SME
Fluid Mechanics · Problem type
A pitot tube: speed from the pressure rise
Use this for “A pitot-static tube in air of 1.2 kg/m³ reads 600 Pa. How fast is the air?”
Diagram
The
stream
stops
at
the
nose:
P₀
−
P
=
1
2
ρV²,
so
V
=
√(2Δ
P
ρ
)
=
√(2
×
600
1.2
)
=
31.62
m/s.
The
gauge
reads
h
=
ΔP
(ρₘ − ρ)g
=
0.06124
m.
This is an example. Type your own values to start a new problem.
ΔP (Pa)
=
½
×
ρ (kg/m³)
×
(
V (m/s)
)
2
Formulas
ΔP
=
1
2
×
ρ
×
V
2
600
=
1
2
×
1.2
×
31.62
2
Assumptions
•
The stream stops at the tube’s nose with no loss; the side ports read the stream’s own pressure.
•
Incompressible: the speed is well under a third of the speed of sound (in air, under about 100 m/s).
•
ΔP is the stagnation pressure minus the static; the U-tube drawn holds water.
Step-by-step
WE KNOW
ΔP = 600 Pa, ρ = 1.2 kg/m³
FIND
stream speed (V)
Step 1 · Find stream speed (V)
Use ΔP = ½ × ρ × V²
Bernoulli
from
the
stream
to
the
nose,
where
it
stops:
ΔP
=
1
2
ρV²,
so
V
=
√(2Δ
P
ρ
).
V
=
√(2
×
ΔP
ρ
)
V
=
√(2
×
600
1.2
)
V
=
√(
1,200
1.2
)
V
=
√
1
,000
V
=
31.62
m/s
Check
Put the numbers back in. Both sides should match.
600
=
1
2
×
1.2
×
31.62
2
✓
Topic check
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