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Choked (Sonic) Mass Flow Rate

Mass flow rate through a nozzle throat when the flow is choked (locally sonic), independent of downstream pressure.

PropulsionRocketryGas Dynamics

Choked (Sonic) Mass Flow Rate Calculator

ṁ = (pc · At / √Tc) · √(γ/R) · (2/(γ+1))(γ+1)/(2(γ−1))
Solve for , pc, At, Tc, γ, or R
pc, At, Tc, γ, R
kg/s
Pa
K
J/kg·K
Solve for:
Result
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Mass Flow Rate vs. Chamber Pressure ṁ ∝ pc (for fixed Tc, γ, R, At)
ṁ(pc) for fixed At, Tc, γ, R Computed point
All values positive • γ > 1 • SI units

Interpretation

Choked mass flow rate: ṁ = (p_c·A_t / √T_c) · √(γ/R)·(2/(γ+1))^((γ+1)/(2(γ−1))). It is the maximum mass flow through a nozzle at sonic condition (throat). Example: p_c=5 MPa, A_t=0.002 m², T_c=3000 K, γ=1.3, R=300 → ṁ ≈ 50 kg/s.

mdot = (p_c * A_t / sqrt(T_c)) * sqrt(γ/R) * (2/(γ+1))^((γ+1)/(2*(γ-1)))
Choked (Sonic) Mass Flow Rate

Variables

SymbolQuantityUnit
mdotMass flow ratekg/s
p_cChamber pressurePa
A_tThroat aream2
T_cChamber temperatureK
γRatio of specific heats
RSpecific gas constantJ/(kg*K)

What it means

When the flow reaches sonic speed at the throat, the mass flow rate is choked and cannot increase further for a given upstream condition. This formula gives the mass flow rate through a nozzle under choked conditions. It is derived from the continuity equation and isentropic relations. This is crucial for rocket and jet engine performance: the mass flow determines thrust. Understanding choked flow is essential for propulsion system design and for analysing nozzle operation.

Worked example

Choked Mass Flow – Two Examples

Real‑World
Scenario: p_c = 7×10⁶ Pa, A_t = 0.05 m², T_c = 3500 K, γ = 1.2, R = 350. Find choked mass flow.
ParameterValue
p_c7×10⁶ Pa
A_t0.05 m²
T_c3500 K
γ1.2
R350
1ṁ = p_c·A_t/√T_c × √(γ/R) × (2/(γ+1))^((γ+1)/(2(γ-1))) = 7e6×0.05/√3500 × √(1.2/350) × (2/2.2)^(2.2/0.4) = 350000/59.16 × 0.05857 × 0.568² = 5916 × 0.05857 × 0.322 = 111.5 kg/s (approx)
Result ≈ 111.5 kg/s ✓ High flow
Scenario: p_c = 1×10⁷, A_t = 0.06, T_c = 3300, γ = 1.22, R = 400. Find ṁ.
ParameterValue
p_c1×10⁷
A_t0.06
T_c3300
γ1.22
R400
1ṁ = 1e7×0.06/√3300 × √(1.22/400) × (2/2.22)^(2.22/0.44) = 600000/57.45 × 0.0552 × 0.522 = 10444 × 0.0552 × 0.522 = 300.8 kg/s (approx)
Result ≈ 301 kg/s ✓ Very high
Key insight: Choked flow occurs at Mach 1 in the throat – mass flow is determined by chamber conditions.

Common mistakes

  • Choked (sonic) mass flow rate: ṁ = (p_c·A_t / √T_c) · √(γ/R) · (2/(γ+1))^((γ+1)/(2(γ−1))).
  • p_c: Chamber pressure (Pa).
  • A_t: Throat area (m²).
  • T_c: Chamber temperature (K).
  • Valid when flow is sonic at throat (M=1).
  • Mass flow is maximum for given p_c and T_c.

Applications

Choked (sonic) mass flow rate, ṁ = (p_c·A_t/√(T_c))·√(γ/R)·(2/(γ+1))^((γ+1)/(2(γ−1))), gives the maximum mass flow through a nozzle throat when the flow is sonic. It is used to size rocket engine throats, to compute chamber pressure, and to design turbopumps. Engineers use this to ensure that the throat area is adequate for the required mass flow, and to set operating conditions. By applying the choked flow equation, aerospace engineers can design stable, predictable rocket engines.

  • Rocket engine throat sizing and design
  • Chamber pressure and propellant flow rate determination
  • Turbopump and feed system design
  • Flow rate measurement and control in engine testing
  • Nozzle design and performance analysis

Frequently Asked Questions

Q01What is the Choked (Sonic) Mass Flow Rate used for?
A01

It gives the mass flow rate through a nozzle throat when the flow is choked (locally sonic), which is independent of downstream pressure.

Q02What do the variables pc, At, Tc, γ, and R represent?
A02

pc = chamber pressure (Pa)
At = throat area (m²)
Tc = chamber temperature (K)
γ = specific heat ratio
R = specific gas constant (J/kg·K)

Q03Why is the choked mass flow rate important?
A03

It determines the maximum mass flow through the nozzle, which sets the thrust capability of the engine.

Q04What are common mistakes when using this formula?
A04

  • Assuming mass flow depends on exit conditions once choked; choked flow depends only on upstream (chamber) conditions.
  • Using the wrong value of γ for combustion gases.
  • Forgetting to use absolute temperature (Kelvin).

Q05Give a worked example.
A05

pc = 10 MPa, At = 0.01 m², Tc = 3500 K, γ=1.2, R=300. ṁ = (10e6×0.01/√3500) × √(1.2/300) × (2/2.2)^((2.2)/(0.4)) = (100000/59.16) × √0.004 × (0.909)^5.5. Compute stepwise: 100000/59.16=1690; √0.004=0.0632; 0.909^5.5≈0.584. ṁ = 1690×0.0632×0.584 ≈ 62.4 kg/s.

Q06How does the chamber pressure affect the mass flow?
A06

ṁ is directly proportional to pc. Doubling chamber pressure doubles the mass flow.

Q07What is the effect of throat area on mass flow?
A07

ṁ is directly proportional to At. A larger throat allows more flow.

Q08How does chamber temperature affect mass flow?
A08

Higher Tc reduces ṁ (since ṁ ∝ 1/√Tc).

Q09What happens if the downstream pressure is lower than the throat pressure?
A09

The flow remains choked; the mass flow is still determined by the upstream conditions.

Q10How is the choked mass flow used in rocket engine design?
A10

It is used to size the throat and to predict the engine’s propellant consumption.