LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
Energy conservation
- Thread starter gar
- Start date
- Status
SAC
Senior Member
- Location
- Massachusetts
I've been quite happy with it. No real surprises about my consumption - I've already been around the house with an ammeter so I know what the baseline is. The (electric) dryer and range are the big ticket items. I've already got it setup so that I save 50w resolution to the second accuracy beyond what the TED itself saves (it is very easy to pull the data down through an html queary directly to the unit). The software is a little flakey, but seems does the job. The history and graphs are good.
The thing that has been most interesting to me has been the voltage. I'm a couple of hundred feet from the transformer, but am the only one on it. I can see the voltage vary quite a bit as my loads switch on and off, but it varies even more significantly over the course of a day regardless of my consumption - more than I would have thought.
winnie
Senior Member
- Location
- Springfield, MA, USA
- Occupation
- Electric motor research
Gar,
I think that you made a small math error when figuring the number of Joules removed from the water when freezing it; I get 273 watt hours.
But a more important point is that the electricity needed to move the heat is not equal to the heat moved. You have to take into account the COP of the heat pump.
A heat pump absorbs a quantity of heat Q1 at temperature T1, and rejects quantity of heat Q2 at temperature T2. (T1 and T2 must be on an absolute temperature scale.) For an _ideal_ heat pump, Q1/T1 = Q2/T2, and since energy is conserved, the difference between Q1 and Q2 must be supplied by an external source. In a _real_ heat pump, real Q2 will always be larger than ideal Q2, and this means more energy consumption. (A heat engine uses the same formula, except that the deviation from ideal goes in the opposite direction. When you are pumping heat it takes more energy to move the heat than the theoretical minimum, when you are converting heat into electricity, then you get less out than the theoretical maximum.)
So an _ideal_ heat pump, moving 210 watt hours of heat from 32F (273K) to 56F (286K) would only need 10 watt hous of external power input.
Of course, fridges don't even get close to this ideal COP
But the COP should be better than 1!
-Jon
I think that you made a small math error when figuring the number of Joules removed from the water when freezing it; I get 273 watt hours.
But a more important point is that the electricity needed to move the heat is not equal to the heat moved. You have to take into account the COP of the heat pump.
A heat pump absorbs a quantity of heat Q1 at temperature T1, and rejects quantity of heat Q2 at temperature T2. (T1 and T2 must be on an absolute temperature scale.) For an _ideal_ heat pump, Q1/T1 = Q2/T2, and since energy is conserved, the difference between Q1 and Q2 must be supplied by an external source. In a _real_ heat pump, real Q2 will always be larger than ideal Q2, and this means more energy consumption. (A heat engine uses the same formula, except that the deviation from ideal goes in the opposite direction. When you are pumping heat it takes more energy to move the heat than the theoretical minimum, when you are converting heat into electricity, then you get less out than the theoretical maximum.)
So an _ideal_ heat pump, moving 210 watt hours of heat from 32F (273K) to 56F (286K) would only need 10 watt hous of external power input.
Of course, fridges don't even get close to this ideal COP
But the COP should be better than 1!-Jon
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
091108-0820 EST
winnie:
I used 2.20462 #/kilogram and thus 5# = 2.268 kilograms = 2268 grams.
For the heat of fusion 79.71 cal/gram. And 4.186798 joules/cal. Thus, heat of fusion is 79.71*4.1868 = 333.7 joules/gram for air-free water.
2268*333.7 = 756,832 joules or watt-sec. Then 756,832/3600 = 210.2 watt-hours.
My conversion factors are from the 40th edition of the Handbook of Chemistry and Physics.
When one looks at the tables for thermal capacity and see resolution of 1 part in 40,000 it is amazing to consider the experimental techniques that were needed to make these measurements.
Where do your calculations differ from mine since there is a very large difference between 210 and 273?
.
winnie:
I used 2.20462 #/kilogram and thus 5# = 2.268 kilograms = 2268 grams.
