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Ventilation standards and fresh-air requirement

Short answer

Using Q = G/(Ci - Co), generation is 0.052 L/s and concentration difference is 600 ppm, giving 0.0867 m³/s or 312 m³/h, which is 8.7 L/s per person. Room volume is 90 m³, so air changes per hour are 3.5.

(a) What is the purpose of ventilation standards? State two ways in which they are specified. (b) A classroom 6 m x 5 m x 3 m is occupied by 10 persons, each releasing 0.0052 L/s of CO2. The outdoor CO2 level is 400 ppm and the indoor level is to be kept at 1000 ppm. Calculate the fresh-air flow required (in m3/h and L/s per person) and the air changes per hour.

(a) Ventilation standards

A ventilation standard (e.g. ASHRAE 62.1) specifies the minimum outdoor air that must be supplied to a space so that the concentration of contaminants stays at a level acceptable to occupants. It protects health and comfort and is the basis for designing HVAC systems.

Ways of specifying:

  1. Ventilation rate procedure: a prescribed airflow per person (L/s per person) plus a component per unit floor area (L/s per m2\text{m}^2) for building-related sources, depending on the use of the space.
  2. Air change rate: the number of room volumes of fresh air supplied per hour (ACH), common for hospitals, kitchens and laboratories.
  3. (Also) Indoor air quality procedure: limiting the concentration of selected contaminants (e.g. a CO2_2 level about 700 ppm above outdoors, or total VOCs).

(b) Numerical

Mass balance at steady state: pollutant generation = removal by ventilation.

G=Q (Ci−Co)⇒Q=GCi−CoG = Q\,(C_i - C_o) \Rightarrow Q = \frac{G}{C_i - C_o}

Generation: G=10×0.0052=0.052G = 10 \times 0.0052 = 0.052 L/s =5.2×10−5 m3= 5.2\times10^{-5}\ \text{m}^3/s.

Concentration difference: Ci−Co=(1000−400)×10−6=6×10−4C_i - C_o = (1000 - 400)\times10^{-6} = 6\times10^{-4}.

Q=5.2×10−56×10−4=0.0867 m3/s=86.7 L/s=312 m3/hQ = \frac{5.2\times10^{-5}}{6\times10^{-4}} = 0.0867\ \text{m}^3/\text{s} = 86.7\ \text{L/s} = 312\ \text{m}^3/\text{h}

Per person: 86.7/10=8.6786.7/10 = 8.67 L/s per person.

Room volume =6×5×3=90 m3= 6\times5\times3 = 90\ \text{m}^3:

ACH=31290=3.47 h−1\text{ACH} = \frac{312}{90} = 3.47\ \text{h}^{-1}

Answer: Fresh air required = 312 m3^3/h (86.7 L/s, i.e. 8.7 L/s per person); ACH = 3.5 per hour.

✓ Answer checked by Claude Opus 5.5 (high effort)

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