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I have an old chest-type freezer (circa 1975) that I would replace if I knew for certian that it was inefficient. It is fairly small and in good condition. Is there a way to determine the efficiency? I have a variety of electricians tools and meters that I could use.

2007-06-02 08:46:41 · 3 answers · asked by DoItRite 3 in Science & Mathematics Engineering

3 answers

I would just use a wrap around amp meter on the existing wires and not wire a new one in series. Turn the freezer off and set up the meter. Then ask someone to turn the thermostat (temperature setting) way up. Take note of the max spike in amps drawn when the compressor starts and the relative duration of the spike. Then watch the amperage "level out". Record both numbers (amp spike and stead state). Next unplug the freezer and take a volt meter reading from the wall outlet. Amps mulitplied by voltage equals watts. You would have to estimate the number of times the freezer starts and multiply the number of starts per year x the voltage x the amp spike at startup x the duration of the spike at startup (probably only a few seconds to get to steady state. Add to this number the steady state amps x the voltage x the estimate of the amount of time the refrigerator runs- (just time a few on-off-on cycles until you get something that looks somewhat consistent). For example, 6 amps x 120 volts x runs 1/3 of every hour x 24 hours in a day x 365 days per year = 2081376 watt hours. this is equal to 2081 kW hours. check your electric bill and get the billing rate. LEts pretend its 8 cents per kilowatt hour, so .08 x 2081 = $166 per year in electricity.

Now if you got one that was twice as efficient it might only run half the time or use half the amperage and it would only cost half as much to operate or $83 per year. The other half of the 166 would be savings in your pocket or $83 per year. It would take many years to save up enough to pay for a new freezer since I doubt you would find one twice as efficient.

NAme plates (usually somewhere on the door edge or door frame are a good source of information as they should state the amperage and voltage ratings of the unit- both the one you currently have and any new one in the store.

2007-06-02 13:38:28 · answer #1 · answered by MrWiz 4 · 0 0

The thermodynamic efficiency of the freezer is not really an important number, unless you need to know how much heat the freezer puts into the surrounding environment. Measuring this would require some huge calorimetry equipment.

What's important is how much energy it takes to keep the interior at a certain temperature.

To measure this, you'd need a computer, thermometer, a voltmeter, and an ammeter that can handle many amps. You'd have to wire the ammeter in series with the freezer (make it part of the circuit), which is rather dangerous, as the typical electrical outlet can kill you fairly easily. Then you have to wire the voltmeter in parallel with the refrigerator, to measure the voltage difference between the terminals.

Set up the computer to log amperage, temperature, and voltage data at regular intervals, and set the freezer to -5 Celsius or whatever you normally run it at. The freezer will likely cycle on and off. After the freezer gets to the correct temperature, record several cycles.

Average the temperature across these cycles. Now multiply all the voltage data times the amperage data to get the power dissipated as a function of time. Average this across several cycles to get the mean power consumption.

Now repeat this for a modern refrigerator of the same capacity.

2007-06-02 09:37:57 · answer #2 · answered by lithiumdeuteride 7 · 0 0

Buying a new higher efficiency freezer would never save enough in operating costs to pay for replacing the old freezer.

I have a really old huge chest type freezer and I keep it in my unheated garage. In the winter (November thru much of March) it costs almost nothing to operate. Even in the Summer when I am working out there, I hardly ever hear it kick in and operate.

These chest type freezers do not cost much to operate to begin with, maybe $5 to $10/m max. An increase in efficiency of 10% would save you $1/m. Not worth the brain damage to think about.

2007-06-02 10:53:17 · answer #3 · answered by gatorbait 7 · 0 0

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