SME
Fluid Mechanics · Problem type
A draining tank: the speed out of a hole
Use this for “Water stands 5 m above a sharp-edged 50 mm hole in a tank’s side. How fast does it leave, and what is the flow rate?”
Diagram
V
=
√
2gh
=
√
2 × 9.81 × 5 m
V = 9.905 m/s
Q
=
Cd
×
πD²
4
×
V
=
0.01186
m³/s.
As the tank drains, h falls and V falls with √h, so it empties ever more slowly.
This is an example. Type your own values to start a new problem.
V (m/s)
=
2
×
9.81
×
h (m)
Q (m³/s)
=
Cd
×
π
4
×
(
D (mm)
)
2
×
V (m/s)
Formulas
V
=
√
2 × 9.81 ×
h
9.905
=
√
2 × 9.81 × 5
Q
=
Cd
×
π
4
×
D
2
×
V
0.01186
=
0.61
×
π
4
×
0.05
2
×
9.905
Formulas are worked in m. The step-by-step shows the conversions.
Assumptions
•
The tank is wide, so its surface falls slowly and its speed is ignored.
•
The surface and the jet are both at atmospheric pressure; h runs from the surface to the hole.
•
Cd = 1 for a smooth rounded hole, about 0.61 for a sharp-edged one; g = 9.81 m/s².
Step-by-step
WE KNOW
h = 5 m, D = 50 mm, Cd = 0.61
FIND
jet speed (V), flow rate (Q)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to m).
D = 50 mm = 0.05 m (1 m = 1000 mm)
Step 2 · Find jet speed (V)
Use V = √(2 × 9.81 × h)
Bernoulli
from
the
still
surface
to
the
jet:
the
drop
h
turns
into
1
2
V
2
,
so
V
=
√
2gh
.
V
=
√
2 × 9.81 × 5
V
=
√
19.62 × 5
V
=
√
98.1
V
=
9.905
m/s
Step 3 · Find flow rate (Q)
Use Q = Cd × π ÷ 4 × D² × V
The hole’s area times the jet speed, cut by Cd for the jet’s narrowing and friction.
Q
=
0.61
×
π
4
×
0.05
2
×
9.905
Q
=
0.61π
4
×
0.0025
×
9.905
Q
=
0.47909
×
0.0025
×
9.905
Q
=
0.0011977
×
9.905
Q
=
0.01186
m³/s
Check
Put the numbers back in. Both sides should match.
9.905
=
√
2 × 9.81 × 5
✓
0.01186
=
0.61
×
π
4
×
0.05
2
×
9.905
✓
Topic check
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