Using a Bath Fan to Equalize Room Temperatures

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Using a Bath Fan to Equalize Room Temperatures

A little math clarifies why this approach usually doesn’t work

Posted on Feb 16 2018 by Martin Holladay

On Green Building Advisor, readers regularly ask questions about room-to-room temperature imbalances — the type of imbalances that may occur when a home has a point-source heater like a ductless minisplit or a wood stove. Here’s a typical question: “I’d like to install a bathroom fan to move air from a warm room to a cool room. Will this approach be enough to equalize the temperatures between the two rooms?”

Scott Gibson did a good job summarizing one such discussion in his Q&A Spotlight article, “Can Bathroom Fans Be Used to Distribute Heat?” But there’s more to say on the issue. In this article, I hope to thoroughly address this topic, once and for all.

We can approach this question by breaking it down into parts:

  • First, some math. We need to know how much heat is lost through the exterior walls and the windows in the room that’s cold. We also need to know how much heat flows through the partition walls (and in some cases, the floor assembly or ceiling assembly) separating the cold room from adjacent warmer rooms.
  • Next, we need to use math to figure out how much heat a bathroom fan can move.
  • Then we can apply what we’ve learned about space heating in winter to the question of summertime cooling.
  • Finally, we can look at different options for solving temperature imbalances between rooms, to determine which approaches make the most sense.

Wall and window specifications matter

How much heat is lost through the exterior walls and windows of any particular room on a cold night? To answer the question for a specific room, we need to know the area of the exterior walls, the area of the windows, and the relevant R-values (or U-factors).

The exterior wall area for a typical bedroom might range from about 80 square feet (for a small bedroom with one exterior wall) to 220 square feet (for a large bedroom with two exterior walls).

The R-value of an exterior wall might range from R-16 (for a poorly insulated wall) to R-40 (for a superinsulated wall). To determine U-factor, we use the formula U=1/R. So an R-16 wall has a U-factor of about 0.062, while an R-40 wall has a U-factor of 0.025.

The window area for a typical bedroom might range from 6 square feet (for a bedroom with one small window) to 45 square feet (for a bedroom with three large windows).

The U-factor of a window might range from 0.40 (for an older window) to 0.17 (for a high-quality triple-glazed window). A U-0.40 window has an R-value of R-2.5, while a U-0.17 window has an R-value of R-5.9.

Calculating heat loss

The heat loss formula for determining transmission losses through floors, roofs, and walls is:
Q = A • U • ΔT
In other words, the rate of heat flow through a building assembly (in BtuBritish thermal unit, the amount of heat required to raise one pound of water (about a pint) one degree Fahrenheit in temperature—about the heat content of one wooden kitchen match. One Btu is equivalent to 0.293 watt-hours or 1,055 joules. /h) is equal to the area of the assembly (in ft²) times the U-factor (in Btu/ft² • hr • F°) of the assembly times the ΔT (in F°). The ΔT (delta-TDifference in temperature across a divider; often used to refer to the difference between indoor and outdoor temperatures.) is the difference in temperature between the interior and the exterior of the building.

For example, let’s calculate the heat loss for a bedroom with two exterior walls, with these assumptions:

  • The bedroom size is 11 feet by 14 feet.
  • There are two windows, each measuring 12.5 square feet (total window area = 25 square feet).
  • The exterior wall area (not including windows) is 175 square feet.
  • The wall R-value is R-24 (U-0.04).
  • The window U-factor is U-0.30.

We’ll look at heat loss under three outdoor temperature conditions: 0°F, 20°F, and 32°F.

  • If the outdoor design temperatureReasonably expected minimum (or maximum) temperature for a particular area; used to size heating and cooling equipment. Often, design temperatures are further defined as the X% temperature, meaning that it is the temperature that is exceeded X% of the time (for example, the 1% design temperature is that temperature that is exceeded, on average, 1% of the time, or 87.6 hours of the year). is 0°F, the delta-T is 70 F°. Under design conditions, the heat loss from the bedroom through the exterior walls and windows is (25 • 0.3 • 70) + (175 • 0.04 • 70) = 1,015 BTU/h.
  • When the outdoor temperature is 20°F, the delta-T is only 50 F°. So at an outdoor temperature of 20°F, the heat loss from the bedroom through the exterior walls and windows is (25 • 0.3 • 50) + (200 • 0.04 • 50) = 775 BTU/h.
  • When the outdoor temperature is 32°F, the delta-T is 38 F°. So at an outdoor temperature of 32°F, the heat loss from the bedroom through the exterior walls and windows is (25 • 0.3 • 38) + (200 • 0.04 • 38) = 551 BTU/h.

