---
title: "Convergent-divergent steam nozzle design"
subject: "Fluid Machines"
chapter: 5
key: "semester_5/fluid_machines:steam-nozzle-numerical"
source: https://www.bctengineer.com/q/semester_5/fluid_machines/steam-nozzle-numerical
site: BCT Engineering
---

# Convergent-divergent steam nozzle design

Fluid Machines · Chapter 5: Steam turbine and Hydraulic machine

- Asked in: year not given, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/fluid_machines/steam-nozzle-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/fluid_machines/chapter-5

## Question

Steam at 10 bar and 250 degrees C (specific volume 0.2327 m³/kg) expands isentropically in a convergent-divergent nozzle to a back pressure of 1.5 bar. The mass flow rate is 1.5 kg/s. Taking the index of expansion n = 1.3 for superheated steam and neglecting the inlet velocity, find (a) the throat pressure, (b) the throat velocity and throat area, and (c) the exit velocity, exit area, and diameters at throat and exit.

## Short answer

Using critical pressure ratio for superheated steam (0.546): throat pressure 5.46 bar. Throat velocity √(2n/(n+1) p1v1) = 512.9 m/s, throat area 10.84 cm² (diameter 37.2 mm). Exit velocity 845.6 m/s, exit area 17.76 cm² (diameter 47.6 mm).

## Answer

**Given:** $p_1=10$ bar $=10^6$ Pa, $v_1=0.2327$ m³/kg, $n=1.3$, $p_2=1.5$ bar, $m=1.5$ kg/s.

$p_1v_1=232.7$ kJ/kg.

#### (a) Throat (critical) pressure

$$
\frac{p_t}{p_1}=\left(\frac{2}{n+1}\right)^{\frac{n}{n-1}}=\left(\frac{2}{2.3}\right)^{4.333}=0.5457
$$

$$
p_t=0.5457\times10=5.46\ \text{bar}
$$

Since $p_2/p_1=0.150<0.5457$, the exit pressure is below the critical pressure, so a convergent-divergent nozzle is needed.

#### (b) Throat velocity and area

$$
V_t=\sqrt{\frac{2n}{n+1}p_1v_1}=\sqrt{\frac{2\times1.3}{2.3}\times10^6\times0.2327}=512.9\ \text{m/s}
$$

Specific volume at throat:

$$
v_t=v_1\left(\frac{p_1}{p_t}\right)^{1/n}=0.2327\times\left(\frac{1}{0.5457}\right)^{1/1.3}=0.3708\ \text{m}^3/\text{kg}
$$

$$
A_t=\frac{m\,v_t}{V_t}=\frac{1.5\times0.3708}{512.9}=10.84\ \text{cm}^2\quad\Big(d_t=37.2\ \text{mm}\Big)
$$

#### (c) Exit conditions

$$
V_2=\sqrt{\frac{2n}{n-1}p_1v_1\left[1-\left(\frac{p_2}{p_1}\right)^{\frac{n-1}{n}}\right]}
=\sqrt{8.667\times10^6\times0.2327\times\left[1-0.6455\right]}=845.6\ \text{m/s}
$$

$$
v_2=v_1\left(\frac{p_1}{p_2}\right)^{1/n}=0.2327\times\left(\frac{10}{1.5}\right)^{1/1.3}=1.0013\ \text{m}^3/\text{kg}
$$

$$
A_2=\frac{m\,v_2}{V_2}=\frac{1.5\times1.0013}{845.6}=17.76\ \text{cm}^2\quad\Big(d_2=47.6\ \text{mm}\Big)
$$

Area ratio $A_2/A_t=1.64$.

**Answer:** (a) $p_t=5.46$ bar; (b) $V_t=512.9$ m/s, $A_t=10.84$ cm²; (c) $V_2=845.6$ m/s, $A_2=17.76$ cm², $d_t=37.2$ mm, $d_2=47.6$ mm.

## Check yourself

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

1. What is the throat pressure for steam at 10 bar with n = 1.3?
   A. About 5.46 bar
   B. About 1.5 bar
   C. About 5.77 bar
   D. About 10 bar
   Answer: A. pt = 0.5457 x 10 bar = 5.46 bar, using (2/(n+1))^(n/(n-1)). (Key point: Throat pressure)

2. Why is a convergent-divergent nozzle needed here?
   A. p2/p1 = 0.15 is below the critical ratio 0.5457
   B. The mass flow is high
   C. The steam is superheated
   D. The inlet velocity is zero
   Answer: A. When the back pressure ratio is below the critical ratio, expansion beyond the throat requires a diverging section. (Key point: Need for convergent-divergent)

3. How is the throat area calculated?
   A. A = m Vt / vt
   B. A = m vt / Vt
   C. A = Vt / (m vt)
   D. A = m / (Vt vt)
   Answer: B. From continuity m = A V / v, so A = m v / V = 1.5 x 0.3708 / 512.9 = 10.84 cm^2. (Key point: Throat area method)

4. How do the exit conditions compare with the throat?
   A. Exit velocity is lower and the area smaller
   B. Exit velocity is higher (845.6 m/s vs 512.9 m/s) and the area larger (17.76 vs 10.84 cm^2)
   C. Same velocity, larger area
   D. Exit area equals throat area
   Answer: B. Steam keeps expanding in the divergent part, speeding up, and its specific volume rises faster, so area grows. (Key point: Exit vs throat)

## Related in Fluid Machines

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