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AC sizing for 600 sq ft room – Does 1.5 tons account for high ceilings & occupancy?

Kotha | Posted in General Questions on

I’m planning a mini-split installation for a 600 sq ft (20’x30′) home office with these specifics:

  • 9 ft ceilings (5400 cubic ft total volume)
  • 3 regular occupants + equipment (~800W heat load)
  • Middle floor of a 3-story building (buffered above/below)
  • South-facing with double-pane windows

An online calculator suggested 1.5 tons (17,820 BTU), but I’m concerned this may not account for:

  1. The additional volume from high ceilings
  2. Occupant/equipment heat gains during peak hours

For those with energy-efficient HVAC experience:

  • Would you adjust the tonnage calculation for these factors?
  • Any rule of thumb for cubic feet vs. square feet in sizing?
  • Does middle-floor placement typically reduce cooling needs noticeably?

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Replies

  1. charlie_sullivan | | #1

    There's no need to account for the volume. But you do need to account for:

    * Solar gain from the south facing windows.

    * The wall area, and how well it's insulated (and the outdoor temperature)

    * The window area and the U-factor of the windows.

    * The 800 W of internal gain.

    * Infiltration or ventilation rate.

  2. walta100 | | #2

    In Central MO the local rule of thumb is 700 SQF per ton for leaky poorly insulated freestanding buildings and seems to provide more than enough cooling.

    Will the space really have 3 occupants in it most of the day?

    How much heat producing equipment will they be operating like computers and printers.

    If this space abuts conditioned spaces on 5 sides 1.5 tons sounds oversized.

    Installing oversized equipment is risky in that it will run a small percentage of the time and if it is a humid climate the system will not remove enough moisture in the short run times. This forces you to set the thermostat lower and you end up being cold and clammy.

    Walta

    1. Kotha | | #4

      Thanks for the detailed MO-specific insights - the 700 sqf/ton rule for leaky buildings is particularly helpful context. You've raised excellent points about my scenario:

      Occupancy: You're right that 3 people won't be constantly present.

      2 occupants daytime (8 hrs)

      1 computer (500W)

      1 printer (occasional use)

      This reduced the load estimate by about 1,500 BTU

      Adjacent conditioned spaces: The tool did account for buffered floors/ceilings (5 sides as you noted) with a 15% load reduction - does that align with your experience?

      Humidity concerns: This is exactly why I'm wary of oversizing.

      1. Expert Member
        BILL WICHERS | | #8

        Volume doesn't matter, but the surface area does, because heat is getting into the space through the surfaces, not from the air enclosed within those surfaces. Your space is a little like the datacenters I design for work, just very much smaller. I can address some of your questions, and note that I work with watts for work, not BTUs (because it's easier for my application doing datacenter work):
        1- The 500w computer is only SENSIBLE load for the A/C. There is no humidity to deal with for this.
        2- Each person is worth about 100w, so another 200w of heat load. Most of this will be sensible load too.

        That's about 700w of heat load, almost all sensible load, which is about 0.2 ton worth of cooling. With a middle floor room, assuming conditioned space above and below on the two adjacent floors, you'll have minimal heat gain from either the floor or the ceiling since I'm assuming the other conditioned spaces will be approximately the same temperature as you are aiming for in your office space. If there is no temperature differential (as is the case if all the spaces are the same temperature), then there is no heat flow in either direction -- no loss, and no gain. You say nothing about what is behind the three walls on the interior of your office, but if we assume conditioned space on the other side of those too, then no appreciable amount of heat gain from those, either.

        That leaves two things to contribute to your sizing that we didn't already cover:
        1- humidity levels (LATENT heat load for the A/C if you're dehumidifying)
        2- Solar gain from the window, and heat gain through the exterior wall.
        This is influenced by the type of window coating, especially SHGC, the area of the windows in square feet, and the area of the exterior wall in square feet along with the R value of the wall. You get into complexity when trying to calculate solar heat gain since the geographic location and time of year come into play. This stuff is usually estimated.

        Again, the enclosed volume of the space really doesn't matter here. Only the area of the surfaces enclosing that volume matter, and in your case, most of the surfaces have conditioned space on the other side, making them not matter much, either. Chances are the equipment will be somewhere around 50% or more of your heat gain in the room here (guessing a bit at the windows), so you're probably around 1/2 ton or so of sensible (solar gain and heat through the wall are SENSIBLE load too), so your 1.5 ton estimate is very much higher than you're likely to actually need.

        Infiltration and exfiltration from other conditioned spaces of similar temperature and humidity levels to your office setpoints doesn't really affect your calculations here. Only outside air has an affect on your space, and anything coming in from non-conditioned adjacent spaces that are significantly warmer than your office. As an example, if your office is built into a warehouse with no A/C, then those "interior" walls will contribute to heat gain in your office. If those other walls just partition your office off from other offices in a big building, we can assume those other offices are kept at "comfy for people" temperatures not too different from your office's setpoint, which means you won't really see much, if any, heat gain through those interior walls (and floor/ceiling too, most likely). That just leaves the exterior wall, and the enclosed equipment.

        In my work, we have no windows, and we minimize the exterior wall areas -- often with a hallway buffering those walls from the outdoors. We can heat gain through the roof in single story buildings (which most dedicated datacenters are), and the heat load of the enclosed equipment so dominates the calculations that it's almost all that matters. ALL the heat put in by the equipment is sensible load, since it doesn't alter the dew point of the interior air. Latent loads are only an issue when you're removing water from the air, which is the case when you're dehumidfying.

        With the information you've provided, and making some assumptions as to the other interior surfaces seperating you from other conditioned spaces, I'm going to assume you are trying to cool this office from the ADDITIONAL heat load of the people and equipment, so your minisplit is probably more of a supplemental A/C, and not handling 100% of the cooling all by itself. If that's the case, I'd probably put in a 1/2 ton unit, which is around double the heat load of your equipment and people in the room. A 3/4 ton will give you more margin for that window wall. If you don't have to do dehumidification, shorter cycles on a 1 ton aren't as much of an issue, but that is probably very oversized for your space. Even the 3/4 ton is probably more than you need if my assumptions are correct.

        Note that you do need to allow for makeup air in the space -- the people in there need to breathe, after all, so you need to know how many CFM that will be, and if it's been conditioned (blowing in from a building HVAC system or an adjacent conditioned office), or if it's unconditioned air. If it's conditioned air, you don't need to worry about it for your purposes, but if it's unconditioned, it will put more load on your minisplit.

        Bill

    2. Expert Member
      DCcontrarian | | #7

      C'mon, Walta. You know you're not doing the OP any favors citing "rules of thumb."

      Kotha: there is a process for calculating the cooling load for a building. It's called Manual J. You need to know the dimensions of the building, what kind of insulation it has, the size and orientation of the windows, the type of windows and doors, and the occupancy. You also need to know the local climate.

      A high ceiling does not by itself affect the cooling load, except that it increases the exterior wall area.

      That online "calculator" you linked to is complete garbage. Houses are like snowflakes, every one is unique. Two houses with similar dimensions can have their cooling load vary by a factor of ten.

  3. gusfhb | | #3

    climate zone/design temp?

    1. Kotha | | #5

      Average Climate zone

      1. Expert Member
        DCcontrarian | | #9

        There's no such thing.

  4. Kotha | | #6

    I don't Know about humidity calculations. Do we have to consider Humidity also in AC sizing because it is not mentioned in this tool.
    https://www.actonnagecalculator.com/p/ac-tonnage-calculator.html

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