SME
Propulsion · Problem type
Rocket thrust and specific impulse
Use this for “An engine burns 250 kg/s at vₑ = 2900 m/s, with pₑ = 70 kPa on a 1 m² exit at sea level. What are F and Iₛₚ?”
Diagram
F
F = 250 × 2,900 + (70,000 − 101,325) × 1
F = 693,700 N
Specific impulse: 282.8 s
This is an example. Type your own values to start a new problem.
F (N)
=
ṁ (kg/s)
×
vₑ (m/s)
+
(
pₑ (Pa)
−
pₐ (Pa)
)
×
Aₑ (m²)
Iₛₚ
Specific impulse
calculated
Formulas
F
=
ṁ
×
vₑ
+
(pₑ
−
pₐ)
×
Aₑ
693,675 = 250 × 2,900 + (70,000 − 101,325) × 1
Iₛₚ
=
F
ṁ
× 9.81
282.844
=
693,675
250 × 9.81
Assumptions
•
Steady flow; the exit pressure pₑ is uniform over the exit area Aₑ.
•
Outside pressure pₐ acts on the whole engine, so only pₑ − pₐ over Aₑ adds thrust.
•
g = 9.81 m/s² defines specific impulse.
Step-by-step
WE KNOW
ṁ = 250 kg/s, vₑ = 2,900 m/s, pₑ = 70,000 Pa, pₐ = 101,325 Pa, Aₑ = 1 m²
FIND
thrust (F), specific impulse (Iₛₚ)
Step 1 · Find thrust (F)
Use F = ṁ × vₑ + (pₑ − pₐ) × Aₑ
Momentum
thrust
ṁvₑ,
plus
the
pressure
term:
Pa
×
m
2
is
N.
F
=
250
×
2,900
+
(70,000
−
101,325)
×
1
F
=
250
×
2,900
+
(−31,325)
F
=
725,000
+
(−31,325)
F
=
693,675
N
Step 2 · Find specific impulse (Iₛₚ)
Use Iₛₚ = F ÷ (ṁ × 9.81)
Thrust over the weight of propellant burned each second.
Iₛₚ
=
693,675
250 × 9.81
Iₛₚ
=
693,675
2,452.5
Iₛₚ = 282.844 s
Check
Put the numbers back in. Both sides should match.
693,675 = 250 × 2,900 + (70,000 − 101,325) × 1 ✓
282.844
=
693,675
250 × 9.81
✓
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