Module 02 · How a Boiler Works5 min read

Boiler Ratings: TPH, F&A 100 °C and the Evaporation Factor

A boiler’s nameplate says how much steam it makes, but under standard conditions, not yours. In India the standard is “from and at 100 °C”. Converting that rating to real output at your pressure and feedwater temperature is the single most useful calculation when buying or checking a boiler.

ByShikhar Singh·Steam Consultant & IT Consultant·Published

You'll learn

  • Explain TPH and the F&A 100 °C rating basis
  • Calculate the evaporation factor
  • Convert a nameplate rating to actual output, and a demand to a rating
  • See how feedwater temperature changes real capacity

Before this: Enthalpy, Latent heat

BeginnerThe idea in plain words, and where you meet it. No maths.

What “5 TPH, F&A 100 °C” means

TPH is tonnes of steam per hour; 5 TPH is 5,000 kg/h. But the same boiler makes different amounts of steam depending on how hot its feedwater is and what pressure it runs at, so a rating needs a reference condition. Indian boilers are rated “from and at 100 °C” (F&A 100 °C): the steam the boiler would make from water already at 100 °C, into steam at 100 °C and atmospheric pressure.

Real plants are never at that condition. Feedwater is usually cooler than 100 °C and steam is at a higher pressure, so each kilogram of real steam needs more heat than a kilogram of F&A steam. The result: a boiler delivers less actual steam than its F&A rating.

IntermediateThe formula, its units, and a worked example on real plant numbers.

The evaporation factor

Evaporation factor
F=hg−hfeed2256.9
Symbols in the formula
SymbolMeaningUnit
hgEnthalpy of steam at working pressurekJ/kg
hfeedEnthalpy of the feedwaterkJ/kg
2256.9Latent heat at 100 °C: the F&A referencekJ/kg
Converting
actual kg/h=F&A ratingFF&A rating=actual kg/h×F
Worked example
What a 5 TPH F&A boiler really delivers
Given
  • Working pressure 10 kg/cm²(g); cold feedwater at 30 °C.
  1. Steam enthalpy 2,780.1 kJ/kg; feedwater 126.7 kJ/kg.
  2. Evaporation factor:
    F=2,780.1−126.72,256.9=1.1757
  3. Actual output:
    5,0001.1757=4,253 kg/h
Answer: About 4,253 kg/h — 15% below the nameplate. With feedwater at 80 °C the same boiler makes 4,617 kg/h.
Chart of evaporation factor against feedwater temperature from 20 to 120 °C at 10 kg/cm²(g), falling from 1.194 with 20 °C feedwater to 1.046 at 100 °C. Pressure between 7 and 17.5 kg/cm²(g) changes it very little.
Hotter feedwater lowers the factor, so the same boiler delivers more real steam. Chart: GyanTosh Fabricators Pvt. Ltd., reusable with credit under CC BY 4.0
ExpertThe fine print: where the simple model breaks, and what it costs.

Sizing and checking with the factor

Sizing runs the other way. A process needing 4,000 kg/h of actual steam at 10 kg/cm²(g) from 30 °C feedwater needs a boiler rated 4,703 kg/h F&A before any margin for distribution losses, blowdown or future load. Buying a “4 TPH” boiler for a 4,000 kg/h demand undersizes it by 18% on day one.

Feedwater temperature is the lever a plant controls. Returning condensate, a deaerator or an economizer raises it, and every degree reduces the evaporation factor and raises real output. That is why a capacity complaint is sometimes solved in the feed tank rather than by buying a bigger boiler.

The reference latent heat, 2,256.9 kJ/kg (539 kcal/kg), is a convention, held fixed so that ratings stay comparable. Feedwater enthalpy is taken at boiler pressure, not at its own saturation pressure; the difference is under 1 kJ/kg but it is what makes published evaporation factor tables reproduce exactly.

Key takeaways

  • Indian boilers are rated F&A 100 °C: steam from 100 °C water into 100 °C steam.
  • Evaporation factor = (hg−hfeed)/2256.9.
  • Actual output = rating ÷ factor: a 5 TPH boiler at 10 kg/cm²(g) from 30 °C feed makes about 4,253 kg/h.
  • Hotter feedwater raises real output; compare quotations on the same basis.

Check your understanding

Questions run from beginner to expert. Try each before opening the answer.

  1. Beginner

    Q1. Why does a boiler usually deliver less steam than its F&A 100 °C rating?

    1. A.Ratings are exaggerated
    2. B.Real feedwater is cooler and real pressure higher than the reference, so each kilogram needs more heat
    3. C.Steam leaks
    4. D.The rating is in a different unit
    Show answer

    B. The F&A basis assumes 100 °C water to 100 °C steam; real conditions need more heat per kilogram.

  2. Intermediate

    Q2. A boiler rated 3,000 kg/h F&A has an evaporation factor of 1.2 on site. What is its actual output?

    1. A.3,600 kg/h
    2. B.3,000 kg/h
    3. C.2,500 kg/h
    4. D.2,400 kg/h
    Show answer

    C. 3,000 ÷ 1.2 = 2,500 kg/h.

  3. Intermediate

    Q3. What happens to the evaporation factor when feedwater gets hotter?

    1. A.It rises
    2. B.It falls
    3. C.No change
    4. D.It becomes negative
    Show answer

    B. Hotter feedwater has higher enthalpy, so less heat is needed per kilogram.

  4. Expert

    Q4. A process needs 4,000 kg/h actual at 10 kg/cm²(g) from 30 °C feed. What F&A rating is needed before margins?

    1. A.4,000 kg/h
    2. B.4,703 kg/h
    3. C.3,402 kg/h
    4. D.8,000 kg/h
    Show answer

    B. F&A rating = actual × factor = 4,000 × 1.1757.

Frequently asked

What does F&A 100 °C mean on a boiler?
From and at 100 °C: the rated steam output assuming feedwater at 100 °C turned into steam at 100 °C. It is a standard reference so that boilers can be compared, not the output at your conditions.
How do you calculate the evaporation factor?
Evaporation factor = (enthalpy of steam at working pressure − enthalpy of feedwater) ÷ 2256.9 kJ/kg. Divide the F&A rating by it to get actual output.
How much steam does a 5 TPH boiler produce?
It depends on pressure and feedwater temperature. At 10 kg/cm²(g) with 30 °C feedwater, a 5 TPH F&A boiler produces about 4,253 kg/h; with 80 °C feedwater, about 4,617 kg/h.

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About the author

Shikhar SinghSteam Consultant & IT Consultant

Shikhar Singh is a Steam Consultant and IT Consultant at GyanTosh Fabricators Pvt. Ltd., where he works on steam system sizing, condensate recovery and the engineering content published on this site.

Full profile and background →