It many ways, it may be simpler to look at this heat loss from a UA perspective (ignoring the delta-T). To calculate the UA, we multiply the area of each component by the U-factor, and add up the various products. Here’s how the UA calculation would work for our example: The UA of the exterior walls is (25 • 0.3) + (175 • 0.04) = 14.5 BTU/hr.

What does this mean? It means that for every 1 F° difference across this wall assembly, there is 14.5 BTU/hr of heat transfer.

What about the floor and the ceiling?

If the bedroom has a floor or a ceiling with unconditioned space on the other side of the assembly, the same calculation method (the one explained for exterior walls) can be used.

For example, an 11 ft. by 14 ft. bedroom has an area of 154 square feet. If the R-value of the ceiling assembly is R-40 (U-0.025), the heat loss at an outdoor design temperature of 0°F is:
154 • 0.025 • 70 = 269.5 BTU/h.

Heat flowing through partitions

Since we’re considering a cold room adjacent to a warm room, we need to consider how much heat flows through the partitions to help heat up the cold room.

A typical partition consists of a 2x4 stud wall with 1/2-inch gypsum wallboard on each side. Most partitions have an R-value of R-2.5 (U-0.4) to R-3 (U-0.33).

Let’s look at a bedroom with R-2.5 partitions measuring 200 square feet. The UA is:
200 • 0.4 = 80 BTU/hr.
That means that for every 1 F° difference across this wall assembly, there is 80 BTU/hr of heat transfer. If the warm room is at 72°F, and the cold room is at 68°F, the delta-T is 4 F°, and heat transfer rate through the partitions will be 4 • 80 = 320 BTU/h.

If this bedroom is on the second floor, and there is a conditioned floor below, we can calculate the heat transfer rate through the floor assembly.

If the floor area is 154 square feet, and the R-value of the floor assembly is R-3 (U-0.33), the UA is:
154 • 0.33 = 51 BTU/h.

If the delta-T between the downstairs room and the upstairs room is 4 F°, the heat transfer rate through the floor assembly is:
4 • 51 = 204 BTU/h.

What about exfiltration?

A rough calculation of exfiltration heat loss can be made using the old I=B=R method. (For more on this method, see How to Perform a Heat-Loss Calculation — Part 2.) Unless your house is extremely leaky, use the method for so-called “tight” homes.

Using this method:

  • A room with one exterior wall is assumed to have an infiltration factor of 0.012, corresponding to an ACH(nat) rate of 0.66.
  • A room with two exterior walls is assumed to have an infiltration factor of 0.018, corresponding to an ACH(nat) rate of 1.0.
  • A room with three exterior walls is assumed to have an infiltration factor of 0.027, corresponding to an ACH(nat) rate of 1.5.
  • The rate of heat loss at the design temperature is calculated using the formula:
    Exfiltration heat loss = Room volume • ΔT • ACH(nat) • Infiltration factor

    So our 11 ft. by 14 ft. bedroom would have exfiltration heat loss at the design temperature of:
    1232 • 70 • 1.0 • 0.012 = 1,034 BTU/h.

    Note that the I=B=R method overestimates air leakage for a home constructed with attention to airtightness.

    Heat generated by people and electrical appliances

    If a room is occupied, the person or people in the room generate heat. So does a light bulb, a TV set, or a computer.

    A sleeping person emits about 250 BTU/h.

    If you know the watt rating of an electrical appliance, you can translate watts to BTU/h using this formula:
    1 watt = 3.413 Btu/h

    A computer and a light might draw 45 watts — equivalent to a heat output of 153 BTU/h.

    Putting it all together

    OK, now let’s see what happens in our bedroom when one person is sleeping. We’ll assume there is no computer, TV, or lights; the outdoor temperature is 0°F (so the delta-T is 70 F°), and the adjacent warm room is at 72°F. It’s a single-story house, so there is no heat rising through the floor assembly into the bedroom. We’ll assume that the space under the bedroom is a crawl space, and that the floor assembly includes R-19 insulation. We’ll assume the space above the ceiling is an attic, with R-38 insulation on the attic floor.

