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Discussion Starter · #1 ·
I purchased a microwave oven for the boat, a 700-watter. I decided that I want to use at at anchorage, too, so I purchased a 3,000-watt peak, 1,500-watt continuous power inverter. The batteries are about 15 feet from the microwave and TV, in a relatively large battery compartment with 4 T105 house batteries.

My question is, should I mount the inverter in the battery compartment and run a 15-foot 110-volt AC line to an outlet on the other side of the boat, or install the inverter next to the microwave on the bulkhead and run a pair of 2 gauge lines from the house batteries to the inverter?

Hopefully, there is someone with this kind of scenario on the forum that can provide some assistance. My biggest concern is battery acid fumes and the possibility of damage to the inverter from those fumes.

Gary :cool:
 

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I agree with the "mount the inverter as close to the battery compartment as possible but not in it". Keep the leads as short and large as possible.
Even with everything done right you may find it less than satisfactory to run the micro wave with the battery bank you are running. ( High load for extended period quickly draws the voltage down.)
The math says a draw on the batteries will be in the range of 69 amps* when starting with fully charged voltage of 12.7 volts. (700/.8)/12.7=68.9 amps
The battery voltage will drop quickly and the amperage draw will rise as that happens.
example
(700/.8)/11.7=74.8 amps
(700/.8)/10.5=83.3 amps - at this voltage the inverter will likely be shutting down.

*assuming 80% efficient conversion
 

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Agree with:
* keep inverter as close as possible to the batteries to reduce voltage drop and to minimise costs of expensive large diameter wiring
* do not keep inverter in the same compartment as the batteries, although sealed AGM have very little acid air vented, but yours' are open (vented) lead acid by Trojan I suppose? The other side of a bulkhead is a good idea.

Do not agree with... or let me check:
* that amperage goes up when voltage goes down. Yes it looks that way when using the formula Amp (I)=watts/volts, but the wattage is variable and depends on the volts!!. In fact there is an resistance that needs to be taken into account and the formula. For instance, if the wattage is 700 at a voltage of say 12.5 Volt (as per specs of the unit). Then the formula is I=W/V, or 700/12.5=56 Amps. At that Voltage!! Then resistance is R=V/I=12.5/56=0.22 Ohm
Then when the voltage drops ie to 12 Volt, resistance remains the same, and therefore the amps go down to I=V/R=12.0/0.22=53.8 Amps..... yes, even if the resting voltage is 12.7 Volt) it will go down under such a load to 12 Volt or less.

In fact inverters are not 100% efficient and 'use' power. A good inverter and the best one I have seen is around 94% efficient (meaning a loss of 6%). Most have an efficiency of around 80-85%, so a loss of 15 to 20 %. So one could add these percentages (ie 6 to 20%) to the amps used.
If one is really pedantic, one can then calculate the real Wattage of the microwave at a lower Voltage. In this case at 12 Volt and 85% efficiency: 549 Watts. All figures rounded.
If you do not believe all the above, think of a lamp. When the battery goes flat (lower voltage) the light goers dim (lower amps).

Of course, if the voltage of the inverted AC voltage (110 Volts) remains the same, then Wattage remains the same, and the amperage has to increase dramatically to make up for the voltage losses. So 'Norah' was right, although intuitively it did not look that way.

Sorry, this was not really the question that was asked.

Clarification:
* previous posters used the term "large" cables. I am sure they mean "large diameter" cable.

Additional info:
* Generally cheap inverters have low efficiency rates, meaning have big losses, or/and use higher amps when in standby mode.
* additional switches or connectors in the 'large current' circuit introduce resistance, and power is lost in form of heat, resulting in reduced voltage or voltage drop.
* Similarly voltage drop goes up substantially by using undersize cabling
* generally when using 100Amps from a battery one has to put approx up to 120 Amps back in again due to battery losses
* Before installing equipment that draws large currents, ensure you have the charging methods all worked out, and can replenish the amp/hrs used.
my 2 cts
 

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If the microwave is 700 watts output that is not the input wattage. My 700 watt microwave consumes a maximum of 1050 watts. Using easy math and allowing for inefficiencies that translates to 105 amps DC at maximum output. I would do as posted and install the inverter close to the battery bank but not in the same compartment and use cables not smaller than 1/0, probably 2/0. Don't forget to fuse the positive cable close to the batteries and don't forget the inverter chassis ground.
 

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Since I am planning a solar and inverter installation, I've been pouring over lots of caluculations and was very interested in this post.

While I'd certainly agree to:
  • Mount the converter close to the battery bank, but
  • Not in the same space as the batteries, and
  • using 1/0 to 4/0 cabling from the battery to inverter, and
  • Understanding that a 700W microcowave uses more than than amount when starting, and
  • You need to factor in battery voltage and efficiency realistically in determining how many amps are being used...

...the critical elements are the installation with the proper size of properly crimped cables, short DC runs and realistic use of the microwave when powering from the inverter.

A draw of even 100A seems devestating until you realized that this is likely being done for only 3-6 minutes. Even running two cycles a day for a full six minutes only comsumes a 20 Amp Hour draw from the battery bank (100 Amp Hour x .2 hours).

