---
title: "Steady state work vs steady flow; turbine, nozzle"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 4
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:steady-state-work-vs-flow"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/steady-state-work-vs-flow
site: BCT Engineering
---

# Steady state work vs steady flow; turbine, nozzle

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

- Asked in: 2075 Baisakh, 6 marks
- Video: [Thermodynamics - Turbines, Compressors, and Pumps in 9 Minutes!](https://www.youtube.com/watch?v=AHVvKzXYIcY) — Less Boring Lectures
- 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/steady-state-work-vs-flow
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-4

## Question

Differentiate between steady state work applications and steady state flow applications with examples. Also write the functions and governing equations for an adiabatic turbine and adiabatic nozzle.

## Short answer

Work devices (turbine, compressor) exchange shaft work; flow devices (nozzle, throttle) do not. Adiabatic turbine: Ẇ_out = ṁ(h₁ − h₂) neglecting KE. Adiabatic nozzle: h₁ + V₁²/2 = h₂ + V₂²/2, so exit velocity V₂ ≈ √[2(h₁ − h₂)] for low inlet speed.

## Answer

In steady flow, the mass flow and the properties at each point do not vary with time (dm_cv/dt = 0, dE_cv/dt = 0), so Σṁ_i = Σṁ_e and Q̇ − Ẇ = ṁ[Δh + ΔKE + ΔPE]. Devices are grouped by whether they exchange shaft work.

| Basis | Steady state work devices | Steady state flow devices (no work) |
|---|---|---|
| Main purpose | Produce or absorb shaft work | Change velocity, pressure or temperature of the stream |
| Work term | Ẇ ≠ 0 | Ẇ = 0 |
| Examples | Turbine, compressor, pump, fan | Nozzle, diffuser, throttle valve, heat exchanger, mixing chamber |
| Heat transfer | Usually negligible (adiabatic) | Nozzle, throttle: negligible; heat exchanger: key effect |
| Governing equation | w = (h₁ − h₂) for turbine (ΔKE, ΔPE neglected) | Nozzle: h₁ + V₁²/2 = h₂ + V₂²/2 |

#### Adiabatic turbine
**Function:** expands high-pressure, high-temperature gas or steam to produce shaft work. Q = 0 and ΔPE ≈ 0:  
Ẇ_out = ṁ [ (h₁ − h₂) + (V₁² − V₂²)/2 ]  
The kinetic energy change is usually small, so w = h₁ − h₂.

#### Adiabatic nozzle
**Function:** accelerates the fluid by decreasing its pressure (increases its kinetic energy). Q = 0, W = 0, ΔPE = 0:  
h₁ + V₁²/2 = h₂ + V₂²/2  
**V₂ = √[ V₁² + 2 (h₁ − h₂) ]**; for inlet velocity ≈ 0, V₂ = √(2 Δh).

Mass balance for both: ṁ₁ = ṁ₂, ṁ = A V / v.

## Check yourself

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

1. Which is a steady-flow device with no shaft work?
   A. Nozzle
   B. Turbine
   C. Compressor
   D. Pump
   Answer: A. Nozzles, diffusers, throttles have Ẇ = 0. (Key point: Work vs flow devices)

2. An adiabatic turbine:
   A. Raises pressure using work input; w = h2 - h1
   B. Expands gas to produce shaft work; w ≈ h1 - h2
   C. Increases velocity with no work
   D. Mixes two streams
   Answer: B. Enthalpy drop becomes work output. (Key point: Turbine function)

3. The nozzle exit velocity is:
   A. V2 = sqrt(V1² - 2(h1 - h2))
   B. V2 = h1 - h2
   C. V2 = sqrt(V1² + 2(h1 - h2))
   D. V2 = V1 + (h1 - h2)
   Answer: C. From h1 + V1²/2 = h2 + V2²/2. (Key point: Nozzle equation)

4. A nozzle works by:
   A. Increasing pressure to cut velocity
   B. Adding heat to the stream
   C. Extracting shaft work
   D. Decreasing pressure to increase kinetic energy
   Answer: D. Pressure drop converts enthalpy into velocity. (Key point: Nozzle purpose)

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