SME
Calculus II · Problem type
Work pumping out a tank
Use this for “A round tank 1.5 m in radius and 3 m deep is full of water. How much work pumps it all out through a pipe 0.5 m above the rim?”
Diagram
The slab at y = 0.5 lifts H + h − y = 2 + 1 − 0.5 = 2.5
W = ∫ ρgπr²(H + h − y) dy from 0 to H = ρgπr²(H²/2 + hH) = 123,276 J
This is an example. Type your own values to start a new problem.
W
=
ρgπr²
∫
0
(H
+
h
−
y)
dy
=
W (J)
ρ
=
ρ (kg/m³)
,
r
=
r (m)
H
=
H (m)
,
h
=
h (m)
y
Height of one slab
example · m
d
Distance that slab rises
calculated · m
J
Lift integral H²/2 + hH
worked out (not typed) · m²
Formulas
d
=
H
+
h
−
y
2.5 = 2 + 1 − 0.5
J
=
H
2
2
+
h
×
H
4
=
2
2
2
+
1
×
2
W
=
ρ
×
9.81
×
π
×
r
2
×
J
123,276
=
1,000
×
9.81
×
π
×
1
2
×
4
Assumptions
•
Cut the liquid into thin flat slabs. The slab at height y above the bottom, dy thick, weighs ρgπr² dy and has to rise H + h − y to the outlet.
•
Adding each slab’s weight times its lift from the bottom to the top gives W = ∫ from 0 to H of ρgπr²(H + h − y) dy = ρgπr²(H²/2 + hH).
•
The tank is an upright cylinder that starts full and is pumped empty; g = 9.81 m/s², water is 1000 kg/m³ and oil about 900 kg/m³.
•
Lengths are in m, density in kg/m³ and work in J (1 J = 1 N·m).
Step-by-step
WE KNOW
r = 1 m, H = 2 m, h = 1 m, ρ = 1,000 kg/m³, y = 0.5 m
FIND
distance that slab rises (d), lift integral H²/2 + hH (J), work to pump the tank empty (W)
Step 1 · Find distance that slab rises (d)
Use d = H + h − y
The
slab
rises
H
−
y
to
the
rim,
then
h
more
to
the
outlet.
d
=
2
+
1
−
0.5
d
=
2.5
m
Step 2 · Find lift integral H²/2 + hH (J)
Use J = H² ÷ 2 + h × H
Each
slab
rises
H
+
h
−
y
.
An
antiderivative
of
H
+
h
−
y
is
(
H
+
h
)
y
−
y
2
2
;
from
0
to
H
it
is
H
2
/2
+
hH.
J
=
2
2
2
+
1
×
2
J
=
4
2
+
2
J
=
2
+
2
J
=
4
m²
Step 3 · Find work to pump the tank empty (W)
Use W = ρ × 9.81 × π × r² × J
Each
slab
weighs
ρgπr²
dy;
times
its
lift
and
added
from
y
=
0
to
y
=
H
,
that
is
ρgπr²
times
the
lift
integral.
W
=
1,000
×
9.81
×
π
×
1
2
×
4
∫
from
0
to
2
of
1,000
×
9.81
×
π
×
1
2
×
(3
−
y
)
dy
=
[1,000
×
9.81
×
π
×
1
2
×
(
3
y
−
y
2
2
)]
from
0
to
2
=
123,276.0957
W
=
123,276
J
Check
Put the numbers back in. Both sides should match.
2.5 = 2 + 1 − 0.5 ✓
4
=
2
2
2
+
1
×
2
✓
123,276
=
1,000
×
9.81
×
π
×
1
2
×
4
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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