For the heat of fusion 79.71 cal/gram. And 4.186798 joules/cal. Thus, heat of fusion is 79.71*4.1868 = 333.7 joules/gram for air-free water.
2268*333.7 = 756,832 joules or watt-sec. Then 756,832/3600 = 210.2 watt-hours.
My conversion factors are from the 40th edition of the Handbook of Chemistry and Physics.
When one looks at the tables for thermal capacity and see resolution of 1 part in 40,000 it is amazing to consider the experimental techniques that were needed to make these measurements.
Where do your calculations differ from mine since there is a very large difference between 210 and 273?
.
winnie
Senior Member
- Location
- Springfield, MA, USA
- Occupation
- Electric motor research
Sorry, I was unclear: I get 273 watt hours for the entire process of cooling 2268g of water from 52F through freezing to 7F. I believe the difference in our results is from the heat removed to cool the water from 52F to 32F.
-Jon
-Jon
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
091108-1031 EST
winnie:
My calculations for the other two phases were:
52 to 32 is 20 F or 11.1 C. I used 4.2 joules per gram per deg C as an approximation of the values 4.2177 @ 0 C to 4.1907 @ 11 C. This calculation was 2268*4.2*11.1 = 105,734 watt-sec or 105734/3600 = 29.4 watt-hours.
32 to 7 is 25 F or 13.9 C. I used 2.09 joules per gram per deg C for ice. This calculation was 2268*2.09*13.9 = 65,888 watt-sec or 65888/3600 = 18.3 watt-hours.
.
winnie:
My calculations for the other two phases were:
52 to 32 is 20 F or 11.1 C. I used 4.2 joules per gram per deg C as an approximation of the values 4.2177 @ 0 C to 4.1907 @ 11 C. This calculation was 2268*4.2*11.1 = 105,734 watt-sec or 105734/3600 = 29.4 watt-hours.
32 to 7 is 25 F or 13.9 C. I used 2.09 joules per gram per deg C for ice. This calculation was 2268*2.09*13.9 = 65,888 watt-sec or 65888/3600 = 18.3 watt-hours.
.
winnie
Senior Member
- Location
- Springfield, MA, USA
- Occupation
- Electric motor research
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
091108-1119 EST
winnie:
I make mistakes like that all the time, and that is why I ask others to check my work.
On the other part of the problem I should not equate the energy pumped to the pumping energy without a lot more basis for the relationship.
By the way the average temperature in the freezer was between about -5 F and about +2 F which are the min and max points of the cycle. Compressor on and off periods are about 1 hour for a 2 hour cycle. Ambient I can not control and over the days I have collected data it has ranged from about 52 F to 65 F.
.
winnie:
I make mistakes like that all the time, and that is why I ask others to check my work.
On the other part of the problem I should not equate the energy pumped to the pumping energy without a lot more basis for the relationship.
By the way the average temperature in the freezer was between about -5 F and about +2 F which are the min and max points of the cycle. Compressor on and off periods are about 1 hour for a 2 hour cycle. Ambient I can not control and over the days I have collected data it has ranged from about 52 F to 65 F.
.
boboelectric
Senior Member
- Location
- Eighty Four,Pa.15330
That's like the ads about Almish heaters. The heater is free.Thee almish wood isn't.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
091108-2046 EST
For now I have finished my 5# water freezing experiment.
The thermistor measuring the water was immersed in the the water and thus in the ice block. Last night about 1815 I removed the jug from the freezer and set it in the room on the floor. At this time the ice block temperature was +2 deg F. The ambient room temperature was about 62 deg F, and remained about that thru the night and today. By 1233 today the ice had not fully melted and the temperature was holding at 31.8 deg F. As a very ball park estimate from this I estimate the thermal resistance from the ice to the room at about 1/3 watt/deg F. Do not conclude that my thermistor thermometer is as accurate as the fractional degree would imply. That value is just what I read.
By 1800 today the ice was all melted and the water temperature was up to 40 deg F.
