Heat rate of thermal power plant
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Jul 02, 2019
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About This Presentation
The book describes the basics of heat rate, how it is to be calculated, the mass balance of the Thermal power station and the requisite data to be collected, the boiler efficiency, turbine efficiency and everything related to the heat rate of the Power Plant.
Slide Content
total output power solutionsManohar Tatwawadi Page1
HEAT AND MASS BALANCE
INDEX
S.N. TOPICS PAGE NOS
1 BASICS OF HEAT RATE 2-5
2
FACTORS AFFECTING THE TURBINE
HEAT RATE 6-12
3
BOILER LOSSES AND EFFICIENCY
13-20
4
210 MW LMW TURBINE CYCLE HEAT
& MASS BALANCE
21 37
5
BOILER EFFICIENCY
CALCULATIONS
38-42
HEAT AND MASS BALANCE
INDEX
S.N. TOPICS PAGE NOS
1 BASICS OF HEAT RATE 2-5
2
FACTORS AFFECTING THE TURBINE
HEAT RATE 6-12
3
BOILER LOSSES AND EFFICIENCY
13-20
4
210 MW LMW TURBINE CYCLE HEAT
& MASS BALANCE
21 37
5
BOILER EFFICIENCY
CALCULATIONS
38-42
total output power solutionsManohar Tatwawadi Page2
BASICS OF HEAT RATE
Generator output= 1 kWH= 860 kCal
Losses in the Generator= 20 kCal
Input to the Generator= 880 kCal
Turbine Exhaust and Regen Losses= 1185 kCal
Heat Input to the Turbine= 2065 kCal(Turbine Heat Rate)
Boiler Output = 2065 kCal
Boiler Losses = 331 kCal
Input Heat Energy to the Boiler= 2396 kCal(Unit Heat rate)
If the Heat Value of Coal is 4000 kCal/kg,= 2396 / 4000 = 0.599kg of coal
then Coal Rate i.e. coal required to generate
oneunit of electricity (1 kWH)
BASICS OF HEAT RATE
Generator output= 1 kWH= 860 kCal
Losses in the Generator= 20 kCal
Input to the Generator= 880 kCal
Turbine Exhaust and Regen Losses= 1185 kCal
Heat Input to the Turbine= 2065 kCal(Turbine Heat Rate)
Boiler Output = 2065 kCal
Boiler Losses = 331 kCal
Input Heat Energy to the Boiler= 2396 kCal(Unit Heat rate)
If the Heat Value of Coal is 4000 kCal/kg,= 2396 / 4000 = 0.599kg of coal
then Coal Rate i.e. coal required to generate
oneunit of electricity (1 kWH)
total output power solutionsManohar Tatwawadi Page3
Plant Heat Rate
If the losses increase, the heat rate will also increase and the coal required will also
increase, thereby increasing the cost of generation.
HEAT RATE OF THERMALPOWER PLANT:
In thermal power plant, Chemical Energy of fuel is converted to electrical energy. The
conversion cycle is based on Thermodynamic Vapor Cycle, called Rankine Cycle.
Conversion takes place through various stages and different processes are involved for
the purpose. Due to the variouslimitations nature has imposed, such as Irreversibility in
the process, heat losses to atmosphere, Friction losses, Heat Transfer losses, to name a
few, efficiency of conversion is always less than 100%. In addition to these losses, some
heat energy is rejected because, steam temperature and pressure drop to such low values
(after doing work in Turbine), that further conversion to useful work is not possible. Due
to all these reasons, energy input is much more for one kWh electrical energy output
from the Generator. If the Chemical / Electrical conversion process should have been
100% efficient, 860 Kcal heat energy input should have given one kWh electrical energy
out put at Generator terminals. ThisratioofElectrical Energy Outputover a certain period
of time toChemical Energy inputto the Plant over the same period is calledHeat rate.
In modern plants, designed for High temperature and pressure Steam admission to
Turbine, efficiency and heat rate can be around 36% and 2400 Kcal/ KWh respectively.
Plant Heat Rate
If the losses increase, the heat rate will also increase and the coal required will also
increase, thereby increasing the cost of generation.
HEAT RATE OF THERMALPOWER PLANT:
In thermal power plant, Chemical Energy of fuel is converted to electrical energy. The
conversion cycle is based on Thermodynamic Vapor Cycle, called Rankine Cycle.
Conversion takes place through various stages and different processes are involved for
the purpose. Due to the variouslimitations nature has imposed, such as Irreversibility in
the process, heat losses to atmosphere, Friction losses, Heat Transfer losses, to name a
few, efficiency of conversion is always less than 100%. In addition to these losses, some
heat energy is rejected because, steam temperature and pressure drop to such low values
(after doing work in Turbine), that further conversion to useful work is not possible. Due
to all these reasons, energy input is much more for one kWh electrical energy output
from the Generator. If the Chemical / Electrical conversion process should have been
100% efficient, 860 Kcal heat energy input should have given one kWh electrical energy
out put at Generator terminals. ThisratioofElectrical Energy Outputover a certain period
of time toChemical Energy inputto the Plant over the same period is calledHeat rate.
In modern plants, designed for High temperature and pressure Steam admission to
Turbine, efficiency and heat rate can be around 36% and 2400 Kcal/ KWh respectively.
total output power solutionsManohar Tatwawadi Page4
The term Heat rate is defined in many ways as follows:
Net Unit Heat rate: It is the ratio of energy input to Boiler in terms of Heat energy of
fuel, for one kWh of electrical energy output atBus Bars, i.e. after UAT. If the out put
and input is considered for a period of an hour, then it is Net Unit Heat rate for one
Hour. Similarly, it can be calculated over a period of a Day, a Week, a Month or a Year.
In this case, it is the sent out energy that is considered, hence, consumption of electrical
energyfor driving the plant s auxiliaries is also accounted for.
Gross Unit Heat rate: It is the ratio of energy input to Boiler in terms of Heat energy of
fuel, for one kWh of electrical energy output at Generator Terminals. In this case,
auxiliary consumptionisNOTaccounted for.
Net Turbine Cycle Heat rate:It is the ratio of heat energy contained in steam admitted
to Turbine for one kWh of electrical energy output at Generator Terminals. In this case,
auxiliary consumption and losses in Boiler areNOTaccounted for.
Operating Heat rate: It is the heat rate calculated by considering the inputs and outputs
from the plant only when it is synchronized with the grid. In this case, the fuel input
required for steam conditioning, from light up to synchronizationis not considered. Also
auxiliaries consumption during the period of plant shut down is not considered.
What information does Heat rate give?
The plant is designed to generate electricity at certain design heat rate. Deviations from
design values give a valuable information regarding the operational and maintenance
practices. Also, by comparison with the historical data, decisions can be taken while
making investments on the maintenance and renovation. Also, problem area can be
identified and analyzed for improvements. A deviation in Gross Turbine Cycle heat rate
tells us about energy conversion scenario in turbine, including condenser and
regenerative feed heating process. If Net average unit heat rate deviates from that of
design, it tells us how much extra amount of energy is put in and how much money is
wasted .
Now a days, tariff for supply of electricity to consumers is fixed by Maharashtra
Electricity Regulatory Commission. While fixing tariff, MERC has given the benchmark
heat rate values for all power plants in MAHAGENCO. If actual heat rate is more than
the benchmark heat rate, the additional expenditure incurred shall not be considered in
Generation cost for fixing tariff. Naturally MAHAGENCOwill have to absorb the cost
of this expenditure. Another important aspect is of conservation of fast depleting natural
resources, such as coal and fuel oil. When power is generated at optimum heat rate,
minimum possible fuel is consumed. Less fuel consumption also leads to lesser extent of
pollutantsadded to the environment. Hence monitoring and controlling the heat rate to
the optimum level has many benefits.
The term Heat rate is defined in many ways as follows:
Net Unit Heat rate: It is the ratio of energy input to Boiler in terms of Heat energy of
fuel, for one kWh of electrical energy output atBus Bars, i.e. after UAT. If the out put
and input is considered for a period of an hour, then it is Net Unit Heat rate for one
Hour. Similarly, it can be calculated over a period of a Day, a Week, a Month or a Year.
In this case, it is the sent out energy that is considered, hence, consumption of electrical
energyfor driving the plant s auxiliaries is also accounted for.
Gross Unit Heat rate: It is the ratio of energy input to Boiler in terms of Heat energy of
fuel, for one kWh of electrical energy output at Generator Terminals. In this case,
auxiliary consumptionisNOTaccounted for.
Net Turbine Cycle Heat rate:It is the ratio of heat energy contained in steam admitted
to Turbine for one kWh of electrical energy output at Generator Terminals. In this case,
auxiliary consumption and losses in Boiler areNOTaccounted for.
Operating Heat rate: It is the heat rate calculated by considering the inputs and outputs
from the plant only when it is synchronized with the grid. In this case, the fuel input
required for steam conditioning, from light up to synchronizationis not considered. Also
auxiliaries consumption during the period of plant shut down is not considered.
What information does Heat rate give?
The plant is designed to generate electricity at certain design heat rate. Deviations from
design values give a valuable information regarding the operational and maintenance
practices. Also, by comparison with the historical data, decisions can be taken while
making investments on the maintenance and renovation. Also, problem area can be
identified and analyzed for improvements. A deviation in Gross Turbine Cycle heat rate
tells us about energy conversion scenario in turbine, including condenser and
regenerative feed heating process. If Net average unit heat rate deviates from that of
design, it tells us how much extra amount of energy is put in and how much money is
wasted .
Now a days, tariff for supply of electricity to consumers is fixed by Maharashtra
Electricity Regulatory Commission. While fixing tariff, MERC has given the benchmark
heat rate values for all power plants in MAHAGENCO. If actual heat rate is more than
the benchmark heat rate, the additional expenditure incurred shall not be considered in
Generation cost for fixing tariff. Naturally MAHAGENCOwill have to absorb the cost
of this expenditure. Another important aspect is of conservation of fast depleting natural
resources, such as coal and fuel oil. When power is generated at optimum heat rate,
minimum possible fuel is consumed. Less fuel consumption also leads to lesser extent of
pollutantsadded to the environment. Hence monitoring and controlling the heat rate to
the optimum level has many benefits.
total output power solutionsManohar Tatwawadi Page5
Calculations of heat rate:
Net Unit Heat rate, for given time period, is calculated by the formula,
(Coal Consumption ? Its Calorific Value)+(Oil Consumption ? Its Calorific Value)
------------------------------------------------------------------------------------------
Generation measured at Bus Bars
To measure coal consumption accurately is very difficult. Also the calorific value of coal
varies and its continuous, on line measurement is not possible.
Hence, in normal practice, unit heat rate is calculated by the simpler method:
Unit Heat rate = Turbine Cycle Heat rate / Boiler Effi.calculated by lossmethod.
Turbine Cycle Heat rate = (Total Heat added to Turbine in Kcal)/Generation in MU
Total Heat added to Turbine Cycle =
((Sp. Enthalpy of S.H. Steam at Boiler Outlet?Total Steam Flow Rate to H.P.T.)
(Sp. Enthalpy of Feed Water at economizer inlet?Feed Water Flowrate at
economiser inlet))
+ (Sp. Enthalpy of R.H. Steam at Reheater outlet Sp. Enthalpy of C.R.H. steam
at Reheater inlet)?Reheat Steam Flow
+ (Sp. Enthalpy of S.H. Steam at Boiler Outlet Sp. Enthalpy of S.H. spray)?
S. H. Attemperator Flow
+ (Sp. Enthalpy of R.H. Steam at Reheater outlet Sp. Enthalpy of Reheat
attemporator)?R. H. Attemperator Flow.
Values of temperature, pressure and flow rate are known from instrumentation and
specific enthalpy can be known from Steam tables. Thevalue of generation is known
from the Energy Meters. If reading of energy meter connected to Generator terminals is
considered in this formula, the heat rate obtained is Gross Heat rate and if that from Bus
Bar energy meter is considered, then it is the net heat rate.
For method of calculation of Boiler efficiency by loss method pl. refer the chapter on the
topic.
Calculations of heat rate:
Net Unit Heat rate, for given time period, is calculated by the formula,
(Coal Consumption ? Its Calorific Value)+(Oil Consumption ? Its Calorific Value)
------------------------------------------------------------------------------------------
Generation measured at Bus Bars
To measure coal consumption accurately is very difficult. Also the calorific value of coal
varies and its continuous, on line measurement is not possible.
Hence, in normal practice, unit heat rate is calculated by the simpler method:
Unit Heat rate = Turbine Cycle Heat rate / Boiler Effi.calculated by lossmethod.
Turbine Cycle Heat rate = (Total Heat added to Turbine in Kcal)/Generation in MU
Total Heat added to Turbine Cycle =
((Sp. Enthalpy of S.H. Steam at Boiler Outlet?Total Steam Flow Rate to H.P.T.)
(Sp. Enthalpy of Feed Water at economizer inlet?Feed Water Flowrate at
economiser inlet))
+ (Sp. Enthalpy of R.H. Steam at Reheater outlet Sp. Enthalpy of C.R.H. steam
at Reheater inlet)?Reheat Steam Flow
+ (Sp. Enthalpy of S.H. Steam at Boiler Outlet Sp. Enthalpy of S.H. spray)?
