SME
Propulsion · Problem type
Ideal exhaust speed from the chamber state
Use this for “Gas with γ = 1.2 and R = 350 J/(kg·K) leaves a 3200 K chamber, expanding to 1% of the chamber pressure. Find vₑ.”
Diagram
To
scale:
exit
width
throat
width
=
√
11.87
=
3.445.
Subsonic before the throat, M = 1 at it, supersonic after.
This is an example. Type your own values to start a new problem.
(
vₑ (m/s)
)
2
=
2
×
γ
×
R (J/(kg·K))
×
Tc (K)
÷
(
γ
−
1)
×
(1
−
(
)
(
−
1)
÷
)
Formulas
vₑ
=
√(2
×
γ
×
R
×
Tc
γ
− 1
×
(1
−
(pₑ/pc)
(
γ
− 1) ÷
γ
))
2,684
=
√(2
×
1.2
×
350
×
3,200
1.2 − 1
×
(1
−
(0.01)
(1.2 − 1) ÷ 1.2
))
Assumptions
•
Isentropic expansion of a perfect gas with constant γ, from rest in the chamber.
•
R is the gas constant of the exhaust: Rᵤ = 8.314 J/(mol·K) over its molar mass.
•
The nozzle is drawn to the exit Mach number that pₑ ÷ pc gives at this γ.
Step-by-step
WE KNOW
γ = 1.2, R = 350 J/(kg·K), Tc = 3,200 K, pₑ/pc = 0.01
FIND
exhaust speed (vₑ)
Step 1 · Find exhaust speed (vₑ)
Use vₑ = √(2 × γ × R × Tc ÷ (γ − 1) × (1 − (pₑ/pc)(γ − 1) ÷ γ))
The enthalpy the gas gives up between the chamber and the exit becomes kinetic energy.
vₑ
=
√(2
×
1.2
×
350
×
3,200
1.2 − 1
×
(1
−
(0.01)
(1.2 − 1) ÷ 1.2
))
2
γ
RTc
γ
− 1
=
2
×
1.2
×
350
×
3,200
0.2
=
1.344 ×
10
7
(pₑ/pc)
(
γ
− 1) ÷
γ
=
0.01
0.1667
=
0.46416
vₑ
=
√(1.344
×
10
7
×
(1
−
0.46416))
=
√
7,201,705
vₑ = 2,684 m/s
Check
Put the numbers back in. Both sides should match.
2,684
=
√(2
×
1.2
×
350
×
3,200
1.2 − 1
×
(1
−
(0.01)
(1.2 − 1) ÷ 1.2
))
✓
Topic check
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