The research paper published by IJSER journal is about Energy Analysis of Thermal Power Plant 1
ISSN 2229-5518
Energy Analysis of Thermal Power Plant
Raviprakash kurkiya, Sharad chaudhary
Abstract — Energy analysis helps designers to find ways to improve the performance of a system in a many way. Most of the conventional energy losses optimization method are iterative in nature and require the interpretation of the designer at each iteration. T ypical steady state plant operation conditions were determined based on available trending data and t he resulting condition of the operation hours. The energy losses from individual components in the plant is calculated based on these operating conditions to determine the true system losses. In this, first law of thermodynamics analysis was performed to evaluate efficiencies and various energy losses. In addition, variation in the per- centage of carbon in coal content increases the overall efficiency of plant that shows the economic optimization of plant.
Keywords — Energy, efficiency, thermal power plant, first law analysis, energy losses, optimization, coal.
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I n boiler, efficiency has a great influence on heating related energy savings. It is therefore important to maximize the
heat transfer to the water and minimize the heat losses in the boiler. The thermal power plant is based on a simple Rankine cycle; steam is used as the working fluid, steam generated from saturated liquid water (feed-water). This saturated steam flows through the turbine, where its internal energy is con- verted into mechanical work to run an electricity generating system. Not all the energy from steam can be utilized for run- ning the generating system because of losses due to friction, viscosity, bend-on-blade, heat losses from boilers i.e. hot flue gas losses, radiation losses and blow-down losses etc [1].
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Raviprakash kurkiya is currently pursuing masters degree program in IET
- DAVV University, India, PH-+91-9630923298. E-mail: raviprakash_kurkiya@yahoo.com
Sharad chaudhary is faculty member at Institute of engineering technolo- gy, DAVV University, India, PH-+91-930323760.
E-mail: sharad.iet@dauniv.ac.in
To optimize the operation of a boiler plant, it is necessary to identify where energy wastage is likely to occur. A significant amount of energy is lost through flue gases as all the heat pro- duced by the burning fuel cannot be transferred to water or steam in the boiler. Since most of the heat losses from the boi- ler appear as heat in the flue gas, the recovery of this heat can result in substantial energy savings. This indicates that there are huge savings potentials of a boiler energy savings by mi- nimizing its losses.
Combustion chamber is the most important part of the boi- ler. The combustor in a boiler is usually well insulated that causes heat dissipation to the surrounding almost zero. It also has no involvement to do any kind of work (w=0). In addition, the kinetic and potential energies of the fluid streams are usually negligible. Then only total energies of the incoming streams and the outgoing mixture remained same for analysis [2]. The conservation of energy principle requires that these two equal each other ’s that is shown in the figure
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The research paper published by IJSER journal is about Energy Analysis of Thermal Power Plant 2
ISSN 2229-5518
Therefore, the total heat released by complete combustion of 1 kg of coal is [1]
HHV = 
(1) In energy efficiency case, we assume that the combustion
chamber there is no heat losses [4]. Therefore, η = 100%
The boiler is considered as a single cross-flow steam pro- duction chamber. The performance of the steam produc- tion chamber plays an important role of the boiler efficien- cy. Heat is transferred from the hot fluid to the cold one through the wall separating them. Heat exchanger is a de- vice where two moving fluid streams exchange heat with- out mixing. A heat exchanger typically involves no work interactions (w=0) and negligible kinetic and potential energy changes for each fluid streams [3]. The outer shell of the heat exchanger is usually well insulated to prevent any heat loss to the surrounding medium.
Energy balance equation for any system
Ein - Eout =
(2) mp {hp – hg} = mw {hs – hw} + Q (3) Heat loss
Q = mp {hp – hg} - mw {hs – hw} (4)
Where,
Efficiency of heat exchanger
= × 
(5)
Energy balance equation for turbine and Work done by turbines actually = 18000 kW
Balancing equation for bleeding mass [5]
Wt = m3 (h3 –h4) + (m3 – m) (h4 – h5) (6) Energy loss
Q = m3h3 – (m3 – m) (h3 – h4) – m (h4 – h5) – W (7) Efficiency
η = 1- 
(8)
Energy balance equation for condenser is-
Heat given to the system
Qgiven = (m3 – m) (h3 – h2) + m (h3 – h7) (9) Heat rejection
Qrej = (m3 – m) (h5 – h1) (10)
So, from the above equation we get,
Qrej = (mc cp) (t5 – t4) (11) Energy balance
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The research paper published by IJSER journal is about Energy Analysis of Thermal Power Plant 3
ISSN 2229-5518
Qloss = m5(h5 – h1) – Qrej (12)
Efficiency
η = 1 - 
(13)
The energy balance equation for feed pump
m2h2 = m1h1 – Wp – Qloss (14)
Heat loss in feed pump
Qloss = m1h1 – Wp- – m2h2 (15) Efficiency
η = 1 – (16)
Basically boiler efficiency can be tested by the following methods:
There are reference standards for Boiler Testing at Site us- ing indirect method namely British Standard, BS 845: 1987 and USA Standard is ‘ASME PTC-4-1 Power Test Code Steam Generating Units’.
