SME
University Physics I: Mechanics
Fluid mechanics
Use this for “Water at 2 m/s and 200 kPa flows from a 10 cm pipe into a 5 cm one. Find the speed and pressure in the narrow part.”
Diagram
The
same
flow
passes
both
sections,
A₁v₁
=
A₂v₂.
v₂
=
v₁
(
D₁
D₂
)
2
=
2
×
4
=
8
m/s.
Δp
=
1
2
ρ(
v₂
2
−
v₁
2
)
=
1
2
×
1,000
×
(
8
2
−
2
2
)
=
30,000
Pa.
Q = A₁v₁ = 0.01571 m³/s.
This is an example. Type your own values to start a new problem.
p₁ (
)
+
½
ρ (kg/m³)
(
v₁ (m/s)
)
2
=
p₂ (
)
+
½
ρ
(
v₂ (m/s)
)
2
D₁
Diameter at point 1
example
D₂
Diameter at point 2
example
Δp
Pressure drop
calculated
Q
Flow rate
calculated
Formulas
D₁
2
×
v₁
=
D₂
2
×
v₂
0.1
2
×
2
=
0.05
2
×
8
Δp
=
1
2
×
ρ
×
(
v₂
2
−
v₁
2
)
30,000
=
1
2
×
1,000
×
(
8
2
−
2
2
)
Δp
=
1
2
×
ρ
×
v₁
2
×
(
(
D₁
D₂
)
4
−
1)
30,000
=
1
2
×
1,000
×
2
2
×
(
(
0.1
0.05
)
4
−
1)
p₂
=
p₁
−
Δp
170,000 = 200,000 − 30,000
Q
=
π
4
×
D₁
2
×
v₁
0.01571
=
π
4
×
0.1
2
×
2
Formulas are worked in m, Pa, m³/s. The step-by-step shows the conversions.
Assumptions
•
Steady, incompressible flow with no friction, along one streamline; water is 1000 kg/m³.
•
The pipe is level; if it is not, add ρg(z₁ − z₂) to p₂ for the drop in height.
•
Point 2 is the narrow part; p₁ and p₂ are absolute, and Δp = p₁ − p₂ is the fall between them.
Step-by-step
WE KNOW
ρ = 1,000 kg/m³, D₁ = 10 cm, D₂ = 5 cm, v₁ = 2 m/s, p₁ = 200 kPa
FIND
speed at point 2 (v₂), pressure drop (Δp), pressure at point 2 (p₂), 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, Pa, m³/s).
D₁ = 10 cm = 0.1 m (1 m = 100 cm)
D₂ = 5 cm = 0.05 m (1 m = 100 cm)
p₁ = 200 kPa = 200,000 Pa (1 kPa = 1000 Pa)
Step 2 · Find speed at point 2 (v₂)
Use D₁² × v₁ = D₂² × v₂
The
same
volume
passes
each
section
every
second,
and
the
area
goes
as
D
2
,
so
the
speed
grows
by
(
D₁
D₂
)
2
.
v₂
=
v₁
×
(
D₁
D₂
)
2
v₂
=
2
×
(
0.1
0.05
)
2
v₂
=
2
×
2
2
v₂
=
2
×
4
v₂
=
8
m/s
Step 3 · Find pressure drop (Δp)
Use Δp = ½ × ρ × (v₂² − v₁²)
Bernoulli
on
a
level
line:
p
+
1
2
ρv²
stays
the
same,
so
the
pressure
falls
by
the
gain
in
1
2
ρv².
Δp
=
1
2
×
1,000
×
(
8
2
−
2
2
)
Δp
=
1
2
×
1,000
×
(64
−
4)
Δp
=
1
2
×
1,000
×
60
Δp = 500 × 60
Δp = 30,000 Pa
Step 4 · Find pressure at point 2 (p₂)
Use p₂ = p₁ − Δp
Where the fluid speeds up, its pressure falls by Δp.
p₂
=
200,000
−
30,000
p₂
=
170,000
Pa
Step 5 · Find flow rate (Q)
Use Q = π ÷ 4 × D₁² × v₁
The
flow
rate
is
the
area
of
the
pipe,
πD²
4,
times
the
speed
through
it.
Q
=
π
4
×
0.1
2
×
2
Q
=
π
4
×
0.01
×
2
Q
=
0.25π
×
0.01
×
2
Q
=
0.0025π
×
2
Q
=
0.01571
m³/s
Step 6 · Convert the answers
Change each answer back to the unit you chose.
Δp = 30,000 Pa = 30 kPa (1 kPa = 1000 Pa)
p₂ = 170,000 Pa = 170 kPa (1 kPa = 1000 Pa)
Q = 0.01571 m³/s = 15.708 L/s (1 m³/s = 1000 L/s)
Check
Put the numbers back in. Both sides should match.
0.1
2
×
2
=
0.05
2
×
8
✓
30,000
=
1
2
×
1,000
×
(
8
2
−
2
2
)
✓
30,000
=
1
2
×
1,000
×
2
2
×
(
(
0.1
0.05
)
4
−
1)
✓
170,000 = 200,000 − 30,000 ✓
0.01571
=
π
4
×
0.1
2
×
2
✓
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