SME
Compressible Flow · Problem type
Mach number after a length of duct
Use this for “Air enters a 5 cm duct at M = 0.5 with f = 0.005. What is the Mach number 2 m downstream?”
Chart
F(0.6716) ≈ 0.2691
F(M) = F₁(M)
The curves cross at M = 0.5.
This is an example. Type your own values to start a new problem.
M₁
Inlet Mach number
example
f
Fanning friction factor
example
D
Diameter
example
L
Duct length
example
(4fL∗/D)₁
Inlet friction parameter
calculated
(4fL∗/D)₂
Exit friction parameter
calculated
M₂
Exit Mach number
calculated
T₂/T₁
Temperature ratio
calculated
p₂/p₁
Pressure ratio
calculated
Formulas
(
4
fL
∗/
D
)₁
=
1 −
M₁
2
1.4 ×
M₁
2
+
(
2.4
2.8
)
×
ln(2.4
×
M₁
2
2 + 0.4 ×
M₁
2
)
1.06906
=
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
)
(
4
fL
∗/
D
)₂
=
(
4
fL
∗/
D
)₁
−
4
×
f
×
L
D
0.26906
=
1.06906
−
4
×
0.005
×
2
0.05
(
4
fL
∗/
D
)₂
=
1 −
M₂
2
1.4 ×
M₂
2
+
(
2.4
2.8
)
×
ln(2.4
×
M₂
2
2 + 0.4 ×
M₂
2
)
0.26906
=
1 −
0.6716
2
1.4 ×
0.6716
2
+
(
2.4
2.8
)
×
ln(2.4
×
0.6716
2
2 + 0.4 ×
0.6716
2
)
T₂/T₁
=
2 + 0.4 ×
M₁
2
2 + 0.4 ×
M₂
2
0.9631
=
2 + 0.4 ×
0.5
2
2 + 0.4 ×
0.6716
2
p₂/p₁
=
(
M₁
M₂
)
×
√
T₂/T₁
0.7306
=
(
0.5
0.6716
)
×
√
0.9631
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 along the duct.
•
Inlet and exit share one choking point, so the exit 4fL∗/D is the inlet one less 4fL ÷ D.
•
The flow stays on the side of M = 1 it entered: friction alone never carries it across.
Step-by-step
WE KNOW
M₁ = 0.5, f = 0.005, D = 0.05 m, L = 2 m
FIND
inlet friction parameter ((4fL∗/D)₁), exit friction parameter ((4fL∗/D)₂), exit Mach number (M₂), temperature ratio (T₂/T₁), pressure ratio (p₂/p₁)
Step 1 · Find inlet friction parameter ((4fL∗/D)₁)
Use (4fL∗/D)₁ = (1 − M₁²) ÷ (1.4 × M₁²) + (2.4 ÷ 2.8) × ln(2.4 × M₁² ÷ (2 + 0.4 × M₁²))
The
Fanno
function
at
the
inlet:
the
friction
parameter
of
the
length
L
∗
to
choking.
(
4
fL
∗/
D
)₁
=
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
(
4
fL
∗/
D
)₁
=
1.06906
Step 2 · Find exit friction parameter ((4fL∗/D)₂)
Use (4fL∗/D)₂ = (4fL∗/D)₁ − 4 × f × L ÷ D
The
duct
uses
up
4
f
L
D
of
the
friction
the
inlet
flow
could
take.
(
4
fL
∗/
D
)₂
=
1.06906
−
4
×
0.005
×
2
0.05
(
4
fL
∗/
D
)₂
=
1.0691
−
0.02
×
2
0.05
(
4
fL
∗/
D
)₂
=
1.0691
−
0.04
0.05
(
4
fL
∗/
D
)₂
=
1.0691
−
0.8
(
4
fL
∗/
D
)₂
=
0.26906
Step 3 · Find exit Mach number (M₂)
Use (4fL∗/D)₂ = (1 − M₂²) ÷ (1.4 × M₂²) + (2.4 ÷ 2.8) × ln(2.4 × M₂² ÷ (2 + 0.4 × M₂²))
No
rearrangement
puts
M₂
on
its
own,
so
it
is
found
by
trial:
each
next
try
is
where
a
straight
line
through
the
last
two
meets
the
target.
Try
M₂
=
0.67:
1 −
0.67
2
1.4 ×
0.67
2
+
(
2.4
2.8
)
×
ln(2.4
×
0.67
2
2 + 0.4 ×
0.67
2
)
=
0.27295
(want
0.26906)
Try
M₂
=
0.68:
1 −
0.68
2
1.4 ×
0.68
2
+
(
2.4
2.8
)
×
ln(2.4
×
0.68
2
2 + 0.4 ×
0.68
2
)
=
0.24978
(want
0.26906)
Try
M₂
=
0.67168:
1 −
0.67168
2
1.4 ×
0.67168
2
+
(
2.4
2.8
)
×
ln(2.4
×
0.67168
2
2 + 0.4 ×
0.67168
2
)
=
0.26894
(want
0.26906)
Try
M₂
=
0.67163:
1 −
0.67163
2
1.4 ×
0.67163
2
+
(
2.4
2.8
)
×
ln(2.4
×
0.67163
2
2 + 0.4 ×
0.67163
2
)
=
0.26906
(want
0.26906)
M₂
=
0.6716
Step 4 · Find temperature ratio (T₂/T₁)
Use T₂/T₁ = (2 + 0.4 × M₁²) ÷ (2 + 0.4 × M₂²)
T₀
stays
the
same
in
adiabatic
flow,
so
T
falls
as
M
rises:
T₀
T
=
1
+
0.2M²
at
each
end.
T₂/T₁
=
2 + 0.4 ×
0.5
2
2 + 0.4 ×
0.6716
2
T₂/T₁
=
2 + 0.4 × 0.25
2 + 0.4 × 0.451
T₂/T₁
=
2 + 0.1
2 + 0.1804
T₂/T₁
=
2.1
2.1804
T₂/T₁ = 0.9631
Step 5 · Find pressure ratio (p₂/p₁)
Use p₂/p₁ = (M₁ ÷ M₂) × √(T₂/T₁)
The
same
mass
flow
through
the
same
area:
pV
T
is
the
same
at
both
ends.
p₂/p₁
=
(
0.5
0.6716
)
×
√
0.9631
p₂/p₁
=
0.5
0.6716
×
0.9814
p₂/p₁ = 0.7445 × 0.9814
p₂/p₁ = 0.7306
Check
Put the numbers back in. Both sides should match.
1.06906
=
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
)
✓
0.26906
=
1.06906
−
4
×
0.005
×
2
0.05
✓
0.26906
=
1 −
0.67163
2
1.4 ×
0.67163
2
+
(
2.4
2.8
)
×
ln(2.4
×
0.67163
2
2 + 0.4 ×
0.67163
2
)
✓
0.9631
=
2 + 0.4 ×
0.5
2
2 + 0.4 ×
0.6716
2
✓
0.7306
=
(
0.5
0.6716
)
×
√
0.9631
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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