SME
Compressible Flow
Fanno and Rayleigh flow
Use this for “Air enters a 5 cm duct at M = 0.5 with f = 0.005. How long can the duct be before the flow chokes?”
Chart
F(0.5) ≈ 1.069
This is an example. Type your own values to start a new problem.
4fL∗/D
=
1
−
M
2
1.4
×
M
2
+
2.4
2.8
×
ln
2.4
×
M
2
2
+
0.4
×
M
2
4
×
×
L∗ (m)
D (m)
=
4fL∗/D
T/T∗
Temperature ratio
calculated
p/p∗
Pressure ratio
calculated
p₀/p₀∗
Stagnation pressure ratio
calculated
V/V∗
Velocity ratio
calculated
Formulas
4fL∗/
D
=
1 −
M
2
1.4 ×
M
2
+
(
2.4
2.8
)
×
ln(2.4
×
M
2
2 + 0.4 ×
M
2
)
1.069
=
1 −
0.5
2
1.4 ×
0.5
2
+
(
2.4
2.8
)
×
ln(2.4
×
0.5
2
2 + 0.4 ×
0.5
2
)
L∗
=
(4fL∗/
D
)
×
D
4 ×
f
2.673
=
1.069
×
0.05
4 × 0.005
T/T∗
=
2.4
2 + 0.4 ×
M
2
1.143
=
2.4
2 + 0.4 ×
0.5
2
p/p∗
=
(
1
M
)
×
√(
2.4
2 + 0.4 ×
M
2
)
2.138
=
(
1
0.5
)
×
√(
2.4
2 + 0.4 ×
0.5
2
)
p₀/p₀∗
=
(
1
M
)
×
(
2 + 0.4 ×
M
2
2.4
)³
1.34
=
(
1
0.5
)
×
(
2 + 0.4 ×
0.5
2
2.4
)³
V/V∗
=
M
×
√(
2.4
2 + 0.4 ×
M
2
)
0.5345
=
0.5
×
√(
2.4
2 + 0.4 ×
0.5
2
)
Assumptions
•
Steady adiabatic flow of air (γ = 1.4) with wall friction in a duct of constant area.
•
f is the Fanning friction factor, a quarter of the Darcy factor, the same all along the duct.
•
Starred values are where the flow reaches M = 1 and chokes, a length L∗ downstream.
•
Friction drives subsonic and supersonic flow alike toward M = 1.
Step-by-step
WE KNOW
M = 0.5, f = 0.005, D = 0.05 m
FIND
friction parameter (4fL∗/D), choking length (L∗), temperature ratio (T/T∗), pressure ratio (p/p∗), stagnation pressure ratio (p₀/p₀∗), velocity ratio (V/V∗)
Step 1 · Find friction parameter (4fL∗/D)
Use 4fL∗/D = (1 − M²) ÷ (1.4 × M²) + (2.4 ÷ 2.8) × ln(2.4 × M² ÷ (2 + 0.4 × M²))
The
Fanno
function:
how
much
friction
the
flow
can
take
before
it
reaches
M
=
1.
4fL∗/
D
=
1 −
0.5
2
1.4 ×
0.5
2
+
(
2.4
2.8
)
×
ln(2.4
×
0.5
2
2 + 0.4 ×
0.5
2
)
0.5
2
=
0.25
1 − 0.25
1.4 × 0.25
=
2.1429
(
2.4
2.8
)
×
ln(2.4
×
0.25
2 + 0.4 × 0.25
)
=
−1.0738
2.1429 − 1.0738 = 1.0691
4fL∗/
D
=
1.069
Step 2 · Find choking length (L∗)
Use L∗ = (4fL∗/D) × D ÷ (4 × f)
Multiply
the
friction
parameter
by
D
,
then
divide
by
4
f
.
L∗
=
1.069
×
0.05
4 × 0.005
L∗
=
1.069
×
0.05
0.02
L∗
=
0.05345
0.02
L∗ = 2.673 m
Step 3 · Find temperature ratio (T/T∗)
Use T/T∗ = 2.4 ÷ (2 + 0.4 × M²)
The
flow
cools
as
it
speeds
up
toward
M
=
1
and
warms
as
it
slows
from
above.
T/T∗
=
2.4
2 + 0.4 ×
0.5
2
T/T∗
=
2.4
2 + 0.4 × 0.25
T/T∗
=
2.4
2 + 0.1
T/T∗
=
2.4
2.1
T/T∗ = 1.143
Step 4 · Find pressure ratio (p/p∗)
Use p/p∗ = (1 ÷ M) × √(2.4 ÷ (2 + 0.4 × M²))
The
square
root
of
T/T∗,
divided
by
M
:
p
falls
as
M
rises.
p/p∗
=
(
1
0.5
)
×
√(
2.4
2 + 0.4 ×
0.5
2
)
p/p∗
=
1
0.5
×
√(
2.4
2 + 0.4 × 0.25
)
p/p∗
=
1
0.5
×
√(
2.4
2 + 0.1
)
p/p∗
=
1
0.5
×
√(
2.4
2.1
)
p/p∗
=
1
0.5
×
√
1.143
p/p∗
=
1
0.5
×
1.069
p/p∗ = 2 × 1.069
p/p∗ = 2.138
Step 5 · Find stagnation pressure ratio (p₀/p₀∗)
Use p₀/p₀∗ = (1 ÷ M) × ((2 + 0.4 × M²) ÷ 2.4)³
Friction
loses
stagnation
pressure:
p₀
is
least
at
M
=
1,
so
this
ratio
is
at
least
1.
p₀/p₀∗
=
(
1
0.5
)
×
(
2 + 0.4 ×
0.5
2
2.4
)³
p₀/p₀∗
=
1
0.5
×
(
2 + 0.4 × 0.25
2.4
)³
p₀/p₀∗
=
1
0.5
×
(
2 + 0.1
2.4
)³
p₀/p₀∗
=
1
0.5
×
(
2.1
2.4
)
3
p₀/p₀∗
=
1
0.5
×
0.875
3
p₀/p₀∗
=
1
0.5
×
0.6699
p₀/p₀∗ = 2 × 0.6699
p₀/p₀∗ = 1.34
Step 6 · Find velocity ratio (V/V∗)
Use V/V∗ = M × √(2.4 ÷ (2 + 0.4 × M²))
M
times
the
square
root
of
T/T∗:
the
speed
over
the
speed
at
M
=
1.
V/V∗
=
0.5
×
√(
2.4
2 + 0.4 ×
0.5
2
)
V/V∗
=
0.5
×
√(
2.4
2 + 0.4 × 0.25
)
V/V∗
=
0.5
×
√(
2.4
2 + 0.1
)
V/V∗
=
0.5
×
√(
2.4
2.1
)
V/V∗
=
0.5
×
√
1.143
V/V∗ = 0.5 × 1.069
V/V∗ = 0.5345
Check
Put the numbers back in. Both sides should match.
1.069
=
1 −
0.5
2
1.4 ×
0.5
2
+
(
2.4
2.8
)
×
ln(2.4
×
0.5
2
2 + 0.4 ×
0.5
2
)
✓
2.673
=
1.069
×
0.05
4 × 0.005
✓
1.143
=
2.4
2 + 0.4 ×
0.5
2
✓
2.138
=
(
1
0.5
)
×
√(
2.4
2 + 0.4 ×
0.5
2
)
✓
1.34
=
(
1
0.5
)
×
(
2 + 0.4 ×
0.5
2
2.4
)³
✓
0.5345
=
0.5
×
√(
2.4
2 + 0.4 ×
0.5
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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