---
title: "Boiler efficiency with superheated steam"
subject: "Applied Thermodynamics"
chapter: 1
key: "semester_3/applied_thermodynamics:boiler-efficiency-superheated-steam"
source: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/boiler-efficiency-superheated-steam
site: BCT Engineering
---

# Boiler efficiency with superheated steam

Applied Thermodynamics · Chapter 1: Boilers

- Asked in: year not given, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/boiler-efficiency-superheated-steam
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_thermodynamics/chapter-1

## Question

A boiler produces 2000 kg/h of superheated steam at 15 bar and 300 °C from feed water at 50 °C. The fuel consumption is 250 kg/h with calorific value 32000 kJ/kg. Find (a) the heat supplied to the water and steam per hour, (b) the boiler efficiency, (c) the factor of evaporation and the equivalent evaporation from and at 100 °C (latent heat 2257 kJ/kg).

## Short answer

Superheated steam at 15 bar, 300°C: h=3038.3 kJ/kg; feedwater at 50°C: 209.4 kJ/kg. Heat supplied = 2000*(3038.3-209.4)=5.66x10^6 kJ/h. Efficiency = 5.66x10^6/(250*32000)=70.7%. Factor = (3038.3-209.4)/2257=1.253. Equivalent evaporation = 2000*1.253=2507 kg/h.

## Answer

#### Enthalpies (steam tables)
- Superheated steam at 15 bar, 300 °C: $h = 3038.3$ kJ/kg.
- Feed water at 50 °C: $h_{fw} = 209.4$ kJ/kg.

#### (a) Heat supplied to water and steam
$$
Q = \dot m_s (h - h_{fw}) = 2000 \times (3038.3 - 209.4) = 5657800\ \text{kJ/h}
$$

#### (b) Efficiency
Heat released by fuel $= 250 \times 32000 = 8000000$ kJ/h.

$$
\eta = \frac{5657800}{8000000} = 0.707
$$

#### (c) Factor and equivalent evaporation
$$
F_e = \frac{h - h_{fw}}{2257} = \frac{3038.3 - 209.4}{2257} = 1.253
$$

$$
E = F_e \times \dot m_s = 1.253 \times 2000 = 2507\ \text{kg/h}
$$

**Answer:** (a) $5.66 \times 10^{6}$ kJ/h; (b) 70.7 %; (c) $F_e = 1.253$, $E \approx 2507$ kg/h.

## Check yourself

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

1. How is the heat supplied to the water and steam per hour found?
   A. ms x h = 2000 x 3038.3
   B. ms (h - hfw) = 2000 x (3038.3 - 209.4)
   C. mf x CV = 250 x 32000
   D. ms x hfw = 2000 x 209.4
   Answer: B. Heat added to the water and steam is steam flow times the enthalpy rise from feed water to superheated steam. (Key point: Heat supplied)

2. What is the boiler efficiency?
   A. About 141 %, 8000000 / 5657706
   B. About 25 %, 2000/8000
   C. About 70.7 %, 5657706 / 8000000
   D. About 90 %, 3038.3/3400
   Answer: C. Efficiency = heat to steam divided by heat released by fuel (250 x 32000 = 8000000 kJ/h). (Key point: Boiler efficiency)

3. What is the factor of evaporation?
   A. 0.798
   B. 1.346
   C. 2.507
   D. 1.253
   Answer: D. Fe = (3038.3 - 209.4)/2257 = 1.253. (Key point: Factor of evaporation)

4. How is the equivalent evaporation from and at 100 °C obtained?
   A. Factor of evaporation x actual steam rate = 1.253 x 2000 = about 2507 kg/h
   B. Actual steam rate divided by factor of evaporation = 1596 kg/h
   C. Fuel rate x factor = 313 kg/h
   D. Heat supplied divided by fuel rate
   Answer: A. E = Fe x ms. (Key point: Equivalent evaporation)

## Related in Applied Thermodynamics

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- [Equivalent evaporation and boiler efficiency](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_thermodynamics%3Aboiler-equivalent-evaporation) — Applied Thermodynamics, chapter 1
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