2
\$\begingroup\$

Apologies for this likely very simple/stupid question.

I have a portable AC (air conditioning) unit that's supposed to be "energy efficient". On the back of the unit, it tells me some details about the unit's power consumption:

Voltage: 115 V-60 Hz
Current consumption: 11.4 A
Cooling capacity: 4100 W (14000 BTU/h)

So here's where I'm a little confused. I see the 4.1 kW and I'm a little confused, because my limited understanding of electricity would be that the power consumption shouldn't exceed the voltage multiplied the current (in this case, 1311 W). Does this mean that my unit is drawing 1.3 kW/h of electricity, or 4.1 kW/h? If it's the former, how does it output more than 3x the power that it draws in?

Also, I was wondering if my power consumption remains constant with this unit plugged in (whether at 1.3 kW/h or 4.1 kW/h), or if it will fluctuate based on the settings? In other words, are the numbers listed on the unit supposed to represent how much power it will draw in no matter what, or does that represent the maximum amount of power it will draw (meaning that it could draw less if I, for example, didn't need as much cooling)? Apologies for my limited understanding here, but just want to make sure I can get a roughly-accurate estimate for my upcoming power bill.

\$\endgroup\$
1
  • \$\begingroup\$ The unit consumes about 1400 watts (the limit in the US for a 15 amp "standard plug outlet" (and often breaker back at the entrance panel) is about 1500 to 1800 watts, max. So this unit is designed for a "standard plug," but will likely consume the entire branch current. Air conditioners "pump" energy. So this unit says it can pump a little more than twice its energy consumption. It's typical for air, which I think at best provides about 2.4X. And that's on a good day. \$\endgroup\$
    – jonk
    Commented May 25, 2020 at 18:38

1 Answer 1

Reset to default
5
\$\begingroup\$

Cooling capacity can exceed power consumption because it's moving heat from one place to another, not creating it.

Those ratings don't actually tell you the power consumption, but it should be less than 1311W, because 1311 is the VA (volts * amperes) and we don't know the power factor.

If you want to know the actual power consumption you can measure it with a device like a Kill a Watt or maybe there are additional specifications available.

The actual power consumption will vary with the settings and the operating conditions (temperature and humidity). The A/C unit turns on and off to try to maintain the temperature. If it is off all the time it draws nothing or maybe a tiny bit for the remote control etc., if it runs all the time it uses the maximum.

\$\endgroup\$
3
  • \$\begingroup\$ Thanks for the answer. I was looking into power meters (like Kill a Watt), but some of them specified a max input of around 1800 W, so I just wanted to make sure I wouldn't overload it. Sounds like I should be okay on that front then! I'm a little confused still about your comment about not knowing the power consumption. I thought VA was equivalent to W? Shouldn't that tell us how much power is being drawn? EDIT: Sorry I just saw your edit. Does this mean that 1311 VA (or W) is the max amount of power that would ever be drawn per hour? \$\endgroup\$
    – Jon Warren
    Commented May 25, 2020 at 18:43
  • \$\begingroup\$ The VA is always >= watts. There's some component to it that involves energy sloshing back and forth between the motor and the mains that is mostly not "real" power consumption. For example, a capacitor or inductor across the mains would draw current but ideally use no power averaged over many cycles. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented May 25, 2020 at 18:45
  • \$\begingroup\$ Yes, basically 1.3kWh of energy per hour of operation should be the maximum. As jonk says it might be a bit more at higher line voltage. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented May 25, 2020 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.