I'm sitting here typing at 11 p.m. by the light of a 100-W incandescent bulb. Not a fluorescent? No, not a fluorescent bulb. Why not? Because I am definitely willing to pay a penny per hour for good quality of light to make my job easier and my life better. Let's look at the return on investment (ROI). If I invest $4 in a new high-efficiency light bulb, what's my ROI? If I replace a cheap incandescent 100-W bulb with a 22-W fluorescent, I can save 1.872 kWh/day, and at 10 cents/kWh, that's 18.7 cents/day. That high-efficiency bulb can pay for itself in 22 days.
That's fine—if I'm running the light 24 hours/day. On the other hand, if I'm running it just 1 hour/week (and many of the lights in my house run only that amount), then the bulb will pay for itself in about 10 years. I'd consider that a lousy payback, a poor ROI, and I wouldn't do that.
In the real world, many other factors exist. If a fluorescent light is turned on and off a lot, its life will be shortened. In such cases, a fluorescent might not save nearly as much as expected, because its life will be degraded. I've heard that a fluorescent light may last 20,000 hours minus five minutes every time the lamp is turned on. A cheap incandescent bulb might last 1000 hours minus 0.1 second every time it's turned on. You figure it out. If a lamp is located in a fixture that makes changing it difficult, the cost of labor for replacing bulbs could be significant. Not all of these computations are simple and objective. Some aspects might be quite subjective.
Lastly, let's get down to Mr. Tipler's proposal (Electronic Design, April 16, p. 109). Is there any motor in my house that runs more than 10 hours/day? No. There aren't any 0.5-hp motors running even 1 hour/day. But if your job in-volves motors that run more than 6 hours/day, such as in the air-conditioning, heating, ventilation, or pumping areas of your company—and there might be dozens or hundreds of them—then this may get interesting.
For example, let's presume that you have several 0.5-hp electrical motors that are several years old. Admittedly, it's not easy to find data on this motor. You may have to take your own data to be sure of the savings factors. If this motor runs around full power for about half the time (about 12 hours/day), it will use around (at least) 500 W × 12 hours, or 6 kWh/day. Let's make some simple assumptions—there's some possibility of: (A) a 10% improvement in efficiency; (B) 12 hours of operation/day (or more); (C) electricity costing 10 cents/kWh.
Under such conditions, a 0.5-hp motor will use 6 kWh, or 60 cents/day, and the savings could be 6 cents/day. Fine. You could save $22 a year, which is enough to pay for a good 25-µF, 400-V ac-rated capacitor that can be used to accomplish these savings.
Is this a good ROI? Well, it's in the right ballpark. If you can invest $22 and get a $22 savings within a year, that's a fairly good ROI. Next year, you may save another $20! So we're headed in the right direction. But if that motor only runs 1 hour/day, then the ROI will be decreased by a factor of 12.
On the other hand, if the cost of energy rises by a factor of two, the time for payback will be decreased by a factor of two. If the motor runs 24 hours/day, there's an additional factor of two. If the real improvement isn't just 10%, but rather 20%, then that's another possible factor of two.
I want all of you readers to grasp the idea that we're working with some fairly crude areas of uncertainty. It would be great if we could make these estimates with 30% uncertainty. Some people like to rely on spreadsheets with 0.01% resolution. Even the tiniest improvement would be great. Yeah, sure. (I don't like computerized spreadsheets, anyhow.*)
In the real world, 10% or 20% im-provements are hard to pin down, and the uncertainties are bigger than that. What if the price of energy doesn't go up by 2:1? What if it does? My main point is that if there's a big percentage of improvement in a very small item, then I agree with my grouchy friend—that isn't significant. But you have to look at the whole picture. What about the cost of labor for connecting the capacitor and installing it in a safe housing? If an electrician had to convert 100 such motors, could he do it for a labor cost of less than $20 each? Maybe.
Another factor is that the motor seems to run cooler, as it draws less power and less current. So the motor may last longer, and that's a good feature. This might be quite important and not easy to quantize.
Based on the data in the report, I wasn't able to see the exact improvement in watts. But it seems to indicate that if a motor runs at less than 100% of rated power, the power savings (by adding the capacitor) is probably about the same 6 cents/day and does not decrease as a fixed percentage of power drawn. So, you don't have to consider this only for motors run at full rated power.
All for now. / Comments invited!
RAP / Robert A. Pease / Engineer
[email protected]—or:
Mail Stop D2597A
National Semiconductor
P.O. Box 58090
Santa Clara, CA 95052-8090
*Refer to "What's All This Spreadsheet Stuff, Anyhow?" (Electronic Design, August 20, 1992, p. 73).