The time required for the saturation of the 210 kg of adsorb

The time required for the saturation of the 210 kg of adsorbent is 4.5 hours. The retentivity fraction is 0.55 and the efficiency of the adsorbent is 0.85. The average molecular weight of the adsorbed vapor is 0.310 kg/mole. The air flow rate is 300 cfm. Calculate the pollutant concentration in the air stream. 130 4600 27600 10000

Solution

Estimated time required for saturation;

           t_s = (2.41 * 10⁷ * R_t * W)/( * q * M * C)

          Where

          t_s   = time of adsorbent service to saturation (hr) = 4.5 hr

          R_t = retentivity (fraction) = 0.55

         W = weight of adsorbent (kg) = 210kg

         = adsorption efficiency (fraction) = 0.85

         M = average molecular weight of adsorbed vapor (g/mole) = 0.310 kg/mole = 310 g/mole

         q = air flow rate (m³/hr) = 300 cfm = 510 m³/hr

         C = entering pollutant concentration (ppm)

          t_s = (2.41 * 10⁷ * R_t * W)/( * q * M * C)

C = (2.41 * 10⁷ * R_t * W)/( * q * M * t_s)

          C= (2.41*10^7*0.55*210) / (0.85* 509.70 * 310 * 4.5 )

         C = 4602.94 ppm