S. H. Attemperator Flow
+ (Sp. Enthalpy of R.H. Steam at Reheater outlet Sp. Enthalpy of Reheat
attemporator)?R. H. Attemperator Flow.
Values of temperature, pressure and flow rate are known from instrumentation and
specific enthalpy can be known from Steam tables. Thevalue of generation is known
from the Energy Meters. If reading of energy meter connected to Generator terminals is
considered in this formula, the heat rate obtained is Gross Heat rate and if that from Bus
Bar energy meter is considered, then it is the net heat rate.
For method of calculation of Boiler efficiency by loss method pl. refer the chapter on the
topic.
total output power solutionsManohar Tatwawadi Page6
FACTORS AFFECTING THE TURBINE HEAT RATE:
1)Main Steam Temperature at H.P.T Inlet
2)Main Steam Pressure at H.P.T Inlet
3)Reheat Steam Temperatureat I.P.T Inlet
4)Reheat Steam Pressure at I.P.T Inlet
5)CondenserVacuum
6)Temperature of Feed Water at Economiser Inlet.
7)Boiler efficiency
8)S.H. and R.H. attemperation flow rate.
The effect of individual parameter is discussed below:
Rankine cycle efficiency,?rankine = 1 (T2/ Tm1) (1)
Where; T2 is temperature of heat rejection,(2)
Tm1 is Mean temperature of steam admission = (h1-h4s) / (s1-s4s). (3)
h1 & s1 are specific enthalpy and entropy of steam at admission temperature and
pressure, h4s and s4s are the Sp. Enthalpy & entropy of feed water at Economiser inlet.
1)Temperature and Pressure of steam admission (M. S. as well as H.R.H): For ?
Rankine to be high, Mean temperature of Steam admission (Tm1 in expression 1
above) should be as high as possible. Metallurgical constrains limit these values for
the given Turbine. However, by maintaining the steam parameters close to the
values specified by the Manufacturer, maximum possible Mean temperature of
Steam admission is achieved thus cycle is operated at design efficiency. Effect on
heat rate due to Deviation from design values for 210 MW LMW plant is as follows:
ParameterExpecte
d Value
Actual
Value
Heat rate
deviation
Kcal/kwh
Excess Coal
Consumption
/KWh ( C.V.
3500 Kcal/Kg)
Excess coal
consumption
over the year,
at 80% PLF
Main Steam
temp.
537? C532? C16.480.00488660 Tons
FACTORS AFFECTING THE TURBINE HEAT RATE:
1)Main Steam Temperature at H.P.T Inlet
2)Main Steam Pressure at H.P.T Inlet
3)Reheat Steam Temperatureat I.P.T Inlet
4)Reheat Steam Pressure at I.P.T Inlet
5)CondenserVacuum
6)Temperature of Feed Water at Economiser Inlet.
7)Boiler efficiency
8)S.H. and R.H. attemperation flow rate.
The effect of individual parameter is discussed below:
Rankine cycle efficiency,?rankine = 1 (T2/ Tm1) (1)
Where; T2 is temperature of heat rejection,(2)
Tm1 is Mean temperature of steam admission = (h1-h4s) / (s1-s4s). (3)
h1 & s1 are specific enthalpy and entropy of steam at admission temperature and
pressure, h4s and s4s are the Sp. Enthalpy & entropy of feed water at Economiser inlet.
1)Temperature and Pressure of steam admission (M. S. as well as H.R.H): For ?
Rankine to be high, Mean temperature of Steam admission (Tm1 in expression 1
above) should be as high as possible. Metallurgical constrains limit these values for
the given Turbine. However, by maintaining the steam parameters close to the
values specified by the Manufacturer, maximum possible Mean temperature of
Steam admission is achieved thus cycle is operated at design efficiency. Effect on
heat rate due to Deviation from design values for 210 MW LMW plant is as follows:
ParameterExpecte
d Value
Actual
Value
Heat rate
deviation
Kcal/kwh
Excess Coal
Consumption
/KWh ( C.V.
3500 Kcal/Kg)
Excess coal
consumption
over the year,
at 80% PLF
Main Steam
temp.
537? C532? C16.480.00488660 Tons
total output power solutionsManohar Tatwawadi Page7
H.R. Steam
temp.
537? C532? C3.33420.00092190Tons
Main Steam
Pressure
140
Kg/cm?
138
Kg/cm?
2.4170.0.00061016 tons
2)CondenserVacuumplays a very important role in efficiency of the Rankine Cycle. If
vacuumis less than design value, i.e. if Condenser absolute pressure is more than
design value, corresponding saturation temperature is more, thus Heat is rejected at
Higher Temperature (T2 in expression 1 isless than design) and cycle efficiency
drops. This increases Heat rate. Also the of the L.P.T. backpressure increases, thus
reducing the conversion of Heat Energy to work in Turbine. This increases specific
steam rate thus increasing fuel consumption. In Condenser, only latent heat is
rejected, hence condensate temperature is always at saturation temperature. If
condenser pressure is less than design value, temperature of condensate shall also be
less. This causes low feed water temperature, thus increasingthe heat rate. Following
table shows effect of deterioration of condenser vacuum on heat rate.
ParameterExpectedActualExcess
Heat
rateKcal
/ KWh
Excess Coal
Consumption /
KWh( C.V.
3500 Kcal/Kg)
Excess coal
consumption
over the year,
at 80% PLF
Condenser
Vacuum
690 mm
Hg.
670 mm
Hg
19 0.0054 7989 Tons
3)Less Temperature of feed water at Economizer inlet causes efficiency of Rankine
Cycle to drop, as Mean temperature of steam admission decreases. Values of h4s and
s4s in expression 3 above are high, thus reducing Mean temperature.
ParameterExpectedActualExcess
Heat
rateKcal
/ kWh
Excess Coal
Consumption /
KWh ( C.V. 3500
Kcal/Kg)
Excess coal
consumption
over the year,
at 80% PLF
Feed Water
Temp
253? C248? C22 0.0063 9261 Tons
H.R. Steam
temp.
537? C532? C3.33420.00092190Tons
Main Steam
Pressure
140
Kg/cm?
138
Kg/cm?
2.4170.0.00061016 tons
2)CondenserVacuumplays a very important role in efficiency of the Rankine Cycle. If
vacuumis less than design value, i.e. if Condenser absolute pressure is more than
design value, corresponding saturation temperature is more, thus Heat is rejected at
Higher Temperature (T2 in expression 1 isless than design) and cycle efficiency
drops. This increases Heat rate. Also the of the L.P.T. backpressure increases, thus
reducing the conversion of Heat Energy to work in Turbine. This increases specific
steam rate thus increasing fuel consumption. In Condenser, only latent heat is
rejected, hence condensate temperature is always at saturation temperature. If
condenser pressure is less than design value, temperature of condensate shall also be
less. This causes low feed water temperature, thus increasingthe heat rate. Following
table shows effect of deterioration of condenser vacuum on heat rate.
ParameterExpectedActualExcess
Heat
rateKcal
/ KWh
Excess Coal
Consumption /
KWh( C.V.
3500 Kcal/Kg)
Excess coal
consumption
over the year,
at 80% PLF
Condenser
Vacuum
690 mm
Hg.
670 mm
Hg
19 0.0054 7989 Tons
3)Less Temperature of feed water at Economizer inlet causes efficiency of Rankine
Cycle to drop, as Mean temperature of steam admission decreases. Values of h4s and
s4s in expression 3 above are high, thus reducing Mean temperature.
ParameterExpectedActualExcess
Heat
rateKcal
/ kWh
Excess Coal
Consumption /
KWh ( C.V. 3500
Kcal/Kg)
Excess coal
consumption
over the year,
at 80% PLF
Feed Water
Temp
253? C248? C22 0.0063 9261 Tons
total output power solutionsManohar Tatwawadi Page8
Reasons forLow Steam temperature and Pressure:
In the Power Plant, there can be many reasons for low temperature of Steam at Boiler
and Reheater outlet. Passing spray water control valves and motorized valves,
inadequately tuned temperature control system, fouled surfaces of the Super Heaters are
some of the reasons. These reasons become more dominant when the plant is operating
at loads below maximum rating. Throttling of steam flow due to partially shut valves is
the major reason for low pressure of steam at Turbine admission.
Reasons for poor vacuum in Condenser:
1)Air ingrace in condenser: Air ejection system of the condenser has the capacity
to remove non-condensable gases present in the steam in normal operation. As
the condenser is operated at less than atmospheric pressure, it is prone for air
leaking in to it. Sealing systems, such as Turbine Gland Sealing, Water sealing of
the evacuation system Valves, are provided to prevent the air ingrace. If Gland
sealing steam pressure and temperature and Valve Gland sealing water
pressure are not maintained properly, atmospheric air enters the condenser in
large quantity. Evacuation system can not remove the excess air and hence
condenser pressure increases. Condensers are also provided with many tapping
points for instrumentation. Many of these tapping points are used only for
carrying out acceptance tests. Once these tests are over, the temporary
instrumentation connected to condenser is removed. If any of such tapping
point remains open by oversight, air enters the condenser. There is also a chance
of cracks developed on the connection between L.P.T. casing and condenser.
Damaged gaskets on flanged joints, leaking vent valves provided on Pressure
gauges, cracked impulse lines, passingvacuumbreaker valves, atmospheric
vent or drain valves on C.E.P. inlet piping, if are open, also cause air ingrace.
Evacuation equipment, such as Steam Ejectors, ElectricalVacuumPumps are
provided with airflow measuring devices. Any increase in the flow rate
indicates air ingrace. Condenser air leaks can be identified by manual inspection
while the plant is on load. Helium Leak Detectors can also check air leaks. When
the unit is shut down, condenser leaks can be detected by filling Condenser with
D.M. Water up to certain high level. Butthis test needs lot of prior preparation.
2) High C.W. Temperature, Insufficient Flow rate or Fouled heat transfer
surface:Condensers are heat exchangers. Heat transfer takes place from steam
to cooling water from the tube surface. Cooling water takes away the Latent
Heat from condensing steam. The heat transfer equation is
Q= U * A * ?Tm (1)
Where Q is heat load on condenser, a function of mass rate of steam condensing
U is the coefficient of heat transfer,
Reasons forLow Steam temperature and Pressure:
In the Power Plant, there can be many reasons for low temperature of Steam at Boiler
and Reheater outlet. Passing spray water control valves and motorized valves,
inadequately tuned temperature control system, fouled surfaces of the Super Heaters are
some of the reasons. These reasons become more dominant when the plant is operating
at loads below maximum rating. Throttling of steam flow due to partially shut valves is
the major reason for low pressure of steam at Turbine admission.
Reasons for poor vacuum in Condenser:
1)Air ingrace in condenser: Air ejection system of the condenser has the capacity
to remove non-condensable gases present in the steam in normal operation. As
the condenser is operated at less than atmospheric pressure, it is prone for air
leaking in to it. Sealing systems, such as Turbine Gland Sealing, Water sealing of
the evacuation system Valves, are provided to prevent the air ingrace. If Gland
sealing steam pressure and temperature and Valve Gland sealing water
pressure are not maintained properly, atmospheric air enters the condenser in
large quantity. Evacuation system can not remove the excess air and hence
condenser pressure increases. Condensers are also provided with many tapping
points for instrumentation. Many of these tapping points are used only for
carrying out acceptance tests. Once these tests are over, the temporary
instrumentation connected to condenser is removed. If any of such tapping
point remains open by oversight, air enters the condenser. There is also a chance
of cracks developed on the connection between L.P.T. casing and condenser.
Damaged gaskets on flanged joints, leaking vent valves provided on Pressure
gauges, cracked impulse lines, passingvacuumbreaker valves, atmospheric
vent or drain valves on C.E.P. inlet piping, if are open, also cause air ingrace.
Evacuation equipment, such as Steam Ejectors, ElectricalVacuumPumps are
provided with airflow measuring devices. Any increase in the flow rate
indicates air ingrace. Condenser air leaks can be identified by manual inspection
while the plant is on load. Helium Leak Detectors can also check air leaks. When
the unit is shut down, condenser leaks can be detected by filling Condenser with
D.M. Water up to certain high level. Butthis test needs lot of prior preparation.
