Question

A design for purifying helium consists of an adiabatic process that splits a helium stream
containing 30-mole-percent methane into two product streams, one containing 97-mole-percent helium
and the other 90-mole-percent methane. The feed enters at 10 bar and 117°C; the methane-rich product
leaves at I bar and 27°C; the heHum-rich product leaves at 50°C and 15 bar. Moreover, wort is
produced by the process. Assuming heHum an ideal gas with Cp = (S/2)R and methane an ideal gas
with Cp = (9/2)R., ealculate the total entropy change of the process on the basis of I mol of feed to
confirm that the process does not violate the second law.

Answer 1

Step:-1 Griven 971. Helium 34. Methane stream 1 Souc (Ti) is bar (P.) Imple feed 30-4. Methane 70% Helium to bar (PP) 117°c (TF) stream 2. 90% Methome Nol Helium 1 bar (P2) 24°C (T2) of Helium CHE Heat .5/2 R Capacity of Meat capacity of Methome que 2

For an Ideal Gas, Enkopy change of System to as) System & lm Tz - R lm Pz Т) en P at constant pressure G = Heat Capacity T2, Ti = final cond Initial temperature Final and Initial Pressure P2, P, R Universal Gas Constant

Entropy change of Helium in steam 1 (AS R lm Pa He G lm Ta 'He Pf 5 504273 R lm R lm is 2 10 117 +273 He - 3.9 Ilmol. K 3.37 I mel. K 2 He (As) -7.27 Il moln K Enkopy change of Methane in stream 1, CACHA lm Ti Rlm PL If Pf R lim 15/10 2 % R lim 504273 117 +373 -7.05 Ilmol. K 3.37 1 mal K

(AS) - 10.42 Ilmol. K Entropy change of Helium in stream 2 He 45) lim Ti = Rlm Pe > CHE TF Anlasts 56 R lm 37+973 R lm 1 2 10 117 +973 He -5.45 5/ mot : K ť 19. 14 Il mol-k 2 He (AS) 13.69 Ilmol. K 2 2 Entropy change of Methane in stream 2, As) lm Tz . R lm Pe Po panx

2 117 + 2+ 3 –9.81 Il meil - K 2 + 19. 14 Il mol. K САҚ leader 9.33 Ilmol. K Basis 1 mole feed (AS) CHA 9% R lm 27+273 - R. lm /10 Moles of Helium 0.7 moles total Entropy change of Methane [(As) "* + (As), *4] 0.3 moles ( -10.42 + 9.33] /malok 301. Methome, of Methane Miche 0.3 moles Fol. Helium, Holes ny 'He 2 NCHA 2 1.-0.327 TIK

total Entropy change of Helium, Mhle [cas), 0.7 miles (-7.27 + 13.69] 5/male-k (AS) He 2 He + (As) *e] 2 (AS) 2 He 4.49 Ilk total Entropy change of the process, As system (As) (as) He 2 + CHA z (-0.327 + 4.49) JIK (AS) System 2 4.163 Ilk This is the Entropy change of the process

Enkopy change of surroumdings, (AS) Sures ♡ Ts Q = Heat transfer Ts - Surrounding / Resewiol temperatūre

Since Adiabatic process, Q-0 (AS) z 0 surroundings total Entropy of Universe (AS) System + AS) Surrol 2 4.163 JIK + 0 JK (As) universe 4.163 JIK & Second haw states thats "Entropy of Universe always wereases Here (As) universe 2 4.163 JA Ilk (AS) > 0 universe so "second haw of thermodynamics ý not violated