    • Heat leaving the room through the exterior walls and windows is 1,015 BTU/hr.
    • Heat leaving the room through the ceiling is 154 • 0.026 • 70 = 280 BTU/hr.
    • Heat leaving the room through the floor is 154 • 0.052 • 70 = 561 BTU/hr.
    • Heat loss from exfiltrating air is 1,034 BTU/hr.
    • Total heat loss is 1015 + 280 + 561 + 1034=2,890 BTU/hr.

    The amount of heat entering the room through the partitions depends on the delta-T between the cold room and the warm room; let’s assume the delta-T is 4 F°. So the heat entering the room through the partitions is 320 BTU/h.

    The heat generated by the sleeping person is 250 BTU/h. We can add that to the 320 BTU/h to calculate the total heat gainIncrease in the amount of heat in a space, including heat transferred from outside (in the form of solar radiation) and heat generated within by people, lights, mechanical systems, and other sources. See heat loss.: 320+250= 570 BTUH/h. So the room is losing heat at a faster rate than it is gaining heat. The net heat loss is 2,570 BTU/h.

    Another calculation method (using UA)

    Earlier in this blog, I mentioned UA calculations. For the bedroom used in our example, the UA of the exterior walls and windows is (25 • 0.3) + (175 • 0.04) = 14.5 BTU/hr.

    That is, for every 1 F° difference across this wall assembly, there is 14.5 BTU/hr of heat transfer.

    So, let’s look at the UA of all of our components:

    • Exterior walls and windows: (25 • 0.3) + (175 • 0.04) = 14.5 BTU/hr
    • Ceiling: 154 • 0.026 = 4 BTU/hr
    • Floor: 154 • 0.052 = 8 BTU/hr
    • Exfiltration (based on 6 cfm): 6 • 1.08 = 6.5 BTU/hr

    So the “room UA to outdoors” is 14.5 + 4 + 8 + 6.5 = 33 BTU/hr.

    What about the heat gain? To keep things simple, we’ll ignore the body heat given off by occupants.

    • Partitions: 200 • 0.4 = 80 BTU/hr
      • So the “warm room UA to cold room” is 80 BTU/hr. That means that for for every 1 F° difference across this wall assembly, there is 80 BTU/hr of heat transfer. If we want the bedroom to be 68°F when the heated room is 72°F (in other words, when the delta-T is 4 F°), the available heat transfer is:
        80 • 4 = 320 BTU/hr.

        If we divide 320 BTU/hr by 33 BTU/hr, we find that (when the room is unoccupied) there is an energy balance when the indoor/outdoor temperature difference is 9.7 F°. With a desired indoor temperature of 68°F, that will occur when the outdoor temperature is 58.3°F. If the outdoor temperature is colder than 58°F, this bedroom (if unoccupied) will be colder than 68°F.

        How much heat can a bath fan move?

        The next question is: Can we supply the missing heat at design conditions, namely 2,570 BTU/h, with a bath fan? The amount of heat that a fan can move is calculated using this formula:
        BTUh = cfm of the fan • delta-T • 1.08

        Expressed another way:
        CFM of the fan = (BTUh) / (delta-T • 1.08)

        Plugging in the numbers, we get:
        CFM of the fan = 2570 / (4 • 1.08) = 595 cfm

        A typical bath fan might move between 50 and 100 cfm. Clearly, a bath fan won’t move enough air to warm up this bedroom at design conditions.

        The bottom line is that it’s really hard to raise the temperature of a 68°F room using 72°F air. It’s a lot easier to do that with a furnace, because the furnace has access to 150°F or 160°F air, as well as a powerful fan moving 500 to 800 cfm.

        In my example, I looked at a bedroom with a net heat loss of 2,570 BTU/h at the design temperature (a particularly cold night). If we change some of the factors — if we consider what happens when the outdoor temperature is warmer, or if we assume that exfiltration heat losses can be lowered by better air sealing measures, or if consider a house with thicker insulation or triple-glazed windows — we might approach a situation where a bath fan or two can make a difference.

        These calculations reinforce a message we repeat often at Pay attention to your thermal envelope, and you won’t need to pay as much attention to your HVAC(Heating, ventilation, and air conditioning). Collectively, the mechanical systems that heat, ventilate, and cool a building. system. Reducing your air leakage rate, increasing the R-value of your insulation, and investing in high-performance windows pay many dividends. One dividend is to reduce room-to-room temperature imbalances in homes with point-source heating.

