A gas scrubber is employed to remove SO_2 from a power plant exhaust stream. The

Question

A gas scrubber is employed to remove SO_2 from a power plant exhaust stream. The exhaust stream is fed to inlet of the scrubber with the following composition: Components enter and leave the scrubber at 1 atm and 340 K. Assuming that all components behave as ideal gases and all SO_2 is removed from the exit scrubber stream, calculate: The molar fractions of the components entering and exiting the scrubber. The minimum theoretical power input required by the scrubber, ignoring kinetic and potential energy effects.

Explanation / Answer

(a): For gas entering the scrubber:

Total moles of all gases entering the scrubber per second = 4.320 mol

Moles of CO2 entering the scrubber per second = 0.733 mol

Hence mole fraction of CO2 entering the scrubber = 0.733 mol / 4.320 mol = 0.1697

Moles of H2O entering the scrubber per second = 0.3 mol

Hence mole fraction of H2O entering the scrubber = 0.3 mol / 4.320 mol = 0.0694

Moles of N2 entering the scrubber per second = 3.284 mol

Hence mole fraction of N2 entering the scrubber = 3.284 mol / 4.320 mol = 0.7602

Moles of SO2 entering the scrubber per second = 0.003 mol

Hence mole fraction of SO2 entering the scrubber = 0.003 mol / 4.320 mol = 0.0007

For gas exiting the scrubber:

Since all SO2 is removed from the exit stream, it will not contain any SO2. Hence the moles of SO2 in the exit stream is 0 and the moles of the rest of the gases in the exit stream is same as the moles of the rest of the gases in the entering stream.

Hence total moles of all gases present in the exit stream per second = 4.320 - 0.003 = 4.317 mol

Moles of CO2 exiting the scrubber per second = 0.733 mol

Hence mole fraction of CO2 exiting the scrubber = 0.733 mol / 4.317 mol = 0.1698

Moles of H2O exiting the scrubber per second = 0.3 mol

Hence mole fraction of H2O exiting the scrubber = 0.3 mol / 4.317 mol = 0.0695

Moles of N2 exiting the scrubber per second = 3.284 mol

Hence mole fraction of N2 exiting the scrubber = 3.284 mol / 4.317 mol = 0.7607

(b): Given temperature, T = 340 K, Pressure, P = 1 atm

Entering moles, ni = 4.320 mol

Hence Volume of entering gas, Vi = nixRT / P = (4.320mol x0.0821 L.atm.mol-1.K-1x340K) / (1 atm)

=   120.588 L

Exiting moles, nf = 4.317

Hence Volume of exiting gas, Vf = nfxRT / P = (4.317mol x0.0821 L.atm.mol-1.K-1x340K) / (1 atm)

=   120.505 L

Now change in volume, dV = Vf - Vi  = 120.505 L - 120.588 L = - 0.083 L

=> dV = - 0.083 L = - 0.083 L x (1 m3 / 1000L) = - 8.3 x 10-5 m3

Now work done W can be calculated as

W = - PxdV = - 1 atm x (- 8.3 x 10-5 m3 ) = - 1 atm x (1.013x105Nm-2 / 1 atm) x (- 8.3 x 10-5 m3 )

= 8.41 N.m = 8.41 J

time, t = 1 sec

Hence theoritical power input = W/t = 8.41J / 1s = 8.41 W (answer)

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