Inverters have improved the sailors lot, no question about it. It is sooo nice to be able to plug in a drill, saw, sander, or soldering pistol. An inverter allows small kitchen appliances to serve equally well in the galley. It can power a computer, recharge portable electronics, and facilitate improved dental hygiene. Just a decade ago, using AC appliances aboard necessitated either an onboard generator or a shorepower connection. Then inverters burst onto the boating scene and today most sailboats with living accommodations are inverter equipped.
Unfortunately the load an inverter places on the house batteries is too often poorly understood. Let’s see if we can remedy that, at least for sailnetters.
Watts is watts Voltage in volts multiplied times current flow in amps gives us power in watts. For example, a 120-volt AC galley appliance that draws 3 amps will be rated at 360 watts. The formula is P = V x I where P is power in watts, V is potential in volts, and I is current in amps. Wattage is useful when talking about inverters because it remains the same regardless of the supply voltage. So if we run this 360-watt appliance on 12 volts—just what we are doing when we use an inverter—the current draw goes up to 30 amps. Why? Because P always equals V times I, so when P is 360 and V is 12 , I must be 30 (12 x 30 = 360).
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We can rewrite the formula as I = P ÷ V to make solving for amps easier. In this example the current draw at 120 volts is 360 ÷ 120, or 3 amps. At 12 volts the current draw becomes 360 ÷ 12, or 30 amps. As a general rule the current draw at 12 volts is 10 times greater than at 120 volts for any given appliance. This “times 10” multiple is why you cannot practically power, for example, an air conditioning unit through your inverter. An efficient 5,000 BTU unit will be rated at 500 watts or more. Dividing watts by 120—the operating voltage of the unit—we find the current draw of a 500 watt AC appliance will be around 4.2 amps. If we operate the same appliance from a 12-volt power supply, the draw on the batteries is 10 times 4.2, or 42 amps.
This level of discharge will flatten an 8D or a pair of golf-cart batteries in about four hours. Even if you had a large enough bank to sustain this load for, say, 12 hours, you would still be faced with needing enough charging capacity to put back 504 amps, plus another 15 percent for battery inefficiency, a significant challenge for most 12-volt charging systems. If you need air conditioning aboard away from the dock, plan on installing a genset.
Power Factor and Other Trolls Some circuit components, most notably coils, oppose any change in the flow of current. This is of little consequence with direct current (DC), but the oscillating nature of alternating current means that a measurable amount of additional power is required to overcome this additional “resistance”—in this case called reactance. AC appliances with motors or other coil components require the inclusion of an appropriate power factor (PF) in the amp-from-wattage calculations. The formula becomes I = P ÷ (V x PF), where PF is one for DC circuits, but less than one for many AC appliances. Sometimes the PF is shown on the appliance faceplate, but knowing the actual value is not essential for our purpose at the moment. Suffice to say that a PF of 0.5 is not unusual, in which case the AC appliance requires twice the amperage the basic I = P÷ V calculation yields for this watt rating. That would make a 500 watt AC appliance draw 8.4 amps at 120 volts, and 84 amps at 12 volts.
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Another consideration, particularly with AC refrigerators and air conditioners, is start-up current. Modern compressors require a start-up current—called inrush—that can be as much as five times the compressor's nominal draw. An inverter must be sized to handle this momentary load. A 500 watt inverter might run a 500 watt refrigerator, but it will not start it. You may need a 1000 or even 1500 watt inverter in order to get the needed surge capacity, in this case perhaps 2500 watts. This high load is momentary, so it likely has no significant implications for the batteries, but the supply wires to the inverter must be capable of carrying the required current. In the case of a 2500 watt load, the 12-volt current will be over 200 amps!
Inverters consume some power in working their magic, as much as 15 percent of the load. In addition some appliances are as much as 30 percent less efficient with the square wave or modified-sine-wave power delivered by less expensive inverters. For example, a 5-amp AC appliance may actually draw 6.5 amps (130 percent) from a modified-sine-wave inverter, and the inverter draws about 75 amps ([6.5 x 10] x 115 percent) from the battery bank. Throw in a Power Factor of 0.5 and the load on the battery increases to 150 amps.
Battery Capacity Four golf-cart batteries—a common house-bank configuration—will have a rated capacity of around 440 amp-hours. However, this number is what is known as a C20 rating, meaning that a steady draw has been placed on the batteries over a 20-hour period to measure their capacity. In this case, if the batteries are new and have a fully charged cell voltage of 2.1 volts, they should be able to supply a continuous current of 22 amps for 20 hours before the cell voltage drops to 1.75 volts (which defines “dead”). But if we discharge the batteries at a higher rate, the true capacity declines. Double the draw to 44 amps, and the bank capacity declines to around 350 Ah. Put a 100-amp load on this bank and the capacity drops to around 260 Ah. Keep in mind that useable capacity is only around 40 percent of total capacity, so a 100 amp continuous load will consume the entire useable capacity of a 440 Ah battery bank in about an hour.
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Bottom Line Before you consider operating any AC appliance from an inverter, you should do the math. Let’s take a 700-watt microwave oven, for example. The PF of microwave ovens is typically around 0.6, so a 700-watt oven will draw around 10 amps (700÷ [120 x 0.6]). If your inverter does not deliver true sine wave power, an appliance inefficiency of around 30 percent comes into play, making the microwave draw around 13 amps (10 x 130 percent). Applying the times-10 rule, your battery bank has to deliver 130 amps at 12 volts, plus another 15 percent to operate the inverter. That makes the load on your battery around 150 amps while the microwave is running.
That might be OK for reheating the Dinty Moore, but try roasting a chicken for 15 minutes and the load on your batteries is similar to cranking your main engine starter motor for that long. The total draw on your battery bank appears to be just 37.5 Ah (150 amps x 1/4 hour), but a 440 Ah battery bank will only deliver 150 amps for around 45 minutes before discharging a 90 percent charged bank to 50 percent. So 15 minutes of microwave operation consumes 1/3 of your useable battery capacity. To put this in perspective, this discharge level is approximately equivalent to what you might expect from operating a 12-volt refrigerator for 24 hours.
A 3000-watt inverter operating at capacity will place a 290-amp load on your battery bank (including inverter inefficiency). Even a 1000-watt inverter draws nearly 100 continuous amps from your batteries. These are big numbers in a 12-volt system designed to deliver a few amps to fans and lights. With realistic expectations, an inverter can deliver an amazing amount of convenience for a modest investment, but high-draw appliances still require a shore cord or a genset.
