Steam stripping of a paraffin (number of stages): 23-Chem-A3 Heat and Mass Transfer, May 2016, Q6

Question 6 — Steam stripping of a paraffin (number of stages) (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Basis 100 kg liquor: 8 kg paraffin ($0.0702$ kmol) + 92 kg non-volatile ($0.6815$ kmol, the carrier $L_s$). Reduce paraffin from 8% to 0.08% by mass. Equilibrium (Raoult): $y=\dfrac{p^{*}}{P}x=\dfrac{53}{101.3}x=0.523\,x$. Steam enters paraffin-free; steam rate $=3\times$ minimum.

Find. Number of theoretical stripping stages.

Counter-current steam stripper: liquor down (carrier $L_s$), steam up (carrier $G_s$); paraffin transfers from liquid to vapour.

Approach. Work in solute/carrier mole ratios so the operating line is straight; find the minimum steam from the top pinch (exit vapour in equilibrium with feed liquor); set the actual steam to $3\times$; then count stages with the Kremser relation / stage stepping.

Terminal mole ratios. $X=\dfrac{\text{mol paraffin}}{\text{mol non-volatile}}$: inlet $X_{0}=\dfrac{0.0702}{0.6815}=0.103$; outlet (0.08% by mass) $X_{N}=9.5\times10^{-4}$. Paraffin removed $=L_s(X_0-X_N)=0.0695$ kmol.
Minimum steam (top pinch). At $G_{s,\min}$ the exit vapour is in equilibrium with the entering liquor: $y_1^{*}=0.523\,x_{0}=0.0489$, i.e. ratio $Y_1^{*}=0.0514$. Since steam enters solute-free, $$G_{s,\min}=\frac{L_s(X_0-X_N)}{Y_1^{*}}=\frac{0.0695}{0.0514}=1.35\ \text{kmol}.$$
Actual steam and stripping factor. $G_s=3G_{s,\min}=4.06$ kmol; the operating slope $L_s/G_s=0.168$. The stripping factor $$S=\frac{m\,G_s}{L_s}=\frac{0.523}{0.168}=3.12\;(>1,\ \text{feasible}).$$
Number of stages (Kremser). With solute-free entering steam, $$N=\frac{\ln\!\Big[\dfrac{X_0}{X_N}\big(1-\tfrac1S\big)+\tfrac1S\Big]}{\ln S}=\frac{\ln[108.6(0.679)+0.321]}{\ln 3.12}=3.79 .$$ $$\boxed{N\approx4\ \text{theoretical stages.}}$$ Stepping stages directly on the (slightly curved) equilibrium line in ratio coordinates gives the same count, confirming four ideal stages suffice.

Quantity	Value
Inlet / outlet ratio $X_0,X_N$	0.103, 9.5 × 10⁻⁴
Minimum steam $G_{s,\min}$	1.35 kmol / 100 kg liquor
Actual steam (3×) / factor $S$	4.06 kmol; $S=3.12$
Theoretical stages	≈ 4

Concept & Theory

Topic: Counter-current steam stripping of a dilute-ish solute; Kremser analysis of a linear-equilibrium cascade.

Key relations: Raoult stripping line $y=(p^{*}/P)x$; carrier balance $G_s(Y_1-Y_{N+1})=L_s(X_0-X_N)$; stripping factor $S=mG_s/L_s$; Kremser stage count.

Why this works: Because both steam and the non-volatile solvent are essentially inert carriers, solute/carrier ratios linearise the operating line and the Kremser geometric-series solution applies. The minimum steam is set by pinching the exit vapour against the feed liquor; multiplying by three lifts $S$ well above unity so only a handful of stages achieve the 100-fold reduction.

Common pitfall: Mixing mole-fraction equilibrium with mole-ratio operating lines without care, and locating the minimum-steam pinch at the wrong end (it is the top, where liquor is richest).

Source: Coulson & Richardson Vol. 2, §12 (stripping); Treybal §8.3 (Kremser equation); Geankoplis §10.3.

Question 6 of 7: Steam stripping of a paraffin (number of stages)

Question 6 — Steam stripping of a paraffin (number of stages) (25 marks)