# Indoor Air Quality Procedure

This procedure limits measured contaminant concentrations.

Must sense specific contaminants.

Ventilation rate procedure predominates.

Pros: Cons:

• Prescriptive • May fail off-design with VAV

• Low capital cost • Energy cost

Perfect dilution (Steady State Ventilation Conditions)

“Well-Mixed Spaces”:

• Perfect mixing of supply/space

• Exhaust at mixed condition

• Constant contaminant generation rate

• Fixed supply flow rate and conditions

Mass balance for contaminants: Mass IN + Generation = Mass OUT

O

V — Supply air flow rate (ma /s)

* * 3 3

Cs — Contaminant concentration of the supplied air (mc /ma )

• # # Q

N — Contaminant sources generated within the space (mc /s)

Cr — Average contaminant concentration in the room (mc3/ma3), (ppm = 10’6xmc3/ma3)

 V, CS Room CR VV/ Ni-*

Example

In a French home, the CO2 concentration in a bedroom was 4000 ppm. The bedroom size is 12 m and room height 2.5 m. Find the air change rate if two occupants were in the bedroom. Suppose that the outdoor CO2 level was 300 ppm and a person breathes out 0.30 L/min. CO2.

Solution:

O

Air change rate — ACH (Air Change rate per Hour):

ACH = V / (Room volume) = 9.72 m3/hr / ( 12 m2 x 2.5 m ) = 0.324 ACH

Ventilation with Recalculation (Single Space)

If outdoor air is mixed with the return air, the supply contaminant concentration has to be determined form the contaminant balance for the mixing process.

Vr

 Room , RVr Cr ^ !s N Vo, Co Vs

Steady state: Vs = Vr, RVs = RVr, Vo = (1-R) Vs

Mass balance: V s Cr = VoCo + RVrCr+ N = (1-R) V s Co + RV s Cr + N

Space concentration:

For a single space: R=>0 minimum concentration

R=>1 concentration =>

System Serving Multiple Spaces

HVAC systems are usually serving multiple spaces that have different requirements for airflow rate of fresh air. The airflow rate requirements for the fresh (outdoor) air VD, are determined from Table 2 (Standard 62) that is presented as Table 4-5 in the textbook (M&P).

A total supply flow rate for each space Vs, is defined from heating/cooling load.

Fraction of outdoor air OA in supply is defined as:

 Vo, A Critical Space has outdoor air fraction “Z”:

 Vs,, Z

Z is the maximum fraction of OA required. Vo Vs

An average fraction of OA for all spaces “X” is: O

Supply OA fraction X Supply OA fraction Z

For over-ventilated space:

=> Some spaces under-ventilated

=> TNo space under-ventilated

-s Most spaces over-ventilated lEnergy wasted

• Airflow rate of fresh air is higher than required

• “Unused” fresh air returns to AHU

Define “corrected” OA fraction “Y” as:

V

Y = — c

V.

Y is OA fraction that accounts for “unused” fresh air.

Fresh air balance:

• Return and supply flows are equal

• Recirculated fraction is R

Critical space is satisfied if: O

Required OA fraction = Z (critical space)

“Unused” OA in return = (Z-X) Vs Recirculated fraction of return:

R = vs — voc = lY Vs

Recirculated OA = R(Z — X)s = (I — Y)(Z — X) Vs Critical space is satisfied if:

Voc+ri-Y)(Z-X)S=ZVS

Y + (1-Y)(Z-X) = Z

“Multiple Spaces Equation”:

O

The equation is used to calculate the corrected fraction of outdoor air that takes into account fraction of the recirculated fresh air. In some cases, the saving are significant if we compare Y and Z.

Example

Four spaces are air-conditioned from a central AHU. The following table gives airflow rates for supply air and fresh air:

 Spaces 1 2 3 4 Total Supply air [cfm] 500 400 600 500 2000 Fresh air [cfm] 200 80 80 75 435 Fresh/Supply 0.4 0.2 0.13 0.15 0.22

Calculate the required OA fraction?

Solution:

Average fraction of OA: O OA fraction for the critical space 1:0

Corrected fraction of OA: O

The required flow rate of OA is O

Notice: X < Y <Z

435 cfm < 540 cfm < 800 cfm

Standard 62-1999 recognizes transients, i. e. high occupancy/low use spaces such as conference rooms. If the peak occupancy is used to determine required flow rate of OA, calculated Y is unnecessary high.

To compensate, may:

• Ventilate based on average occupancy

• Lag ventilation start

• Control OA with CO2 concentration

Removal of Contaminants

Air Cleaner can take place in ducts or in spaces.

Gas and odor removal

Absorption (solid/liquid absorbers, air washer)

Chemisorption (by chemical reaction)

E. g.: UV+Ti02 photocatalytic oxidization of VOCs

Particulate removal by filtering

Particles:

Size and shape Specific gravity Concentration Electrical properties

Air cleaners Types:

Fibrous media unit filter Renewable media filter Electronic air cleaners Combination air cleaners

Filter Efficiency

O

Air-flow resistance (loss of total P at a given flow rate) — operating costs AP ~ V2

Filtration can reduce OA (outdoor air) requirement.

System with recirculation should use filters.

 Media Filtration HEPA Filters Activated Carbon Adsorption UV Photocatalytic Oxidation Odors X X VOCs X X Bio-Aerosols X X Dust X X

Example

GivenN, Co, Vo, VR Find Cr

 V o+V r. V o, Co Cm -»4 V 0, Cr V o+V r, Cr

Solution:

For the room (mass conservation):

 O (1)

 (2) For the filter:

O

 (3) For the mixing:

O

(2)+(3):

 C _ V0C0 + vrcr V0+VR
 (1-8) (4)

 (l)+(4): ^ VuC0(l-s) + N

 (5)

 V0+VRe