2) High C.W. Temperature, Insufficient Flow rate or Fouled heat transfer
surface:Condensers are heat exchangers. Heat transfer takes place from steam
to cooling water from the tube surface. Cooling water takes away the Latent
Heat from condensing steam. The heat transfer equation is
Q= U * A * ?Tm (1)
Where Q is heat load on condenser, a function of mass rate of steam condensing
U is the coefficient of heat transfer,
total output power solutionsManohar Tatwawadi Page9
A is the surface area of tubes
?Tm is Log Mean Temperature Difference,
?Ti-?Tf
Where ?Tm =---------------------- (2 )
Ln (?Ti / ?Tf)
?Ti = (saturation temperature of steam C.W. inlet temperature)(3)
?Tf = (saturation temperature of steam C.W. outlet temperature)(4)
Also called Terminal Temperature Difference or TTD
Relationship between Water flow rate and heat load is given by
mw = Q / ( cp * (T2 T1) ) (5)
(T2-T1) = (mw * cp) / Q (6)
Where mw is Mass flow rate of Water
cp is specific heat of water = 4.18Kj/ Kg / ?C,
T2 is Temperature of Water at condenser outlet
T1 is Temperature of Water at condenser inlet,
In the installed system, Mass flow rate of water (depends on the C.C.W pumping
capacity) and Heat Load (Mass of steam from LPT exhaust) becomes constant. And as
per equation 3 above, heat removal capacity solely depends on (T2-T1). Temperature of
Cooling Water, T2, at Condenser outlet can increase only up to the value decided by
design T.T.D. for the condenser,
Design value for T.T.D. in Condensers is generally 2.5? C, as designing condenser for
TTDbelow this is not viable. Hence, ultimately, the heat removal becomes directly
dependent on Cooling Water Inlet temperature (assuming other factors to be constant
for the given case). Increase in this temperature will cause reduction in mass of steam
getting condensed. In such cases, some steam remains in vapour form, causing
Condenser Pressure to increase. Similarly, even if Cooling Water temperature is within
design limits, but its mass flow rate reduces, same scenario can be expected.
If heat transfercoefficient deteriorates, it again lead to increased Condenser Pressure,
as all the steam do not condense because of insufficient cooling.
A is the surface area of tubes
?Tm is Log Mean Temperature Difference,
?Ti-?Tf
Where ?Tm =---------------------- (2 )
Ln (?Ti / ?Tf)
?Ti = (saturation temperature of steam C.W. inlet temperature)(3)
?Tf = (saturation temperature of steam C.W. outlet temperature)(4)
Also called Terminal Temperature Difference or TTD
Relationship between Water flow rate and heat load is given by
mw = Q / ( cp * (T2 T1) ) (5)
(T2-T1) = (mw * cp) / Q (6)
Where mw is Mass flow rate of Water
cp is specific heat of water = 4.18Kj/ Kg / ?C,
T2 is Temperature of Water at condenser outlet
T1 is Temperature of Water at condenser inlet,
In the installed system, Mass flow rate of water (depends on the C.C.W pumping
capacity) and Heat Load (Mass of steam from LPT exhaust) becomes constant. And as
per equation 3 above, heat removal capacity solely depends on (T2-T1). Temperature of
Cooling Water, T2, at Condenser outlet can increase only up to the value decided by
design T.T.D. for the condenser,
Design value for T.T.D. in Condensers is generally 2.5? C, as designing condenser for
TTDbelow this is not viable. Hence, ultimately, the heat removal becomes directly
dependent on Cooling Water Inlet temperature (assuming other factors to be constant
for the given case). Increase in this temperature will cause reduction in mass of steam
getting condensed. In such cases, some steam remains in vapour form, causing
Condenser Pressure to increase. Similarly, even if Cooling Water temperature is within
design limits, but its mass flow rate reduces, same scenario can be expected.
If heat transfercoefficient deteriorates, it again lead to increased Condenser Pressure,
as all the steam do not condense because of insufficient cooling.
total output power solutionsManohar Tatwawadi Page10
Reasons for High C.W. temperature:
In Cooling Towers, evaporative cooling of Hot water takes place. Air, sucked bythe
C.T. Fan, flows in cross flow direction to water flow, comes in contact with air, causing
evaporation of water. The heat energy required is taken from Water, thus cooling it. The
rate of evaporation is dependent on Relative Humidity of air and its drybulb
temperature C.T. design is made considering yearly average value of R.H. found from
historical data.
If the R.H. and Dry bulb temperature of ambient air is high, evaporation is low and
hence Water temperature does not drop to the design values. Thissituation may arise
during some periods of the year and is not controllable. The controllable reasons are;
1.Non availability of some of the C.T. fans,
2.Unequal distribution of water to individual cell of the cooling tower,
3.Some of the water not coming in contact with air stream,
4.Reduced surface are of mass of water due to damaged or plugged nozzles,
5.Sensible heat gain by cold water when it flows from C.T. to C.W. Pump sump.
Reasons for Low C.W. Flow rate;
1)C.W. Flow rate required for maintaining CondenserVacuumat rated generation
from the plant are calculated by designers. Accordingly C.W. Pump rating is calculated.
Velocity of cooling water through condenser tubes is the controlling factor. The pumps
selection is based on calculated values of Hydraulic Resistance of the C.W. Lines,
Condenser tubes, elevation to which hot water should reach etc. Hydraulic resistance
of the C.W. circuit increases due to following reasons:
i.Number of Plugged condenser tubes more than considered while designing the
system
ii.Reduction in Tube cross sectional area due to scaling in the tube or deposit of
mud, algae or organic growth within the tubes
iii.Throttling of Flow distribution valves at C.T. Cells
iv.Throttled isolating valves in the system
v.Deterioration of pump performance due to eroded or corroded impeller.
vi.Heavy and undetected leakage from the under ground piping.
Reasons for High C.W. temperature:
In Cooling Towers, evaporative cooling of Hot water takes place. Air, sucked bythe
C.T. Fan, flows in cross flow direction to water flow, comes in contact with air, causing
evaporation of water. The heat energy required is taken from Water, thus cooling it. The
rate of evaporation is dependent on Relative Humidity of air and its drybulb
temperature C.T. design is made considering yearly average value of R.H. found from
historical data.
If the R.H. and Dry bulb temperature of ambient air is high, evaporation is low and
hence Water temperature does not drop to the design values. Thissituation may arise
during some periods of the year and is not controllable. The controllable reasons are;
1.Non availability of some of the C.T. fans,
2.Unequal distribution of water to individual cell of the cooling tower,
3.Some of the water not coming in contact with air stream,
4.Reduced surface are of mass of water due to damaged or plugged nozzles,
5.Sensible heat gain by cold water when it flows from C.T. to C.W. Pump sump.
Reasons for Low C.W. Flow rate;
1)C.W. Flow rate required for maintaining CondenserVacuumat rated generation
from the plant are calculated by designers. Accordingly C.W. Pump rating is calculated.
Velocity of cooling water through condenser tubes is the controlling factor. The pumps
selection is based on calculated values of Hydraulic Resistance of the C.W. Lines,
Condenser tubes, elevation to which hot water should reach etc. Hydraulic resistance
of the C.W. circuit increases due to following reasons:
i.Number of Plugged condenser tubes more than considered while designing the
system
ii.Reduction in Tube cross sectional area due to scaling in the tube or deposit of
mud, algae or organic growth within the tubes
iii.Throttling of Flow distribution valves at C.T. Cells
iv.Throttled isolating valves in the system
v.Deterioration of pump performance due to eroded or corroded impeller.
vi.Heavy and undetected leakage from the under ground piping.
total output power solutionsManohar Tatwawadi Page11
Reasons for deterioration of Heat transfer coefficient:
Scaling and fouling, corrosion, and organic growth on condenser tubes reduces the
ability of heat transfer between Steam and cooling water. Ingrace of ambient of air in to
the condenser, which blankets the tube surface. Air has very low thermal conductivity
and it causes drop in Heat Transfer coefficient.
To minimize the problems of scaling, it is extremely necessary that cooling water
softness be maintained. Calcium and Magnesium salt precipitates stick to the metal
surface forming hard and difficult to remove scales. These salts have very poor thermal
conductivity. Commonly encountered scales are
i.Calcium Carbonate
ii.Calcium Sulphate
iii.Silicate Scales
iv.Calcium Orthophosphate
v.Magnesium salts
vi.Iron salts
Fouling is caused by deposition of suspended matter, insoluble in water. Foulants are
Mud and silt, Natural Organics, Microorganisms, Air borne Dust, Vegetation etc.
Preventive Measures:
The concentration of salts takes place because of evaporation of water in the cooling
towers. Even if softened water is used, concentration of these salts increases in closed
circulation system. One of the ways to reduce the concentration is taking fresh water in
to the cooling pond to make up for the evaporated water. But by this method, huge
quantity of make up water is required. Another way is to softening. But soft water has
greater tendency for corrosion. Maintaining pH of waterbetween 6.0 to 8.0 by feeding
acid in the system. But there are many disadvantages such as control of pH, safety in
handling huge quantity of acid etc. On line circulation of sponge balls through
condenser tubes, and occasional acid cleaning of the condenser tubes are other ways to
prevent scaling.
Microbial Growth: Microorganisms enter cooling towers through air, make up water and
dust. The major problems are Algae, Fungi and Bacteria. Chlorine is usually adequate to
prevent the growth. But, it is effective only if pH is 8.3 or below. Free chlorine of 0.2 to
0.5 ppm is sufficient. Beyond 8.3 pH Chlorination does not satisfactory results.
Reasons for deterioration of Heat transfer coefficient:
Scaling and fouling, corrosion, and organic growth on condenser tubes reduces the
ability of heat transfer between Steam and cooling water. Ingrace of ambient of air in to
the condenser, which blankets the tube surface. Air has very low thermal conductivity
and it causes drop in Heat Transfer coefficient.
To minimize the problems of scaling, it is extremely necessary that cooling water
softness be maintained. Calcium and Magnesium salt precipitates stick to the metal
surface forming hard and difficult to remove scales. These salts have very poor thermal
conductivity. Commonly encountered scales are
i.Calcium Carbonate
ii.Calcium Sulphate
iii.Silicate Scales
iv.Calcium Orthophosphate
v.Magnesium salts
vi.Iron salts
Fouling is caused by deposition of suspended matter, insoluble in water. Foulants are
Mud and silt, Natural Organics, Microorganisms, Air borne Dust, Vegetation etc.
Preventive Measures:
The concentration of salts takes place because of evaporation of water in the cooling
towers. Even if softened water is used, concentration of these salts increases in closed
circulation system. One of the ways to reduce the concentration is taking fresh water in
to the cooling pond to make up for the evaporated water. But by this method, huge
quantity of make up water is required. Another way is to softening. But soft water has
greater tendency for corrosion. Maintaining pH of waterbetween 6.0 to 8.0 by feeding
acid in the system. But there are many disadvantages such as control of pH, safety in
handling huge quantity of acid etc. On line circulation of sponge balls through
condenser tubes, and occasional acid cleaning of the condenser tubes are other ways to
prevent scaling.
Microbial Growth: Microorganisms enter cooling towers through air, make up water and
dust. The major problems are Algae, Fungi and Bacteria. Chlorine is usually adequate to
prevent the growth. But, it is effective only if pH is 8.3 or below. Free chlorine of 0.2 to
0.5 ppm is sufficient. Beyond 8.3 pH Chlorination does not satisfactory results.
total output power solutionsManohar Tatwawadi Page12
Temperature of feed water at Economizer inlet:
Feed water temperature is another factor, which decides the efficiency of Rankine Cycle,
as is evident from expression 1 above. Tm1 decreases if temperature of feed water at
Boiler outlet is low. High availability of feed water heating system and also its optimum
performance are important factors. Reasons for poor performance of feed heaters are:
1.Scaling of the tubes
2.Inadequate venting of Feed waters before cutting those in service
3.Passing and leaking heater bypass valves
4.Heater getting bypassed frequently due to High water level of because of inefficient
heater level control instrumentation
*********
Temperature of feed water at Economizer inlet:
Feed water temperature is another factor, which decides the efficiency of Rankine Cycle,
as is evident from expression 1 above. Tm1 decreases if temperature of feed water at
Boiler outlet is low. High availability of feed water heating system and also its optimum
performance are important factors. Reasons for poor performance of feed heaters are:
1.Scaling of the tubes
2.Inadequate venting of Feed waters before cutting those in service
3.Passing and leaking heater bypass valves
4.Heater getting bypassed frequently due to High water level of because of inefficient
heater level control instrumentation
*********
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BOILER LOSSES AND EFFICIENCY:
Boilers are designed to operate at certain efficiency. Typical figures of the losses in the
Boiler (designed values) are:
Loss taking place % loss
Dry Flue Gas loss 4.64
H?2Oand H2 in fuel 5.60
H2O in air 0.18
Unburnt Carbon 0.60
Radiation 0.19
Unaccounted 0.40
Manufacturers Margin 0.50
Total Losses 12.11
Efficiency 87.9
Controllable losses are 1) Dry Flue Gas loss and 2) Unburnt Carbon. Losses due to
Moisture in fuel and air are uncontrollable. Ambient air, when introduced in the boiler,
also carries with it water vapors. Hydrogen in Coal reacts with Oxygen in air and forms
moisture. Along with flue gas, water vapors also receive heat energy produced from
combustionof fuel. This energy is lost to atmosphere through Chimney.