Indirect method is also called as heat loss method. A de- tailed procedure for calculating boiler efficiency by indirect method is given below. However, it may be noted that the practicing energy mangers in industries prefer simpler cal- culation procedures.
The principle losses that occur in a boiler are:
L1- Loss of heat due to dry flue gas (Sensible heat)
L2- Loss due to hydrogen in fuel (H2)
L3- Loss of heat due to moisture in fuel (H2O) L4- Loss of heat due to moisture in air (H2O)
L5- Loss of heat due to carbon monoxide (CO)
L6- Loss of heat due to radiation and unaccounted
L7- Unburnt losses in fly ash (Carbon)
L8- Unburnt losses in bottom ash (Carbon) Boiler Efficiency by indirect method
= 100 – (L1 + L2 + L3 + L4 + L5 + L6 + L7 +L8)
In the above, loss due to moisture in fuel and the loss due to combustion of hydrogen are dependent on the fuel, and cannot be controlled by design.
Theoretical (stoichiometric) air fuel ratio and excess air supplied are to be determined first for computing the boi- ler losses. The equation is given below for the same.
a) Theoretical air required for combustion
= [(11.6 × C) + {34.8 × (H2 – O2/8)} + (4.35 × S)] / 100 (17)
b) % Excess air supplied (EA)
= 
(18) Normally O2 measurement is recommended. If O2 mea-
surement is not available, use CO2 measurement
= 
(from flue gas analysis) (19) Where, = 
Moles of N2 = 
Moles of C = 
c) Actual mass of air supplied / kg of fuel (AAS)
= {1 + EA / 100} × theoretical air (20)
1. Heat loss due to dry flue gas
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The research paper published by IJSER journal is about Energy Analysis of Thermal Power Plant 4
ISSN 2229-5518
L1 = (21)
2. Heat loss due to evaporation of water formed due to H2 in fuel (%):- The combustion of hydrogen causes a heat loss because the product of combustion is water. This water is converted to steam and this carries away heat in the form of its latent heat.
L2 = 
(22)
3. Heat loss due to moisture present in fuel: - This moisture loss is made up of the sensible heat to bring the moisture to boiling point, the latent heat of evaporation of the moi s- ture, and the superheat required to bring this steam to the temperature of the exhaust gas. This loss can be calculated with the following equation
L3 = 
(23)
4. Heat loss due to moisture present in air
L4 = 
(24)
5. Heat loss due to incomplete combustion:- Products formed by incomplete combustion could be mixed with oxygen and burned again with a further release of energy
L5 = 
(25)
6. Heat loss due to radiation and convection: - The other heat losses from a boiler consist of the loss of heat by radiation and convection from the boiler casting into the surround- ing boiler house.
Normally surface loss and other unaccounted losses is 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%
L8 = 
(28)
The study is based on the energy analysis and energy savings of an industrial boiler. Data has been collected from various thermal power plant. Fuel consumption, excess air, steam production rate, pressure and temperature, air temperature, inlet and outlet temperature of water and steam, inlet and out- let temperature of flue gas of boiler etc. are collected for the analysis.
Parameters used for efficiency calculation
Parameters | Unit | Quantity |
Steam temperature | 0C | 525 |
Steam pressure | kg/cm2 | 88 |
Steam flow | Tph | 75 |
Mass flow rate of fuel | kg/s | 4.167 |
Mass flow rate of water | kg/s | 23.88 |
Turbine power | kW | 18000 |
Ambient temperature | 0C | 35 |
Temperature of flue gases | 0C | 275 |
The proximate analysis and ultimate analysis of lignite coal are given in Table 2
L6 = 0.548 x [ (Ts / 4
1.25
– (Ta / 55.5 4 + 1.957 x
(Ts – Ta)
x √*(196.85 Vm + 68.9)/68.9] (26)
7. Heat loss due to unburnt in fly ash (%)
L7 = (27)
8. Heat loss due to unburnt in bottom ash (%)
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The research paper published by IJSER journal is about Energy Analysis of Thermal Power Plant 5
ISSN 2229-5518
The boiler analysis results of this study are summarized in Tables 4, in which the various main components of Rankine cycle are introduced. The overall boiler energy efficiency is
tabulated-
Some parameters obtain by calculated for losses
Higher heating value = 18.856 MJ/kg Lower heating value = 7.4949 MJ/kg Stoichiometric ratio (a/f ) = 6.1260
Mass flow rate of water feed in boiler drum = 23.89 kg/s Mass flow rate of hot product in boiler drum = 41.77 kg Mass flow rate of steam bleeding = 6.7286 kg/s
Mass of water supply for condensation in condenser = 38.41 kg/s
By using above values, the various losses are calculated. The calculated values are shown in the following table3.
