into.

Your cook top should have a UL sticker somewhere that indicates normal voltage and maximum amperage. Look for the sticker. Failing that get the model number and go online to see what the published specs are.

[Caribdis]

OP posted pictures of his cooktop specs in post #37. It's a 120v unit at 1800 watts at full power, so doesn't use ~200 watts. OP acknowledged his mis-measurement of the AC amps used by the hob in earlier posts....

Originally Posted by Adelie

Ok
I see 2 problems.

1. I bet you cooktop is really 220v. That doubles your wattage and the current draw on the 12v side. Not a big deal, just get a bigger inverter. I have a 1,000w inverter under the seat of my car to run a laser printer for work that’s under the other front seat. $100 or so, big. The problem is getting a 220v inverter. Fine in the EU, doable but more expensive in the US.
2. The mid range inverters supply a modified sine wave AC. More expensive inverters ($400ish) will give you something approaching a true sine wave. I don’t know if an induction cooktop needs any better than a modified sine wave but it’s something to look into.

Your cook top should have a UL sticker somewhere that indicates normal voltage and maximum amperage. Look for the sticker. Failing that get the model number and go online to see what the published specs are.

[john61ct]

Originally Posted by Caribdis

OP posted pictures of his cooktop specs in post #37. It's a 120v unit at 1800 watts at full power, so doesn't use ~200 watts. OP acknowledged his mis-measurement of the AC amps used by the hob in earlier posts....

And the OP by now has a much better understanding of the initial units conversion issues.

As well as likely an appreciation of the fact that running this big a consumer load will stress even a very big expensive DC electric system, compared to what they were originally planning.

[Trawlerman]

Is it only me that finds some of these discussions intensely frustrating? It seems to me that it is always due to posters not taking time to really read and think about the question.



For example, in this one, the OP clearly had an error in the givens. There was no way he was boiling water at 1.7 amps regardless of whether is was with 120, or 12 volts, and that should have been the clue to what needed to be the prime topic. We went on and on about how to calculate amp-hours which must surely have made his eyes cross. The rest of the stuff was good stuff, but it did not address the fundamental fault of the original premise. More than once, there was a post that pointed out that the 1.7 amps was impossible, but the conversions carried on.