        More on the possible contribution of a bathroom fan

        A GBA reader with programming skills, Nick Welch, wrote a program to calculate how fast a room cools off and to determine whether a bath fan can help reduce room-to-room temperature imbalances.

        Welch wrote, “Given a 10 x 10 x 10 room with two R-15 walls, an R-50 ceiling, an R-20 floor (facing unconditioned air below), and two R-1.5 indoor walls; indoor temperature of 72°F and outdoor temperature of of 25°F: The room cools off logarithmically, approaching about 65.8°F and basically reaching that within 20 to 25 minutes.

        “Add a transfer fan that blows in 50 cfm of 72°F air, and the room still cools off with a similar curve, but it levels off at about 67.4°F, and it takes a few less minutes to get to the leveling off point. Make the fan 100 cfm, and it levels off at 68.3°F.

        “I experimented with other R-values, other cfm rates, and other temperatures, and basically the theme is the same: the fan does help, but maybe not enough to be worth the hassle. The trend seems to be that you need around 100 cfm to cut the temperature differential roughly in half, and doing much better than that gets into ridiculous cfm numbers — the cfm-to-temperature-differential doesn’t scale linearly.

        “Here is the code if anyone wants to scrutinize it:

        What about an open door?

        I’ll be frank: I don’t know how much air typically moves through an open door. At least three GBA readers have posted opinions on the issue, though.

        • A GBA reader Jon R. claims that “A 150 cfm fan will move roughly half the heat of an open door [in other words, an open door equals a 300 cfm fan].”
        • According to GBA reader Kevin Dickson, “An open door can swap over 100 cfm, and it doesn’t even need a thermostat. The heat knows where to go and how fast to go.”
        • Energy consultant Marc Rosenbaum wrote in an article published on GBA, “I tend to think of an open door as being equal to a decent bath fan — it’s worth 50-100 cfm.” He also wrote, “It’s different depending on where the door is. If it is up a floor, the open door has more air flow through it than if it is on the same level — the difference in height between the heated room and the door opening helps that movement of air and energy.”

        My gut tells me to go with Marc Rosenbaum’s assessment.

        Summertime cooling

        Everyone agrees that room-to-room temperature imbalances are often worse in summer than in winter. In other words, a strategy that works for heating may not work for cooling.

        The main reason is that cooling loads are often dictated by solar gain through windows. A room with a window that gets direct sunlight can have a significantly higher cooling load than an adjacent room without such a window.

        Marc Rosenbaum wrote, “Cooling is harder [than heating]. A cooling source on the main floor won’t cool the upstairs.”

        David Butler, a building systems engineer, posted the following thoughts in a comment on GBA: “You can sometimes get away with indirect heat distribution in cold (heat-only) climates because folks are more tolerant or even prefer lower temperatures in bedrooms. But when you have a cooling load, especially in bedrooms, you’ve got to deliver supply air to where the load is. A ducted minisplit would seem to be the best solution...”

        Butler goes on to provide helpful hints on ducted minisplit system design: “With metal ducts, length isn’t much of an issue, which is why it’s misleading to say a ducted mini can only handle x feet of duct. It’s the elbows and other fittings that disrupt airflow. The Fujitsu ducted minisplit can handle a whole house with a carefully thought-out layout…. The Fujitsu can handle 0.35 IWC external static pressure — more than enough for a well-designed metal duct system. For a hallway drop ceiling scenario, my preference is a fabricated ‘extended plenum’ reduction trunk with curved branch take-offs, where the branch duct serves as the boot for over-the-door supply registers. Thus no elbows. On the return side, in lieu of factory filter, I specify an oversized filter box at the butt end of the blower with filter sized to about 175 cfm/ft2.

        “I find the biggest challenge is meeting the required left-side service access clearance. Most halls aren’t wide enough for unit width plus side clearance. However, if there’s an adjacent laundry or closet on the left side of the hall, you can shift the unit all the way to that side and put a drop ceiling and access panel on the immediate other side of the wall.”