Losses | % |
Dry flue gas loss (L1) | 08.86 |
Heat loss due to formation of water from hydrogen in fuel (L2) | 05.54 |
Losses due to the moisture in fuel (L3) | 03.91 |
Losses due to the moisture in air (L4) | 0.341 |
Losses due to incomplete combustion of fuel (L5) | 02.89 |
Radiation losses (L6) | 01.37 |
Losses due to un-burnt in fly ash (L7) | 0.241 |
Losses due to un-burnt in bottom ash (L8) | 03.42 |
Total losses = L1 + L2 + L3 + L4 + L5 + L6 + L7 +L8
= 8.86 + 5.54+3.91+.341+2.89+1.37+.241+3.42
= 26.57%
Boiler Efficiency by indirect method
= 100 – (L1 + L2 + L3 + L4 + L5 + L6 + L7 +L8)
= 100 – 26.57
η = 73.43%
In addition, if the variation occurs in coal composition by the increases in carbon percentage of coal and other constitutes being decrease in same ratio.
Due to this, the results obtain are show in figure below:-
1. Fig.7 shows. As the carbon percentage, increase the value of efficiency is also increase. This could be attri- buted in higher heating value. So the increases in car- bon percentage is directly proportional to efficiency.
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The research paper published by IJSER journal is about Energy Analysis of Thermal Power Plant 6
ISSN 2229-5518
77.7
77.6
77.5
77.4
77.3
77.2
77.1
77
4. Fig.10 shows. Increase in efficiency as the moisture percentage is decrease.
Efficiency vs. Moisture
77.7
77.6
77.5
77.4
76.9
45 50 55 60 65 70 75
Carbon (%)
2. This fig.8 shows that the variation in carbon percen- tage is increase in higher heating value through the consumption of fuel. Therefore, it decreases mass flow rate of fuel. That shows the increase in carbon percentage is inversely proportional to mass flow rate of fuel.
4
77.3
77.2
77.1
77
76.9
76.8
12 13 14 15 16 17 18 19 20 21
Moisture (%)
2.4
2.3
2.2
x 10
Mass Flow Rate of Fuel vs. Higher heating value
5. Fig.11 shows. Increase in efficiency as the ash percen- tage decrease.
2.1
2
1.9
1.8
1.7
1.6
3.6 3.8 4 4.2 4.4 4.6 4.8 5 5.2 5.4 5.6
Mass Flow Rate of Fuel (kg/s)
3. Fig.9 shows. Increase in efficiency as the oxygen per- centage is decrease.
Efficiency vs. oxygen
77.7
77.7
77.6
77.5
77.4
77.3
77.2
77.1
77
76.9
76.8
5 5.5 6 6.5 7
ASH (%)
77.6
77.5
77.4
77.3
77.2
77.1
77
76.9
76.8
10 11 12 13 14 15 16 17 18 19 20
oxygen (%)
Then, we obtain there is increase in the boiler efficiency by
0.61% (approximately) with the help of MATLAB software is given below:-
1. Increment in the carbon percentage increases efficien- cy.
2. Decrement is ash percentage that observed increases
efficiency.
3. In the case of moisture, decrement it has been ob-
served that the rise in efficiency.
Energy analysis of a thermal power plant is reported in this paper. It provides the basis to understand the performance of
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The research paper published by IJSER journal is about Energy Analysis of Thermal Power Plant 7
ISSN 2229-5518
a fluidized bed coal fired boiler, feed pump, turbine and con- denser. MATLAB 2008a computer programming is used for the analysis. The energy balance sheet shows that theoretical
losses in various component of boiler. It provides information for selection of the components which has maximum losses so, that optimization techniques could be used to make it more efficient. The various energy losses of plant, through different components are calculated which indicates that maximum energy losses occur in turbine.
Following conclusions can be drawn from this study:
The coal type affects the first law efficiency of the sys- tem considerably.
It has been also analysed that a part of energy loss oc-
curs through flue gases.
The carbon content in the coal has to be proper.
The presence of moisture has a detrimental effect on overall efficiency.
If we use the heat recovery system to recover the heat
losses through flue gases then it will be more useful for us.
With the growing need of the coal, which is an non renewable source of energy and depleting with a very fast pace, it is de- sirable to have such optimal techniques (better quality of coal) which can reduce the energy losses in the coal fired boiler and improves its performance these create impact on production and optimizations uses of energy sources. In addition this study shows the better quality of coal giving the high per- formance of plant and even though the consumption of coal is been reduced that creates economic condition for overall plant
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c – Carbon
h – Hydrogen
o – Oxygen
s – Sulphur
E – Rate of energy
dE system/ dt – Rate of change
C p = Specific heat capacity, kJ/kg0C E = Rate of energy
m =Mass flow rate, kg/s
Q = Energy losses
s = Specific entropy, kJ/kg
T = Temperature0C
η = Energy efficiency
in – input
out – Output p – Product
a – Air
g – Flue gases s – Steam
w – Water
B – Boiler
f – Fuel
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