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May 2009
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Say we have two days in a row, identical conditions (temp/humidity/thermostat set points, etc.) From purely an electrical consumption standpoint, which situation is better (assume "sleep" and "wake" was set at 73 degrees in both scenarios):

1) have leave setback to 78 degrees begin at 5:30 am knowing that at x:xx time that afternoon, a/c will begin cycling on.

2) have the leave setback to 78 degrees begin at 4:30 am knowing that at x:xx time that afternoon (presumably earlier than in case 1) a/c will begin cycling on.

In case 2, I would assume the temperature difference between indoors and outdoors, although not greater at any identical time between cases, the a/c would run longer. In case 1, the a/c would likely short cycle more in the morning, making it less efficient.

Is that correct thinking? Is it different for say a single stage a/c vs. a dual stage a/c? Could a dual stage a/c actually use less power in case 2 conditions than in case 1?

Most importantly, where is my logic flawed?

2. If I read your post correctly, the "leave" start time is only one hour apart. The outdoor temperature isn't likely to rise or fall all that much between four and five in the morning.

Personally my thermostat does not go into "leave" mode until I leave the house in the morning. The house is often still occupied after my departure, but the occupants like it warmer than I, and the house does not rapidly warm up that time of day after the "leave" mode kicks in. Reason I delay the "leave" mode until my departure, in addition to hinting at personal comfort, is so morning moisture generation via showering, breakfast, etc. can be removed from the air prior to the "leave" mode kicking in, where the unit may not run for a few hours before the interior temp warms up enough to kick it back on. Longer than I want higher humidity hanging around uncontrolled.

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