SME
Compressible Flow · Problem type
Area ratio A/A∗ and Mach number
Use this for “What area ratio A/A∗ gives M = 2? What subsonic M has A/A∗ = 1.6875?”
Diagram
T₀
T
=
1
+
0.2
×
2
2
=
1.8;
p₀
p
=
7.824.
Drawn
to
scale:
A
A
∗
=
1.688,
so
the
duct
has
passed
a
throat
and
widens
again.
This is an example. Type your own values to start a new problem.
A/A∗
=
1
M
×
1
+
0.2
×
M
2
1.2
3
B
Branch
calculated
Formulas
A/A∗
=
(
1
M
)
×
(
1 + 0.2 ×
M
2
1.2
)³
1.688
=
(
1
2
)
×
(
1 + 0.2 ×
2
2
1.2
)³
B
=
subsonic
if
M
<
1,
else
supersonic
2 ≥ 1, so supersonic
Assumptions
•
Isentropic flow of air, γ = 1.4, so 2 ÷ (γ + 1) = 1 ÷ 1.2 and (γ + 1) ÷ (2(γ − 1)) = 3.
•
A∗ is the throat area where the flow would reach M = 1, so A/A∗ is never below 1.
•
Each A/A∗ above 1 has two Mach numbers: subsonic before the throat, supersonic after it.
Step-by-step
WE KNOW
M = 2
FIND
branch (B), area ratio (A/A∗)
Step 1 · Find branch (B)
Use B = subsonic if M < 1, else supersonic
Below
M
=
1
the
duct
narrows
toward
the
throat;
at
or
above
it,
the
flow
is
supersonic
and
widens.
2 ≥ 1: supersonic
B
=
subsonic
if
2
<
1,
else
supersonic
B
=
supersonic
Step 2 · Find area ratio (A/A∗)
Use A/A∗ = (1 ÷ M) × ((1 + 0.2 × M²) ÷ 1.2)³
The
area–Mach
relation:
the
area
the
flow
needs
at
M
,
in
units
of
the
throat
area.
2 ≥ 1, on the supersonic branch: supersonic
A/A∗
=
(
1
2
)
×
(
1 + 0.2 ×
2
2
1.2
)³
A/A∗
=
1
2
×
(
1 + 0.2 × 4
1.2
)³
A/A∗
=
1
2
×
(
1 + 0.8
1.2
)³
A/A∗
=
1
2
×
(
1.8
1.2
)
3
A/A∗
=
1
2
×
1.5
3
A/A∗
=
1
2
×
3.375
A/A∗ = 0.5 × 3.375
A/A∗ = 1.688
Check
Put the numbers back in. Both sides should match.
1.688
=
(
1
2
)
×
(
1 + 0.2 ×
2
2
1.2
)³
✓
2 ≥ 1, so supersonic ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.