MDI Emissions Reporting Guidelines for the ... - Polyurethanes

K MDI = 0.69
C f = 15%
MDI Emissions Reporting Guidelines for the Polyurethanes Industry
359 = the molar volume of an ideal gas in ft 3 /lb-mole @ 0 o
C and 1-atmosphere
VP MDI = 0.08788 mm
C mdi = (0.08788/760) x 10 6 = 115.6 ppmv
R MDI = e^(-k * t R )
k = the first order reaction rate constant in min -1
t R = the reaction time in minutes. (Total residence time in oven)
The cord passes through 30 feet of oven at a temperature of 280 o
F and then another 30 feet where
the cords are cooled before going through a second vat solution. Though the reaction is complete
after 30 feet, a worst-case scenario is to assume completion is done after the cord exits the oven:
Solving for k:
t R = (30 ft x 2)/175 ft/hr = 0.34 hr = 21 min
R MDI = 0.001 (assuming complete reaction is 99.9%)
k = 0.329 min -1
If we substitute the k constant in the equation, we will find the free amount of MDI at the midpoint
of the reaction is 0.0316 or that the reaction is 96.84% complete.
Substituting the values into the equation the emissions per hour are:
L = (V air /359) * (273.15/T sp ) * 60 * (C mdi /1000000) * M W * k MDI * C f * R MDI
L = (8500/359)*(273.15/394.3)* (60)* (115.6/1000000)*(250.26)*(0.69)*(0.15)*(0.0316)
L = 0.093 lbs/hr
This represents the maximum emission rate at steady state conditions prior to the exhaust stream
being treated.
The entire concentration profile of the free MDI can be calculated and plotted assuming a first order
reaction mechanism. Plotting the un-reacted profile and taking the area under the curve, we find
that the effective unreacted fraction is 0.146 or 85.4% completion. Therefore:
L = the emissions in lb/hr
V air = 8500 ft 3 /min
T sp = 250 o
F = 394.3 K
M W = 250.26
K MDI = 0.69
C f = 15%
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