Boiler Efficiency Improvement through Analysis of Losses

IJSRD - International Journal for Scientific Research & Development| Vol. 1, Issue 3, 2013 | ISSN (online): 2321-0613
All rights reserved by www.ijsrd.com 801
Abstract— Thermal is the main source for power generation
in India. The percentage of thermal power generation as
compare to other sources is 65 %. The main objective of
thermal power plant is to fulfill the energy demands of the
market and to achieve these demands; plant requires
technical availability with the parts reliability and
maintenance strategy. This paper deals with the
determination of current operating efficiency of Boiler and
calculates major losses for Vindhyachal Super thermal
power plant (India) of 210 MW units. Then identify the
causes of performance degradation. Also find the major
causes of heat losses by Fault Tree Analysis (FTA) and
recommends its appropriate strategy to reduce major losses.
The aim of performance monitoring is continuous evaluation
of degradation i.e. decrease in performance of the steam
boiler. These data enable additional information which is
helpful in problem identification, improvement of boiler
performance and making economic decisions about
maintenance schedule.
Keywords: Boiler Efficiency, Fault Tree Analysis (FTA),
Maintenance
I. INTRODUCTION
Today, most of the electricity produced throughout the
world is from steam power plants. Looking to the statically
data, rate which power demand increases is extremely very
high compared to the rate at which generation capacity
increase. Efficient operation of power plants has always
been important to utilities. The heat rate of a conventional
coal fired power plant is a measure of how efficiently it
converts the chemical energy contained in the fuel into
electrical energy. This conversion is accomplished in four
major steps. First, the chemical energy in the fuel is
converted into thermal energy, then the thermal energy is
converted into kinetic energy, then the kinetic energy is
converted in mechanical energy, and last the mechanical
energy is converted to electrical energy. In each of these
sub-processes, some energy is lost to the environment. Some
of the fuel is not burned completely, some of the thermal
energy is lost out through the stack and also rejected to the
cooling water, some of the kinetic and mechanical energy
produces heat instead of electricity, and last, some of the
electricity that is produced is used by these sub-processes.
A boiler is an enclosed vessel that provides a
means for combustion heat to be transferred into water until
it becomes heated water or steam [2]. The hot water or
steam under pressure is then usable for transferring the heat
to a process. So Boiler is the main equipment to produce
power but every equipment does not runs with full
efficiency regularly/continuously, it requires regular
maintenance and condition monitoring to it reliable in
future. Therefore to calculate boiler heat loss and their
causes makes easy to apply appropriate maintenance
strategy for increasing efficiency of power plant [1].
Objective
 To evaluate operating efficiency of Boiler
 Determine various types of losses related to boiler
operation
 Identification the causes of performance degradation
and its performance analysis
 Identifying heat rate gaps and then implementing
corrective actions to eliminate the efficiency loss.
II. METHODOLOGY
Thermal efficiency of boiler is defined as the percentage of
heat input that is effectively utilized to generate steam.
There are two methods of assessing boiler efficiency.
A. The Direct Method:
Where the energy gain of the working fluid (water and
steam) is compared with the energy content of the boiler
fuel [4].
B. The Indirect Method:
Where the efficiency is the difference between the losses
and the energy input.
In indirect method the efficiency can be measured
easily by measuring all the losses occurring in the boilers
using the principles to be described.
The disadvantages of the direct method can be
overcome by this method, which calculates the various heat
losses associated with boiler. The efficiency can be arrived
at, by subtracting the heat loss fractions from 100.An
important advantage of this method is that the errors in
measurement do not make significant change in efficiency.
