---
title: "Convergent-divergent nozzle with efficiency"
subject: "Turbo Machines"
chapter: 4
key: "semester_7/turbo_machines:cd-nozzle-efficiency-exit-area-numerical"
source: https://www.bctengineer.com/q/semester_7/turbo_machines/cd-nozzle-efficiency-exit-area-numerical
site: BCT Engineering
---

# Convergent-divergent nozzle with efficiency

Turbo Machines · Chapter 4: Gas Turbine Nozzles

- Asked in: year not given, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/turbo_machines/cd-nozzle-efficiency-exit-area-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/turbo_machines/chapter-4

## Question

Air at 800 kPa and 600 K (stagnation values) expands in a convergent-divergent nozzle to a back pressure of 100 kPa. The nozzle efficiency is 92% (ratio of actual to isentropic enthalpy drop) and the mass flow is 2 kg/s. Find (a) the throat area, (b) the exit temperature and velocity, (c) the exit Mach number and (d) the exit area. Take $\gamma = 1.4$, $c_p = 1.005$ kJ/kg K and $R = 0.287$ kJ/kg K. Assume the throat flow is isentropic.

## Short answer

Throat flow chokes at p* = 422.6 kPa, giving throat area A* = 15.15 cm^2. Isentropic exit temperature T_es = 331.2 K; with 92% efficiency actual T_e = 352.7 K, exit velocity C_e = 705.0 m/s. Exit Mach number Ma_e = 1.87, exit area A_e = 28.72 cm^2.

## Answer

**Data:** $p_0 = 800$ kPa, $T_0 = 600$ K, $p_e = 100$ kPa, $\eta_n = 0.92$, $\dot m = 2$ kg/s.

#### (a) Throat area
Pressure ratio $100/800 = 0.125 < 0.528$, so the nozzle is choked and a divergent part is needed.
$$
\begin{aligned}
p^* &= 0.5283 \times 800 = 422.6\ \text{kPa}, \quad T^* = \frac{2 \times 600}{2.4} = 500.0\ \text{K} \\
C^* &= \sqrt{1.4 \times 287 \times 500.0} = 448.2\ \text{m/s}, \quad \rho^* = \frac{422.6}{0.287 \times 500.0} = 2.945\ \text{kg/m}^3 \\
A^* &= \frac{\dot m}{\rho^* C^*} = \frac{2}{2.945 \times 448.2} = 15.15\ \text{cm}^2
\end{aligned}
$$

#### (b) Exit temperature and velocity
Isentropic exit temperature:
$$
T_{es} = 600\left(\frac{100}{800}\right)^{0.2857} = 331.2\ \text{K}, \qquad \Delta h_s = 1.005 \times (600 - 331.2) = 270.1\ \text{kJ/kg}
$$
Actual enthalpy drop:
$$
\Delta h = \eta_n \Delta h_s = 0.92 \times 270.1 = 248.5\ \text{kJ/kg}
$$
$$
T_e = 600 - \frac{248.5}{1.005} = 352.7\ \text{K}, \qquad C_e = \sqrt{2 \times 1000 \times 248.5} = 705.0\ \text{m/s}
$$
(The isentropic velocity would be 735.0 m/s.)

#### (c) Exit Mach number
$$
a_e = \sqrt{1.4 \times 287 \times 352.7} = 376.5\ \text{m/s}, \qquad Ma_e = \frac{705.0}{376.5} = 1.87
$$

#### (d) Exit area
$$
\rho_e = \frac{100}{0.287 \times 352.7} = 0.988\ \text{kg/m}^3, \qquad A_e = \frac{\dot m}{\rho_e C_e} = \frac{2}{0.988 \times 705.0} = 28.72\ \text{cm}^2
$$

| Section | $T$ (K) | $p$ (kPa) | $C$ (m/s) | Area (cm²) |
|---|---|---|---|---|
| Throat | 500.0 | 422.6 | 448.2 | 15.15 |
| Exit | 352.7 | 100 | 705.0 | 28.72 |

**Answer:** $A^* = 15.15$ cm²; $T_e = 352.7$ K, $C_e = 705.0$ m/s; $Ma_e = 1.87$; $A_e = 28.72$ cm².

## Check yourself

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

1. Why is a divergent section needed for p0 = 800 kPa and back pressure 100 kPa?
   A. The pressure ratio 0.125 is below 0.528, so the throat is choked and further expansion is supersonic
   B. The temperature is above 500 K
   C. The efficiency is below 100%
   D. The mass flow is too small
   Answer: A. A pressure ratio below the critical value means sonic flow at the throat and supersonic expansion beyond it needs increasing area. (Key point: Need for divergent part)

2. The throat area is found from
   A. A* = m/(rho_e C_e)
   B. A* = m/(rho* C*) with p* = 0.5283 p0 and T* = 2T0/(gamma+1)
   C. A* = m/(rho_0 C_0)
   D. A* = m p_e
   Answer: B. At the throat Ma = 1, so p* = 422.6 kPa, T* = 500 K, C* = 448.2 m/s, giving A* = 15.15 cm^2. (Key point: Throat area method)

3. How is nozzle efficiency used to get the exit state?
   A. Multiply the isentropic exit pressure by 0.92
   B. Divide the isentropic enthalpy drop by 0.92
   C. Actual enthalpy drop = 0.92 x isentropic drop (270.1 kJ/kg), giving T_e = 352.7 K higher than T_es = 331.2 K
   D. Multiply the throat area by 0.92
   Answer: C. Delta h = eta_n Delta h_s = 248.5 kJ/kg; the exit is hotter than isentropic because of friction, and C_e = sqrt(2 x 248.5 x 1000) = 705 m/s. (Key point: Use of nozzle efficiency)

4. The exit Mach number is
   A. 1.00
   B. 1.87
   C. 2.50
   D. 0.94
   Answer: B. Ma_e = C_e/a_e = 705.0/376.5 = 1.87, supersonic as expected after the throat. (Key point: Exit Mach number)

5. The exit area is larger than the throat area because
   A. density at exit is higher
   B. the exit velocity is lower than at the throat
   C. density falls more than velocity rises, so A = m/(rho C) increases
   D. the nozzle efficiency is below one
   Answer: C. Density drops from 2.945 to 0.988 kg/m^3 while velocity rises only from 448 to 705 m/s, so A rises from 15.15 to 28.72 cm^2. (Key point: Exit area)

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