---
title: "Convergent nozzle: choking and mass flow"
subject: "Turbo Machines"
chapter: 4
key: "semester_7/turbo_machines:convergent-nozzle-choking-mass-flow"
source: https://www.bctengineer.com/q/semester_7/turbo_machines/convergent-nozzle-choking-mass-flow
site: BCT Engineering
---

# Convergent nozzle: choking and mass flow

Turbo Machines · Chapter 4: Gas Turbine Nozzles

- Asked in: year not given, 8 marks
- Video: [Fluid Mechanics Lesson 15B: Compressible Flow and Choking in Converging Ducts](https://www.youtube.com/watch?v=YuQgcbvYiIc) — John Cimbala
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/turbo_machines/convergent-nozzle-choking-mass-flow
- Whole chapter: https://www.bctengineer.com/md/semester_7/turbo_machines/chapter-4

## Question

Air at a stagnation pressure of 500 kPa and a stagnation temperature of 500 K flows isentropically through a convergent nozzle of exit area 20 cm². Find the exit velocity, exit pressure and mass flow rate when the back pressure is (a) 350 kPa and (b) 200 kPa. Take $\gamma = 1.4$, $R = 0.287$ kJ/kg K.

## Short answer

Critical pressure p* = 264.1 kPa. (a) At 350 kPa back pressure the nozzle is unchoked: exit pressure 350 kPa, exit velocity 312.0 m/s, mass flow 1.685 kg/s. (b) At 200 kPa the nozzle is choked: exit pressure 264.1 kPa, exit velocity 409.2 m/s, mass flow 1.808 kg/s.

## Answer

Critical pressure ratio for air:
$$
\frac{p^*}{p_0} = \left(\frac{2}{\gamma + 1}\right)^{\gamma/(\gamma-1)} = 0.5283 \;\Rightarrow\; p^* = 0.5283 \times 500 = 264.1\ \text{kPa}
$$

#### (a) Back pressure 350 kPa
Since $350 > 264.1$ kPa, the nozzle is **not choked** and the exit pressure equals the back pressure, $p_e = 350$ kPa.
$$
T_e = T_0\left(\frac{p_e}{p_0}\right)^{0.2857} = 500 \times 0.7^{0.2857} = 451.6\ \text{K}
$$
$$
C_e = \sqrt{2c_p(T_0 - T_e)} = \sqrt{2 \times 1005 \times 48.4} = 312.0\ \text{m/s}
$$
$$
\rho_e = \frac{p_e}{RT_e} = \frac{350}{0.287 \times 451.6} = 2.701\ \text{kg/m}^3
$$
$$
\dot m = \rho_e C_e A_e = 2.701 \times 312.0 \times 0.0020 = 1.685\ \text{kg/s}
$$
Exit Mach number $= 312.0/\sqrt{1.4 \times 287 \times 451.6} = 0.73$.

#### (b) Back pressure 200 kPa
Since $200 < 264.1$ kPa, the nozzle is **choked**. The exit pressure is $p^* = 264.1$ kPa (the remaining drop to 200 kPa takes place outside the nozzle) and $Ma_e = 1$.
$$
T^* = \frac{2T_0}{\gamma + 1} = \frac{2 \times 500}{2.4} = 416.7\ \text{K}
$$
$$
C^* = \sqrt{\gamma R T^*} = \sqrt{1.4 \times 287 \times 416.7} = 409.2\ \text{m/s}
$$
$$
\rho^* = \frac{p^*}{RT^*} = \frac{264.1}{0.287 \times 416.7} = 2.209\ \text{kg/m}^3
$$
$$
\dot m_{max} = \rho^* C^* A = 2.209 \times 409.2 \times 0.0020 = 1.808\ \text{kg/s}
$$

Lowering the back pressure further does not change the mass flow; it can only be raised by increasing $p_0$ or $A$.

| Case | $p_e$ (kPa) | $C_e$ (m/s) | $\dot m$ (kg/s) |
|---|---|---|---|
| (a) 350 kPa | 350 | 312.0 | 1.685 |
| (b) 200 kPa (choked) | 264.1 | 409.2 | 1.808 |

**Answer:** (a) $C_e$ = 312.0 m/s, $\dot m$ = 1.685 kg/s; (b) choked, $p_e$ = 264.1 kPa, $C_e$ = 409.2 m/s, $\dot m$ = 1.808 kg/s.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For p0 = 500 kPa air, how do you decide whether the convergent nozzle is choked?
   A. Compare back pressure with p* = 0.5283 p0 = 264.1 kPa; choked if back pressure is lower
   B. Compare back pressure with p0
   C. Check if the exit area is above 20 cm^2
   D. Choked whenever the back pressure is above p*
   Answer: A. If the back pressure is below the critical pressure, the exit stays at p* and the nozzle is choked. (Key point: Critical pressure test)

2. At a back pressure of 350 kPa the exit pressure is
   A. 264.1 kPa
   B. 500 kPa
   C. 200 kPa
   D. 350 kPa
   Answer: D. 350 kPa exceeds p* = 264.1 kPa, so the nozzle is unchoked and the exit pressure equals the back pressure. (Key point: Case (a) not choked)

3. For the choked case (b), the exit temperature and velocity are found from
   A. T_e = T0 (200/500)^0.2857 and the isentropic velocity
   B. T* = 2T0/(gamma+1) = 416.7 K and C* = sqrt(gamma R T*) = 409.2 m/s
   C. T_e = T0 and C = 0
   D. The back pressure of 200 kPa directly
   Answer: B. Choked flow has Ma = 1 at the exit, so the sonic temperature and speed apply rather than the back-pressure values. (Key point: Case (b) choked exit conditions)

4. Once the nozzle is choked, lowering the back pressure further causes
   A. higher mass flow
   B. lower mass flow
   C. no change in mass flow; it increases only by raising p0 or A
   D. flow reversal
   Answer: C. Mass flow is fixed at its maximum (1.808 kg/s) by throat conditions; only higher p0 or larger area can raise it. (Key point: Effect of lowering back pressure)

## Related in Turbo Machines

- [Convergent-divergent nozzle with efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Acd-nozzle-efficiency-exit-area-numerical) — Turbo Machines, chapter 4

## Related in other subjects

- [Steam nozzle 4 MPa: mass flow, exit velocity, area](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Asteam-nozzle-4mpa-450c-to-2mpa) — Fundamental of Thermodynamics & Heat Transfer, chapter 4
- [Air nozzle: exit temperature and exit area](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Anozzle-air-exit-temperature-area) — Fundamental of Thermodynamics & Heat Transfer, chapter 4
- [Convergent-divergent steam nozzle design](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Asteam-nozzle-numerical) — Fluid Machines, chapter 5
- [Adiabatic air turbine: mass flow and areas](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Aturbine-1000k-to-400k-areas) — Fundamental of Thermodynamics & Heat Transfer, chapter 4, asked 2×