        Strategies to address room-to-room temperature imbalances

        I’ll summarize what we know about bedrooms in homes with point-source heating:

        • If you’re building a new house with point-source heating, an excellent thermal envelope — one with a low rate of air leakage, insulation that exceeds minimum code requirements, and high-performance windows — will help provide comfort in bedrooms without any heat source.
        • Unless you like your bedroom to be several degrees colder than the rest of the house, you should leave your bedroom door open during the day to help equalize temperatures.
        • Many owners of new homes built with attention to airtightness are satisfied with the comfort of their bedrooms even when the bedroom lacks a source of heat. To be satisfied, it's best if you don't mind the fact that your bedroom will occasionally be 5 F° colder than the living room, and it's best if you don't mind leaving your bedroom open during the day. It's also best if you live in a climate that doesn't require a lot of air conditioning.
        • If you prefer a warm bedroom during the winter, you may want to install an electric-resistance heater in your bedroom for cold nights.
        • A bathroom fan is unlikely to equalize temperatures between adjacent rooms, although you can always do the math to look at your particular case.
        • A bathroom fan costs money to install, requires electricity to operate, and is unlikely to satisfy performance expectations. In most cases, it makes more sense to take the money you would have spent on a bathroom fan and invest the money in one of two possible solutions: (1) installing an electric resistance heater, or (2) upgrading from a ductless minisplit in the living room to a ducted minisplit that serves multiple rooms.
        • When it comes to cooling, the two most important variables are (1) whether the bedroom has windows that get direct sunlight in hot weather, and (2) the quality of the home’s thermal envelope. If the bedroom is subject to significant solar gain, it will probably need some type of air conditioning.

        Marc Rosenbaum provides some rules of thumb: “My guideline is that if people will tolerate 4 F° lower than the heated space (which in my mind means 72°F heated space, 68°F bedroom) and they leave the doors mostly open, then a point-source heater is viable when the heat loss is 1,000 BTU/hour and the room is occupied; 1,500 BTU/hr is kind of my soft cut-off for considering it. Beyond that, I’ll provide some electric resistance backup in those rooms.”

        In a comment posted on GBA, certified Passive HouseA residential building construction standard requiring very low levels of air leakage, very high levels of insulation, and windows with a very low U-factor. Developed in the early 1990s by Bo Adamson and Wolfgang Feist, the standard is now promoted by the Passivhaus Institut in Darmstadt, Germany. To meet the standard, a home must have an infiltration rate no greater than 0.60 AC/H @ 50 pascals, a maximum annual heating energy use of 15 kWh per square meter (4,755 Btu per square foot), a maximum annual cooling energy use of 15 kWh per square meter (1.39 kWh per square foot), and maximum source energy use for all purposes of 120 kWh per square meter (11.1 kWh per square foot). The standard recommends, but does not require, a maximum design heating load of 10 W per square meter and windows with a maximum U-factor of 0.14. The Passivhaus standard was developed for buildings in central and northern Europe; efforts are underway to clarify the best techniques to achieve the standard for buildings in hot climates. Consultant John Semmelhack notes, “I’m always a little baffled by the transfer fan concept, since it adds back about half of the cost difference between a ductless and ducted minisplit, while solving much less than half of the problem.”

        Peter Yost echoes Semmelhack’s conclusion: “By the time you purchase two quality fans that are quiet enough, energy-efficient enough, and easily adjusted, you might as well go with a dropped-soffit, ducted hallway solution.” In other words, just install a ducted minisplit.

        Martin Holladay’s previous blog: “Things You Do Not Need.”

        Click here to follow Martin Holladay on Twitter.

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  1. Carl Seville

Feb 16, 2018 5:29 PM ET

Edited Feb 16, 2018 5:33 PM ET.

So, why does leaving the door open work?
by Stephen G

I really wouldn't expect leaving the door open to have higher air / heat movement than a dedicated fan, but this does seem to make a significant difference. Why?

Nevermind. I see discussion above, but it seems crazy to me that an open door pushes 300 CFM...

Feb 16, 2018 6:02 PM ET

Edited Feb 16, 2018 7:31 PM ET.

open doors and infiltration
by Jon R

Nice article.

Here is a source for open door flow calculation. I get 322 CFM at 5F.

A 20" box fan assisting air movement through an open doorway can efficiently move 1000+ CFM.

If a ducted fan blows into a closed door room and slightly pressurizes it (say 3 pascals), in most conditions, there is no heat loss due to outdoor air in-filtration. I assume the values used reflect fairly windy design conditions.

Be careful with "supplemental" heat - if the bedroom is kept 1 degree warmer than the rest of the house, it will always supply more than 100% of the bedroom's heat.

