SME
University Physics I: Mechanics · Problem type
Speed of water from a hole
Use this for “Water stands 2 m above a 2 cm² hole in a tank. How fast does it leave, and at what flow rate?”
Diagram
v
=
√
2gh
=
√
2 × 9.81 × 2 m
v = 6.264 m/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)
a
Area of the hole
example
Q
Flow rate
worked out (not typed)
Formulas
v
=
√
2 × 9.81 ×
h
6.2642
=
√
2 × 9.81 × 2
Q
=
a
×
v
0.001253 = 0.0002 × 6.2642
Formulas are worked in m², m³/s. The step-by-step shows the conversions.
Assumptions
•
The tank is wide, so its surface falls slowly and the speed of the surface is ignored.
•
The surface and the hole are both open to the air, so the pressures cancel; h runs from the surface down to the hole.
•
No losses at the hole (a sharp edge passes only about 0.61 of Q); g = 9.81 m/s².
Step-by-step
WE KNOW
h = 2 m, a = 2 cm²
FIND
speed of the jet (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 m², m³/s).
a = 2 cm² = 0.0002 m² (1 m² = 10⁴ cm²)
Step 2 · Find speed of the jet (v)
Use v = √(2 × 9.81 × h)
Bernoulli
from
the
still
surface
to
the
jet,
both
at
air
pressure:
the
drop
h
turns
into
1
2
v
2
,
so
v
=
√
2gh
.
v
=
√
2 × 9.81 × 2
v
=
√
19.62 × 2
v
=
√
39.24
v
=
6.2642
m/s
Step 3 · Find flow rate (Q)
Use Q = a × v
The flow rate is the area of the hole times the speed through it.
Q
=
0.0002
×
6.2642
Q
=
0.001253
m³/s
Step 4 · Convert the answers
Change each answer back to the unit you chose.
Q = 0.001253 m³/s = 1.2528 L/s (1 m³/s = 1000 L/s)
Check
Put the numbers back in. Both sides should match.
6.2642
=
√
2 × 9.81 × 2
✓
0.001253 = 0.0002 × 6.2642 ✓
Topic check
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