SME
Fluid Mechanics · Problem type
Drag on a bluff body and the power to push it
Use this for “A car with CD = 0.3 and a frontal area of 2.2 m² drives at 30 m/s in air of 1.2 kg/m³. Find the drag and the power it takes.”
Table
Speed (V)
(m/s)
Drag force (FD)
(N)
Power against drag P = 10.69 kW
This is an example. Type your own values to start a new problem.
FD (N)
=
½
×
ρ (kg/m³)
×
(
V (m/s)
)
2
×
CD
×
A (m²)
P (kW)
=
FD (N)
×
V (m/s)
Formulas
FD
=
1
2
×
ρ
×
V
2
×
CD
×
A
356.4
=
1
2
×
1.2
×
30
2
×
0.3
×
2.2
P
=
FD
×
V
10,692 = 356.4 × 30
Formulas are worked in W. The step-by-step shows the conversions.
Assumptions
•
CD is measured for the body’s shape and taken as constant over these speeds, well under the speed of sound.
•
A is the frontal area, the body’s shadow seen from straight ahead.
•
Still air: V is the body’s speed through it, and P = FDV is the power against drag alone.
Step-by-step
WE KNOW
V = 30 m/s, CD = 0.3, A = 2.2 m², ρ = 1.2 kg/m³
FIND
drag force (FD), power against drag (P)
Step 1 · Find drag force (FD)
Use FD = ½ × ρ × V² × CD × A
Drag
is
CD
times
the
dynamic
pressure
1
2
ρV²
times
the
frontal
area.
FD
=
1
2
×
1.2
×
30
2
×
0.3
×
2.2
FD
=
1
2
×
1.2
×
900
×
0.3
×
2.2
FD = 0.6 × 900 × 0.3 × 2.2
FD = 540 × 0.3 × 2.2
FD = 162 × 2.2
FD = 356.4 N
Step 2 · Find power against drag (P)
Use P = FD × V
Power
is
force
times
speed;
drag
grows
as
V
2
,
so
the
power
grows
as
V
3
.
P
=
356.4
×
30
P
=
10,692
W
Step 3 · Convert the answers
Change each answer back to the unit you chose.
P = 10,692 W = 10.69 kW (1 kW = 1000 W)
Check
Put the numbers back in. Both sides should match.
356.4
=
1
2
×
1.2
×
30
2
×
0.3
×
2.2
✓
10,692 = 356.4 × 30 ✓
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