Flue gas loss and Unburnt Carbon loss are the controllable losses.Effect of deviation of
some of the parameters on Heat rate :
ParameterExpectedActualExcess
Heat rate
Kcal /
kWh
Excess Coal
Consumption /
KWh ( C.V. 3500
Kcal/Kg)
Excess coal
consumption
over the year, at
80% PLF
Excess
Oxygen
3.5 %4.0%3.4670.001 1600 Tons
BOILER LOSSES AND EFFICIENCY:
Boilers are designed to operate at certain efficiency. Typical figures of the losses in the
Boiler (designed values) are:
Loss taking place % loss
Dry Flue Gas loss 4.64
H?2Oand H2 in fuel 5.60
H2O in air 0.18
Unburnt Carbon 0.60
Radiation 0.19
Unaccounted 0.40
Manufacturers Margin 0.50
Total Losses 12.11
Efficiency 87.9
Controllable losses are 1) Dry Flue Gas loss and 2) Unburnt Carbon. Losses due to
Moisture in fuel and air are uncontrollable. Ambient air, when introduced in the boiler,
also carries with it water vapors. Hydrogen in Coal reacts with Oxygen in air and forms
moisture. Along with flue gas, water vapors also receive heat energy produced from
combustionof fuel. This energy is lost to atmosphere through Chimney.
Flue gas loss and Unburnt Carbon loss are the controllable losses.Effect of deviation of
some of the parameters on Heat rate :
ParameterExpectedActualExcess
Heat rate
Kcal /
kWh
Excess Coal
Consumption /
KWh ( C.V. 3500
Kcal/Kg)
Excess coal
consumption
over the year, at
80% PLF
Excess
Oxygen
3.5 %4.0%3.4670.001 1600 Tons
total output power solutionsManohar Tatwawadi Page14
Unburnt
Carbon
1.0%1.5 %3.7820.0011 1700 Tons
Flue Gas
Temp
135 14518.670.00533 7853 Tons
Moisture
in coal
9% 11%2.75.00078 1156 Tons
Flue Gas Loss:
Combustion of fuel produces flue gas. Its major constituents are
1.Carbon Di Oxide produced by Carbon & Oxygen reaction,
2.Nitrogen from air,
3.Fly ash,
4.Oxygen,
5.Water Vapours.
Temperature of flue gas leaving air pre heaters is maintained at 135 to 140? C. Total Heat
content in the flue gas is =
(Volume of flue gas in m?/sec?Sp. Heat of the flue gas?Specific Weight?Flue gas
temperature)
?Specific Heat of the flues gas is 30.6 kJ / Kgmole/? C.
?Specific weightof the flue gas is 0.796 Kg/ m?.
When boiler is operated with Optimum air supply and temperature of flue gas at APH
outlet must is maintained within the design limits, flue gas loss is at its minimum.
Primary Air + Secondary air is the total Combustion air supplied to Boiler. Depending
on the Coal Analysis and required velocity of air + coal mixture through coal pipes,
manufacturers specify P.A. Flow through coal mill in relation to Coal Feeding.
Combustion Air requirement for the Boiler:Requirement ofair for combustion of coal
varies as per the constituents of coal being fired. If it is less than required, incomplete
combustion takes place leading to high unburnt carbon loss. If it is more than required,
combustion can be complete but Flue gas quantityincrease leading to higher flue gas
losses.
Unburnt
Carbon
1.0%1.5 %3.7820.0011 1700 Tons
Flue Gas
Temp
135 14518.670.00533 7853 Tons
Moisture
in coal
9% 11%2.75.00078 1156 Tons
Flue Gas Loss:
Combustion of fuel produces flue gas. Its major constituents are
1.Carbon Di Oxide produced by Carbon & Oxygen reaction,
2.Nitrogen from air,
3.Fly ash,
4.Oxygen,
5.Water Vapours.
Temperature of flue gas leaving air pre heaters is maintained at 135 to 140? C. Total Heat
content in the flue gas is =
(Volume of flue gas in m?/sec?Sp. Heat of the flue gas?Specific Weight?Flue gas
temperature)
?Specific Heat of the flues gas is 30.6 kJ / Kgmole/? C.
?Specific weightof the flue gas is 0.796 Kg/ m?.
When boiler is operated with Optimum air supply and temperature of flue gas at APH
outlet must is maintained within the design limits, flue gas loss is at its minimum.
Primary Air + Secondary air is the total Combustion air supplied to Boiler. Depending
on the Coal Analysis and required velocity of air + coal mixture through coal pipes,
manufacturers specify P.A. Flow through coal mill in relation to Coal Feeding.
Combustion Air requirement for the Boiler:Requirement ofair for combustion of coal
varies as per the constituents of coal being fired. If it is less than required, incomplete
combustion takes place leading to high unburnt carbon loss. If it is more than required,
combustion can be complete but Flue gas quantityincrease leading to higher flue gas
losses.
total output power solutionsManohar Tatwawadi Page15
For Pulverized coal fired Boilers, 20% Excess air supplied under specific
conditions, ensure complete combustion. By maintaining 3.5 % Oxygen in flue gas (On
dry flue gas basis) at Economizer outlet ensures, that the Boiler if being fed with 20%
excess air. It needs to be emphasized that Specific Conditions must be metto ensure
minimum losses. These conditions are:
1.Fuel particle size must confirm to specified dimensions.
2.All the coal nozzles must admit equalmass of fuel in furnace and hence , primary air
velocity through pipes must be equal and as P.A. flow to mill should be
proportional to mill loading as specified by the manufacturer
3.Coal / Air mixture temperature at Pulveriser outlet must be 77? C.
4.Secondary air must enter combustion chamber from pre determined places only.
5.Secondary air must enter the furnace at predetermined velocity from all elevations.
6.Diffusers on the coal nozzles must be in proper condition to ensure that the jet of
air/coal mixture, emanating from nozzle, is well distributed.
7.Furnace must be air tight to eliminate possibility of entry of ambient air.
When all these conditions are satisfied, then only efficient combustion in the
furnace, supplied with 20 % excess air is ensured. Fuel admission and combustion
system has following equipment to ensure these conditions.
1.Oxygen Analyzers: In situ, Zirconia probe Oxygen Analyzers, installed on
Economiser outlet ducts, continuously monitor Oxygen in flue gas. Automatic air
flow control loop regulates F.D. Fan Inlet Guide Vanes in such a way that 3.5%
Oxygen in flue gas is maintained through out the operation of Boiler.
2.Fuel air dampers (named after the coal elevations i.e. A, B, C, D etc) on all the Four
Corners should be open only for theelevations that are in service. Position of these
dampers must be equal for all the corners. Regulation of these dampers is as per the
quantity of coal feeding measured as Coal Feeder speed. Dampers of the elevations
AA. FF, BC and DE should open equallyfor all Four Corners. These dampers are
regulated to maintain Furnace Windbox D.P. to the value specified by the
manufacturer. Dampers AB, CD and EF are regulated as per Fuel Oil pressure for Oil
elevation in service. For the oil elevation not in service, dampers regulate as per the
Furnace Windbox D.P.
3.Orifice plates in Coal Pipes: To ensure that all burners (nozzles) at all coal elevations
admit equal mass per sec in the furnace, two requirements should be fulfilled.
Primary air flow velocity in each of the pipe must be equal and fuel/ air ratio in all
pipes should be the same. Inserting the Orifice plates, thus equalizing the hydraulic
For Pulverized coal fired Boilers, 20% Excess air supplied under specific
conditions, ensure complete combustion. By maintaining 3.5 % Oxygen in flue gas (On
dry flue gas basis) at Economizer outlet ensures, that the Boiler if being fed with 20%
excess air. It needs to be emphasized that Specific Conditions must be metto ensure
minimum losses. These conditions are:
1.Fuel particle size must confirm to specified dimensions.
2.All the coal nozzles must admit equalmass of fuel in furnace and hence , primary air
velocity through pipes must be equal and as P.A. flow to mill should be
proportional to mill loading as specified by the manufacturer
3.Coal / Air mixture temperature at Pulveriser outlet must be 77? C.
4.Secondary air must enter combustion chamber from pre determined places only.
5.Secondary air must enter the furnace at predetermined velocity from all elevations.
6.Diffusers on the coal nozzles must be in proper condition to ensure that the jet of
air/coal mixture, emanating from nozzle, is well distributed.
7.Furnace must be air tight to eliminate possibility of entry of ambient air.
When all these conditions are satisfied, then only efficient combustion in the
furnace, supplied with 20 % excess air is ensured. Fuel admission and combustion
system has following equipment to ensure these conditions.
1.Oxygen Analyzers: In situ, Zirconia probe Oxygen Analyzers, installed on
Economiser outlet ducts, continuously monitor Oxygen in flue gas. Automatic air
flow control loop regulates F.D. Fan Inlet Guide Vanes in such a way that 3.5%
Oxygen in flue gas is maintained through out the operation of Boiler.
2.Fuel air dampers (named after the coal elevations i.e. A, B, C, D etc) on all the Four
Corners should be open only for theelevations that are in service. Position of these
dampers must be equal for all the corners. Regulation of these dampers is as per the
quantity of coal feeding measured as Coal Feeder speed. Dampers of the elevations
AA. FF, BC and DE should open equallyfor all Four Corners. These dampers are
regulated to maintain Furnace Windbox D.P. to the value specified by the
manufacturer. Dampers AB, CD and EF are regulated as per Fuel Oil pressure for Oil
elevation in service. For the oil elevation not in service, dampers regulate as per the
Furnace Windbox D.P.
3.Orifice plates in Coal Pipes: To ensure that all burners (nozzles) at all coal elevations
admit equal mass per sec in the furnace, two requirements should be fulfilled.
Primary air flow velocity in each of the pipe must be equal and fuel/ air ratio in all
pipes should be the same. Inserting the Orifice plates, thus equalizing the hydraulic
total output power solutionsManohar Tatwawadi Page16
resistance of all the pipes equalizes pipe velocity. Cold air flow tests are conducted
on coal mills at regular intervals. Results from these tests give valuable information
of condition of Orifice plates and partially or fully choked up pipes. If coal mill is
operated with Primary air flow rate less than that specified, velocity of coal air
mixture drops below 20 mtr/sec, causing separation of coal particles from stream
and consequent settlement in pipes, resulting partial choke up. If the temperature of
coal / air mixture at coal mill outlet drops below 60? C, there is a possibility of
condensation of water vapor which also result in separation of coal particles and its
settlement.
4.Mill air flow control dampers: For ensuring the coal / air ratio equal, P.A. flow rate to
mill should be as per mill loading and hence regulated by feeder speed. Coal mill
manufactures givethe P.A. Flow rate and mill loading characteristics.
5.Mill temperature control system: By ensuring coal air mixture at 77? C, adequate
dryness of coal is ensured, which is one of the important requirements for proper and
efficient combustion.
6.Furnace Windbox DP Control system: Velocity at which secondary air enters the
furnace is determined by Furnace Wind box differential pressure. For every boiler,
value of Furnace Wind box differential pressure is specified for different loading
conditions. By sticking to the specified values, it is ensured that velocity of secondary
air is as per the combustion reaction requirement. For this purpose, opening of
Secondary Air dampers of the wind box is controlled by automatic control loop for
Furnace Windbox DP.Set point for this loop is generated as per the boiler load as
indicated in the enclosed Fig.1.
7.Corner Firing: For achieving efficient and sustainedcombustion at desired rate,
Oxygen in Air must reach the Coal particles at that rate. Oxygen molecule reach
burning coal particles by a process called Diffusion. Ratio of Concentration of
Oxygen at particle surface to that in surrounding gas mixture decides rate of
diffusion. This rate is highest when Coal particle is surrounded by air which contains
21 % Oxygen. Furnace atmosphere is made of mixture of Coal, Air, Flue Gases and
Ash particles. To ensure that coal particles will always remain surrounded by air,
place of air admission, velocity at which air is admitted and turbulence in the furnace
are of primeimportance. First two requirements are fulfilled as discussed above.
Tangential firing fulfills requirement of turbulence.
8.Air tight Furnace: Furnace pressure is always maintained at 4 5 mm W.C. below
atmosphere. If furnace is not air tight, ambient airwill enter furnace. But, the velocity
of this air is very low. This air can not mix with the jets of Secondary air and Primary
air / Fuel mixture admitted at very high velocities and hence does not take part in
combustion. But, it travels with flue gas, and distorts the Oxygen reading, thus
resistance of all the pipes equalizes pipe velocity. Cold air flow tests are conducted
on coal mills at regular intervals. Results from these tests give valuable information
of condition of Orifice plates and partially or fully choked up pipes. If coal mill is
operated with Primary air flow rate less than that specified, velocity of coal air
mixture drops below 20 mtr/sec, causing separation of coal particles from stream
and consequent settlement in pipes, resulting partial choke up. If the temperature of
coal / air mixture at coal mill outlet drops below 60? C, there is a possibility of
condensation of water vapor which also result in separation of coal particles and its
settlement.
4.Mill air flow control dampers: For ensuring the coal / air ratio equal, P.A. flow rate to
mill should be as per mill loading and hence regulated by feeder speed. Coal mill
manufactures givethe P.A. Flow rate and mill loading characteristics.
5.Mill temperature control system: By ensuring coal air mixture at 77? C, adequate
dryness of coal is ensured, which is one of the important requirements for proper and
efficient combustion.
