SPP2601 Assignment 1 2026 Minor Test | Due 8 May 2026

SPP2601 MINOR TEST


QUESTION 1 – Distillation with side stream

1.1. Mass balances

Distillate D=1560 kmol/hD=1560 kmol/h, xD=0.96xD=0.96

Bottoms xW≤0.09xW≤0.09

Side stream S=0.20 FS=0.20F, xS=0.70xS=0.70

64% of ethanol in the feed is recovered in the distillate

Reflux ratio R=2.5R= 2.5

Let FF = feed rate, zFzF = ethanol mole fraction in feed, WW = bottoms rate.

Ethanol recovery:



0.64 (FzF)=DxD=1560×0.96=1497.6 kmol/h0.64(FzF)=DxD=1560×0.96=1497.6 kmol/h
⇒ FzF=1497.60.64=2340 kmol/h⇒ FzF=0.641497.6 = 2340 kmol/h.

Overall balance:



F=D+W+S=1560+W+0.2FF=D+W+S=1560+W+0.2F → 0.8F=1560+W0.8F =
1560+W → W=0.8F−1560W=0.8F−1560.

Component balance:



FzF=DxD+WxW+SxSFzF=DxD+WxW+SxS
2340=1497.6+(0.8F−1560)×0.09+0.2F×0.702340=1497.6+(0.8F−1560)×0.09+0.2F×0.7

, 0
2340=1497.6+0.072F−140.4+0.14F2340 = 1497.6+0.072F−140.4+0.14F
2340=1357.2+0.212F2340=1357.2+0.212F → F=982.80.212=4635.8 kmol/hF=0.21298
2.8 = 4635.8 kmol/h.




zF=2340/4635.8=0.505zF=2340/4635.8=0.505

W=0.8×4635.8−1560=3708.6−1560=2148.6 kmol/hW=0.8×4635.8−1560=3708.6−1560
= 2148.6 kmol/h

S=0.2×4635.8=927.2 kmol/hS=0.2×4635.8= 927.2 kmol/h

1.2. Liquid and vapour flows

L=RD=2.5×1560=3900 kmol/hL=RD=2.5×1560=3900 kmol/h
V=L+D=3900+1560=5460 kmol/hV=L+D=3900+1560 = 5460 kmol/h (constant molar
overflow)

Above side stream:
L=3900L=3900, V=5460V = 5460

Between side stream and feed:
LS=L−S=3900−927.2=2972.8 kmol/hLS=L−S=3900−927.2=2972.8 kmol/h
VS=V=5460VS=V= 5460

Below feed (saturated liquid, q=1q=1):
Lˉ=LS+F = 2972.8+4635.8 = 7608.6 kmol/hLˉ= LS+F=2972.8+4635.8 = 7608.6 kmol/h
Vˉ=Lˉ−W = 7608.6−2148.6=5460 kmol/hVˉ=Lˉ−W=7608.6−2148.6 = 5460 kmol/h