SME
Material & Energy Balances · Problem type
An evaporator with a bypass
Use this for “Juice at 12% solids is concentrated to 58%, then blended back with bypassed juice to 42%. How much bypasses?”
Diagram
This is an example. Type your own values to start a new problem.
F (kg/h)
×
xf
=
P (kg/h)
×
xₚ
F (kg/h)
=
W (kg/h)
+
P (kg/h)
=
B (kg/h)
+
E (kg/h)
W (kg/h)
=
E (kg/h)
×
1
−
xf
xc
The first two lines balance the whole process, solids then total. The evaporator keeps its solids, E × xf = (E − W) × xc, which gives the third line; the fourth splits the feed.
Formulas
F
×
xf
=
P
×
xₚ
100 × 0.12 = 28.5714 × 0.42
F
=
W
+
P
100 = 71.4286 + 28.5714
W
=
E
×
(1
−
xf
xc
)
71.4286
=
90.0621
×
(1
−
0.12
0.58
)
F
=
B
+
E
100 = 9.9379 + 90.0621
Assumptions
•
Steady state. The evaporator removes pure water; the solids all stay in the liquid.
•
The bypass has the fresh feed’s composition; it joins the concentrate at the mixing point.
•
Balances: the whole process (solids and total), the split, and the evaporator’s water.
Step-by-step
WE KNOW
F = 100 kg/h, xf = 0.12, xc = 0.58, xₚ = 0.42
FIND
product (P), water evaporated (W), evaporator feed (E), bypass (B)
Step 1 · Find product (P)
Use F × xf = P × xₚ
All the solids fed leave in the product.
P
=
F
×
xf
xₚ
P
=
100
×
0.12
0.42
P
=
12
0.42
P
=
28.5714
kg/h
Step 2 · Find water evaporated (W)
Use F = W + P
What doesn’t leave as product leaves as vapour.
W
=
F
−
P
W
=
100
−
28.5714
W
=
71.4286
kg/h
Step 3 · Find evaporator feed (E)
Use W = E × (1 − xf ÷ xc)
The
evaporator’s
solids
go
from
E
at
xf
to
E
×
xf
xc
;
the
difference
is
W
.
E
=
W
×
xc
xc − xf
E
=
71.4286
×
0.58
0.58 − 0.12
E
=
71.4286
×
0.58
0.46
E
=
41.43
0.46
E
=
90.0621
kg/h
Step 4 · Find bypass (B)
Use F = B + E
The feed not sent to the evaporator bypasses it.
B
=
F
−
E
B
=
100
−
90.0621
B
=
9.9379
kg/h
Check
Put the numbers back in. Both sides should match.
100 × 0.12 = 28.5714 × 0.42 ✓
100 = 71.4286 + 28.5714 ✓
71.4286
=
90.0621
×
(1
−
0.12
0.58
)
✓
100 = 9.9379 + 90.0621 ✓
Topic check
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