---
title: "Steam turbine: entropy generation per kg"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 5
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:steam-turbine-entropy-generation-500k"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/steam-turbine-entropy-generation-500k
site: BCT Engineering
---

# Steam turbine: entropy generation per kg

Fundamental of Thermodynamics & Heat Transfer · Chapter 5: Second Law of Thermodynamics

- Asked in: 2069 Bhadra, 8 marks
- 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-turbine-entropy-generation-500k
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-5

## Question

The conditions of steam at entrance and exit of a turbine are: h1 = 3456.5 kJ/kg, S1 = 7.2338 kJ/kgK, V1 = 150 m/s; and h2 = 2792.8 kJ/kg, S2 = 7.4665 kJ/kgK, V2 = 100 m/s respectively. The work output per kg of steam flow is 600 kJ. Heat transfer between of 500K. Determine the entropy generation per kg steam flow.

## Short answer

The entropy generation per kg of steam is 0.373 kJ/kg·K. Heat loss of 69.95 kJ/kg at 500 K contributes an entropy transfer of –0.1399 kJ/kg·K, while the steam's entropy increases by 0.2327 kJ/kg·K.

## Answer

**Reading of the problem.** The statement "Heat transfer between of 500K" is read as: the turbine casing loses heat to surroundings at a boundary temperature T_b = 500 K.

**Given:** h₁ = 3456.5 kJ/kg, s₁ = 7.2338 kJ/kg·K, V₁ = 150 m/s; h₂ = 2792.8 kJ/kg, s₂ = 7.4665 kJ/kg·K, V₂ = 100 m/s; work w = 600 kJ/kg.

#### Step 1: Heat transfer per kg (steady-flow energy equation)
q − w = (h₂ − h₁) + (V₂² − V₁²)/2

(V₂² − V₁²)/2 = (10000 − 22500)/2 = −6250 J/kg = −6.25 kJ/kg
h₂ − h₁ = 2792.8 − 3456.5 = −663.7 kJ/kg

q = 600 + (−663.7) + (−6.25) = **−69.95 kJ/kg**

The negative sign means heat is lost by the steam to the surroundings.

#### Step 2: Entropy balance for the steady-flow device (per kg)
s_gen = (s₂ − s₁) − q/T_b

s₂ − s₁ = 7.4665 − 7.2338 = 0.2327 kJ/kg·K
q/T_b = −69.95/500 = −0.1399 kJ/kg·K

s_gen = 0.2327 + 0.1399

**Answer: s_gen = 0.3726 kJ/kg·K ≈ 0.373 kJ/kg·K** (positive, so the process is irreversible and possible).

## Check yourself

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

1. Heat transfer per kg is found from:
   A. q = w + (h2 − h1) + (V2^2 − V1^2)/2
   B. q = h2 − h1
   C. q = w
   D. q = w − (h2 − h1)
   Answer: A. The steady-flow energy equation gives q = −69.95 kJ/kg. (Key point: Heat transfer from energy equation)

2. q = −69.95 kJ/kg means:
   A. heat is added to the steam
   B. the process is adiabatic
   C. work is input
   D. heat is lost by steam to surroundings
   Answer: D. Negative q indicates heat leaving. (Key point: Sign of heat)

3. The entropy generation per kg is:
   A. s_gen = s2 − s1 + q/T_b
   B. s_gen = (s2 − s1) − q/T_b
   C. s_gen = s2 − s1
   D. s_gen = q/T_b
   Answer: B. Subtract the entropy transfer by heat from the entropy change. (Key point: Entropy balance)

4. The result s_gen ≈ 0.373 kJ/kg·K is:
   A. negative
   B. zero
   C. positive, so irreversible and possible
   D. impossible
   Answer: C. 0.2327 + 0.1399 = 0.3726. (Key point: Result)

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