Average KW/hr calculation on Trane Voyager 25 Tons
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  1. #1
    Join Date
    Feb 2011
    Posts
    7

    Average KW/hr calculation on Trane Voyager 25 Tons

    Hello all,

    I want to calculate the average KW/hr that this RTU have.

    The unit is a Trane Voyager 25 Tons:

    Cooling Performance(a)
    Gross Cooling Capacity 282,000
    EER (Downflow/Horizontal)(b) 10/10
    Nominal CFM / ARI Rated CFM 10000 / 8750
    ARI Net Cooling Capacity 272,000
    Integrated Energy Efficiency Ratio (IEER)(c) 10.4
    Percent Capacity @ part load
    (Stage 1/Stage 2)
    50/100
    System Power (kW) 27.20

    I live in a very hot weather and i have 16 units of this kind.

    My question is do you know what is the average KW/hr, because i know not always the dual compressors are on. How can i get this calculation? some one have an average?

    Thanks so much for your help!!

    best regards!!

  2. #2
    Join Date
    Feb 2011
    Posts
    7
    Any help here? or i must post on another section?

    Thanks!!

  3. #3
    Join Date
    Sep 2010
    Location
    Wa.
    Posts
    119
    That should be done while unit is on so that proper voltage at the unit can be measured.

  4. #4
    Join Date
    Sep 2010
    Location
    Wa.
    Posts
    119
    Is this a heat pump or A/C unit? Does it have electric strip heat?

  5. #5
    Join Date
    Feb 2011
    Posts
    7
    Hello,

    Its a heat pump, and have a resistor to heat the air.

    Thanks!!

  6. #6
    Join Date
    Dec 2002
    Posts
    603
    I worked this out
    EER = energy efficient ratio (Btu/Wh) there-for 10 EER = to 12/10 =1.2 btu/wh times 1000 W or 1 KW =1,200 BTU/H and for IEER=12/10.4=1.15 W/H x 1kw=1150 btu/h so lets split it to 1175 BTU/H/ KW now then if we devide12,000 BTU/TON by 1175/h we get 10.2 KW/H

  7. #7
    Join Date
    Dec 2002
    Posts
    603
    If we do it this way I get this
    The gross btu = 300,000 btu/h and the net = 272,000 btu/h which is ~91% efficiency now if we find the btu/h we devide by 272,000 by 24Hours I get 11,333 btu/h and if I devide by 3.412 btu/watt I get 3.322Kw/h I think this one is correct where am I wrong?

  8. #8
    Join Date
    Sep 2002
    Location
    Hampton Roads, Virginia
    Posts
    1,593
    Quote Originally Posted by atamai View Post
    Hello all,

    I want to calculate the average KW/hr that this RTU have.

    The unit is a Trane Voyager 25 Tons:

    Cooling Performance(a)
    Gross Cooling Capacity 282,000
    EER (Downflow/Horizontal)(b) 10/10
    Nominal CFM / ARI Rated CFM 10000 / 8750
    ARI Net Cooling Capacity 272,000
    Integrated Energy Efficiency Ratio (IEER)(c) 10.4
    Percent Capacity @ part load
    (Stage 1/Stage 2)
    50/100
    System Power (kW) 27.20

    I live in a very hot weather and i have 16 units of this kind.

    My question is do you know what is the average KW/hr, because i know not always the dual compressors are on. How can i get this calculation? some one have an average?

    Thanks so much for your help!!

    best regards!!
    How can anyone give an average KW/hr if a figure for runtime of each compressor is not available?

    Kevin
    "Profit is not the legitimate purpose of business. The legitimate purpose of business is to provide a product or service that people need and do it so well that it's profitable."

    James Rouse

  9. #9
    Join Date
    Dec 2002
    Posts
    603
    Quote Originally Posted by klrogers View Post
    How can anyone give an average KW/hr if a figure for runtime of each compressor is not available?

    Kevin
    I am able to plot the KW/h with changing the load as I show below
    I can tell you with 272,000BTU/h with R-22 the KW/h = 27.2 with 131*F SCT and 42*F SST, EER = 10
    At 195,000 Btu/h, with R-22 =19KW/h, with SST= 42*f and SCT = 129*F, EER = 10.32
    At 158,000 BTU/h, R-22 = 13.75KW/H, with SST = 45*F and SCT = 125*F, EER = 11,53
    If I went on you could plot the curve to give you the various KW/h at various loads
    All of the above were plotted with a volumetric efficiency of 90% because of the difference in the beginning of the gross refrigeration effect and the net refrigeration effect

  10. #10
    Join Date
    Feb 2011
    Posts
    7
    Hello thanks for the information, but for example do you know what is the average time for each load?

    You tell me at first with 272,000BTU/h KW/h = 27.2, but you know what is the average time that the HVAC stay with this load?

    After maybe the HVAC stay with 195,000 Btu/h, with R-22 =19KW/h but how long?

    Im asking on averages i think it varies depending on conditions but do you have any average? or how can i get this information?

    Im preparing a executive presentation to talk about saving between control this HVAC manually vs IP Thermostats, and i need the average cost for this HVAC.

    Thanks a lot for your help

  11. #11
    Join Date
    Dec 2002
    Posts
    603
    T0 find the average I would have to take a series of loads lets say 10 and find an average and thats a lot of work but if I have the time I will see if I can do that any way i find this problem interesting. Of course if I could find the run time for each compressor it would make it a better prediciction

  12. #12
    Join Date
    Feb 2011
    Posts
    7
    Thanks!!

    Anyone have an idea to find the run time for each compressor?

  13. #13
    Join Date
    Dec 2002
    Posts
    603
    Average KW/H of 25 ton Air Conditioning with R22
    Heat load BTU/H KW/H EER SCT
    1. 272,000 27.18 10 130
    2.242,000 23,80 10.16 129
    3.212,000 20.44 10.48 128
    4.182,000 17.20 10.64 127
    5.152,000 13.84 10.98 126
    6.122,000 10.94 11.16 125
    7.92,000 8.12 11.34 124
    8.62,000 5.33 11.64 123
    ----------------------------------------------------------------------
    15.86 =Average. In your case if you are in a warm climate I would use the top 4 KW/H because of the run time would most likely be at those conditions. EX OSA = 95+ for most of the run time but you would know that so adjust the numbers according to that knowledge. So let’s say we take the top 4 and add them = 88.64 devide by 4 = 22.16 KW/H If you really want to get a handle on this I would suggest installing a watt meter on say 20% of the units or better yet install a DDC contol system for watt meter readouts. let us know what you decide.

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