 Boiler Efficiency by Indirect Method: Calculation
Procedure and Formula
The data required for calculation of boiler efficiency using
indirect method are [2]:
1) Ultimate analysis of fuel (H2, O2, S, C, moisture
content, ash content)
2) Percentage of Oxygen or CO2 in the flue gas
3) Flue gas temperature in 0
C (Tf)
Boiler
Efficiency
Evalution
Method
Direct Method Indirect Method
Boiler Efficiency Improvement through Analysis of Losses
Virendra Nagar1
Dr. V. K. Soni2
Dr. V. K. Khare3
1, 2, 3
Mechanical Engineering Department, M.A.N.I.T Bhopal (M. P.) India
S.P.B.Patel Engineering College, Mehsana, Gujarat

(IJSRD/Vol. 1/Issue 3/2013/0099)
4) Ambient temperature in 0
C (Ta) & humidity of air in
kg/kg of dry air.
5) GCV of fuel in kcal/kg
6) Percentage combustible in ash (in case of solid fuels)
7) GCV of ash in kcal/kg (in case of solid fuels)
Unit Load = 210MW
Steam Flow = 615 T/hr
Total Coal Flow
Wet Bulb Temp
Dry bulb Temp
=
=
=
140 T/hr
24
o
C
30
o
C
%CO
2
in Flue gas = 14.3
%CO in flue gas = 0.5
Average flue gas temperature = 180
o
C
Ambient temperature = 30
o
C
Humidity in ambient air = 0.014 kg/ kg dry air
Carbon in Ash / kg of coal = 0.002 kg/kg coal
Ratio of bottom ash to fly ash = 90:10
Fuel Analysis (in %)
Ash content in fuel = 40
Moisture in coal = 12.2
Carbon content = 39.71
Hydrogen content = 2.58
Nitrogen content = 0.76
Oxygen content
Sulphur
=
=
4.15
0.6
GCV of Coal
CV of Carbon
=
=
3501 kCal/kg
8077.8 kCal/kg
Fig. 1 : Fuel content
Step – 1 Find theoretical air requirement
Theoretical air required for complete combustion =
= [(11.6 x C) + {34.8 x (H2 – O2/8)} + (4.35 x
S)]/100 kg/kg of fuel
= [(11.6 x 39.71) + {34.8 x (2.58 -4.15/8)} +
(4.35 x 0.6)]/100 kg/kg of fuel
= [460.636 +71.73 +2.61]/100 kg/kg of fuel
= 5.34 kg/kg of fuel
Step – 2 Find theoretical CO2 %
% CO2 at theoretical condition (CO2)t = Moles of C/ (Moles
of C + Moles of N2)
Where,
Moles of N2 = (Wt of N2 in theoretical air/Mol. wt of N2) +
(Wt of N2 in fuel/ Mol. wt of N2)
Moles of N2 = [(5.34 x 77/100)/28] + .0076/28
= 0.1471
Where moles of C = 0.3971/12 = 0.033
(CO2) t= 0.033/ (0.1471+0.033) = 18.33%
Step – 3 to find Excess air supplied
Actual CO2 measured in flue gas = 14.3%
Excess Air supplied (EA) %=7900x [(CO2%)t-
(CO2%)a]/(CO2%)ax[100-[(CO2%)t]
Excess Air supplied (EA) % = 7900 x [18.33-14.3]/14.3 x
[100-18.33]
Excess Air supplied (EA) % = 27.26%
Step – 4 to find actual mass of air supplied
Actual mass of air supplied = {1 + EA/100} x theoretical air
Actual mass of air supplied = {1 +27.26/100} x 5.34
= 6.795 kg/kg coal
Step –5 to find actual mass of dry flue gas
Mass of dry flue gas = (Mass of CO2 +Mass of N2 content in
the fuel+ Mass of N2 in the combustion air supplied + Mass
of oxygen in flue gas)
Mass of dry flue gas = [(0.3971 x 44)/12] + 0.0076 +
[(6.795 x77)/100] + [(6.795-
5.34) x 23]/100
= 1.463 + 5.23 + 0.334