6.Furnace Windbox DP Control system: Velocity at which secondary air enters the
furnace is determined by Furnace Wind box differential pressure. For every boiler,
value of Furnace Wind box differential pressure is specified for different loading
conditions. By sticking to the specified values, it is ensured that velocity of secondary
air is as per the combustion reaction requirement. For this purpose, opening of
Secondary Air dampers of the wind box is controlled by automatic control loop for
Furnace Windbox DP.Set point for this loop is generated as per the boiler load as
indicated in the enclosed Fig.1.
7.Corner Firing: For achieving efficient and sustainedcombustion at desired rate,
Oxygen in Air must reach the Coal particles at that rate. Oxygen molecule reach
burning coal particles by a process called Diffusion. Ratio of Concentration of
Oxygen at particle surface to that in surrounding gas mixture decides rate of
diffusion. This rate is highest when Coal particle is surrounded by air which contains
21 % Oxygen. Furnace atmosphere is made of mixture of Coal, Air, Flue Gases and
Ash particles. To ensure that coal particles will always remain surrounded by air,
place of air admission, velocity at which air is admitted and turbulence in the furnace
are of primeimportance. First two requirements are fulfilled as discussed above.
Tangential firing fulfills requirement of turbulence.
8.Air tight Furnace: Furnace pressure is always maintained at 4 5 mm W.C. below
atmosphere. If furnace is not air tight, ambient airwill enter furnace. But, the velocity
of this air is very low. This air can not mix with the jets of Secondary air and Primary
air / Fuel mixture admitted at very high velocities and hence does not take part in
combustion. But, it travels with flue gas, and distorts the Oxygen reading, thus
total output power solutionsManohar Tatwawadi Page17
replacing the Secondary air. It is therefore extremely important that tramp air entry
be prevented.
9.Pulverization of coal for design particle size: The above discussions deal with the
importance of Fuel firing equipment and air supply to boiler. Role of particle size is
as important as that of proper supply and distribution of air in the furnace. As
explained, care is taken that coal particles will always be surrounded by air in the
furnace. In furnace, very small sizeair Packets are interspersed in the
homogeneous mixture of gases. Total oxygen required for complete combustion of
the individual particle depends on mass of particle, which in turn depends on the
size to which particle is pulverized. Smaller is the sizeof particle, smaller the
quantity of Oxygen required for its complete combustion. Hence, by ensuring that
70% of Coal passes through 200 Mesh, it will always remain surrounded by air
packet which will contain enough Oxygen. But, it is also important that size
distribution of balance 30 % coal should be:
Passing through 100 mesh; 85% and above
Retained by 50 Mesh: Less than 0.5%
Resident time of particles in the furnace is generally 1 to 2 seconds. Bigger particles will
not burn completely due to lack ofOxygen, within this time and leave the furnace as
unburnt Carbon, thus increasing losses. Coarser particles also lead to increase in
slagging.
Optimization of Combustion Process:Supplying 20% excess air ensures that
combustion will be complete. However,there is always a possibility that in certain type
of Coal and combustion conditions, Excess Oxygen requirements can even go below
20%. It may also be possible that in some conditions, excess Oxygen requirements may
be more than 20%. In power plants, wherecoal from different mines is fired regularly,
such conditions may arise very frequently. To ensure that combustion remain efficient in
varying condition and Optimum air is supplied to Boiler in all conditions, Carbon Mono
Oxide monitoring in flue gas is done. If combustion is not complete, concentration of CO
in flue gases increases. Complete combustion is indicated by 100 ppm Co in flue gas at
Economizer outlet. If combustion is incomplete due to insufficient air, Co level shot up
immediately to very concentration values. Fig. 2 shows the variations in Co with ref to
Air supplied to Boiler.
Other Factors:
Following factors also cause deterioration of plant performance, thus increasing heat
rate. Many times, these factors are not measurable directly by plant s instrumentation.
But, their effect can be known from regular tests.
replacing the Secondary air. It is therefore extremely important that tramp air entry
be prevented.
9.Pulverization of coal for design particle size: The above discussions deal with the
importance of Fuel firing equipment and air supply to boiler. Role of particle size is
as important as that of proper supply and distribution of air in the furnace. As
explained, care is taken that coal particles will always be surrounded by air in the
furnace. In furnace, very small sizeair Packets are interspersed in the
homogeneous mixture of gases. Total oxygen required for complete combustion of
the individual particle depends on mass of particle, which in turn depends on the
size to which particle is pulverized. Smaller is the sizeof particle, smaller the
quantity of Oxygen required for its complete combustion. Hence, by ensuring that
70% of Coal passes through 200 Mesh, it will always remain surrounded by air
packet which will contain enough Oxygen. But, it is also important that size
distribution of balance 30 % coal should be:
Passing through 100 mesh; 85% and above
Retained by 50 Mesh: Less than 0.5%
Resident time of particles in the furnace is generally 1 to 2 seconds. Bigger particles will
not burn completely due to lack ofOxygen, within this time and leave the furnace as
unburnt Carbon, thus increasing losses. Coarser particles also lead to increase in
slagging.
Optimization of Combustion Process:Supplying 20% excess air ensures that
combustion will be complete. However,there is always a possibility that in certain type
of Coal and combustion conditions, Excess Oxygen requirements can even go below
20%. It may also be possible that in some conditions, excess Oxygen requirements may
be more than 20%. In power plants, wherecoal from different mines is fired regularly,
such conditions may arise very frequently. To ensure that combustion remain efficient in
varying condition and Optimum air is supplied to Boiler in all conditions, Carbon Mono
Oxide monitoring in flue gas is done. If combustion is not complete, concentration of CO
in flue gases increases. Complete combustion is indicated by 100 ppm Co in flue gas at
Economizer outlet. If combustion is incomplete due to insufficient air, Co level shot up
immediately to very concentration values. Fig. 2 shows the variations in Co with ref to
Air supplied to Boiler.
Other Factors:
Following factors also cause deterioration of plant performance, thus increasing heat
rate. Many times, these factors are not measurable directly by plant s instrumentation.
But, their effect can be known from regular tests.
total output power solutionsManohar Tatwawadi Page18
Low efficiency of H.P. Turbines, I.P. Turbine and L.P. Turbine: Turbine cylinder
isentropic efficiency is the measure of how efficiently turbine has converted input heat
energy in tomechanical work. Isentropic efficiency of Turbine Cylinder is given by:
Actual Enthalpy of steam at Inlet Actual Enthalpy of steam at exhaust
Efficiency =------------------------------------------------------------------------------
Actual Enthalpyof steam at Inlet Ideal Enthalpy of steam at exhaust
Actual Enthalpy is known from steam parameters at Inlet and Exhaust. If steam
expands in turbine without change of Entropy, then it is called ideal expansion. By
finding out Temperature for Actual exhaust pressure and actual entropy of steam at
Turbine inlet, value of ideal enthalpy is known. Turbine manufacturers give the
expected Efficiencies. Any subsequent deviation from expected values indicate
deterioration of Turbine and can be corrected in theplanned outages.
Air Heater leakage:
In Trisector Airheaters, air leakage, through seals, in to flue gas takes place. Due to
rotating rotor, the air side and flue gas side sectors are sealed by radial as well as axial
seal plates. Deterioration of sealing arrangement increases air leakage increasing I.D.
Fans loading. Leakage of ambient air in to flue gas through damaged ducts and through
E.S.P. Hoppers is another reason of increased loading fo the I.D. Fans. The extent of both
the leakages can be so high that I.D. Fan loading reaches its maximum, leading to either
restriction on Generation or in worst case, purposeful reduction of Secondary air. By
measuring Oxygen at Air Heater outlet and ESP outlet monitoring of extent of air
leakage is possible.
Make up water consumption:
Consumption of make up water is because of following reasons:
1.Soot blowing
2.Steam ejectors
3.Opening of C.B.D.
4.Passing of drain valves
5.Leakages of steam or feed water.
6.Steam used for Oil heating and steam tracing of oil lines.
7.Operation of auto drain traps to remove condensate from steam pipelines.
To certain extent, steam consumed for Soot Blowing, Oil heating and Ejectors and
Low efficiency of H.P. Turbines, I.P. Turbine and L.P. Turbine: Turbine cylinder
isentropic efficiency is the measure of how efficiently turbine has converted input heat
energy in tomechanical work. Isentropic efficiency of Turbine Cylinder is given by:
Actual Enthalpy of steam at Inlet Actual Enthalpy of steam at exhaust
Efficiency =------------------------------------------------------------------------------
Actual Enthalpyof steam at Inlet Ideal Enthalpy of steam at exhaust
Actual Enthalpy is known from steam parameters at Inlet and Exhaust. If steam
expands in turbine without change of Entropy, then it is called ideal expansion. By
finding out Temperature for Actual exhaust pressure and actual entropy of steam at
Turbine inlet, value of ideal enthalpy is known. Turbine manufacturers give the
expected Efficiencies. Any subsequent deviation from expected values indicate
deterioration of Turbine and can be corrected in theplanned outages.
Air Heater leakage:
In Trisector Airheaters, air leakage, through seals, in to flue gas takes place. Due to
rotating rotor, the air side and flue gas side sectors are sealed by radial as well as axial
seal plates. Deterioration of sealing arrangement increases air leakage increasing I.D.
Fans loading. Leakage of ambient air in to flue gas through damaged ducts and through
E.S.P. Hoppers is another reason of increased loading fo the I.D. Fans. The extent of both
the leakages can be so high that I.D. Fan loading reaches its maximum, leading to either
restriction on Generation or in worst case, purposeful reduction of Secondary air. By
measuring Oxygen at Air Heater outlet and ESP outlet monitoring of extent of air
leakage is possible.
Make up water consumption:
Consumption of make up water is because of following reasons:
1.Soot blowing
2.Steam ejectors
3.Opening of C.B.D.
4.Passing of drain valves
5.Leakages of steam or feed water.
6.Steam used for Oil heating and steam tracing of oil lines.
7.Operation of auto drain traps to remove condensate from steam pipelines.
To certain extent, steam consumed for Soot Blowing, Oil heating and Ejectors and
total output power solutionsManohar Tatwawadi Page19
Water lost through C.B.D. can be calculated. If this data is monitored regularly, extent of
leakage fromsystem can be guessed. Any leakage from system indicates heat lost and
lead to increased heat rate.
Spray Water Flow rate for Steam temperature Control.
There is no direct effect of attemperation flow in heat rate deviation. But increased
sprey flow rate indicates deterioration of Boiler Conditions.
Auxiliary Consumption:
Increased Auxiliary Consumption indicates more energy consumed by auxiliaries. It also
makes less energy available for distribution to consumers. Closely monitoring these
values helpsin monitoring of health of the auxiliary. Regular energy audit gives
valuable information on repairs to be carried out and planned maintenance.
Conclusions:From above discussions, it can be concluded that, operation of the
Thermal Power Plant at optimum conditions reduces Gross Unit heat rate. The factors
that affect heat rate are:
1)Parameters of steam at HPT, IPT inlets,
2)Condenser Performance
3)Cooling Tower Performance,
4)Combustion of fuel in Boiler with Optimum air supply, thus reducing Dry
Flue Gas lossand Unburnt Carbon loss.
5)Auxiliary Consumption
6)Air heater leakage
7)Duct Leakage
8)Ingrace of tramp air in Boiler
9)Make up water consumption
10)Turbine Cylinder Efficiency
11)Feed Water temperature at Economizer Inlet.
Water lost through C.B.D. can be calculated. If this data is monitored regularly, extent of
leakage fromsystem can be guessed. Any leakage from system indicates heat lost and
lead to increased heat rate.
Spray Water Flow rate for Steam temperature Control.
There is no direct effect of attemperation flow in heat rate deviation. But increased
sprey flow rate indicates deterioration of Boiler Conditions.
Auxiliary Consumption:
Increased Auxiliary Consumption indicates more energy consumed by auxiliaries. It also
makes less energy available for distribution to consumers. Closely monitoring these
values helpsin monitoring of health of the auxiliary. Regular energy audit gives
valuable information on repairs to be carried out and planned maintenance.
Conclusions:From above discussions, it can be concluded that, operation of the
Thermal Power Plant at optimum conditions reduces Gross Unit heat rate. The factors
that affect heat rate are:
1)Parameters of steam at HPT, IPT inlets,
2)Condenser Performance
3)Cooling Tower Performance,
4)Combustion of fuel in Boiler with Optimum air supply, thus reducing Dry
Flue Gas lossand Unburnt Carbon loss.
5)Auxiliary Consumption
6)Air heater leakage
7)Duct Leakage
8)Ingrace of tramp air in Boiler
9)Make up water consumption
10)Turbine Cylinder Efficiency
11)Feed Water temperature at Economizer Inlet.
total output power solutionsManohar Tatwawadi Page20
Fig. 1, Variation in Furnace Windbox DP control ckt. set point with load
140 mm Wcl.
Furnace
Windbox DP
40 mm Wcl.
