---
title: "Steam turbine 10 MPa, 550°C: isentropic efficiency"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 5
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:steam-turbine-10mpa-550c-isentropic-eff"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/steam-turbine-10mpa-550c-isentropic-eff
site: BCT Engineering
---

# Steam turbine 10 MPa, 550°C: isentropic efficiency

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

- Asked in: 2069 Poush, 8 marks
- Video: [Steam Turbine Isentropic Efficiency |Thermodynamics](https://www.youtube.com/watch?v=5uRyos0qdeA) — Engineering Hack
- 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-10mpa-550c-isentropic-eff
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-5

## Question

Steam enters an adiabatic turbine at 10 MPa and 550°C. Exit conditions are 0.06 MPa and a quality of 96%. Determine the isentropic efficiency and actual work output for a mass flow rate of 10 kg/s.

## Short answer

The isentropic efficiency is 83.5%. Actual work per kg of steam is 940.9 kJ/kg, giving a power output of 9.41 MW for 10 kg/s.

## Answer

**Steam table values (read from tables)**

Inlet, 10 MPa, 550 °C: h₁ = 3502.0 kJ/kg, s₁ = 6.7584 kJ/kg·K.

At 0.06 MPa (T_sat = 85.9 °C): h_f = 359.84, h_fg = 2293.0, h_g = 2652.9 kJ/kg; s_f = 1.1452, s_fg = 6.3859, s_g = 7.5311 kJ/kg·K.

#### Actual exit (x₂ = 0.96)
h₂ = h_f + x₂ h_fg = 359.84 + 0.96 × 2293.0 = 2561.1 kJ/kg

#### Isentropic exit (s₂s = s₁)
x₂s = (6.7584 − 1.1452)/6.3859 = 0.8790
h₂s = 359.84 + 0.8790 × 2293.0 = 2375.4 kJ/kg

#### Isentropic efficiency
η_T = (h₁ − h₂)/(h₁ − h₂s) = (3502.0 − 2561.1)/(3502.0 − 2375.4) = 940.9/1126.6

**η_T = 0.835 (83.5 %)**

#### Actual power
w_actual = h₁ − h₂ = 940.9 kJ/kg (adiabatic, kinetic energy neglected)

Ẇ = ṁ w = 10 × 940.9 = 9409 kW

**Answer: η_T ≈ 83.5 %, actual work = 940.9 kJ/kg, power ≈ 9.41 MW.**

## Check yourself

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

1. How is the isentropic efficiency of this turbine defined?
   A. (h2 - h2s)/(h1 - h2)
   B. (s2 - s1)/s1
   C. (h1 - h2)/(h1 - h2s)
   D. (h1 - h2s)/(h1 - h2)
   Answer: C. Turbine efficiency is actual work over isentropic work: (h1 - h2)/(h1 - h2s). (Key point: Isentropic efficiency definition)

2. How is the isentropic exit state 2s found?
   A. Use s2s = s1 at 0.06 MPa to get x2s, then h2s
   B. Use T2s = T1 at 0.06 MPa
   C. Use x2s = 0.96
   D. Use h2s = h1 at 0.06 MPa
   Answer: A. Isentropic means constant entropy, so s2s = s1 fixes the quality x2s and then h2s = hf + x2s*hfg. (Key point: Isentropic exit state)

3. Which expression gives the actual exit enthalpy h2?
   A. h2 = hf + 0.96 hfg
   B. h2 = hf + 0.96 sfg
   C. h2 = hg x 0.96
   D. h2 = h1 x 0.96
   Answer: A. With quality 0.96 at 0.06 MPa, h2 = hf + x*hfg = 359.84 + 0.96 x 2293.0 = 2561.1 kJ/kg. (Key point: Actual exit enthalpy)

4. What is the actual power output for 10 kg/s?
   A. 1126.6 kW
   B. 9409 kW
   C. 940.9 kW
   D. 11 266 kW
   Answer: B. Actual work is h1 - h2 = 940.9 kJ/kg (adiabatic), times 10 kg/s gives about 9409 kW. (Key point: Actual power)

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