---
title: "Air nozzle: exit temperature and exit area"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 4
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:nozzle-air-exit-temperature-area"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/nozzle-air-exit-temperature-area
site: BCT Engineering
---

# Air nozzle: exit temperature and exit area

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

- Asked in: 2072 Magh, 8 marks
- Video: [ch4-TEST-nozzle-IG-PG](https://www.youtube.com/watch?v=5iNkuxLGwOQ) — Subrata Bhattacharjee
- 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/nozzle-air-exit-temperature-area
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-4

## Question

Air enters a nozzle steadily at 300 KPa, 127°C and with a velocity of 40 m/s and leaves at 100 KPa and with a velocity of 300 m/s. The heat lost from nozzle surface is 20 KJ/Kg of the air. The inlet area of nozzle is 100 cm². Determine:
i) Exit temperature of the air and
ii) Exit area of the nozzle
[Take R = 287J/kgK and Cp = 1005 J/kgK]

## Short answer

Exit temperature T₂ = 336 K (63 °C), obtained from energy equation with heat loss q = (h₂-h₁)+(V₂²-V₁²)/2. Mass flow is 1.045 kg/s; exit area A₂ = 33.6 cm².

## Answer

**Given:** air, steady flow: P₁ = 300 kPa, T₁ = 127 °C = 400 K, V₁ = 40 m/s, A₁ = 100 cm² = 0.01 m²; P₂ = 100 kPa, V₂ = 300 m/s; heat lost q = 20 kJ/kg. W = 0, ΔPE = 0.

**Energy equation per kg:**  
−q_loss = (h₂ − h₁) + (V₂² − V₁²)/2

c_p (T₂ − T₁) = −20 − (300² − 40²)/2000 = −20 − 44.2 = −64.2 kJ/kg

#### i) Exit temperature
T₂ = T₁ − 64.2/1.005 = 400 − 63.88 = **336.1 K (≈ 63 °C)**

#### ii) Exit area
Inlet specific volume: v₁ = RT₁/P₁ = 0.287 × 400/300 = 0.3827 m³/kg  
Mass flow: ṁ = A₁V₁/v₁ = 0.01 × 40/0.3827 = 1.045 kg/s

Exit: v₂ = RT₂/P₂ = 0.287 × 336.1/100 = 0.9647 m³/kg  
A₂ = ṁ v₂/V₂ = 1.045 × 0.9647/300 = 0.003361 m² = **33.6 cm²**

**Answer: T₂ = 336 K (63 °C); A₂ = 33.6 cm².**

## Check yourself

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

1. For the nozzle with heat loss:
   A. c_p(T2 - T1) + (V2² - V1²)/2 = -q_loss
   B. Δh = 0
   C. Δh + ΔKE = +q_loss
   D. ΔKE = 0
   Answer: A. W = 0 and heat leaves. (Key point: Energy equation)

2. T2 is:
   A. ≈ 400 K
   B. ≈ 336 K, lower than T1
   C. ≈ 464 K
   D. ≈ 300 K
   Answer: B. c_p ΔT = -20 - 44.2 = -64.2 kJ/kg. (Key point: Exit temperature)

3. Mass flow is:
   A. ṁ = A1 v1/V1
   B. ṁ = 0.01 kg/s
   C. ṁ = A1V1/v1 = 1.045 kg/s
   D. ṁ = A2V2
   Answer: C. v1 = RT1/P1 = 0.3827. (Key point: Mass flow)

4. Exit area:
   A. 100 cm²
   B. A2 = ṁ V2/v2
   C. 3.36 cm²
   D. A2 = ṁ v2/V2 = 33.6 cm²
   Answer: D. v2 = 0.9647 m³/kg. (Key point: Exit area)

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