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Thermodynamics Application

  1. Oct 20, 2007 #1
    im having a hard time explaining this, so im going to do my best. please feel free to ask questions on things that arent clear.

    i want to find out how much one would save on their electricity bill (from the lights and powering the AC to cool the room). ive already calculated the money im saving from the lights, but what about powering the AC?

    lets say i leave my lights on 24/7. the total power consumption by the lights is 1kW.
    would I consider that the rate of heat transfer (Qdot)? if i want the heat transfer (Q), i would multiply 8760 hr/yr * 3600 s/hr and get 31,536 kJ. is that correct so far? now if thats correct, this is where i reach a dead end.

    what other information am i missing to calculate how much the AC system is using per year?

    would i treat this as a refrigeration cycle, the room being the system?
     
  2. jcsd
  3. Oct 20, 2007 #2
    please forgive me for posting this in the wrong section mods :( could you guys please move it?
     
  4. Oct 20, 2007 #3
    Save money by doing...what though?

    What is the actual, you know "treatment"?

    Turning the AC off? using a different type of AC? leavings the lights off in the room?
     
  5. Oct 20, 2007 #4
    yea thats what im not understanding either, i will have to ask my professor. on the paper it does say you will have to make some assumptions hmm...
     
    Last edited: Oct 20, 2007
  6. Oct 20, 2007 #5

    stewartcs

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    Power companies charge by the kilowatt-hour. For example mine charges me 13 cents per kilowatt-hour. So, in order to determine how much you are spending on AC or whatever, you need to determine how many kilowatt-hours your AC or whatever is consuming.

    A 1 ton AC consumes approximately 3,517 watts. Multiply that times however many hours you run it per day to determine the watt-hours. Divide that by 1000 to get the kilowatt-hours.

    For example, if you run a 1 ton AC for 6 hours per day you end up with 3,517 x 6 / 1000 = 21.102 kilowatt-hours. Multiply that times the rate of 0.13 per kilowatt-hour and you get $2.74 per day or about $82.29 per month. That's for just a 1 ton AC. Most are larger than that. I have two 4 ton AC's in my house.

    Does that answer your question?
     
  7. Oct 22, 2007 #6
    i thought responded here already...what the hell

    yes that helps out, thanks.
     
  8. Oct 23, 2007 #7
    How much heat

    I think the essence of the problem is to remove the heat generated by the light bulbs. I once had a problem similar to this, where a large computer was in a small air-conditioned room with a couple of technicians (one tech puts out 100 watts of heat).

    To solve this, you need to know the amount of heat to be removed, the temperature of the room, the temperature of the exhaust, and the efficiency of the refrigerator. Assume that ALL of the power to the lights becomes heat. Figure out from this how much energy you need to run the refrigerator, and add that to the energy savings from turning out the lights.

    I hope you enjoy unit conversion problems, you are going to get a lot of practice!
     
  9. Oct 23, 2007 #8

    russ_watters

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    Sorry I missed this before:
    Actually, small air conditioners tend to run more like 1.5 kW/ton.
     
  10. Oct 23, 2007 #9

    stewartcs

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    Opps!

    Russ is correct! The 3,517 was already in watt-hours (1 ton = 12,000 BTU's = 3,517 watt-hours), not watts. Sorry!

    The rest of calculation is the same though.

    Moral of the story: Make sure you look at your unit to find how many watts it consumes.

    If it doesn't state the watts specifically, you can find it by dividing the BTU's by the EER (energy efficiency rating).
     
    Last edited: Oct 23, 2007
  11. Oct 23, 2007 #10

    russ_watters

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    No, that's not it (tons is a power, not an energy*). The 1 ton is the rate of heat absorption air conditioner. It doesn't really have anything to do with the energy input.

    So another way to say this is that a 1 ton air conditioner provides 3.5kW of cooling and requires 1.5 kW of electricity to do it.

    As a convention, people tend to say "BTU" even when they mean BTU/h - which is usually shown (confusingly) as BTUH. They are treated interchangeably, so you need to know the context.

    *This also has a caveat: from what I understand, the original definition of a "ton" was based on the cooling capacity of a ton of ice, for a day. I guess then it was used interchangeably as a rate and a total heat.
     
    Last edited: Oct 23, 2007
  12. Oct 23, 2007 #11

    stewartcs

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    Yeah, don't know what I was thinking...again, Russ had it right with the cooling capacity in watts and the required energy to perform the cooling.

    At any rate once you know the electrical energy in watts required by the AC to perform the cooling (1 ton of cooling or whatever) then you can find out how much it will cost you based on the rate the utility provider is charging.

    Thanks Russ...
     
  13. Oct 23, 2007 #12

    russ_watters

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    No prob. FYI, I have a 3 ton air conditioner and I measured its power input to be almost exactly 4.5kW. My highest electric bill in the summer was $110 and I pay $.13 per kWh. Figure about $60 of that usage was a/c. That means....

    $60/$.13=461 kWh

    461 kWh/4.5 kW = 102.5 hr, or I used my a/c an average of 3.3 hours a day in the summer.
     
  14. Oct 24, 2007 #13

    stewartcs

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    I remember having cheap power bills like that back when I lived in North Carolina. I live in Texas now and my average power bill in the summer is about $400. That's with two 4 ton AC's for both cases (plus a pool pump running 4 hours a day).
     
  15. Oct 24, 2007 #14
    thanks guys, ill have another go at the problem sometime soon :)
     
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