40 % 70 %
Boiler Load
Fig.2, Change in CO in flue gas with combustion air supply
CO in flue gas
In ppm
Deficient airsupply
100 ppm
Optimum Air Supply Air supply to Boiler
********
Fig. 1, Variation in Furnace Windbox DP control ckt. set point with load
140 mm Wcl.
Furnace
Windbox DP
40 mm Wcl.
40 % 70 %
Boiler Load
Fig.2, Change in CO in flue gas with combustion air supply
CO in flue gas
In ppm
Deficient airsupply
100 ppm
Optimum Air Supply Air supply to Boiler
********
total output power solutionsManohar Tatwawadi Page21
210 MWLMWTURBINE CYCLE HEAT & MASS BALANCE
Heat & Mass Balance
Calculation of Heat Rate of 210 MW BHEL Turbine Cycle with Regenerative Feed Water
Heating consisting of seven feed water heaters shown in the schematic flow diagram
attached.
From station to station the C.W. Temp for the condenser cooling changes and therefore
the vacuum conditions and the exhaust temperature also changes accordingly. A
specimen calculation shown has been done for the values at 652000 kgs of steam per
hour flow at 130 kg/cm
2
absolute pressure and 555
0
C superheat Temperature. Similar
such calculations can be done at different loads for calculation of heat rate of the turbine
regenerative feed water heating cycle at those particular loads.
While calculating the Turbine Heat Rate following assumptions have been made in the
original drawings, giving the values of Enthalpy and other Properties.
1.Under cooling in the drain coolers of Heater No 6 and heater No 7 has been taken
as 10
0C. The effect of drain cooler of Heater No 5 has been ignored.
2.Wherever a dash has been given in place of reading, the equipment is inoperative
in that range.
3.The values of temperatures mentioned are approximate. Whenever the steam is
wet (especially at the LP Exhaust), instead of giving the temperature, dryness
fraction has been mentioned in the temperature column.
4.When press in Heater No 5 approaches 10.5 Kg/cm
2absolute, drain from Heater
No 6 is diverted to Deaerator bypassing Heater No 6 and drain of Heater No 5 is
diverted to Heater No 4.
5.The Specific Heat Rate does not include the consumption of steam in the main
Ejector.
Similar such calculations for the turbine heat rate can also be carried out for over load
working of TA set at 215.78 MW and at 211.05 MW Loads, at 670 T/hr steam flow and
with 3% makeup water added to the Hotwell of Condenser and at different cooling
water temperatures ranging from 30
0C to 36
0C at Condenser inlet. A separate calculation
can also be done to calculate the turbine heat rate when HP Heaters are out of service at
206.135 MW Load.
These figures should however, be not taken as guaranteed performance figures and will
vary from station to station depending upon the actual working conditions, ageof
stations etc.
210 MWLMWTURBINE CYCLE HEAT & MASS BALANCE
Heat & Mass Balance
Calculation of Heat Rate of 210 MW BHEL Turbine Cycle with Regenerative Feed Water
Heating consisting of seven feed water heaters shown in the schematic flow diagram
attached.
From station to station the C.W. Temp for the condenser cooling changes and therefore
the vacuum conditions and the exhaust temperature also changes accordingly. A
specimen calculation shown has been done for the values at 652000 kgs of steam per
hour flow at 130 kg/cm
2
absolute pressure and 555
0
C superheat Temperature. Similar
such calculations can be done at different loads for calculation of heat rate of the turbine
regenerative feed water heating cycle at those particular loads.
While calculating the Turbine Heat Rate following assumptions have been made in the
original drawings, giving the values of Enthalpy and other Properties.
1.Under cooling in the drain coolers of Heater No 6 and heater No 7 has been taken
as 10
0C. The effect of drain cooler of Heater No 5 has been ignored.
2.Wherever a dash has been given in place of reading, the equipment is inoperative
in that range.
3.The values of temperatures mentioned are approximate. Whenever the steam is
wet (especially at the LP Exhaust), instead of giving the temperature, dryness
fraction has been mentioned in the temperature column.
4.When press in Heater No 5 approaches 10.5 Kg/cm
2absolute, drain from Heater
No 6 is diverted to Deaerator bypassing Heater No 6 and drain of Heater No 5 is
diverted to Heater No 4.
5.The Specific Heat Rate does not include the consumption of steam in the main
Ejector.
Similar such calculations for the turbine heat rate can also be carried out for over load
working of TA set at 215.78 MW and at 211.05 MW Loads, at 670 T/hr steam flow and
with 3% makeup water added to the Hotwell of Condenser and at different cooling
water temperatures ranging from 30
0C to 36
0C at Condenser inlet. A separate calculation
can also be done to calculate the turbine heat rate when HP Heaters are out of service at
206.135 MW Load.
These figures should however, be not taken as guaranteed performance figures and will
vary from station to station depending upon the actual working conditions, ageof
stations etc.
total output power solutionsManohar Tatwawadi Page22
1)INTRODUCTION
Before calculating the Turbine Heat Rate we shall first consider the stage by stage
efficiency of the Turbine Cycle.
A Complete Turbine Cycle single line schematic diagram is as per attached drawing,
wherein the parameters for Steam, Condensate and Feed water are marked for full load.
Similar such chart can be worked out by inserting the values for the parameters
calculated from BHEL Drawing No C-210-130/TDC-210-60.2. The Turbine Cycle
Diagram indicates the direction of flow of various fluids involved.
2)STAGE BY STAGE EFFICIENCY
2.1Ejector:-
Ejector is supplied with steam from Deaerator (d4) at 4.5Kg/cm
2having 155
0C Temp
and the condensate from hotwell passes through the same, which is pumped by the
condensate extraction pump to the further feed water regenerative heating cycle. The
ejector drain is recovered back to the flash chamber on the condenser. Calculations have
been made assuming the efficiency of 99.8% for the Ejector from which the Enthalpy /
Temperature forthe drain from the Ejector to the Flash Chamber is derived.
P-4.5, T-155
F-1500, E-659.83
Steam from Deaerator
Condensate
to G.C.1
T-47.27
F-1500
*T-45.07
F-480106
T-43.16
F-480106
Condensate
from CEP
Discharge
Drain to Flash
Chamber
*Assuming the Effy. to be 99.8%
Ejector Outlet Temp T Calculated
Output Heat taken by condensate
Input heat given by steam
480106(45.07-43.16)*100
1500 (659.83-T)
= =T
45.07
0
C
1)INTRODUCTION
Before calculating the Turbine Heat Rate we shall first consider the stage by stage
efficiency of the Turbine Cycle.
A Complete Turbine Cycle single line schematic diagram is as per attached drawing,
wherein the parameters for Steam, Condensate and Feed water are marked for full load.
Similar such chart can be worked out by inserting the values for the parameters
calculated from BHEL Drawing No C-210-130/TDC-210-60.2. The Turbine Cycle
Diagram indicates the direction of flow of various fluids involved.
2)STAGE BY STAGE EFFICIENCY
2.1Ejector:-
Ejector is supplied with steam from Deaerator (d4) at 4.5Kg/cm
2having 155
0C Temp
and the condensate from hotwell passes through the same, which is pumped by the
condensate extraction pump to the further feed water regenerative heating cycle. The
ejector drain is recovered back to the flash chamber on the condenser. Calculations have
been made assuming the efficiency of 99.8% for the Ejector from which the Enthalpy /
Temperature forthe drain from the Ejector to the Flash Chamber is derived.
P-4.5, T-155
F-1500, E-659.83
Steam from Deaerator
Condensate
to G.C.1
T-47.27
F-1500
*T-45.07
F-480106
T-43.16
F-480106
Condensate
from CEP
Discharge
Drain to Flash
Chamber
*Assuming the Effy. to be 99.8%
Ejector Outlet Temp T Calculated
Output Heat taken by condensate
Input heat given by steam
480106(45.07-43.16)*100
1500 (659.83-T)
= =T
45.07
0
C
total output power solutionsManohar Tatwawadi Page23
2.2Gland Steam Condenser GC1:-
The Gland Steam Condenser GC1, which is next to the Ejector in the regenerative feed
water heating cycle, sucks steam and air mixture from outermost HP, IP andLP Turbine
glands. The necessary required vacuum of about 100 mm is established in the Gland
Steam Condenser by an Ejector, which is provided with steam (d2) from the Deaerator.
The condensate (gd1) is recovered back to the flash chamber on the condenser.From the
calculations it is seen that the efficiency of the Gland Steam Condenser No 1 is 99.7%.
2.3LP Heater No 1:-
T-100
F-1660
Drain to Cond
Flash Tank
T-45.07
F-480106
Condensate
from Ejector
P-4.5, T-155
F-400, E-659.83
Steam from Deaerator
Condensate
to LPH 1
T-47.18
F-480106
Ejector
GLAND STEAM
COOLER (GC1)
Gland Steam
Condenser Effy.
480106(47.18-45.07)*100
1260(728.20-100)+400(659.83-100)
= =99.7%
Steam Air
Mixture from
glands
P-0.97, T-280
F-1260, E-728.20
2.2Gland Steam Condenser GC1:-
The Gland Steam Condenser GC1, which is next to the Ejector in the regenerative feed
water heating cycle, sucks steam and air mixture from outermost HP, IP andLP Turbine
glands. The necessary required vacuum of about 100 mm is established in the Gland
Steam Condenser by an Ejector, which is provided with steam (d2) from the Deaerator.
The condensate (gd1) is recovered back to the flash chamber on the condenser.From the
calculations it is seen that the efficiency of the Gland Steam Condenser No 1 is 99.7%.
2.3LP Heater No 1:-
T-100
F-1660
Drain to Cond
Flash Tank
T-45.07
F-480106
Condensate
from Ejector
P-4.5, T-155
F-400, E-659.83
Steam from Deaerator
Condensate
to LPH 1
T-47.18
F-480106
Ejector
GLAND STEAM
COOLER (GC1)
Gland Steam
Condenser Effy.
480106(47.18-45.07)*100
1260(728.20-100)+400(659.83-100)
= =99.7%
Steam Air
Mixture from
glands
P-0.97, T-280
F-1260, E-728.20
total output power solutionsManohar Tatwawadi Page24
The Feed water Heater 1 (LP Heater No 1) is fed with Extraction Steam (e1) from the
Extraction No 1 on LP Turbine and the heater drain(hd1) is fed to the flash chamber on
condenser. The condensate after GC1 flows through Heater No 1. From the calculations
shown in the sketch the efficiency comes to 99.80%.
2.4Gland Steam Condenser No 2:-
Condensate
to GC 2
T-61.05
F-480106
T-47.18
F-480106
Condensate
from GC 1
T-65.93
F-11989
Drain to flash
chamber in Cond.
(hd1)
P-0.89, T-85
F-11989, E-622.05
Steam from
Extraction e1
from LPT
Effy of LPH No 1
480106(61.05-47.18)*100
11989(622.05-65.93)
= =99.8%
LPH 1
The Feed water Heater 1 (LP Heater No 1) is fed with Extraction Steam (e1) from the
Extraction No 1 on LP Turbine and the heater drain(hd1) is fed to the flash chamber on
condenser. The condensate after GC1 flows through Heater No 1. From the calculations
shown in the sketch the efficiency comes to 99.80%.
2.4Gland Steam Condenser No 2:-
Condensate
to GC 2
T-61.05
F-480106
T-47.18
F-480106
Condensate
from GC 1
T-65.93
F-11989
Drain to flash
chamber in Cond.
(hd1)
P-0.89, T-85
F-11989, E-622.05
Steam from
Extraction e1
from LPT
Effy of LPH No 1
480106(61.05-47.18)*100
11989(622.05-65.93)
= =99.8%
LPH 1
total output power solutionsManohar Tatwawadi Page25
Gland Steam Condenser No 2 is fed with Leakage steam from HP and IP Turbine Glands
and the drain from GC2 is recovered back to flash tank on the condenser. As per the
calculations shown in the sketch, the efficiency comes to 80%, because not all the
condensate at full load passes through GC2. However, actually a part of the condensate
is bypassed over GC2. The efficiency of GC2 drops accordingly. That is to say that if 80%
of the condensate is passing through GC2, the efficiency drops to 80%.
2.5LP Heater No 2:-
Condensate
to LPH 2
T-68.10
F-480106
T-61.5
F-480106
Condensate
from LPH 1
T-72.94
F-4949
Drain to flash
chamber on Cond.
(gd2)
P-0.50, T-325
F-4949, E-756.90
Steam from HP
& IP Glands g2
Effy of Gland
Steam
Condenser GC 2
0.8*480106(68.10-61.05)*100
4949(756.80-72.94)
= =80% (20% of
condensate is
bypassed)
80%
20%
Flow through GC2
GC 2
Gland Steam Condenser No 2 is fed with Leakage steam from HP and IP Turbine Glands
and the drain from GC2 is recovered back to flash tank on the condenser. As per the
calculations shown in the sketch, the efficiency comes to 80%, because not all the
condensate at full load passes through GC2. However, actually a part of the condensate
is bypassed over GC2. The efficiency of GC2 drops accordingly. That is to say that if 80%
of the condensate is passing through GC2, the efficiency drops to 80%.
