Answer to Question #109552 in Mechanics | Relativity for Moneeb

Question

Model equation for contaminant concentration in a space and apply to the following. A person breathes out CO2 at the rate of 0.3L/min. The concentration of CO2 in the incoming ventilation air is 400 ppm. It is desired to hold the concentration in the room below 1000ppm. Assuming that the air in the room is perfectly mixed, what is the minimum rate of air flow is required to maintain the desired level as per ASHRAE standard 62-1999?. If the same room is designed to occupy 6 persons determine the total amount of CO2 per second breath out and amount of incoming air to meet the ASHRAE standard 62-1999 desired comfortable ambient.

Answer 1

Ppm stands for parts per million by volume. We will express ppm in units of mass per unit volume.

Transform the words into equations and convert all time units to hours.

(1) A person breathes out CO₂ at the rate of 0.3 L/min. After time t, the volume of CO₂ will be

$$V_P(t)=0.3t\cdot60=18t\text{ L}.$$

This volume will correspond to a mass of the carbon dioxide of

$$m_P(t)=\Mu_{\text{CO}_2}\frac{V_P}{V_\Mu}=44\frac{18t}{22.4}=\\ \space\\=35.4t\text{ g}.$$

It means that each hour a person breathes out 35.4 grams of carbon dioxide.

(2) The concentration of CO₂ in the incoming ventilation air is 400 ppm. Convert ppm to g/L:

$$C_I=ppm\cdot\frac{\Mu_{\text{CO}_2}}{V_\Mu}=400\frac{44}{22.4}=786\text{ g/L}.$$

It means that each liter of incoming air contains 786 g of the oxide per 8 hours. Per one our it is

$$C_I=98.3\text{ g/L}.$$

How fast the incoming air is blown into the room? Assume the rate is Q liters per hour. Therefore, the mass of CO₂ in the room air will increase each second at the rate of

$$m_I(t)=98.3Qt\text{ g}.$$

(3) Assume that the output air flow G takes $C_O$ grams of CO₂ a second:

$$m_O(t)=C_OGt.$$

After time t, the mass of the oxide in the room must comply with the following equation:

$$m_P(t)+m_I(t)=m_O(t),\\ \space\\ 35.4t+98.3Qt=C_OGt\\ \space\\ C_OG>35.4+98.3Q.$$

Six persons breath out

$$m=0.3/60\cdot6\cdot\frac{44}{22.4}=0.059\text{ g/s}.$$

The air flow:

$$G=\frac{35.4+98.3Q}{C_O}.$$