Using Nick Welch's nice program and simply changing to 0F outside, it yields 62F, 66F and 69F at 0, 100 and 300 CFM.

Feb 17, 2018 7:08 AM ET

Response to Stephen G
by Martin Holladay

Q. "Why does leaving the door open work?"

A. On the molecular level, the molecules in the air -- mostly nitrogen, but also oxygen and other molecules -- are always bouncing around, as long as they are warmer than absolute zero. The molecules near the open doorway are capable of bouncing through the open doorway. This is the mechanism that allows the molecules rising from a smelly pair of tennis shoes to reach your nose 10 feet away.

Air also moves by convection. Air tends to stratify in a still room, with warmer air near the ceiling and cooler air near the floor. (Small air leaks in the ceiling and floor in winter accentuate this stratification.) If the hallway is 4 F° warmer than the bedroom, then strata at different temperatures are encountering each other at the open doorway. At any given stratum, the warmer hallway air will tend to move upward diagonally through the doorway to the cooler bedroom. As air moves through the doorway into the bedroom, the cool air on the bedroom floor moves horizontally through the open doorway into the hallway, setting up a convection current.

Feb 17, 2018 8:32 AM ET

Edited Feb 17, 2018 8:38 AM ET.

Response to Jon R
by Martin Holladay

Jon R,
You wrote, "If a ducted fan blows into a closed door room and slightly pressurizes it (say 3 pascals), in most conditions, there is no heat loss due to outdoor air infiltration."

But you can't have infiltration without the same amount of exfiltration. Pressurizing a room increases exfiltration; the makeup air would be the air pushed into the room by the pressurizing fan, of course. Is that good or bad? You decide.

Clearly, if the pressurizing fan is pushing indoor air from the hallway or living room into a bedroom, then outdoor air is entering cracks in the thermal envelope -- perhaps cracks in the hallway or living room -- to replace the exfiltrating air leaving the bedroom due to pressurization.

Feb 17, 2018 8:33 AM ET

Another strategy (maybe)
by Reid Baldwin

I was thinking about the issue of heating bedrooms with closed doors and it occurred to me that a Minotair, CERV, or other heat-pump based ventilator would have sufficient capacity to handle these rooms in many houses. Of course, there is more to the issue than having enough capacity. I haven't figured out exactly how it would work, but I will share my half-baked ideas here in case others are inclined to help finish baking them.

Obviously, the fresh air supplies would need to be in these bedrooms. The Minotair controls sense temperature of the stale air (I don't know about the CERV). If the stale air returns are in bathrooms, the unit wouldn't call for heat until the bathrooms got cold and then wouldn't stop producing heat when the bedrooms get to the setpoint. To make sure the unit provides heating and cooling at the right times, you either need to put the stale air returns in the bedrooms or modify the unit to use a remote temperature sensor. If the returns and supplies are both in the bedrooms, does the rest of the house get ventilated properly? Is it practical to modify the unit to use a remote temperature sensor?

Feb 17, 2018 8:46 AM ET

Response to Reid Baldwin
by Martin Holladay

If the living room is heated by a ductless minisplit, you are basically suggesting that the homeowner could purchase and install an additional heat pump -- in this case, a heat pump manufactured by CERV or Minotair -- to heat the cooler bedroom or bedrooms.

That approach would work, of course, with the right ducts, controls, and thermostat settings. But your suggestion still means that the house needs a second heat pump to handle the heating load of the bedrooms. I fail to see why this suggestion is better than simply installing a ducted minisplit to handle both the living room and the bedrooms.

Feb 17, 2018 10:53 AM ET

"If a ducted fan blows into a
by Jon R

"If a ducted fan blows into a closed door room and slightly pressurizes it (say 3 pascals), in most conditions, there is no heat loss due to outdoor air infiltration."

Martin is correct that this would be more precisely worded as " closed door room heat loss...". There is makeup infiltration elsewhere, but it isn't relevant to the subject of this article.

A non-thermostatically-zoned ducted system isn't going to regulate well as loads change non-proportionally (eg, solar gain in one room and not the other).

Feb 18, 2018 9:17 AM ET

Response to Martin
by Reid Baldwin

I labelled this a half-baked idea because I hadn't yet convinced myself it is a good idea nor convinced myself it is a bad idea. I agree with you that when the bedrooms are close to the family room, a single ducted mini-split would be preferable. For the original article, close proximity is a fair assumption.

