Dear All, I have a scientific question which I hope you help me through: There is a class room (Volume = 210m3), which had Co2 value of 420 ppm and temperature 17°C when nobody existed in it and all windows were closed, after 30 minutes there was some students in the room, where co2 value reached 1200 ppm and temperature increased 1°C while windows are still closed, There is an equation in the german code which applies the variables: volume flow rate, co2 and number of people. I have tried to apply this equation but it doesn't work if volume flow rate or number of occupants = 0. What I know is that a student generates heat = 70 W. how do you think such a problem could be solved ? Best regards
The formula is probably equivalent to:
Ep x n = Ve [1000 ppm - Co]
where
Ep=per capita CO2 emission
n = number of people inside
Ve = air exchange rate with the outside (m3/s)
Co(t) = CO2 concentration of outdoor air at time "t"
This formula is a static, equilibrium one, and does not consider the transient from empty to occupied room.
If there is no CO2 production in the room, there is no need for ventilation.
If there is no ventilation, no one should occupy the room.
What you need to solve this problem is a material balance of the form: accumulation+in-out=0. You can find examples of such balances in bachelor-level text books on transport phenomena & heat and mass transfer.
Please find attached a short chapter in which you will find the basic equations. Assuming that sitting students produce about 18 liter CO2 per hour, and knowing the CO2 concentration versus time and infiltration rate, you may use equation 1 to find the number of students versus time.
The same equation 1 can be used for heat, replacing the concentration by temperature, air density by its heat capacity (Wh/m³K) and contaminant source by heat generation (70 W * number students) with air flow rate in m³/h.
1. Initial CO2 levels at 420 ppm, close to atmospheric levels. Implies indoors was at equilibrium with atmospheric CO2 levels. This does not add extra information.
2. Using heat balance to get the answer is quite difficult since students are not the only heat source in that room
3. Assuming 1200 ppm is what you get at equilibrium (that is, the CO2 value plateaus out), outdoors is 420 ppm, and students are engaged in typical class work (near sedentary) gives their CO2 emission rate of about 0.0052 L per second (for every student)
4. This will give the ventilation per student is ~6.7L/s [= 0.0052/(1200-420)] - remember 1200 and 420 are in parts per million, so a factor of 10e6
5. If you had a count of students this time, you can estimate total outdoor air entering into the room.
6. Next time you have students in the classroom, you can estimate from the measured CO2 value and the outdoor air ventilation rte you estimated in step 5 the number of students there are.
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