From measurements during a power outage earlier this year I concluded that about 3 hours was all that I wanted for a freezer warm up time before power needed to be reapplied. This related to how long I could keep the generator off during the night.
Frozen water jugs are not much help in maintaining a low temperature in a freezer without power. This is because the greatest heat storage capability occurs at 32 deg F. Dry ice is far better, -109.3 deg F. Having a siphon tank of liquid CO2 is another means.
.
For now I have finished my 5# water freezing experiment.
The thermistor measuring the water was immersed in the the water and thus in the ice block. Last night about 1815 I removed the jug from the freezer and set it in the room on the floor. At this time the ice block temperature was +2 deg F. The ambient room temperature was about 62 deg F, and remained about that thru the night and today. By 1233 today the ice had not fully melted and the temperature was holding at 31.8 deg F. As a very ball park estimate from this I estimate the thermal resistance from the ice to the room at about 1/3 watt/deg F. Do not conclude that my thermistor thermometer is as accurate as the fractional degree would imply. That value is just what I read.
By 1800 today the ice was all melted and the water temperature was up to 40 deg F.
From measurements during a power outage earlier this year I concluded that about 3 hours was all that I wanted for a freezer warm up time before power needed to be reapplied. This related to how long I could keep the generator off during the night.
Frozen water jugs are not much help in maintaining a low temperature in a freezer without power. This is because the greatest heat storage capability occurs at 32 deg F. Dry ice is far better, -109.3 deg F. Having a siphon tank of liquid CO2 is another means.
.
glene77is
Senior Member
- Location
- Memphis, TN
Gar,
Thanks for publishing your experiment
on Water Freezing Experiment.
This posting is a good example
of the expert information available on these forums.
About 3 hours, that is good to know. I have had people ask me about this info.
Today, I will replace a IR led with a red one (with diffuse cover), in old remote controller for TV. May have to try slowing down the clock/timer by adding a cap.
The little Kids can zap themselves!
That is a far fetch from my years at the university,
but thats life!
Thanks for publishing your experiment
on Water Freezing Experiment.
This posting is a good example
of the expert information available on these forums.
About 3 hours, that is good to know. I have had people ask me about this info.
Today, I will replace a IR led with a red one (with diffuse cover), in old remote controller for TV. May have to try slowing down the clock/timer by adding a cap.
The little Kids can zap themselves!
That is a far fetch from my years at the university,
but thats life!
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
091109-0911 EST
glene77is:
The freezer that I am currently monitoring is an upright and has the condensing coil on the rear and not in the side walls. It is fairly full and contains maybe 2 gallons of water in jugs, a shelf full of bread, two shelves of berries and greens, plus some miscellaneous items. Most of these items are not high density water containing product. Even though quite full there is not a lot of water mass in the freezer.
In particular the berry and greens products should not get too warm. At 60 deg F ambient the freezer cools from +2 F to -5 F in about 0.66 hours, and warms to +2 F in about 1.16 hours. The curve is approximately an exponential, but not really. The initial slope is about 4 deg in 0.33 hours, 12 deg/hour. Further up, between -1 and +2.5 the slope looks more linear, and is about 4 deg/hour. Suppose I ball park the time constant at 6 deg/hour to 63% ( (60+5)*0.63 = 41 deg change ). Then I have 6.8 hours to rise 41 deg from -5, or to +36. If instead the ambient is 100 deg F, then the rise is 105*0.63 = 66 deg in 6.8 hours or about 4.5 hours to reach about 32 deg F. But I do not want to get close to 32 deg.
My other freezer has cool time of about 0.33 hour from +4.5 deg F to 0 F and a rise time of about 0.5 hours. Both rise and fall are much straighter than the other freezer. This freezer is about 42 years old. It has less peak inrush current than the other.
There is a substantial thermal gradient in the freezer. So the meaning of the temperature measurement is dependent upon location.
I have no idea what the design criteria are for freezers and refrigerators. My goal in these experiments is to see what one can learn with simple instrumentation that may be useful to the average consumer in evaluating their energy consumption.
