---
title: "Equivalent evaporation and boiler efficiency"
subject: "Applied Thermodynamics"
chapter: 1
key: "semester_3/applied_thermodynamics:boiler-equivalent-evaporation"
source: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/boiler-equivalent-evaporation
site: BCT Engineering
---

# Equivalent evaporation and boiler efficiency

Applied Thermodynamics · Chapter 1: Boilers

- Asked in: year not given, 8 marks
- Video: [How to find out boiler efficiency and equivalent evaporation - SPPU paper solutions](https://www.youtube.com/watch?v=uQdT7Fk3wjo) — CHINMAY ACADEMY
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/boiler-equivalent-evaporation
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_thermodynamics/chapter-1

## Question

A boiler generates 4500 kg of steam per hour at 10 bar with dryness fraction 0.95 from feed water at 40 °C. The coal burnt is 600 kg/h and its calorific value is 29000 kJ/kg. Determine (a) the equivalent evaporation from and at 100 °C, (b) the factor of evaporation, (c) the boiler efficiency, and (d) the equivalent evaporation per kg of coal. Take the latent heat of steam at 100 °C as 2257 kJ/kg.

## Short answer

Using steam tables, wet steam enthalpy at 10 bar with x=0.95 is 2676.4 kJ/kg. Heat added per kg = 2508.8 kJ/kg. Equivalent evaporation = 4500*2508.8/2257 = 5002 kg/h; factor of evaporation = 1.112; boiler efficiency = (4500*2508.8)/(600*29000) = 64.9%; equivalent evaporation per kg coal = 5002/600 = 8.34 kg steam/kg coal.

## Answer

#### Data from steam tables
At 10 bar: $h_f = 762.7$ kJ/kg, $h_{fg} = 2014.4$ kJ/kg. Feed water at 40 °C: $h_{fw} = 167.6$ kJ/kg.

Enthalpy of wet steam:

$$
h = h_f + x\,h_{fg} = 762.7 + 0.95 \times 2014.4 = 2676.4\ \text{kJ/kg}
$$

Heat added per kg of steam $= h - h_{fw} = 2676.4 - 167.6 = 2508.8$ kJ/kg.

#### (a) Equivalent evaporation
$$
E = \frac{\dot m_s\,(h - h_{fw})}{2257} = \frac{4500 \times 2508.8}{2257} = 5002\ \text{kg/h}
$$

#### (b) Factor of evaporation
$$
F_e = \frac{h - h_{fw}}{2257} = \frac{2508.8}{2257} = 1.112
$$

#### (c) Boiler efficiency
$$
\begin{aligned}
\eta &= \frac{\dot m_s (h - h_{fw})}{\dot m_f \times CV} = \frac{4500 \times 2508.8}{600 \times 29000}\\
&= \frac{11289600}{17400000} = 0.649
\end{aligned}
$$

#### (d) Equivalent evaporation per kg of coal
$$
\frac{E}{\dot m_f} = \frac{5002}{600} = 8.34\ \text{kg steam/kg coal}
$$
(Actual evaporation is $4500/600 = 7.5$ kg/kg coal.)

**Answer:** (a) about 5002 kg/h; (b) 1.112; (c) about 64.9 %; (d) about 8.34 kg/kg of coal.

## Check yourself

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

1. How is the enthalpy of the wet steam found here?
   A. h = hf + hfg = 2777.1 kJ/kg
   B. h = x hf + hfg = 2738.2 kJ/kg
   C. h = hg - x hfg = 762.7 kJ/kg
   D. h = hf + x hfg = 762.7 + 0.95 x 2014.4 = 2676.4 kJ/kg
   Answer: D. For wet steam the enthalpy is hf plus x times hfg. (Key point: Enthalpy of wet steam)

2. What is the factor of evaporation and how is it found?
   A. (h - hfw)/2257 = 2508.8/2257 = 1.112
   B. (h - hfw)/2014.4 = 1.245
   C. 2257/(h - hfw) = 0.90
   D. h/2257 = 1.186
   Answer: A. Factor of evaporation is heat added per kg of steam divided by latent heat at 100 °C (2257 kJ/kg). (Key point: Factor of evaporation)

3. What is the boiler efficiency?
   A. About 77.5 %, using 4500 x 2676.4 over fuel heat
   B. About 64.9 %, steam heat 4500 x 2508.8 divided by fuel heat 600 x 29000
   C. About 5002 %, using equivalent evaporation
   D. About 11.1 %, using the factor of evaporation
   Answer: B. Efficiency = heat to steam / heat from coal = 11289480/17400000 = 0.649. (Key point: Boiler efficiency)

## Related in Applied Thermodynamics

- [Boiler efficiency with superheated steam](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_thermodynamics%3Aboiler-efficiency-superheated-steam) — Applied Thermodynamics, chapter 1
- [Boiler trial heat balance sheet](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_thermodynamics%3Aboiler-heat-balance-sheet) — Applied Thermodynamics, chapter 1

## Related in other subjects

- [Steam turbine: entropy generation per kg](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Asteam-turbine-entropy-generation-500k) — Fundamental of Thermodynamics & Heat Transfer, chapter 5
- [Calorific value and air from ultimate analysis](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Adulong-calorific-value-fuel-oil) — Energy Resources, chapter 1
- [Steam turbine 10 MPa, 550°C: isentropic efficiency](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Asteam-turbine-10mpa-550c-isentropic-eff) — Fundamental of Thermodynamics & Heat Transfer, chapter 5
- [Ideal Brayton cycle: net work per kg and efficiency](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Abrayton-work-per-kg-efficiency) — Fundamental of Thermodynamics & Heat Transfer, chapter 6