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Dilution Ventilation Calc - Sanity Check

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Engineer6512

Mechanical
Nov 4, 2002
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I'm calculating required outdoor air for a vehicle test facility where the vehicles will be running continuously in the space. Here are my calcs - please feel free to comment on anything I might be missing.

CO is the dominant contaminant. Allowable max concentration is 29 mg/m3.

CO generation rate: 11.67 g/min = 11670 mg/min
4 vehicles running: 4 * 11670 = 46680 mg/min
(source ASHRAE applications, chp. 13, table 6 avg of winter emmissions)

Molecular weight, CO: 28
Moles CO per mg: = 0.001/28 = 3.57E-05 mol/mg
T = 298 K
P = 1 atm
R = 0.08206 L*atm/(K*mol)
Volume CO generated per min: = 46680 * (3.57E-05 * 0.08206 * 298 / 1) * (0.001 L/m3) = 0.040768 m3/min

K = 5 (imperfect mixing)
Dilution Air: V = (G/C)*K = (0.040768 m3/min / 29E-06 ppm) * 5 = 7028 m3/min ~ 250000 CFM

Did I miss something here? Or is this really the flow rate I need?

Thanks!
 
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You are assuming constant tailpipe CO emissions and perfect tailpipe / room air mixing. CO will be a function of engine size, rpm and load. You will buildup pockets of CO then it will mix in with room air. I believe should also look at air volume changes per hours and cfm/ft2 and these will be a function of room size. The IMC code of 1 ½ CFM/ft2 may be very short for a very small space.
 
Hi walkes, yes its an indoor arena, but size of the space does not matter with a dilution calc... only the rate of generation and the allowable concentration, per ACGIH.

 
My main issue that I am not sure about is if I am handling the units and the conversion of CO from mg/min to a volume of gas correctly. I am using ACGIH dilution calcs as a reference, but the only examples they give are for evaporation of a liquid in pints per hour and the associated constants for that system of units. ASHRAE 62.1-2004 has basically the same equations in an appendix but again does not provide any examples.
 
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