Answer to Question #109542 in Mechanical Engineering 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

Person breathes out at the rate 0.3 L/min = $\frac{0.3 L}{60s}= 0.005 L/s$ for 6 person $CO_2$ expels

out = $6 \times 0.005=0.030 \frac{L}{s}$

The concentration of CO₂ is 400 ppm and it has to maintain upto 1000 ppm.

AT STP volume of CO2 is 22.4 L which is equal to 1 mole = 44 g

so mass of CO₂ expells out = $0.03\times 44= 1.32 g$ /s

400ppm means 400 g of CO2 present in 1000000 particles of air

concentration of incoming air = $400 \times \frac{44}{22.4} = 98.3 \frac{g}{L}$

six person breaths out = $\frac{0.3}{60}\times6\times \frac{44}{22.4}=0.059 g/s$

s per the standards

air flows= $\frac{35.3+98.3Q}{C_o}$