from Wiki
". . .consider a 5000 BTU/h air-conditioning unit, with a SEER of 10, operating for a total of 1000 hours during an annual cooling season (e.g., 8 hours per day for 125 days).
The annual total cooling output would be:
5000 BTU/h * 8 h/day * 125 days = 5,000,000 BTU
With a SEER of 10, the annual electrical energy usage would be about:
5,000,000 BTU / 10 BTU/W·h = 500,000 W·h
The average power usage may also be calculated more simply by:
Average power = (BTU/h) / (SEER, BTU/W·h) = 5000 / 10 = 500 W
If your electricity cost is 20¢/kW·h, then your operating cost is:
0.5 kW * 20¢/kW·h = 10¢/h"
Seems like you need to first specify the size of the unit. Also, in general, more complexity = less reliability (= higher SEER).
If your yearly cooling is 50 Therms then the 14.3 would use 350 kwh, with the 21.0 using 238 kwh.
If you pay $0.15/kwh then your cost for the 14.3 is $53/yr and for the 21 is about $36/yr.
53-36 = $17/yr, so it would take $400/($17/yr) = 24 yrs to make up your $400.
At 500 Therms/yr it'd take 2.4 yrs.
Do you know how many kwh you use for cooling each year? You may be able to figure it out using the cooling degree day data on this site
http://www.degreedays.net/
plus your yearly elec. bill record.
You'll also need
http://www.google.com/search?client=safari&rls=en-us&q="heat+gain"+house&ie=UTF-8&oe=UTF-8