2.5LP Heater No 2:-
Condensate
to LPH 2
T-68.10
F-480106
T-61.5
F-480106
Condensate
from LPH 1
T-72.94
F-4949
Drain to flash
chamber on Cond.
(gd2)
P-0.50, T-325
F-4949, E-756.90
Steam from HP
& IP Glands g2
Effy of Gland
Steam
Condenser GC 2
0.8*480106(68.10-61.05)*100
4949(756.80-72.94)
= =80% (20% of
condensate is
bypassed)
80%
20%
Flow through GC2
GC 2
total output power solutionsManohar Tatwawadi Page26
The steam from IP Turbine Extraction No 2 (e2) is fed to the feed water heater LP Heater
No. 2. The drain (hd2) is pumped back with the help of Drip Pump into the main
condensate flow path before the condensate entry to Heater no 3 as shown. This also
increases the temperature of condensatefrom 100.86
0
C to 1o1.52
0
C at the entry in Heater
No 3. From the calculations as above the efficiency of LP Heater No 2 comes to 99.8%.
2.6LP Heater No 3:-
Condensate
to LPH 3
T-101.52
F-555281
T-68.10
F-480106
Condensate
from GC 2
T-105.78
F-75175
Drain to
Condensate line
(hd2)
P-1.369, T-183
F-25369, E-678.41
Steam from IPT
out e2
Effy of LPH 2
480106(100.86-68.10)*100
25369(678.41-105.78)+49806(130.26-105.78)
= =99.8%
LPH 2
Drip Pump
Drain from
LPH3 (hd3)
T-
F-49806
T-100.86, F-480106
The steam from IP Turbine Extraction No 2 (e2) is fed to the feed water heater LP Heater
No. 2. The drain (hd2) is pumped back with the help of Drip Pump into the main
condensate flow path before the condensate entry to Heater no 3 as shown. This also
increases the temperature of condensatefrom 100.86
0
C to 1o1.52
0
C at the entry in Heater
No 3. From the calculations as above the efficiency of LP Heater No 2 comes to 99.8%.
2.6LP Heater No 3:-
Condensate
to LPH 3
T-101.52
F-555281
T-68.10
F-480106
Condensate
from GC 2
T-105.78
F-75175
Drain to
Condensate line
(hd2)
P-1.369, T-183
F-25369, E-678.41
Steam from IPT
out e2
Effy of LPH 2
480106(100.86-68.10)*100
25369(678.41-105.78)+49806(130.26-105.78)
= =99.8%
LPH 2
Drip Pump
Drain from
LPH3 (hd3)
T-
F-49806
T-100.86, F-480106
total output power solutionsManohar Tatwawadi Page27
Heater No 3 is provided with steam from IP Turbine Extraction No 3 (e3) and the drain
from heater no 4 (hd4) is also fed back to heater No 3. It may also be noted that the
amount of condensate passing through heater No 3 increases in quantity by the amount
of drain from heater No 2 (hd2). Heater no 3 drain is fed back to LP Heater No 2. The
calculated efficiency of LP Heater No 3 comes to 99.8%.
2.7LP Heater no 4:-
Condensate
to LPH 4
T-125.27
F-555281
T-101.52
F-555281
Condensate
from LPH2 &
Drip Pump
T-130.26
F-49806
Drain to LPH 2
(hd3)
P-2.974, T-264
F-21033, E-715.68
Steam from IPT
Extr (e3)
Effy of LPH 3
555281(125.27-101.52)*100
21033(715.68-130.26)+28773(161.73-130.26)
= =99.8%
LPH 3
Drain from
LPH4 (hd4)
T-
F-28773
Heater No 3 is provided with steam from IP Turbine Extraction No 3 (e3) and the drain
from heater no 4 (hd4) is also fed back to heater No 3. It may also be noted that the
amount of condensate passing through heater No 3 increases in quantity by the amount
of drain from heater No 2 (hd2). Heater no 3 drain is fed back to LP Heater No 2. The
calculated efficiency of LP Heater No 3 comes to 99.8%.
2.7LP Heater no 4:-
Condensate
to LPH 4
T-125.27
F-555281
T-101.52
F-555281
Condensate
from LPH2 &
Drip Pump
T-130.26
F-49806
Drain to LPH 2
(hd3)
P-2.974, T-264
F-21033, E-715.68
Steam from IPT
Extr (e3)
Effy of LPH 3
555281(125.27-101.52)*100
21033(715.68-130.26)+28773(161.73-130.26)
= =99.8%
LPH 3
Drain from
LPH4 (hd4)
T-
F-28773
total output power solutionsManohar Tatwawadi Page28
LP Heater No 4 is fed with Extraction steam from IP Turbine (e4). Leakage steam from
HP and IP Glands (g3) is also fed to the LP Heater No 4. Drain from LP Heater No 4 is
fed back to LP Heater No 3. HP Heater No 5 drain (hd5) is cascaded to LP Heater No 4
below 150 MW load. For loads more than 150 MW HP Heater No 5 drain normally goes
to Deaerator. In the calculations at full load, HP Heater No 5 drain (hd5) is not
considered as fed toHeater No 4. The calculated Efficiency comes to 99.7%.
2.8Deaerator:-
Condensate
to
Deairator
T-156.52
F-555281
T-125.27
F-555281
Condensate
from LPH3
T-161.73
F-28773
Drain to LPH 3
(hd4)
P-6.911, T-364
F-23838, E-762.69
Steam from IPT
Extr (e4)
Effy of LPH 4
555281(156.52-125.27)*100
23838(762.69-161.73)+4935(782.31-161.73)
= =99.7%
LPH 4
Leak off
steam HP
& IP (g3)
P-6.9, T-161.73
F-4935, E-
LP Heater No 4 is fed with Extraction steam from IP Turbine (e4). Leakage steam from
HP and IP Glands (g3) is also fed to the LP Heater No 4. Drain from LP Heater No 4 is
fed back to LP Heater No 3. HP Heater No 5 drain (hd5) is cascaded to LP Heater No 4
below 150 MW load. For loads more than 150 MW HP Heater No 5 drain normally goes
to Deaerator. In the calculations at full load, HP Heater No 5 drain (hd5) is not
considered as fed toHeater No 4. The calculated Efficiency comes to 99.7%.
2.8Deaerator:-
Condensate
to
Deairator
T-156.52
F-555281
T-125.27
F-555281
Condensate
from LPH3
T-161.73
F-28773
Drain to LPH 3
(hd4)
P-6.911, T-364
F-23838, E-762.69
Steam from IPT
Extr (e4)
Effy of LPH 4
555281(156.52-125.27)*100
23838(762.69-161.73)+4935(782.31-161.73)
= =99.7%
LPH 4
Leak off
steam HP
& IP (g3)
P-6.9, T-161.73
F-4935, E-
total output power solutionsManohar Tatwawadi Page29
Dearator is supplied with heating steam from Extraction tapped from either Extraction
no 5 or Extraction No 6 (ed). Extraction No 5 is taken from IP Turbine and Extraction No
6 is taken from HP Turbine at HP Exhaust. For loads below 150 MW, deaerator heating
steam supply (ed) is derived from Extraction No 6, which is changed over to Extraction
No 5 at loads above 150 MW. Auxiliary steam supply to the main Ejector, Gland steam
ejectorand sealing steam of Turbine Glands is changed over from auxiliary PRDS to
Deaerator.
The Deaerator pegging steam supply is initially given from auxiliary PRDS, which is
changed over from auxiliary source to Extraction No 6 at about 90 to 100 MW Load.
Thegland leakage from ESV and IV is fed to the deaerator. Heater No 5 and Heater No 6
Drains (hd5 or hd6) are also fed to the deaerator. Actually HP Heaters are taken in
service after 60 to 70 MW Load is taken on the turbine. Heater No 5 Drain (hd5) goes to
Deaerator
pegging steam
From PRDS or
Extr.6(ed)
Drain from HPH
5/6 (hd5 or hd6)
Gland leakage
from ESV/IV
Steam supply to
glands (d1)
Steam supply to
Ejectors (main
& GC1) d2+d4
Condensate
from LPH4
P-7.5, T-479
F-2300, E-820.28
P-1.03, T-141
F-2930, E-659.83
P-4.5, T-155
F-1900, E-659.83
T-181.51
F-93256
E-802.51
F-5984
T-156.52
F-555281
Feed water
to HPH5
T-173.12
F-652000
(555281(166.71-156.52)+4830(820.28-659.83))*100
2300(820.28-166.71)+93256(181.51-166.71)+5984(802.92-
166.71)
Deaerator Effy
= =
96.8%
T-166.71
F-652000
Dearator is supplied with heating steam from Extraction tapped from either Extraction
no 5 or Extraction No 6 (ed). Extraction No 5 is taken from IP Turbine and Extraction No
6 is taken from HP Turbine at HP Exhaust. For loads below 150 MW, deaerator heating
steam supply (ed) is derived from Extraction No 6, which is changed over to Extraction
No 5 at loads above 150 MW. Auxiliary steam supply to the main Ejector, Gland steam
ejectorand sealing steam of Turbine Glands is changed over from auxiliary PRDS to
Deaerator.
The Deaerator pegging steam supply is initially given from auxiliary PRDS, which is
changed over from auxiliary source to Extraction No 6 at about 90 to 100 MW Load.
Thegland leakage from ESV and IV is fed to the deaerator. Heater No 5 and Heater No 6
Drains (hd5 or hd6) are also fed to the deaerator. Actually HP Heaters are taken in
service after 60 to 70 MW Load is taken on the turbine. Heater No 5 Drain (hd5) goes to
Deaerator
pegging steam
From PRDS or
Extr.6(ed)
Drain from HPH
5/6 (hd5 or hd6)
Gland leakage
from ESV/IV
Steam supply to
glands (d1)
Steam supply to
Ejectors (main
& GC1) d2+d4
Condensate
from LPH4
P-7.5, T-479
F-2300, E-820.28
P-1.03, T-141
F-2930, E-659.83
P-4.5, T-155
F-1900, E-659.83
T-181.51
F-93256
E-802.51
F-5984
T-156.52
F-555281
Feed water
to HPH5
T-173.12
F-652000
(555281(166.71-156.52)+4830(820.28-659.83))*100
2300(820.28-166.71)+93256(181.51-166.71)+5984(802.92-
166.71)
Deaerator Effy
= =
96.8%
T-166.71
F-652000
total output power solutionsManohar Tatwawadi Page30
Heater No 4 for loads less than 150 MW. For Turbine loads higher than 170 MW, Heater
No 6 Drain (hd6) goes to Heater No 5. And heater no 5 drain (hd5) is diverted to
Deaerator instead of Heater No 4. From the heat balance calculations, the deaerator
efficiency comes to 96.8%.
2.9Boiler Feed Pump:-
It may be noted that the total quantity of feed water supplied from feed water tank to
boiler feed water pump is equal to main steam flow to the HP Turbine.
In this connection, it may be noted that heat added by Boiler Feed water pump due to
Mechanical churning of water has been also taken into consideration, which is equal to
4179320 Kcal.
2.10HP Heater No 5:-
T=166.71
Flow=
= 652000(173.12-166.71) = 4179320 Kcal.
Heater No 4 for loads less than 150 MW. For Turbine loads higher than 170 MW, Heater
No 6 Drain (hd6) goes to Heater No 5. And heater no 5 drain (hd5) is diverted to
Deaerator instead of Heater No 4. From the heat balance calculations, the deaerator
efficiency comes to 96.8%.
2.9Boiler Feed Pump:-
It may be noted that the total quantity of feed water supplied from feed water tank to
boiler feed water pump is equal to main steam flow to the HP Turbine.
In this connection, it may be noted that heat added by Boiler Feed water pump due to
Mechanical churning of water has been also taken into consideration, which is equal to
4179320 Kcal.
2.10HP Heater No 5:-
T=166.71
Flow=
= 652000(173.12-166.71) = 4179320 Kcal.
total output power solutionsManohar Tatwawadi Page31
HP Heaters are taken into service in block after a load of 60 to 70 MW is taken on the
turbogenerator. HP Heater No 5 is supplied with steam from Extraction No 5 (e5) taken
from IP Turbine. Heater No 5 Drain (hd5) goes to Heater No 4 below 150 MW, which is
changed over to Deaerator at loads more than 170 MW. Heater No 6 Drain (hd6) is
cascaded to Heater no 5. From the heat balance calculations the efficiency of HP Heater
No 5 comes to 93.7%.
2.11HP Heater No 6:-
Feed Water
to HPH 6
T-189.69
F-652000
T-173.12
F-652000
Feed water
from BFP
Discharge
T-181.51
F-93256
Drain to Deaeator (hd5)
P-12.96, T-444
F-16640, E-802.92
Steam from IPT
Extr (e5)
Effy of HPH 5
652000(189.69-173.12)*100
16640(802.92-181.51)+76616(198.05-181.51)
= =93.7%
HPH 5
Drain from
HPH6 (hd6)
T-198
F-76616, E-
HP Heaters are taken into service in block after a load of 60 to 70 MW is taken on the
turbogenerator. HP Heater No 5 is supplied with steam from Extraction No 5 (e5) taken
from IP Turbine. Heater No 5 Drain (hd5) goes to Heater No 4 below 150 MW, which is
changed over to Deaerator at loads more than 170 MW. Heater No 6 Drain (hd6) is
cascaded to Heater no 5. From the heat balance calculations the efficiency of HP Heater
No 5 comes to 93.7%.