What about a situation where the bedrooms are on a second floor? In that case, the ducted mini-split for the bedrooms would also be an incremental unit and probably more costly than the incremental cost of a heat-pump ventilator over a conventional ERV. (I am presuming that a fully ducted, balanced ventilation system is already in the budget.) In that case, would having most of the fresh air supplies in the bedrooms make the ventilation system bad at its primary job?

Feb 18, 2018 11:36 AM ET

The advantage of a HRV
by Jon R

The advantage of a HRV connected heat pump is avoiding additional ducts. But I doubt they are designed to be able to independently control the amount of fresh air and heat going to a specific room. For example, someone closing the register to create a cool bedroom would get no ventilation air.

Feb 18, 2018 12:21 PM ET

Response to Reid Baldwin
by Martin Holladay

I'm siding with Jon on this one. The Minotair and CERV controls basically aim to achieve good ventilation performance -- the heat recovery function (via the air source heat pump) is a benefit, but these appliances aren't controlled by a thermostat, and they aren't intended to provide space heating as a primary function.

Feb 18, 2018 2:25 PM ET

Magic Box
by Andy Kosick

First, Bravo for answering a controversial question with a lesson in room by room load calculation.

Second, while recognizing that the Minotair and CERV are designed to provide ventilation, I've spent a lot of time thinking about these things lately. When I consider the fact that my local Habitat affiliate is building a house that only requires about 400cfm or less to heat and cool, I hope somebody somewhere is trying to design a single box with a single set of ducts that can heat, cool, and ventilate a high performance home. It would seem a shame if someone is not, because it's entirely possible, and, I think, essential to scaling affordable high performing housing.

Feb 18, 2018 3:13 PM ET

I agree - it is possible to
by Jon R

I agree - it is possible to build something that would produce a tailored mix of heat/cool, ventilation and humidity for each room and deliver it via a dedicated duct. It would provide more comfort than anything being discussed.

A Chiltrix heat pump with a fan coil in each room + HRV ducts is the closest thing I can think of. And it isn't crazy expensive.

Feb 19, 2018 8:23 AM ET

Response to Andy Kosick (Comment #11)
by Martin Holladay

It's possible to design a "magic box" that provides space heating, cooling, ventilation, and even domestic hot water. Several companies have worked on developing such a device, and a few have even hit the market, in Canada, Europe, and the U.S., over the past 20 years.

It's a little like selling a TV with a built-in video tape player, DVD player, and intercom. You can do it. But when one of the items in the device breaks, you either have to take it in to the repair shop, or throw the whole thing away.

Sometimes separate devices make more sense.

Feb 21, 2018 4:04 PM ET

Great Information Here
by Kevin Camfield

It seems to me that this may also speak to the importance of getting the ducting right and using a room by room heat loss analysis for ducted heat pump systems. We are looking at using ducted Mitsubishi setup in our house using a multi-position and horizontal ducted indoor units. I don't have a lot of experience with what came before, but it appears that the temperature and the volume of the heated air is much less with these systems than say with a ducted gas furnace. It seems like that then may require more precision with getting the right air flow to each room when using a ducted multi-split system with a single thermostat for multiple rooms.

Feb 21, 2018 4:21 PM ET

Response to Kevin Camfield
by Martin Holladay

If you are designing a heating system -- regardless of the type of heating system -- the first step is to perform a room-by-room heat loss calculation.

Whether the rooms have ducts connected to a furnace, or ducts connected to a ductless minisplit, or baseboard radiation connect by tubing to a boiler, you can't design the system without an accurate understanding of each room's design heat load.

If you do a bad job with the room-by-room heat load calculations, occupants can be uncomfortable -- whether the heat comes from a furnace, boiler, or heat pump.

Feb 22, 2018 9:41 PM ET

Response to Martin
by Kevin Camfield


Thanks. I'm working through that now with a local energy consultant. I haven't found an HVAC company that willing to go to that detail. I agree with you that without the analysis one is just guessing.

Feb 23, 2018 6:25 AM ET

Response to Kevin Camfield
by Martin Holladay

Your report -- "I haven't found an HVAC company that willing to go to that detail" -- is common but infuriating. Building codes require that a room-by-room Manual J calculation be performed before installing a heating system, but the provision is widely ignored and rarely enforced.

For more information, see Saving Energy With Manual J and Manual D.

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