Question

In the production of alternative and natural sweeteners from Stevia, the cleaned leaves at a rate of 875 kg/h are first passed through a mechanical sorter. 23.0% of the fresh feed are small leaves and can go straight to the drier. They are dried to 8.00% moisture, from the original 46.7% moisture. After drying, they are ground to a fine powder. The remaining large leaves are first fed to a grinder where the leaves are ground to a pulp. They are then fed to an extractor where hexane is also added to extract steviol glycosides, the “sweet component” of the stevia leaves. Fresh hexane at 27.2 kg/h is added to a recycled condensed hexane stream and is then fed to the extractor at 2.75 kg hexane per kg of ground stevia leaves. The exit stream of the extractor then passes through a filter where the filter cake (predominantly containing pulp) is separated from the liquid part of the stream (filtrate). The filtrate is then fed to a decanter where the hexane stream is separated from the aqueous stream. The aqueous stream is virtually pure water. The hexane stream is fed to an evaporator where all of the hexane is evaporated, leaving pure steviol glycosides. The evaporated hexane is then condensed and recycled back to the extractor. Some information for the composition of streams and materials are: Fresh Stevia leaves contain 46.7% moisture, 3.80% steviol glycosides, and the remainder is pulp; The filter cake is 92.4% pulp, the remainder is a mixture of hexane and steviol glycoside.; Two final products are produced in the process: dried ground Stevia leaves and pure steviol glycoside.
Find the following:
a. Complete diagram

b. Rate of production of the two final products in tons/year (assume that 1 yr. operation = 300 days of continuous operation)

c. Rate of hexane recycle (kg/h) and recycle ratio (recycled hexane/fresh hexane)

Answer 1

the flowsheet is drawn schematically and is given by

the streams are labelled for convenience.

Given feed F = 875 kg/h

large leaves flowrate L = 0.77 x 875 = 673.75 kg/h

small leaves flowrate S = 0.23 x 875 = 201.25 kg/h

small leaves stream :

mass balance :

S = M + D

201.25 = M + D ----------------eq-1

moisture content in the feed stream = 0.467

moisture content in leaving stream D = 0.08

by water balance :

0.467 S = 0.08 D + M

0.467 x 201.25 = 0.08 D + M ------------------eq-2

on solving eq-1 and eq-2, we get

D = 116.6 kg/h

M = 84.65 kg/h

the product stream flowrate P₂ = D = 116.6 kg/h

Large leaves stream :

grinded stream flowrate = G = L (since no mass is lost in grinder)

G = 673.75 kg/h

balance at the extractor :

G + K = E

673.75 + K = E

K = H + R

H is the fresh hexane stream = 27.2 kg/h

K = 27.2 + R

K is the mixed stream of Hexane. ( recycle + fresh)

mass ratio of hexane to leaves = 2.75

for 1 kg of leaves, hexane fed = 2.75 kg

for 673.75 kg, hexane fed in K = 673.75 x 2.75 = 1852.8125 kg/h

K = 1852.8125 kg/h

E = 673.75 + 1852.8125 = 2526.5625 kg/h

mass balance at the filter :

E = C + U

2526.5625 = C + U

C is the cake formed. U is the filtrate flowrate.

C contains 92.4 % of pulp.

Overall Pulp balance :

pulp present in large leaves streams L = pulp separated in C

0.495 x L = 0.924 x C

C = 0.495 x 673.75 / 0.924 = 360.9375 kg/h

water/moisture balance :

moisture present in large leaves stream = aqueous stream leaving in separator. ( no water is present in cake or the exit product pure steviol glycoside stream)

0.467 x L = W

W = 0.467 x 673.75

W = 314.64 kg/h

hexane balance :

hexane entering the process in H = hexane leaving the process in C

H = hexane in C

hexane in C = 27.2 kg/h

steviol glycoside in C = C - (pulp + hexane)

stevial glycoside in C = 360.9375 - ( 333.506 + 27.2 ) = 0.2315 kg/h

steiviol glycoside balance :

stevial glycoside entering in L = stevial glycoside collected as final product P₁ + stevil glycoside lost in C

0.038 x L = P₁ + 0.2315

P₁ = 0.038 x 673.75 - 0.2315 = 25.371 kg/h

balance at the decanter :

U = V + W

U = E - C

E = 2526.5625 kg/h

C = 360.9375 kg/h

U = 2526.5625 - 360.9375 = 2165.625 kg/h

W = 314.64 kg/h

V = U - W

V = 2165.625 - 314.64 = 1850.985 kg/h

V is the stream fed to the evaporator.

V = P₁ + R

R = 1850.985 - 25.371 = 1825.614 kg/h

the recycle hexane stream flowrate R = 1825.614 kg/h

b)

production rate of pure steviol glycoside = P₁ = 25.371 kg/h = 25.371 kg/h x 10^-3 x 300 days/year x 24 hours/day = 182.6712 tonnes/year

production rate of dried ground stevia = P₂ = 116.65 kg/h = 116.65 x 10^-3 x 300 x 24 = 839.88 tonnes/year

c)

recycle rate of hexane R = 1825.614 kg/h

recycle ratio = R / H = 1825.614 / 27.2 = 67.12