One kmol/s of a gas consisting of 75 mol% methane and 25% n-pentane at 300 K and

Question

One kmol/s of a gas consisting of 75 mol% methane and 25% n-pentane at 300 K and 1 atm is to be scrubbed with 2 kmol/s of a nonvolatile paraffin oil entering the absorber free of pentane at 308 K. Estimate the number of ideal trays for adiabatic absorption of 98.6% of the pentane. Neglect the solubility of methane in the oil, and assume operation to be at constant pressure. The pentane forms ideal solutions with the paraffin oil. The average molecular weight of the oil is 200 and the heat capacity is 1.884 kJ/kg-K. The heat capacity of methane over the range of temperatures to be encountered is 35.6 kJ/kmol-K; for liquid pentane, is 177.5 kJ/kmol-K; for pentane vapor, is 119.8 kJ/kmol-K. The latent heat of vaporization of npentane at 273 K is 27.82 MJ/kmol (Treybal, 1980).

Explanation / Answer

The number of ideal stages is given by the Kremser equation

N = (Xo - XN)/(XN - (YN+1/))

where Xo is the concentration of the input liquid stream = 0.25

XN is the concentration of the exit liquid stream = 0.25 - 0.986 * 0.25

= 0.0035

Applying mole balance over liquid and gas stream

Ls (Xo - XN) = Gs (YN+1 - YN)

1 * (0.25 - 0.0035) = 2 * (YN+1 - 0)

YN+1 = 0.123

YN+1 = *XN

0.123 = *0.0035

= 35.14

Therefore ideal number of stages, N =   (Xo - XN)/(XN - (YN+1/))

= (0.25 - 0.0035)/(0.0035 - 0.123/35.14)

= 0.2465/0.03

= 8.21 stages which is approximately 9 stages

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