For power and voltage monitoring I am using a Kill-A-Watt EZ and a TED system. Temperature is measured with thermistors. Temperature monitoring may change to Dallas Semiconductor one-wire devices.
.
.
glene77is:
The freezer that I am currently monitoring is an upright and has the condensing coil on the rear and not in the side walls. It is fairly full and contains maybe 2 gallons of water in jugs, a shelf full of bread, two shelves of berries and greens, plus some miscellaneous items. Most of these items are not high density water containing product. Even though quite full there is not a lot of water mass in the freezer.
In particular the berry and greens products should not get too warm. At 60 deg F ambient the freezer cools from +2 F to -5 F in about 0.66 hours, and warms to +2 F in about 1.16 hours. The curve is approximately an exponential, but not really. The initial slope is about 4 deg in 0.33 hours, 12 deg/hour. Further up, between -1 and +2.5 the slope looks more linear, and is about 4 deg/hour. Suppose I ball park the time constant at 6 deg/hour to 63% ( (60+5)*0.63 = 41 deg change ). Then I have 6.8 hours to rise 41 deg from -5, or to +36. If instead the ambient is 100 deg F, then the rise is 105*0.63 = 66 deg in 6.8 hours or about 4.5 hours to reach about 32 deg F. But I do not want to get close to 32 deg.
My other freezer has cool time of about 0.33 hour from +4.5 deg F to 0 F and a rise time of about 0.5 hours. Both rise and fall are much straighter than the other freezer. This freezer is about 42 years old. It has less peak inrush current than the other.
There is a substantial thermal gradient in the freezer. So the meaning of the temperature measurement is dependent upon location.
I have no idea what the design criteria are for freezers and refrigerators. My goal in these experiments is to see what one can learn with simple instrumentation that may be useful to the average consumer in evaluating their energy consumption.
For power and voltage monitoring I am using a Kill-A-Watt EZ and a TED system. Temperature is measured with thermistors. Temperature monitoring may change to Dallas Semiconductor one-wire devices.
.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
091111-1318 EST
glene77is:
My Admiral freezer, the one currently instrumented, the first one referenced in the previous post, was disconnected from power at 0823 this morning. The ambient was 56 deg F and remained at that value until now, 1304, when I reapplied power.
This is a period of about 4.6 hours and the rise was 13 deg F, or about 2.8 deg per hour.
If the ambient temperature was 100 deg F, then the rise could have been in the range of 25 deg.
Both the freezer thermometer and the thermistor sensor were on the second shelf from the top, not the warmest location. These were sensing the shelf and air temperature. Probably realistic for the surface temperature of items in the freezer.
I believe my other freezer has a higher rate of rise.
One should run experiments on their own freezer to determine a realistic time for storage with no power available. Obviously one needs an understanding of what any measurements mean and there needs to be a criteria for maximum temperature to prevent thawing.
In the past I have heard reference to 24 hours for a closed freezer as a criteria relative to how long one could be without power. I think this is far too long.
.
glene77is:
My Admiral freezer, the one currently instrumented, the first one referenced in the previous post, was disconnected from power at 0823 this morning. The ambient was 56 deg F and remained at that value until now, 1304, when I reapplied power.
This is a period of about 4.6 hours and the rise was 13 deg F, or about 2.8 deg per hour.
If the ambient temperature was 100 deg F, then the rise could have been in the range of 25 deg.
Both the freezer thermometer and the thermistor sensor were on the second shelf from the top, not the warmest location. These were sensing the shelf and air temperature. Probably realistic for the surface temperature of items in the freezer.
I believe my other freezer has a higher rate of rise.
One should run experiments on their own freezer to determine a realistic time for storage with no power available. Obviously one needs an understanding of what any measurements mean and there needs to be a criteria for maximum temperature to prevent thawing.
In the past I have heard reference to 24 hours for a closed freezer as a criteria relative to how long one could be without power. I think this is far too long.
.
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