help w/ engr calcs
re: How much electricity to bake a loaf and then also to cool the house from the heat generated.
(from Kwh i can compute cost via the marginal $/kwh price.)
I want to estimate the electricity (KWH) needed to bake a loaf _and_to cool the house from the added heat.
Math/engineering comments/corrections are invited.
Guesses/assumptions about the oven:
.75 hours at average 2,000 watts to preheat oven and cast iron dutch oven. 1.5 Kwh.
1.25 hrs (75 min, it's a 1200 g ww boule) at average 1,000 watts to maintain heat. 1.25 kwh.
2.75 kwh total.
It's resistive heating, so I assume it all goes to heating the house (worst case scenario): 2750 watt-hours * 3.4 btu/watt-hour = 9350 btus.
I guess I could turn off everything else in the house and take a meter-reading before and after baking to get a better kwh number.
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My question, how many Kwh will be consumed by my central air conditioner to provide 9350 btu of cooling? What is the likely efficiency of that conversion?
The outdoor unit (compressor) looks 20 years old. The indoor unit looks at least 30 years old.
Anyone have any estimates on the ac's efficiency percent on the kwh/btu conversion?
I suspect you already know that you’re asking a question that can only be answered generally and that answer is dependent on a number of a assumptions about:
Sooo... how badly do you really want to know?
Best wishes,
Phil
So far...
my marginal cost per Kwh is $.11 incl tax. And we don't have time-of-day adjustments.
Through search-engine-ing, I found that the main effective ratio of Btu to watt-hour for AC is the EER. My inside unit is too old to have that on the unit. And I dont want to walk through poison ivy to check out the name-plate data on the outside unit.
Unlike resistive heating, the AC is a "heat pump", so instead of the "raw" factor of 3.4 btu/watt-hour, modern systems can get up into the 13 range.
So I may ask a contractor what a typical EER was for this size unit in 2000.
As a very rough guess, I may go with 5, and assume my system is not tuned/optimized, and then, as a worst case scenario, use a 3.4 figure, just because it's the "neutral efficiency" (or whatever the technical term is) conversion factor.
I realize it's possible for an AC to have an actual EER lower than 3.4 if it's in bad enough shape.
So... the cost to run the oven, and the AC, to bake a loaf of my bread in the summer: at worst, likely 61 cents.
Dave, I am with Phil that it will be extremely difficult to determine the cooling cost. I also would add in that your system comes on and off based on a thermostat - and if it is located far from the kitchen, it is possible that the impact is not as linear as you oven consumed x watts, so you AC must cool off that many watts.
On the cost of the oven, I think you need to reconsider your math. If in fact the bake element is 2,000 watts, while it will be on most of the time during the preheat, it won't be on constantly - it will go a little higher, then turn off as the temp coasts down, then repeat. Once you are cooking, it will do the same dance, though since it is also raising the temp of the bread, the intervals that it stays on will be longer. So you can either turn off every thing else and check the meter reading - or you can watch the oven like a hawk and try to determine what percentage of the time the element is on during baking.
Measure the approximate consumption for baking a loaf.
You could turn off all loads, except oven and cooling, and measure the kWh difference directly from your meter by taking readings at start and end of your experiment. If you have critical loads such as fridge and freezer that you don't want to turn off, you could still deduce their average kWh consumption from the result. Manufacturers typically give information about appliance's power consumption. That would give you a rough idea of how much energy goes into baking and cooling your home.
It should be sufficient to take a meter reading at,say, 2 hours before beginning the oven pre-heat/bake cycle, then a reading when you start that cycle, and again when you complete the cycle and turn off the oven, and yet again when you are satisfied the "cool down the house" phase is completed. Assume that the initial 2 hour period is your "baseline usage" off of which you can compute the incremental usage in the next measured phases. Do this a couple of times and you will get a reasonably good estimate. It's bound to be pretty close compared to all the unknowns in the "calculated" approach. If you repeat the cycle measurements two or three times on comparable weather days, you can tighten it up some, also. I grant it is not an exact approach, and will be impacted by weather differences and such, but the proposed calculations will not be perfect either, for the reasons already offered.
