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  1. #1
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    Nov 2005
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    I wanted some opinions on this.

    I am doing a replacement of equipment at a bank job. There are 2 existing water cooled AC units, one (1) 40-ton and one (1) 5-ton. There is an exsting cooling tower there of 55-tons.

    The 40-ton AC and 55-ton cooling tower are over 20 years old and are shot, so we are replacing them. The 5-ton was replaced a few years ago and works fine.

    The question I have, should I use a smaller tower for this application or keep the tower oversized?

  2. #2
    If that tower is always running at below 80% you may have some growth issues. It *should* be ok, with the fan running at a lower speed on a VFD to match the load, but there may be more specifics to the project site.

    Since the system is over 20 years old, I'd get an engineer to take a look at the piping system & chem treatment to see if you can optimize it. Plus, when that engineer releases signed & stamped drawings all the responsibility is on him and not you when something gets mucked up.

  3. #3
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    yes the engineer had already inspected it and determined all pipes and duct can stay just the units need replacement.


  4. #4
    Join Date
    Nov 2003
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    Philadelphia
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    its hard to tell because you are talking nominal numbers here. depends on your local design conditions and the water temperatures you are looking for. dont forget the cooling tower has to reject the heat of compression as well as the building load. rule of thumb is an additional 25% of the building load for heat of compression.

    45 tons x 1.25 = 56.25 tons

    so i wouldnt decrease the tower. i also agree with marauder, get some signed and sealed drawings from an engineer.

  5. #5
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    I agree with "du mech eng". Sizing based on tonnage is an iffy proposition at best. Find out the GPM and Delta T on the equipment and the design conditions of your location and go from there.

  6. #6
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    it's 165 GPM

    Water from 95F to 85F with 78F entering air wet bulb temp.


  7. #7
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    Is that 165 GPM the design of the existing tower or the requirements of the condensing units? Just curious, where are you located? Your design temps are the same as ours here in ATL.

  8. #8
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    NYC area.

    165 is the existing I suppose. The layout never changed since the inittal installation.


  9. #9
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    Philadelphia
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    those numbers sound like nominal numbers for the tower. 95/85/78 is exactly what CTI uses for nominal rating of towers. it is possible however that those are the numbers that the system was designed for and in that case no, i would definately not downsize that tower.

    Cooling Tower Heat Rejected (Btu/hr) = 500 x GPM x Range
    1 Cooling Tower Ton = 15,000 Btu/hr

    So...

    500 x 165 x (95-85)/15,000 = 55 tons

    and incase anyone is not familar with towers, yes that is 15,000 not 12,000. 1 cooling tower ton is defined as cooling 3 GPM by 10 deg F.

    3 GPM x 500 x 10 deg F = 15,000 Btu/hr

    i still think that the engineer that looked at the system should provide signed and sealed drawings before doing this changeout. he/she would know exactly what's out there as opposed to guessing over the internet.

  10. #10
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    How did you figure the 500 btu/hr?


  11. #11
    Join Date
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    Location
    Philadelphia
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    Originally posted by klabkebash
    How did you figure the 500 btu/hr?
    the 500 is a conversion factor, not btu/hr.

    the units are: (Btu x min)/(gal x hr x degF)

    here's a quick derivation:

    the basic equation for heat transfer is:

    q = mass flow x specific heat x delta T

    the specific heat of water is 1 Btu/(lb x degF)

    since we are used to dealing with volumetric flow and not mass flow in this industry, we need to convert GPM of water to mass flow. to convert from volumetic flow to mass flow you simply multiply by the density. the density of water is about 8.34 lbs/gallon.

    so now we have the following equation:

    q = 8.34 x GPM x delta T

    this equation would give us heat transfer in Btu/min. since we typically deal with Btu/hr we must multiply this equation by 60 minutes per hour and we end up with...

    q = 8.34 x 60 x GPM x delta T

    or

    q = 500 x GPM x delta T

    since the specific heat and density of water actually change with the temperature of the water this equation only holds true for certain water conditions. it is however pretty accurate at the given temperatures.

  12. #12
    Join Date
    Nov 2003
    Location
    Philadelphia
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    by the way, you could also derive the following air side equations in the same manner as i did the water side above. just different units.

    Qtotal = 4.5 x CFM x Enthalpy Change
    Qsensible = 1.08 x CFM x Temperature Change
    Qlatent = 0.68 x CFM x Humidity Ratio Change (grains/lb)
    Qlatent = 4760 x CFM x Humidity Ration Change (lb/lb)

  13. #13
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    Location
    New York
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    yea they made sense

    yea they made sense. The converstion factors and U values brought me back to when I learned load calcs in design classes.

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