---
title: "Steam nozzle 4 MPa: mass flow, exit velocity, area"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 4
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:steam-nozzle-4mpa-450c-to-2mpa"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/steam-nozzle-4mpa-450c-to-2mpa
site: BCT Engineering
---

# Steam nozzle 4 MPa: mass flow, exit velocity, area

Fundamental of Thermodynamics & Heat Transfer · Chapter 4: First Law of Thermodynamics

- Asked in: 2070 Magh, 8 marks
- Video: [4.31 Steam enters a nozzle operating at steady state at 20 bar, 2808°C,  with a velocity of 80 m/s.](https://www.youtube.com/watch?v=VCc8OLxYd9Q) — GR Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/steam-nozzle-4mpa-450c-to-2mpa
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-4

## Question

Steam at 4 MPa, 450°C enters a nozzle operating at steady state with a velocity of 50 m/s. Steam leaves the nozzle at 2 MPa and 300°C. The inlet area of the nozzle is 80 cm² and heat loss from the nozzle surface occurs at the rate of 100 kW. Determine:
i) The mass flow rate of steam
ii) The exit velocity of the steam and
iii) The exit area of the nozzle
[Refer the attached table for the properties of steam]

## Short answer

Using steam table data, mass flow rate ṁ = A₁V₁/v₁ = 5.0 kg/s. Exit velocity from energy equation is 759 m/s. Exit area A₂ = 8.26 cm².

## Answer

**Given:** steam nozzle, steady flow: P₁ = 4 MPa, T₁ = 450 °C, V₁ = 50 m/s, A₁ = 80 cm² = 0.008 m²; exit 2 MPa, 300 °C; heat loss Q̇ = −100 kW; W = 0, ΔPE = 0.

**Steam table values:**
- State 1 (4 MPa, 450 °C): v₁ = 0.08004 m³/kg, h₁ = 3331.0 kJ/kg
- State 2 (2 MPa, 300 °C): v₂ = 0.12550 m³/kg, h₂ = 3024.2 kJ/kg

#### i) Mass flow rate
ṁ = A₁V₁/v₁ = 0.008 × 50/0.08004 = **4.997 kg/s ≈ 5.0 kg/s**

#### ii) Exit velocity
Energy equation: Q̇ = ṁ [ (h₂ − h₁) + (V₂² − V₁²)/2 ]

V₂²/2 = V₁²/2 + (h₁ − h₂) + Q̇/ṁ  
 = 50²/2 + (3331.0 − 3024.2) × 1000 + (−100 × 1000)/4.997  
 = 1250 + 306 800 − 20 012 = 288 038 J/kg

V₂ = √(2 × 288 038) = **759 m/s**

#### iii) Exit area
A₂ = ṁ v₂/V₂ = 4.997 × 0.1255/759 = 8.26 × 10⁻⁴ m² = **8.26 cm²**

**Answer: ṁ = 5.0 kg/s; V₂ = 759 m/s; A₂ = 8.26 cm².**

## Check yourself

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

1. Mass flow:
   A. ṁ = A1V1/v1 ≈ 5.0 kg/s
   B. ṁ = A1 v1/V1
   C. 0.5 kg/s
   D. 50 kg/s
   Answer: A. 0.008 × 50 / 0.08004. (Key point: Mass flow)

2. Exit velocity from:
   A. Δh = 0
   B. Q̇ = ṁ[(h2 - h1) + (V2² - V1²)/2]
   C. V2² = 2Δh only
   D. Ẇ = ṁΔh
   Answer: B. W = 0, heat loss included. (Key point: Energy equation)

3. Heat loss makes V2:
   A. Higher
   B. Unchanged
   C. Slightly lower than without loss
   D. Zero
   Answer: C. Energy leaves, less KE. (Key point: Heat loss effect)

4. Exit area:
   A. 80 cm²
   B. A2 = ṁ V2/v2
   C. 826 cm²
   D. A2 = ṁ v2/V2 ≈ 8.26 cm²
   Answer: D. Using v2 = 0.1255 and V2 = 759. (Key point: Exit area)

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