While this can't be done indefinitely without battery bank replenishment via charging, it is no diferent than a 1A draw for 20 hours. Yes, there will be a voltage drop when you are drawing 100A, but after 6 minutes the battery will "recover" slightly.

If you make your connections low resistance and use the microwave judiciously when off the charging grid, you shouldn't have any problems.

In my planned installation, the microwave is the largest AC draw (also 700W) and I'm going to be very careful with the installation.

Murph

S/V Amalia
1965 Cal 30
Muskegon, MI
 

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Do not agree with... or let me check:
* that amperage goes up when voltage goes down.
That depends on the load. If it is a resistive load like a heater current goes down with voltage in accordance with E=IR. For active loads like inverters, radios, and most motors current goes up as voltage goes down in accordance with P=IE.
 

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If the microwave is 700 watts output that is not the input wattage. My 700 watt microwave consumes a maximum of 1050 watts. Using easy math and allowing for inefficiencies that translates to 105 amps DC at maximum output. I would do as posted and install the inverter close to the battery bank but not in the same compartment and use cables not smaller than 1/0, probably 2/0. Don't forget to fuse the positive cable close to the batteries and don't forget the inverter chassis ground.
This is an excellent point!! I was assuming he was using the max wattage from the specifications sheet. Never assume..... D'oh...! Output wattage is NOT input wattage.

We have an Emerson 700W microwave. It draws a measured (2 minutes ago) 1190W at 119V !!!!

1190W at 11.5V = 103.5A X 1.20 (for approx inverter inefficiency) = 124.2A

If the battery voltage sags to 10.7V we are now at approx 133.A....

Battery temperature, type and the Peukerts constant will also dictate how long the bank can support a 100+A load before dropping out the inverter at 10.5V or so....

Keep in mind that even a 3% voltage drop equates to the inverter dropping out at a terminal voltage of approx 10.8V and the wire voltage drop does not include terminals, fuses, terminations or switches... This means an inverter sized for a wire drop of 3% could in actuality drop out at 11V or more as installed.....

With short term microwave use most good sized flooded deep cycle house banks can easily survive this load for the duration the microwave is run, if they are healthy. If they are sulfated, and not healthy, you'll know it pretty quickly...
 

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That depends on the load. If it is a resistive load like a heater current goes down with voltage in accordance with E=IR. For active loads like inverters, radios, and most motors current goes up as voltage goes down in accordance with P=IE.
+1... Plus we are talking about inverters here so as battery voltage drops the current goes up. The output voltage of most inverters remains quite steady even as battery voltage drops...

I've yet to see an inverter that did not do this, but I am always open to someone showing me one..;);)
 

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My thanks to those who mentioned the power loss within the micro wave oven and inverter power circuit. My good friend Al Zheimers helped me with my first post.

On one of my own systems I have inverter shut down when the battery voltage is still showing 12.0 volts when I draw 100 or more amps from a 750 amp hour bank of Surrette/Rolls AGM's.
 

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Discussion Starter · #14 ·
Well, I tested the system today, both ways, and the best method turned out to be the short battery to inverter leads with a heavy gauge 15-foot run of 110-VAC line to the microwave and TV. The distance from the battery to the inverter is just 10 inches, and I constructed a special connection using Automobile battery cables that attached directly to the batteries and 2-gauge wire that is connected to the inverter using screw compression connections. I tinned the ends of the wires where they connect to the inverter just to be safe. I monitored the voltage drop while boiling a coffee cup of water in the microwave. I started out at 13.6-VDC and dropped to 12.5 during a 4 minute microwave on high. I did not look at the current drain. Maybe next time. I purchased the battery cables at Pep Boys and they were on sale at about $7 each. (Got an old farts discount.)

When I tested the system with 15 foot, 2 gauge wires, the voltage drop was down to 11.7, which I thought was just too much.

My house batteries are a series parallel rig of 4 T-105 lead acid golf cart batteries. All the batteries are just 2 years old and in perfect condition. I keep them fully charged with a 100-watt solar panel. My boat's biggest drain, like any cruising boat, is the refrigerator/freezer, which draws 6 amps when running. Fortunately, it's very well insulated and doesn't run constantly like some I've come across. I insulated it myself with 6-inches of styrofoam on the outside of the compartment. This cut the running time of the compressor by more than 70 percent over the half-inch of insulation that was originally installed.

Thanks everyone for your advice and expertise,

Gary :cool:
 

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..... 2-gauge wire that is connected to the inverter using screw compression connections. I tinned the ends of the wires where they connect to the inverter just to be safe.
If the battery cable connection at the inverter is the type where a plate is pressed onto the wire by a threaded machine screw the wires should not be tinned - tinning hardens the cable end and doesn't allow for the required compression for a good connection.
 

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Discussion Starter · #16 ·
Yes, tinning can make it more difficult to compress the cable, but this connection is not with a small machine screw, it's a stainless steel 1/4-20 bolt and I had no trouble compressing it whatsoever.

Gary :cool:
 
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