SME
Fluid Mechanics · Problem type
From the model’s force to the prototype’s
Use this for “A 1 m car model feels 200 N of drag at 120 m/s. What drag does the 4 m car feel at 30 m/s in the same air?”
This is an example. Type your own values to start a new problem.
Fₘ (N)
=
CF
×
ρₘ (kg/m³)
×
(
Vₘ (m/s)
)
2
×
(
Lₘ (m)
)
2
Fₚ (N)
=
CF
×
ρₚ (kg/m³)
×
(
Vₚ (m/s)
)
2
×
(
Lₚ (m)
)
2
Formulas
Fₘ
=
CF
×
ρₘ
×
Vₘ
2
×
Lₘ
2
200
=
0.01157
×
1.2
×
120
2
×
1
2
Fₚ
=
CF
×
ρₚ
×
Vₚ
2
×
Lₚ
2
200
=
0.01157
×
1.2
×
30
2
×
4
2
Assumptions
•
The model is similar in shape and its Reynolds number matches the prototype’s.
•
Then the force coefficient CF = F ÷ (ρV²L²) is the same on both.
•
So Fₚ = Fₘ(ρₚ ÷ ρₘ)(Vₚ ÷ Vₘ)²(Lₚ ÷ Lₘ)².
Step-by-step
WE KNOW
Fₘ = 200 N, ρₘ = 1.2 kg/m³, Vₘ = 120 m/s, Lₘ = 1 m, ρₚ = 1.2 kg/m³, Vₚ = 30 m/s, Lₚ = 4 m
FIND
force coefficient (CF), prototype force (Fₚ)
Step 1 · Find force coefficient (CF)
Use Fₘ = CF × ρₘ × Vₘ² × Lₘ²
Divide
the
model’s
force
by
its
ρV²
L
2
:
this
coefficient
is
the
same
on
both.
CF
=
Fₘ
ρₘ ×
Vₘ
2
×
Lₘ
2
CF
=
200
1.2 ×
120
2
×
1
2
CF
=
200
1.2 × 14,400
CF
=
200
17,280
CF = 0.01157
Step 2 · Find prototype force (Fₚ)
Use Fₚ = CF × ρₚ × Vₚ² × Lₚ²
The
shared
force
coefficient
times
the
prototype’s
own
ρV²
L
2
.
Fₚ
=
0.01157
×
1.2
×
30
2
×
4
2
Fₚ = 0.01157 × 1.2 × 900 × 16
Fₚ = 0.013884 × 900 × 16
Fₚ = 12.4956 × 16
Fₚ = 200 N
Check
Put the numbers back in. Both sides should match.
200
=
0.01157
×
1.2
×
120
2
×
1
2
✓
200
=
0.01157
×
1.2
×
30
2
×
4
2
✓
← Previous lesson
A ship model: matching the Froude number
Next lesson →
Which number must the model match?
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Dimensional analysis
Main lesson
A ship model: matching the Froude number
Use this for “A 100 m ship sails at 10 m/s. How fast should its 4 m model (1:25) be towed?”
Which number must the model match?
Use this for “A model of a spillway is tested in a lab. Should it match the Reynolds, Froude, Mach or Weber number?”