SME
Fluid Mechanics
Bernoulli equation
Use this for “Water flows through a venturi that narrows from 100 mm to 50 mm, and the pressure falls 30 kPa. Find the flow rate.”
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.
ΔP (kPa)
=
½
×
ρ (kg/m³)
×
(
V₁ (m/s)
)
2
×
(
D₁ (mm)
D₂ (mm)
4
−
1)
Q (m³/s)
=
π
4
×
(
D₁ (mm)
)
2
×
V₁ (m/s)
Continuity gives V₂ = V₁(D₁ ÷ D₂)², which in ΔP = ½ρ(V₂² − V₁²) gives the first line; Q = A₁V₁ with A₁ = πD₁² ÷ 4 gives the second.
V₂
Throat speed
calculated · m/s
Formulas
ΔP
=
1
2
×
ρ
×
V₁
2
×
(
(
D₁
D₂
)
4
−
1)
30,000
=
1
2
×
1,000
×
2
2
×
(
(
0.1
0.05
)
4
−
1)
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
)
Q
=
π
4
×
D₁
2
×
V₁
0.01571
=
π
4
×
0.1
2
×
2
Formulas are worked in Pa, m. The step-by-step shows the conversions.
Assumptions
•
Steady, incompressible flow with no friction losses, along one streamline.
•
The meter is level, so the heights cancel from Bernoulli’s equation.
•
A real venturi passes about 0.98 of this ideal flow (its discharge coefficient).
Step-by-step
WE KNOW
ΔP = 30 kPa, D₁ = 100 mm, D₂ = 50 mm, ρ = 1,000 kg/m³
FIND
inlet speed (V₁), throat speed (V₂), 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 Pa, m).
ΔP = 30 kPa = 30,000 Pa (1 kPa = 1000 Pa)
D₁ = 100 mm = 0.1 m (1 m = 1000 mm)
D₂ = 50 mm = 0.05 m (1 m = 1000 mm)
Step 2 · Find inlet speed (V₁)
Use ΔP = ½ × ρ × V₁² × ((D₁ ÷ D₂)⁴ − 1)
Put
V₂
=
V₁
(
D₁
D₂
)
2
from
continuity
into
ΔP
=
1
2
ρ
(
V₂
2
−
V₁
2
),
then
solve
for
V₁
.
V₁
=
√(2
×
ΔP
÷
(
ρ
×
(
(
D₁
D₂
)
4
−
1)))
V₁
=
√(2
×
30,000
÷
(1,000
×
(
(
0.1
0.05
)
4
−
1)))
V₁
=
√(2
×
30,000
1,000 × (
2
4
− 1)
)
V₁
=
√(2
×
30,000
1,000 × (16 − 1)
)
V₁
=
√(2
×
30,000
1,000 × 15
)
V₁
=
√(2
×
30,000
15,000
)
V₁
=
√(
60,000
15,000
)
V₁
=
√
4
V₁
=
2
m/s
Step 3 · Find throat speed (V₂)
Use D₁² × V₁ = D₂² × V₂
The
same
flow
passes
each
section:
the
area
shrinks
by
(
D₂
D₁
)
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 4 · Find flow rate (Q)
Use Q = π ÷ 4 × D₁² × V₁
The flow rate is the inlet’s area times its speed.
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
Check
Put the numbers back in. Both sides should match.
30,000
=
1
2
×
1,000
×
2
2
×
(
(
0.1
0.05
)
4
−
1)
✓
0.1
2
×
2
=
0.05
2
×
8
✓
30,000
=
1
2
×
1,000
×
(
8
2
−
2
2
)
✓
0.01571
=
π
4
×
0.1
2
×
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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