Boiler performance (Part 1) - Equivalent evaporation - Notes

In these notes Equivalent evaporation related with Boiler performance is explained

1. Boiler performance (Part 1) -
Equivalent evaporation
Presentation by:

Avdhesh Tyagi
Asst. Prof.
Department of Mechanical Engineering
G. L. Bajaj Institute of Technology & Management, Greater Noida
Disclaimer: The materials provided in this presentation and any comments or information
provided by the presenter are for educational purposes only. Nothing conveyed or provided
should be considered legal, accounting or tax advice.

2. Name of Subject: Applied Thermodynamics
Subject Code: KME401
Department of Mechanical Engineering
G. L. Bajaj Institute of Technology and Management,
Greater Noida, Uttar Pradesh
College code: 192
Affiliated to Dr. A.P.J. Abdul Kalam Technical University (Formerly UPTU Lucknow).
Unit 3
Topic covered: Equivalent evaporation

3. Boiler performance
The performance of any machine reduces with
time and boilers are no exception. The reasons
are poor combustion, fouling and poor
operation and maintenance. Deterioration of
fuel quality and water quality also leads to
poor performance of boiler. It is important to
understand that how we can measure the
performance a boiler and compare it with
performance of other boilers.

4. Boiler performance
Some of the boiler performance measuring
parameters are:
1.Evaporative capacity
2.Evaporation ratio
These 2 parameters measure the performance of
a particular boiler.
3. Equivalent evaporation
4. Boiler efficiency
These 2 parameters measure and compare the
performance of different boilers.

5. Evaporative capacity
The “Evaporative capacity” of a boiler is given
as its capacity to generate steam in unit time
(generally 1 hour).
It is generally be expressed in terms of:
i.kg of steam/hour
ii.kg of steam/hour/m
2
of heating surface

6. Evaporation ratio
The “Evaporation ratio” of a boiler is the ratio of
amount of steam generated to the amount of
fuel burnt by boiler in unit time (generally 1
hour).
Mathematically,
Evaporation ratio=
Amount of steam generated
Amount of fuel burnt

It is expressed in terms of kg of steam/kg of fuel.

7. Comparison of performances
of different boilers
Both “Evaporative capacity” and “Evaporation
ratio” good assessment tool for evaluation of
performance of a particular boiler. But here is
a question.
Can we use “Evaporative capacity” and
“Evaporation ratio” for the comparison of
performance of different boilers?
NO.
Why?

8. Performance of Boilers
Different boilers generate steam at different
pressures and temperatures by using feed water
at different temperatures. It means that input
conditions are not comparable and output
results are also quite for different boilers.
For justified comparison of performance of
different boilers, the input and output conditions
must be as per some standard conditions. Also,
the assessment tool must incorporate these
standard conditions.

9. Performance of Boilers
“Evaporative capacity” and “Evaporation ratio” have
no exact means for comparison of performance of
different boilers because it does not deal with
standard conditions for a justified comparison.
In such conditions the selection of “Evaporative
capacity” and “Evaporation ratio” for the
assessment of performance of different boilers will
not be a rational decision.
Hence we need a assessment tool that incorporates
some standard conditions for justified comparison
of performance of different boilers.

10. Performance of Boilers
The adopted standard conditions are as following:
1. Inlet conditions of feed water (1 kg):
Saturated water at 100℃ and 1 atm. pr.
2. Outlet conditions of produced steam (1 kg):
Dry and saturated steam at 100℃ and 1 atm. pr.
“Equivalent evaporation” is an assessment tool
that deals with abovementioned standard
conditions. Now we should discuss in detail about
“Equivalent evaporation”.

11. Equivalent evaporation
It is defined as the amount of water evaporated
(or steam generated) per unit time during
conversion of feed water at 100℃ into dry and
saturated steam at 100℃.
It is frequently expressed as the “amount of
water evaporated from and at 100℃.”
Now we will try to derive a mathematical
expression for equivalent evaporation.

12. Equivalent evaporation
Let
m = Actual mass of water evaporated or steam
generated/hour,
m
f = Actual mass of fuel burnt/hour,
m
a = Actual mass of water evaporated or steam
generated/kg of fuel (m/m
f)
h = Sp. Enthalpy of steam generated
h
w = Sp. Enthalpy of feed water

13. Equivalent evaporation
Actual heat gained by m
a kg of feed water (or
steam generated)/kg of fuel = m
a(h-h
w) kJ
Heat required by 1 kg of feed water at 100℃ and
1 atm. pr. for its conversion into dry and
saturated steam at 100℃ and 1 atm. pr. = 2257 kJ
Using above two heat amounts as per definition
of equivalent evaporation (m
e), we get its
mathematical expression

14. Equivalent evaporation
Equivalent evaporation ??????
�=
??????
??????ℎ−ℎ
??????
2257

Putting the value of m
a we get
Equivalent evaporation ??????
�=
?????? ℎ−ℎ
??????
??????
� x 2257

Above expression is expressed in terms of kg of
steam/kg of fuel. It may also be expressed in
terms of kg of steam/m
2
of heating surface by
suitable conversion.

15. Factor of evaporation
Actual heat gained by 1 kg of feed water (or
steam generated) per unit time = (h-h
w) kJ
Heat required by 1 kg of feed water at 100℃ and
1 atm. pr. for its conversion into dry and
saturated steam at 100℃ and 1 atm. pr. = 2257 kJ
Now the comparison of above two heat amounts
gives us factor of evaporation (F
e).
Factor of evaporation ??????
�=
ℎ−ℎ
??????
2257

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