Question 10.1: Toluene is known to be a HAP in a contaminated gaseous emiss...

The following contaminated air emission stream was obtained from an industrial plant:

HAP                                                                                                 = toluene
Stream temperature, T_{e}                                                             = 90ºF
HAP emission stream concentration, HAP_{e}                       = 1000 ppmv

Because the adsorption isotherms for toluene are presented in Table 1, the default value for a working capacity of 0.100 is not needed in this example. Table 1 presents the isotherms for toluene at an adsorption temperature of 77ºF, which is lower than the stream temperature of 90ºF in this example. Assume that this difference in temperature does not significantly affect the calculation results. The working capacity value, W_{c}, is usually 50\% of the equilibrium capacity (W_{e}). Using Eq. (1) and values from Table 1, the W_{e} is calculated as follows:

Note: _{}^{a}\textrm{Each} isotherm is of the form: W_{e} = kP^{m}. (See text for definition of terms).
Data are for adsorption on Calgon-type “BPL” carbon (4 \times 10 mesh).
_{}^{b}\textrm{Equations }should not be extrapolated outside of these ranges

W_{e} = k  {(P_{partial})}^{m}                                                    (1)

where

P_{partial} = (HAP_{e}) (14.696 \times 10^{−6})                      (2)

P_{partial} = (HAP_{e}) (14.696 \times 10^{−6}) = 1,000 ppmv (14.696 \times 10^{−6}) = 0.0147 psia

From Table 1, k = 0.551, and m = 0.110. Then, substituting into Eq. (1) yields

W_{e} = (0.551) (0.0147  psia)^{0.110}

W_{e} = 0.346 lb toluene/lb carbon

Because the working capacity, W_{c}, is usually 50\% of equilibrium capacity (W_{e}),

W_{c} = 0.50  W_{e} = 0.50 (0.346 lb toluene / lb carbon) = 0.173 lb toluene/lb carbon

Because the adsorption isotherm for toluene was available (Table 1), the default value for working capacity of 0.100 is not needed in this example.