= 7.027 kg/kg coal
Step – 6 to find all losses
% Heat loss due to dry flue gas = [m x Cp x (Tf – Ta) x 100]
/ GCV of fuel
m = mass of dry flue gas in kg/kg
Cp = Specific heat of flue gas (0.23 kcal/kg)
% Heat loss due to dry flue gas = 7.027x0.23x (180-30)
x100/3320
% Heat loss due to dry flue gas (L1) = 7.302%
% Heat loss due to formation of water from H2 in fuel (L2)
Where,
H2 percentage of H2 in 1 kg of fuel
Cp – Specific heat of superheated steam (0.45 kcal/kg)
L2 = 9 x 0.0258x {584+0.45(180-30)} x 100/3320
L2 = 0.2322 x 651.5 x 100/3320
L2 = 4.55%
% heat loss due to evaporation of moisture present in fuel
(L3)
L3= M x {584 + Cp (Tf-Ta)} x 100/GCV of fuel
Where,
M – % moisture in 1kg of fuel
Cp – Specific heat of superheated steam (0.45 kcal/kg)
L3= 0.122 x {584 + 0.45(180-30)} x 100/3320
L3= 2.39%
% heat loss due to moisture present in air (L4)
= AAS x humidity x Cp x (Tf-Ta) x100/GCV
Ash content
in fuel 40%
Moisture in
coal
12.20%
Carbon
content
39.71%
Hydrogen
content
2.58%
Nitrogen
content
0.76%
Oxygen
content
4.15%
Sulphur
0.60%
9 x H2 {584 + Cp (Tf – Ta)} x 100
GCV of fuel

L4 = 6.795 x0.014x 0.45x (180-30) x 100/3320
L4 = 0.193%
% Heat loss due to partial conversion of C to CO (L5) =
L5=
=
=
L5 = 2.32%
%heat loss due to radiation and other unaccounted loss (L6)
are assumed based on the type and size of the boiler as given
below
For industrial fire tube / packaged boiler = 1.5 to 2.5%
For industrial water tube boiler = 2 to 3%
For power station boiler = 0.4 to 1%
% Loss due to Unburnt Carbon (L7) = UxCVcx100/Gcv
Where,
U = Carbon in Ash / kg of coal
CVc = CV of Carbon
L7 = 0.002 x 8077.8 x 100/3320
L7 = 0.48%
Boiler efficiency by indirect method
= 100 – (L1+ L2+ L3+ L4+ L5+ L6+ L7)
=100-(7.302+4.55+2.39+0.193+2.32+0.3+0.48)
= 82.465%
Input/output Parameter kCal / kg of
coal
% loss
Heat Input = 3320 100
Losses in boiler
1. Dry flue gas, L
1
= 242.42 7.302
2. Loss due to hydrogen in fuel, L
2
= 151.06 4.55
3. Loss due to moisture in fuel, L
3
= 79.34 2.39
4. Loss due to moisture in air, L
4
= 6.407 0.193
5. Partial combustion of C to CO,
L
5
= 77.02 2.32
6. Surface heat losses, L
6
= 9.96 0.3
7. Loss due to Unburnt carbon, L
7
= 15.93 0.48
Boiler Efficiency = 100 – (L
1
+ L
2
+ L
3
+ L
4
+ L
5
+ L
6
+ L
7
) =
82.447%
III. ANALYSIS
The above mathematical calculation to determining the
actual heat losses in boiler by using indirect method, So the
major heat losses in boiler occurs due to Dry heat gas loss
(7.302%),Loss due to hydrogen in fuel (4.55%),Loss due to
moisture in fuel (2.39%) and Partial combustion of C to CO
(2.32%). To find causes behind above losses in detail by
using Fault Tree Analysis (FTA). And also recommend
appropriate maintenance strategy to reduce the causes of
degradation boiler performance. Analysis shows different
types of heat losses, Energy waste in Kcal/Kg of coal,
Contents of fuel etc.
