# Humidity, Air Intake, and Heating Requirements

Could someone explain or point me to the math for calculating the kind of heating and equipment necessary to maintain <x ppm CO2 in a roughly passive house (e.g., 0.5 ACH) with y occupants in, e.g., zone 6, while maintaining at least 40% indoor RH at 70 F in winter?

From a public/clinical health perspective, it’s important to maintain low CO2 and moderate humidity. This offsets some of the energy savings that an airtight house would otherwise enjoy.

The relationship between ambient ppm, ACH, and the number of occupants (exhaling ~50k ppm in their breath) is relatively straightforward, but I’m unclear on how to factor in the demands for humidity, and also on how such high air intake rates might increase heating needs.

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## Replies

Tight houses need to be ventilated.

The 2018 IRC requires continuous mechanical ventilation, the amount depends on the number of bedrooms and the square footage of the house. It ranges from 30 CFM for a house that is under 1500 square feet with one bedroom, to 165 CFM for a house with 7 or more bedrooms and over 7,500 square feet. The whole thing is in table M1505.4.3(2).

There's a lot of criticism of those numbers and they're pretty arbitrary. You could probably come up with a better number for your house analytically. High-end systems use C02 sensors and humidity sensors to control ventilation.

We've had some debates here about whether it's more efficient to ventilate or run a dehumidifier if you need to remove humidity from your house in the winter. Factors include how cold it is outside, how humid it is outside and inside, whether you have a heat recovery ventilator, how efficient your dehumidifier is and what your heating system is and how efficient it is. It's not a simple question.

Here's an article with a formula for the ventilation rate needed for a given CO2 level:

https://www.abe.iastate.edu/extension-and-outreach/indoor-air-quality-carbon-monoxide-and-carbon-dioxide-aen-125/

That of course varies greatly with the individuals' metabolism and their activity level.

If you want to provide that ventilation with a minimum of energy expended for heating the ventilation air, and you want to maintain higher humidity inside that you'd get with simple ventilation strategies, the best way to ventilate is with an energy-recovery ventilator. Different models will have different performance, and the performance will also depend on the air flow rate. Here's a spec sheet for a high performance unit that has between 75% and 89% heat recovery, and 52 to 82% humidity recovery. (see p. 10)

https://www.zehnderamerica.com/wp-content/uploads/2021/03/Zehnder_ComfoAir_Q450_Specification_2021.03.18.pdf

Given the flow and the temperature difference inside to out, you can calculate the heat energy needed without heat recovery and then multiply by (1 - heat recovery efficiency). The handy formula for that is Q = 1.08 x CFM x ΔT , where ΔT is the inside-to-out temperature difference and Q is the heat load in BTU/h. So with heat recovery, it's Q = 1.08 x CFM x ΔT x (1 - heat recovery efficiency).

Humidity is much harder to predict, because of the many moisture sources in the house. You might be maintain 40% with just he ERV and no active humidification.

Here's an analytic take on ventilation requirements:

Atmospheric air is about 420 PPM of CO2. Indoor air starts to feel stuffy at about 1000 PPM. So replacing indoor air with outdoor nets you 580 PPM. A typical human produces about 3.2 lbs of CO2 per day. To transport 3.2 lbs of CO2 at 580 PPM requires about 5,500 lbs of air. Air at standard conditions is about 0.0765 lb/cu ft so 5,500 lbs of air is about 72,000 cubic feet. That's per day, or 3,000 cubic feet per hour or 50 cfm.

So 50 cfm per occupant. Including infiltration.

Continuing the analysis, how much dehumidification do you need for occupant activity? A quick google tells me that you should assume 185 BTU/hr per occupant for latent heat removal. Latent heat of water is 970 BTU/lb so that's 0.19 pounds of water per hour that needs to be removed, or 4.5 pounds per day. How much ventilation it takes to do that depends on the moisture content of the replacement air.

There are 7000 grains in a pound so we need to remove 31,500 grains of water per day. With the same 5500 lbs of air ventilation that we need for CO2 removal, that means we need to remove 5.72 grains per pound of air. Interior air at 72F and 50% RH has 59 grains of water per pound of air, so we need to replace it with air that has about 53 grains per pound.

That could be 72F air at 45% RH

80F air at 34% RH

60F air at 68% RH

50F air at 100% RH

In most circumstances the ventilation that you need for CO2 removal will also take care of humidity. If the replacement air is any temperature below 50F you will be removing too much moisture.

If you have an energy recovery ventilator you have to adjust for its efficiency.

Final part: how much energy does all this ventilation require?

BTU/hr= 1.08* CFM * temperature difference

So if inside is 72F and outside is 20F, that 50 CFM takes (1.08)*(50)*(52) =2808 BTU/hr.

If you have a heat recovery ventilator or energy recovery ventilator you can adjust for its efficiency.

Thanks very much. I meant to frame my question in the context of loads assumed for ERVs. I've been wondering whether additional humidification might be required in a very dry climate in winter with an ERV, especially given high air exchange rates and ~70% humidity recovery, and if this all amounts to a much more energy-intensive system than people traditionally assume (higher exchange rates = more heating of the air, and additional humidification = more energy). Have also been wondering about "surge" capacity if, e.g., a dozen people are over for four hours, or if the only practical solution there is to open windows. I will attempt to work through some of the numbers this weekend. It's helpful to know how others approach this. I'm not aware of what kinds of integrated systems exist to balance humidity and CO2 but will look into that also.

I saw in 2019 Martin wrote "A Passive House shouldn't need a humidifier. By reducing random air leaks through the building envelope, you should have adequate indoor humidity levels during the winter." I need to convince myself of the math on that. This is an intermittently occupied house at 8600', and so it will frequently be starting from low humidity (and obviously low CO2). I should figure out how many hours to reach ~40% RH.

Intermittent occupation makes it more complicated. The moisture buffering of the building materials and furnishings will make it hard to bring up to 40% humidity quickly. Unless you can drastically drop the ventilation rate and temperature when it's not occupied, so that it stays close to 40% humidity.

Check out https://www.healthyheating.com/.