SME
Process Design · Problem type
Drum size from holdup time
Use this for “A drum must hold 10 minutes of a 0.5 m³/min liquid flow while half full, with L = 3D. Find D and L.”
Diagram
V = π × r² × L
V = π × 0.8095² × 4.8572 ≈ 10 m³
This is an example. Type your own values to start a new problem.
V (m³)
×
φ
=
F (m³/min)
×
t (min)
V (m³)
=
(π
÷
4)
×
k
×
(
D (m)
)
3
L (m)
=
k
×
D (m)
F is the liquid flow and φ the share of the drum the liquid fills (0.5 when half full); D is the diameter, twice the radius drawn.
r
Radius
worked out (not typed) · m
Formulas
V
×
φ
=
F
×
t
10 × 0.5 = 0.5 × 10
L
=
k
×
D
4.8572 = 3 × 1.6191
V
=
(
π
4
)
×
D
2
×
L
10
=
(
π
4
)
×
1.6191
2
×
4.8572
V
=
(
π
4
)
×
k
×
D
3
10
=
(
π
4
)
×
3
×
1.6191
3
r
=
D
2
0.8095
=
1.6191
2
Assumptions
•
The drum is a plain cylinder; the heads’ extra volume is left out, which errs on the safe side.
•
Holdup time is how long the liquid flow takes to fill the liquid share of the drum.
•
The ratio k = L ÷ D is chosen first, often about 3 (2.5 to 5 is common).
Step-by-step
WE KNOW
F = 0.5 m³/min, t = 10 min, φ = 0.5, k = 3
FIND
drum volume (V), diameter (D), radius (r), length (L)
Step 1 · Find drum volume (V)
Use V × φ = F × t
The
liquid
held
is
flow
×
holdup
time;
it
fills
only
the
share
φ
,
so
divide
by
φ
.
V
=
F
×
t
φ
V
=
0.5
×
10
0.5
V
=
5
0.5
V
=
10
m³
Step 2 · Find diameter (D)
Use V = (π ÷ 4) × k × D³
With
L
=
kD
the
volume
is
(
π
4
)kD³,
so
take
the
cube
root.
D
=
∛(4
×
V
π ×
k
)
D
=
∛(4
×
10
π × 3
)
D
=
∛(
40
3π
)
D
=
∛4.244
D
=
1.61906
m
Step 3 · Find radius (r)
Use r = D ÷ 2
The radius is half the diameter.
r
=
1.61906
2
r
=
0.8095
m
Step 4 · Find length (L)
Use L = k × D
The
length
is
the
chosen
ratio
k
=
L
D
times
the
diameter.
L
=
3
×
1.61906
L
=
4.8572
m
Check
Put the numbers back in. Both sides should match.
10 × 0.5 = 0.5 × 10 ✓
4.8572 = 3 × 1.61906 ✓
10
=
(
π
4
)
×
1.61906
2
×
4.8572
✓
10
=
(
π
4
)
×
3
×
1.61906
3
✓
0.8095
=
1.61906
2
✓
Topic check
See what you know: 5 questions on this topic
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