Fig. 3: Percentage losses in boiler
Fault Tree Analysis (FTA) The heat rate fault tree is used to
identify areas in the plant where heat rate degradation may
be occurring without conducting expensive tests. The fault
tree is structured to provide a process by which decisions
can be determined that narrow down the cause of the
problem based on available information [3].
0
1
2
3
4
5
6
7
8
% Loss
% Loss

 Maintenance Recommendation
On the basis of FTA & computing the heat losses there are
three major losses and their sub losses that affect the boiler
efficiency. The corrective actions should be taken for
minimizing the heat losses are shown below:
Type of losses Corrective Actions
Dry gas losses
Boiler casing air
in-leakage
O2 reading should be taken at several
stages
Air preheater
leakage
Condition monitoring of inlet and
outlet temperature of gases.
Incorrect fuel-to-
air ratio
Proper O2 monitoring system
Improper burner
damper settings
 Adjust burner tilts
 Continuous check burners
setting.
Fouled heat
transfer surfaces
Periodic cleaning of Fouled heat
transfer surfaces
Moisture losses
Excessive soot
blowing
Optimize blow shoot selectively
Change in
ambient
conditions
Adjust the primary air temperature at
air Preheater
Change in coal
quality
Periodically check the coal quality
Increase in coal
surface moisture
 Check coal surface moisture
before entering in mill
 Supply the proper amount of
primary air
Tube leaks
 Acoustic condition monitoring
for leakage
 Take corrective when boiler
shutdown
Incomplete combustion
Incorrect fuel-to-
air ratio
Proper O2 monitoring system
Burner tips
plugged
Use bypass burner
Decrease in mill
fineness
 Adjust and control the fineness
of pulverized coal
 Collect coal sample from
pulverize mills and analyze for
fineness
 Set proper classifier settings
 Set proper mill journal and
spring tension.
Burner damper
settings
Adjust damper setting properly
Change in coal
quality
Collect coal sample from pulverize
mills and analyze the coal quality
periodically.
IV. CONCLUSION
In this paper we focused on the heat losses of boiler and find
that the actual major heat losses are:-
Dry heat gas loss (7.302%), Loss due to hydrogen in fuel
(4.55%), Loss due to moisture in fuel (2.39%) and Partial
combustion of C to CO (2.32%) And also find the reason
behind the heat losses. On the basis of causes of heat losses,
we suggest the corrective actions to increase the boiler
performance.
ACKNOWLEDGEMENT
We are thankful to Department of Mechanical Engineering
of MANIT Bhopal and BMD department of NTPC
Vindhyachal India for providing the required facilities
needed for the successful completion of this paper.
BOILER LOSSES
Moisture
losses
EXCESSIVE
SOOT
BLOWING
CHANGE IN
AMBIENT
CONDITIONS
CHANGE IN COAL
QUALITY
INCREASE IN
COALSURFACE
MOISTURE
TUBE LEAKS
Incomplete
combustion A
Radiation
losses
Dry Gas
losses
BOILER CASING
AIR IN-LEAKAGE
AIR PREHEATER
LEAKAGE
INCORRECT FUEL-
TO-AIR RATIO
IMPROPER
BURNER DAMPER
SETTINGS
FOULED HEAT
TRANSFER
SURFACES
BOILER
WATERWALLS
ECONOMIZER
AIR PREHEATER
SUPERHEATER
REHEATER
A
INCORRECT
FUEL-TO-AIR
RATIO
BURNER TIPS
PLUGGED
DECREASE IN MILL
FINENESS
CLASSIFIER VANES
IMPROPERLY
ADJUSTED
RING OR ROLLER
WEAR
LOSS OF ROLLER
TENSION
EXCEEDING MILL
CAPACITY
CLASSIFIER VANE
WEAR
BURNER DAMPER
SETTINGS
CHANGE IN COAL
QUALITY

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Boiler Efficiency Improvement through Analysis of Losses

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Boiler Efficiency Improvement through Analysis of Losses