If it were me, I'd try to do both, and use them for a "reality check" on each other. When results are in obvious and gross disagreement, one or both are off. When in closer agreement confidence in both results increases, as it is less likely, although not impossible, that both are off in the same direction by the same amount so as to end up in "accidental" agreement.
Good luck with it, and let us know what you find out. I find it an interesting question that I'd like to know the answer to. I can't investigate it for myself: no air conditioning!
OldWoodenSpoon
A fun calculation for sure. Do you really want to know the answer? LOL. It's a bit like calculating the cost of a home grown tomato. The cost/benefit ratio will differ for everyone.
That said, I caution that whatever number you derive will only be good for a certain exterior temperature, sun angle and humidity. In other words, heating up the kitchen in January is beneficial, not so much in July.
Extremely difficult to calculate ac efficiency without knowing the last time your unit was recharged with a refrigerant. Have you changed the filter to allow for good airflow inside? Is a bookshelf sitting in front of one of the cold air ducts?
A good compressor has a life span of 30 years, probably not linear in efficiency. I would take whatever efficiency it may have had when new and cut it by at least half, more if it is low on refrigerant.
Seems the easiest, but sadly the least fun method, would be to take a meter reading before and after?
As always on this forum, those wishing to help with a problem struggle to have correct input information. In that regard, go with the meter and pay attention to exterior temperature. We won't get into how well insulated your house is, we want to grow the size of your loaf and your enjoyment of baking, not the cost. In the end, if you love baking and your bread, it's worth it. Come on, isn't it really? Enjoy your hobby!
I appreciate the feedback/comments to help think this through.
@barry: My guessimates of 2,000/1,000 watts are averages (over the time period) after taking the intermitency (is that the right word?) into account. I'm supposing the "on" wattage to be higher. During warm-up, the bottom element is on continuously, and the top/broiler element is intermittent. Once up to temp, the bottom element is then intermittent, and I am not sure if the top element comes on at all. (Dan did an experiment in his electric oven in which he could not visually see the top element come on, but a temp probe gave good evidence that it did come on, just not long enough to turn red and be visible.)
@pul: thank-you.
@OWS: I like your 2 hour baseline idea.
@grainbrain: Yes, this exercise is definitely seasonal. The "driver" for the question is the normal thought of many bakers "I don't want to heat up the house and spend extra money on running the AC more than I have to."
So the AC part of this calculation is only for "cooling season" where the natural and constant daytime/early-evening heat-flow is going to be from outside to inside. Hence, the AC is going to be on intermitently anyway, and logically speaking, all the BTUs from the oven is our "delta" for which we need to compensate with the AC. That "delta", in BTUs, appears to me to be regardless of outside temp and household insulation efficiency, since the windows remain closed/etc the whole time, baking or not. With having closed windows anyway, I don't "vent" the oven heat, I have to run the AC to get rid of it.
In winter/heating season, of course the BTUs from the oven "help". And in the fall/spring, we're straddling the calculations of net-help, versus venting the heat through open windows, versus extra-cost in terms of AC cooling.
Hence, if the "complaint/excuse" is having to run the AC more in the summer to compensate for the bake, let's put a figure/cost on it.
My first order guesstimate is $.31 extra AC cost for a bake. That extra cost seems cheaper than the "extra" cost of buying an artisan loaf at a store/bakery, compared to a home-made loaf.
Uh-oh. Now I need to figure out the home-made loaf cost. ;-P
Grainbrain, you're right. I just remembered that outside temp _does_ affect efficiency ( BTUs/kwh) of the outside unit. So, yeah, the hotter it is outside, the more it costs to get rid of "X" BTUs from the oven.