SME
Compressible Flow · Problem type
Rayleigh flow ratios
Use this for “Air flows at M = 0.4 in a heated duct. Find p/p∗, T/T∗, T₀/T₀∗ and p₀/p₀∗ for Rayleigh flow.”
Chart
R(0.4) ≈ 0.529
This is an example. Type your own values to start a new problem.
p/p∗
=
2.4
1
+
1.4
×
M
2
T₀/T₀∗
=
2.4
×
M
2
×
(2
+
0.4
×
M
2
)
(1
+
1.4
×
M
2
)
2
T/T∗
Temperature ratio
calculated
p₀/p₀∗
Stagnation pressure ratio
calculated
V/V∗
Velocity ratio
calculated
Formulas
p/p∗
=
2.4
1 + 1.4 ×
M
2
1.961
=
2.4
1 + 1.4 ×
0.4
2
T/T∗
=
M
2
×
(p/p∗)
2
0.6151
=
0.4
2
×
(1.961)
2
T₀/T₀∗
=
2.4
×
M
2
×
(2
+
0.4
×
M
2
)
÷
(1 + 1.4 ×
M
2
)
2
0.529
=
2.4
×
0.4
2
×
(2
+
0.4
×
0.4
2
)
÷
(1 + 1.4 ×
0.4
2
)
2
p₀/p₀∗
=
(p/p∗)
×
(
2 + 0.4 ×
M
2
2.4
)
3.5
1.157
=
1.961
×
(
2 + 0.4 ×
0.4
2
2.4
)
3.5
V/V∗
=
2.4
×
M
2
1 + 1.4 ×
M
2
0.3137
=
2.4
×
0.4
2
1 + 1.4 ×
0.4
2
Assumptions
•
Steady frictionless flow of air (γ = 1.4) in a duct of constant area, heat added or taken away.
•
Heating drives subsonic and supersonic flow toward M = 1; cooling drives it away from M = 1.
•
Starred values are at M = 1, where heating has raised T₀ as far as it can go.
Step-by-step
WE KNOW
M = 0.4
FIND
pressure ratio (p/p∗), temperature ratio (T/T∗), stagnation temperature ratio (T₀/T₀∗), stagnation pressure ratio (p₀/p₀∗), velocity ratio (V/V∗)
Step 1 · Find pressure ratio (p/p∗)
Use p/p∗ = 2.4 ÷ (1 + 1.4 × M²)
Momentum:
p
+
ρV²
=
p(1
+
γM²)
stays
the
same
along
the
duct,
so
the
pressure
falls
as
M
rises.
p/p∗
=
2.4
1 + 1.4 ×
0.4
2
p/p∗
=
2.4
1 + 1.4 × 0.16
p/p∗
=
2.4
1 + 0.224
p/p∗
=
2.4
1.224
p/p∗ = 1.96078
Step 2 · Find temperature ratio (T/T∗)
Use T/T∗ = M² × (p/p∗)²
The
same
mass
flow
through
the
same
area:
pM
÷
√T
is
the
same
everywhere,
and
M
=
1
at
the
star.
T/T∗
=
0.4
2
×
(1.96078)
2
T/T∗ = 0.16 × 3.845
T/T∗ = 0.6151
Step 3 · Find stagnation temperature ratio (T₀/T₀∗)
Use T₀/T₀∗ = 2.4 × M² × (2 + 0.4 × M²) ÷ (1 + 1.4 × M²)²
How
far
the
heat
added
so
far
has
taken
T₀
toward
the
most
it
can
be,
at
M
=
1.
T₀/T₀∗
=
2.4
×
0.4
2
×
(2
+
0.4
×
0.4
2
)
÷
(1 + 1.4 ×
0.4
2
)
2
T₀/T₀∗
=
2.4
×
0.4
2
×
(2
+
0.4
×
0.16)
÷
(1 + 1.4 × 0.16)
2
T₀/T₀∗
=
2.4
×
0.4
2
×
(2
+
0.064)
÷
(1 + 0.224)
2
T₀/T₀∗
=
2.4
×
0.4
2
×
2.064
1.224
2
T₀/T₀∗
=
2.4
×
0.16
×
2.064
1.498
T₀/T₀∗
=
0.384
×
2.064
1.498
T₀/T₀∗
=
0.7926
1.498
T₀/T₀∗ = 0.529
Step 4 · Find stagnation pressure ratio (p₀/p₀∗)
Use p₀/p₀∗ = (p/p∗) × ((2 + 0.4 × M²) ÷ 2.4)3.5
Bring
each
end
to
rest:
p₀
p
=
(1 + 0.2
M
²)
3.5
,
which
is
1.2
3.5
at
the
star.
p₀/p₀∗
=
1.96078
×
(
2 + 0.4 ×
0.4
2
2.4
)
3.5
p₀/p₀∗
=
1.9608
×
(
2 + 0.4 × 0.16
2.4
)
3.5
p₀/p₀∗
=
1.9608
×
(
2 + 0.064
2.4
)
3.5
p₀/p₀∗
=
1.9608
×
(
2.064
2.4
)
3.5
p₀/p₀∗
=
1.9608
×
0.86
3.5
p₀/p₀∗ = 1.9608 × 0.58985
p₀/p₀∗ = 1.157
Step 5 · Find velocity ratio (V/V∗)
Use V/V∗ = 2.4 × M² ÷ (1 + 1.4 × M²)
The
same
mass
flow
through
the
same
area:
V
V
∗
=
ρ∗
÷
ρ.
V/V∗
=
2.4
×
0.4
2
1 + 1.4 ×
0.4
2
V/V∗
=
2.4
×
0.4
2
1 + 1.4 × 0.16
V/V∗
=
2.4
×
0.4
2
1 + 0.224
V/V∗
=
2.4
×
0.4
2
1.224
V/V∗
=
2.4
×
0.16
1.224
V/V∗
=
0.384
1.224
V/V∗ = 0.3137
Check
Put the numbers back in. Both sides should match.
1.96078
=
2.4
1 + 1.4 ×
0.4
2
✓
0.6151
=
0.4
2
×
(1.96078)
2
✓
0.529
=
2.4
×
0.4
2
×
(2
+
0.4
×
0.4
2
)
÷
(1 + 1.4 ×
0.4
2
)
2
✓
1.157
=
1.96078
×
(
2 + 0.4 ×
0.4
2
2.4
)
3.5
✓
0.3137
=
2.4
×
0.4
2
1 + 1.4 ×
0.4
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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