SME
Compressible Flow · Problem type
Heat added in a duct (Rayleigh)
Use this for “Air at M = 0.3 and T₀ = 400 K takes in 300 kJ/kg in a duct. Find the exit Mach number and the heat that chokes it.”
Chart
R(0.445) ≈ 0.6057
R(M) = R₁(M)
The curves cross at M = 0.3.
This is an example. Type your own values to start a new problem.
q (kJ/kg)
=
c
p
(
T₀₂ (K)
−
T₀₁ (K)
)
(T₀/T₀∗)₂
=
(T₀/T₀∗)₁
×
T₀₂ (K)
T₀₁ (K)
On the graph R is T₀/T₀∗: the level line R₁ is the inlet’s ratio, met at M₁; heating multiplies it by T₀₂ ÷ T₀₁, up to the dot at the exit.
M₁
Inlet Mach number
example
M₂
Exit Mach number
calculated
qₘₐₓ
Choking heat
calculated · kJ/kg
Formulas
q
=
1.005
×
(
T₀₂
−
T₀₁
)
300 = 1.005 × (698.5 − 400)
(T₀/T₀∗)₁
=
2.4
×
M₁
2
×
(2
+
0.4
×
M₁
2
)
÷
(1 + 1.4 ×
M₁
2
)
2
0.34686
=
2.4
×
0.3
2
×
(2
+
0.4
×
0.3
2
)
÷
(1 + 1.4 ×
0.3
2
)
2
(T₀/T₀∗)₂
=
((T₀/T₀∗)₁)
×
T₀₂
T₀₁
0.605711
=
0.34686
×
698.5
400
(T₀/T₀∗)₂
=
2.4
×
M₂
2
×
(2
+
0.4
×
M₂
2
)
÷
(1 + 1.4 ×
M₂
2
)
2
0.605711
=
2.4
×
0.445
2
×
(2
+
0.4
×
0.445
2
)
÷
(1 + 1.4 ×
0.445
2
)
2
qₘₐₓ
=
1.005
×
T₀₁
×
(
1
(T₀/T₀∗)₁
−
1)
757
=
1.005
×
400
×
(
1
0.34686
−
1)
Assumptions
•
Steady frictionless flow of air in a duct of constant area: cₚ = 1.005 kJ/(kg·K), γ = 1.4.
•
Heat q per kilogram raises T₀ alone: q = cₚ(T₀₂ − T₀₁).
•
Heating drives the flow toward M = 1 on the side it entered; qₘₐₓ brings it to M = 1.
Step-by-step
WE KNOW
M₁ = 0.3, T₀₁ = 400 K, q = 300 kJ/kg
FIND
exit stagnation temperature (T₀₂), inlet ratio ((T₀/T₀∗)₁), exit ratio ((T₀/T₀∗)₂), exit Mach number (M₂), choking heat (qₘₐₓ)
Step 1 · Find exit stagnation temperature (T₀₂)
Use q = 1.005 × (T₀₂ − T₀₁)
Divide
q
by
cₚ,
then
add
it
to
T₀₁
.
T₀₂
=
T₀₁
+
q
1.005
T₀₂
=
400
+
300
1.005
T₀₂
=
400
+
298.5
T₀₂
=
698.5075
K
Step 2 · Find inlet ratio ((T₀/T₀∗)₁)
Use (T₀/T₀∗)₁ = 2.4 × M₁² × (2 + 0.4 × M₁²) ÷ (1 + 1.4 × M₁²)²
The Rayleigh T₀ ratio at the inlet: how near T₀ already is to its most, at M = 1.
(T₀/T₀∗)₁
=
2.4
×
0.3
2
×
(2
+
0.4
×
0.3
2
)
÷
(1 + 1.4 ×
0.3
2
)
2
(T₀/T₀∗)₁
=
2.4
×
0.3
2
×
(2
+
0.4
×
0.09)
÷
(1 + 1.4 × 0.09)
2
(T₀/T₀∗)₁
=
2.4
×
0.3
2
×
(2
+
0.036)
÷
(1 + 0.126)
2
(T₀/T₀∗)₁
=
2.4
×
0.3
2
×
2.036
1.126
2
(T₀/T₀∗)₁
=
2.4
×
0.09
×
2.036
1.268
(T₀/T₀∗)₁
=
0.216
×
2.036
1.268
(T₀/T₀∗)₁
=
0.4398
1.268
(T₀/T₀∗)₁ = 0.34686
Step 3 · Find exit ratio ((T₀/T₀∗)₂)
Use (T₀/T₀∗)₂ = ((T₀/T₀∗)₁) × T₀₂ ÷ T₀₁
Both stations share one T₀∗, so the ratio grows with T₀.
(T₀/T₀∗)₂
=
0.34686
×
698.5075
400
(T₀/T₀∗)₂
=
242.3
400
(T₀/T₀∗)₂ = 0.605711
Step 4 · Find exit Mach number (M₂)
Use (T₀/T₀∗)₂ = 2.4 × M₂² × (2 + 0.4 × M₂²) ÷ (1 + 1.4 × M₂²)²
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
=
0.44:
2.4
×
0.44
2
×
(2
+
0.4
×
0.44
2
)
÷
(1 + 1.4 ×
0.44
2
)
2
=
0.59748
(want
0.605711)
Try
M₂
=
0.45:
2.4
×
0.45
2
×
(2
+
0.4
×
0.45
2
)
÷
(1 + 1.4 ×
0.45
2
)
2
=
0.61393
(want
0.605711)
Try
M₂
=
0.445:
2.4
×
0.445
2
×
(2
+
0.4
×
0.445
2
)
÷
(1 + 1.4 ×
0.445
2
)
2
=
0.60574
(want
0.605711)
Try
M₂
=
0.44498:
2.4
×
0.44498
2
×
(2
+
0.4
×
0.44498
2
)
÷
(1 + 1.4 ×
0.44498
2
)
2
=
0.60571
(want
0.605711)
M₂
=
0.445
Step 5 · Find choking heat (qₘₐₓ)
Use qₘₐₓ = 1.005 × T₀₁ × (1 ÷ ((T₀/T₀∗)₁) − 1)
The
heat
that
takes
T₀
from
T₀₁
up
to
T₀∗
=
T₀₁
T₀/T₀∗
₁,
where
the
flow
reaches
M
=
1.
qₘₐₓ
=
1.005
×
400
×
(
1
0.34686
−
1)
qₘₐₓ = 1.005 × 400 × (2.883 − 1)
qₘₐₓ = 1.005 × 400 × 1.883
qₘₐₓ = 402 × 1.883
qₘₐₓ = 757 kJ/kg
Check
Put the numbers back in. Both sides should match.
300 = 1.005 × (698.5075 − 400) ✓
0.34686
=
2.4
×
0.3
2
×
(2
+
0.4
×
0.3
2
)
÷
(1 + 1.4 ×
0.3
2
)
2
✓
0.605711
=
0.34686
×
698.5075
400
✓
0.605711
=
2.4
×
0.44498
2
×
(2
+
0.4
×
0.44498
2
)
÷
(1 + 1.4 ×
0.44498
2
)
2
✓
757
=
1.005
×
400
×
(
1
0.34686
−
1)
✓
Topic check
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