SME
Compressible Flow
Nozzle flow
Use this for “A nozzle with Aₑ/Aₜ = 1.6875 runs from 1000 kPa and 600 K through a 0.002 m² throat. Find Mₑ, pₑ, Tₑ and ṁ.”
Diagram
To
scale:
exit
width
throat
width
=
√
1.688
=
1.299.
Subsonic before the throat, M = 1 at it, supersonic after.
This is an example. Type your own values to start a new problem.
Aₑ/Aₜ
=
1
Mₑ
×
1
+
0.2
×
Mₑ
2
1.2
3
pₑ (kPa)
=
p₀ (kPa)
(1
+
0.2
×
Mₑ
2
)
3.5
T₀
Stagnation temperature
example · K
Aₜ
Throat area
example
Tₑ
Exit temperature
calculated · K
ṁ
Mass flow
calculated · kg/s
Formulas
Aₑ/Aₜ
=
(
1
Mₑ
)
×
(
1 + 0.2 ×
Mₑ
2
1.2
)³
1.6875
=
(
1
2
)
×
(
1 + 0.2 ×
2
2
1.2
)³
pₑ
=
p₀
÷
(1 + 0.2 ×
Mₑ
2
)
3.5
127,800
=
1,000,000
÷
(1 + 0.2 ×
2
2
)
3.5
Tₑ
=
T₀
1 + 0.2 ×
Mₑ
2
333.3
=
600
1 + 0.2 ×
2
2
ṁ
=
0.6847
×
p₀
×
Aₜ
÷
√
287 ×
T₀
3.3
=
0.6847
×
1,000,000
×
0.002
÷
√
287 × 600
Formulas are worked in Pa. The step-by-step shows the conversions.
Assumptions
•
Isentropic flow of air from the reservoir to the exit, γ = 1.4 and R = 287 J/(kg·K).
•
The design (shock-free) solution: the throat is choked and the exit is supersonic.
•
The back pressure equals pₑ, so the jet leaves without shocks or expansion fans.
Step-by-step
WE KNOW
p₀ = 1,000 kPa, T₀ = 600 K, Aₜ = 0.002 m², Aₑ/Aₜ = 1.6875
FIND
exit Mach number (Mₑ), exit pressure (pₑ), exit temperature (Tₑ), mass flow (ṁ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to Pa).
p₀ = 1,000 kPa = 1,000,000 Pa (1 kPa = 1000 Pa)
Step 2 · Find exit Mach number (Mₑ)
Use Aₑ/Aₜ = (1 ÷ Mₑ) × ((1 + 0.2 × Mₑ²) ÷ 1.2)³
Supersonic branch, Mₑ on both sides: guess Mₑ = 2, put it in on the right, and repeat until it settles. The last round is shown.
Mₑ
=
√(5
×
(1.2
×
((Aₑ/Aₜ) × 2)
1 ÷ 3
−
1))
Mₑ
=
√(5
×
(1.2
×
(1.6875 × 2)
1 ÷ 3
−
1))
Mₑ = √(5 × (1.2 × ∛3.375 − 1))
Mₑ = √(5 × (1.2 × 1.5 − 1))
Mₑ = √(5 × (1.8 − 1))
Mₑ
=
√
5 × 0.8
Mₑ
=
√
4
Mₑ = 2
Step 3 · Find exit pressure (pₑ)
Use pₑ = p₀ ÷ (1 + 0.2 × Mₑ²)3.5
The isentropic pressure ratio at the exit Mach number.
pₑ
=
1,000,000
÷
(1 + 0.2 ×
2
2
)
3.5
pₑ
=
1,000,000
÷
(1 + 0.2 × 4)
3.5
pₑ
=
1,000,000
÷
(1 + 0.8)
3.5
pₑ
=
1,000,000
1.8
3.5
pₑ
=
1,000,000
7.824
pₑ = 127,804.5255 Pa
Step 4 · Find exit temperature (Tₑ)
Use Tₑ = T₀ ÷ (1 + 0.2 × Mₑ²)
The isentropic temperature ratio at the exit Mach number.
Tₑ
=
600
1 + 0.2 ×
2
2
Tₑ
=
600
1 + 0.2 × 4
Tₑ
=
600
1 + 0.8
Tₑ
=
600
1.8
Tₑ = 333.3 K
Step 5 · Find mass flow (ṁ)
Use ṁ = 0.6847 × p₀ × Aₜ ÷ √(287 × T₀)
The
throat
is
choked
at
M
=
1:
it
passes
0.6847
p₀
÷
√
RT₀
per
unit
area.
ṁ
=
0.6847
×
1,000,000
×
0.002
÷
√
287 × 600
ṁ
=
0.6847
×
1,000,000
×
0.002
÷
√
172
,200
ṁ
=
0.6847
×
1,000,000
×
0.002
415
ṁ
=
684,700
×
0.002
415
ṁ
=
1,369.4
415
ṁ
=
3.3
kg/s
Step 6 · Convert the answers
Change each answer back to the unit you chose.
pₑ = 127,800 Pa = 127.8 kPa (1 kPa = 1000 Pa)
Check
Put the numbers back in. Both sides should match.
1.6875
=
(
1
2
)
×
(
1 + 0.2 ×
2
2
1.2
)³
✓
127,800
=
1,000,000
÷
(1 + 0.2 ×
2
2
)
3.5
✓
333.3
=
600
1 + 0.2 ×
2
2
✓
3.3
=
0.6847
×
1,000,000
×
0.002
÷
√
287 × 600
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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