2.11HP Heater No 6:-
Feed Water
to HPH 6
T-189.69
F-652000
T-173.12
F-652000
Feed water
from BFP
Discharge
T-181.51
F-93256
Drain to Deaeator (hd5)
P-12.96, T-444
F-16640, E-802.92
Steam from IPT
Extr (e5)
Effy of HPH 5
652000(189.69-173.12)*100
16640(802.92-181.51)+76616(198.05-181.51)
= =93.7%
HPH 5
Drain from
HPH6 (hd6)
T-198
F-76616, E-
total output power solutionsManohar Tatwawadi Page32
Heater No 6 is supplied with steam from Extraction No 6 (e6) from HPT Exhaust. It is
cold reheat steam. The spindle leakage steam from HPT (g5) is also fed to Heater No 6.
Heater No 7 Drain (hd7) is cascaded to Heater no 6 and Heater no 6 drain (hd6) is
cascaded to Heater No 5. From the calculations the efficiency is 99.7%.
2.12HP Heater No 7:-
Feed Water
to HPH 7
T 228.80
F-652000
T 189.69
F-
Feed water
from HPH 5
T-198.01
F-76616
Drain to HPH 5 (hd6)
P-28.07, T-327
F-40391, E-732.99
Steam from
HPT Exh. (e6)
Effy of HPH 6
652000(228.80-189.69)*100
40391(732.99-198.05)+31844(238.65-198.05)+4381(806.50-198.05)
= =99.7%
HPH 6
Drain from
HPH7 (hd7)
T-238
F-31844, E-
Spindle
Leakage from
HPT (g5)
P-28.07, T-464
F-4381, E-806.50
Heater No 6 is supplied with steam from Extraction No 6 (e6) from HPT Exhaust. It is
cold reheat steam. The spindle leakage steam from HPT (g5) is also fed to Heater No 6.
Heater No 7 Drain (hd7) is cascaded to Heater no 6 and Heater no 6 drain (hd6) is
cascaded to Heater No 5. From the calculations the efficiency is 99.7%.
2.12HP Heater No 7:-
Feed Water
to HPH 7
T 228.80
F-652000
T 189.69
F-
Feed water
from HPH 5
T-198.01
F-76616
Drain to HPH 5 (hd6)
P-28.07, T-327
F-40391, E-732.99
Steam from
HPT Exh. (e6)
Effy of HPH 6
652000(228.80-189.69)*100
40391(732.99-198.05)+31844(238.65-198.05)+4381(806.50-198.05)
= =99.7%
HPH 6
Drain from
HPH7 (hd7)
T-238
F-31844, E-
Spindle
Leakage from
HPT (g5)
P-28.07, T-464
F-4381, E-806.50
total output power solutionsManohar Tatwawadi Page33
Heater No 7 is provided with Extractionsteam from HP Turbine (e7). The drain (hd7) is
cascaded to Heater No 6. Feed water at the outlet of the heater is finally fed to back to
the boiler through the feed water regulating valve. From the calculations the efficiency
comes out to be 99.5%.
2.13Condenser:-
Feed Water to
Economiser
T 253.94
F-652000
T 228.80
F-
Feed water
from HPH 6
T 238.65
F-31844
Drain to HPH 6 (hd7)
P-42.18, T-381
F-31844, E-755.97
Ext. Steam
from HPT (e7)
Effy of HPH 7
652000(253.94-228.80)*100
31844(755.97-238.65)
= =99.5%
HPH 7
Heater No 7 is provided with Extractionsteam from HP Turbine (e7). The drain (hd7) is
cascaded to Heater No 6. Feed water at the outlet of the heater is finally fed to back to
the boiler through the feed water regulating valve. From the calculations the efficiency
comes out to be 99.5%.
2.13Condenser:-
Feed Water to
Economiser
T 253.94
F-652000
T 228.80
F-
Feed water
from HPH 6
T 238.65
F-31844
Drain to HPH 6 (hd7)
P-42.18, T-381
F-31844, E-755.97
Ext. Steam
from HPT (e7)
Effy of HPH 7
652000(253.94-228.80)*100
31844(755.97-238.65)
= =99.5%
HPH 7
total output power solutionsManohar Tatwawadi Page34
210 MW Turbine has two condensers connected in parallel. The total quantity of the
condensed steam exhausted from the LP Turbine is mixed with drains from Ejector
consisting of condensate of steam from Deaerator(d4) and GC1 condensate of steam
from HPT and IPT.
The steam from LPT at a Pressure of 0.0889Kg/cm2 and quantity 460008 kg/hr is
condensed by cooling water quantity 270000 m3 which equals to 27000000 kg/hr and
hence the circulating water required for cooling 1 kg of steam is 27000000/460008 = 58.6
kg.
From the heat balance equation of condenser:-
The total heat loss to cooling water = Steam from LPT Exhaust * Enthalpy of water at
Hotwell
= 2 X 230004 * (589.18 43.16) = 251173568 Kcal/hr.
Cooling
Water out
to Cooling
Towers
Cooling
Water from
CW Pumps
T 30
F 270000 m
3
CEP
CEP
Disc. To
Ejector
P-22, T 43
F 480106, E-43.1
Make up
20 T
Drains to cond
Flash Tank.
gd1-1660
gd2-4949
hd1-11989
ejd-1500
Steam from LPTP-0.0889, F-2*230004, E-589.18
Steam to circ.water Ratio= 0.00178
T. T. D.. =11
0
C Effy = 80%
210 MW Turbine has two condensers connected in parallel. The total quantity of the
condensed steam exhausted from the LP Turbine is mixed with drains from Ejector
consisting of condensate of steam from Deaerator(d4) and GC1 condensate of steam
from HPT and IPT.
The steam from LPT at a Pressure of 0.0889Kg/cm2 and quantity 460008 kg/hr is
condensed by cooling water quantity 270000 m3 which equals to 27000000 kg/hr and
hence the circulating water required for cooling 1 kg of steam is 27000000/460008 = 58.6
kg.
From the heat balance equation of condenser:-
The total heat loss to cooling water = Steam from LPT Exhaust * Enthalpy of water at
Hotwell
= 2 X 230004 * (589.18 43.16) = 251173568 Kcal/hr.
Cooling
Water out
to Cooling
Towers
Cooling
Water from
CW Pumps
T 30
F 270000 m
3
CEP
CEP
Disc. To
Ejector
P-22, T 43
F 480106, E-43.1
Make up
20 T
Drains to cond
Flash Tank.
gd1-1660
gd2-4949
hd1-11989
ejd-1500
Steam from LPTP-0.0889, F-2*230004, E-589.18
Steam to circ.water Ratio= 0.00178
T. T. D.. =11
0
C Effy = 80%
total output power solutionsManohar Tatwawadi Page35
If the condenser efficiency is assumed to be 80% then the gain in temp of cooling water
= (251173568/0.8)/27000000 = 11
0
C
From the calculations shown it will be seen that if 270000 m
3
/h water flow is maintained
for the circulating water through the condenser having 80% efficiency the terminal
temperature difference for the circulating water will be of the order of 11
0
C.
3)TURBINE CYCLE HEAT RATE
The turbine cycle heat rate is calculated as per the formula:-
Total Heat Supplied to the Turbine in kcalQM*(MSE-FWE) + QR*(RHE-CRHE)
Heat Rate =
Total Generation in KWHW
QM= Main Steam Flow = 652000 Kg/hr
MSE= Main Steam Enthalpy = 819.94 Kcal/kg
FWE= Feed Water Enthalpy at HP7 Outlet= 253.94 kcal/kg
QR =Reheat Steam Flow = 566369 kg/hr
RHE =Enthalpy of Reheat Steam= 845.83 Kcal/kg
CRHE=Enthalpy of Cold Reheat Steam= 732.99 Kcal/kg
W =Total Power Developed at Generator output in KWH= 211902 KWH
652000 * (819.94-253.94) + 566369 * (845.83 732.99)
HR at 210 MW=
211902
652000 * 566 + 566369 * 112.84432941078
= = = 2043.11 Kcal/KWH
211902 211902
Calculated Heat Rates at different loads are :-
a)2043.11 Kcal/KWH at 210 MW
b)2068 Kcal/KWH at 176 MW,
c)2070 Kcal/KWH at 150 MW and
d)2156 Kcal/KWH at 100 MW
4)EXTRACTIONS
4.1Total Heat given to Turbine Cylinders
If the condenser efficiency is assumed to be 80% then the gain in temp of cooling water
= (251173568/0.8)/27000000 = 11
0
C
From the calculations shown it will be seen that if 270000 m
3
/h water flow is maintained
for the circulating water through the condenser having 80% efficiency the terminal
temperature difference for the circulating water will be of the order of 11
0
C.
3)TURBINE CYCLE HEAT RATE
The turbine cycle heat rate is calculated as per the formula:-
Total Heat Supplied to the Turbine in kcalQM*(MSE-FWE) + QR*(RHE-CRHE)
Heat Rate =
Total Generation in KWHW
QM= Main Steam Flow = 652000 Kg/hr
MSE= Main Steam Enthalpy = 819.94 Kcal/kg
FWE= Feed Water Enthalpy at HP7 Outlet= 253.94 kcal/kg
QR =Reheat Steam Flow = 566369 kg/hr
RHE =Enthalpy of Reheat Steam= 845.83 Kcal/kg
CRHE=Enthalpy of Cold Reheat Steam= 732.99 Kcal/kg
W =Total Power Developed at Generator output in KWH= 211902 KWH
652000 * (819.94-253.94) + 566369 * (845.83 732.99)
HR at 210 MW=
211902
652000 * 566 + 566369 * 112.84432941078
= = = 2043.11 Kcal/KWH
211902 211902
Calculated Heat Rates at different loads are :-
a)2043.11 Kcal/KWH at 210 MW
b)2068 Kcal/KWH at 176 MW,
c)2070 Kcal/KWH at 150 MW and
d)2156 Kcal/KWH at 100 MW
4)EXTRACTIONS
4.1Total Heat given to Turbine Cylinders
total output power solutionsManohar Tatwawadi Page36
High Pressure Turbine
Flow EnthalpyHeat in Kcal
Input Steam652000 819.94 534600880
Output Steam566369 732.99 415142813
Heat Given to HP Turbine119458067
Intermediate Pressure Turbine
Flow EnthalpyHeat in Kcal
Input Steam566369 845.83 479051891
Output Steam470857 678.41 319434097
Heat Given to IP Turbine159617794
Low Pressure Turbine
Flow EnthalpyHeat in Kcal
Input Steam470857 678.41 319434097
Output Steam460008 589.18 271027514
Heat Given to LP Turbine48406583
Total input to the Turbine Cylinders = Heat given to HP + IP + LP Cylinders
= 119458067 + 159617794 + 48406583 = 327482444 Kcal for generating211902 kWH of
Power.
A smallquantity of Leakage steam from ESV & IV spindle leaks to Deaerator and gland
steam condenser, which is also utilized for feed water regenerative heating. The
condensate derived thereby is also added in the system. The heat content of this
auxiliary steamis neglected.
4.2The total quantity of steam extracted for regeneration
SymbolWork PressureTemp.FlowEnthalpyTotal Heat
e1 Extraction 1LP Heater 10.88954511989622.057,457,757
e2 Extraction 2LP Heater 21.36918325369678.4117,210,583
e3 Extraction 3LP Heater 32.97426421033715.6815,052,897
High Pressure Turbine
Flow EnthalpyHeat in Kcal
Input Steam652000 819.94 534600880
Output Steam566369 732.99 415142813
Heat Given to HP Turbine119458067
Intermediate Pressure Turbine
Flow EnthalpyHeat in Kcal
Input Steam566369 845.83 479051891
Output Steam470857 678.41 319434097
Heat Given to IP Turbine159617794
Low Pressure Turbine
Flow EnthalpyHeat in Kcal
Input Steam470857 678.41 319434097
Output Steam460008 589.18 271027514
Heat Given to LP Turbine48406583
Total input to the Turbine Cylinders = Heat given to HP + IP + LP Cylinders
= 119458067 + 159617794 + 48406583 = 327482444 Kcal for generating211902 kWH of
Power.
A smallquantity of Leakage steam from ESV & IV spindle leaks to Deaerator and gland
steam condenser, which is also utilized for feed water regenerative heating. The
condensate derived thereby is also added in the system. The heat content of this
auxiliary steamis neglected.
4.2The total quantity of steam extracted for regeneration
SymbolWork PressureTemp.FlowEnthalpyTotal Heat
e1 Extraction 1LP Heater 10.88954511989622.057,457,757
e2 Extraction 2LP Heater 21.36918325369678.4117,210,583
e3 Extraction 3LP Heater 32.97426421033715.6815,052,897
total output power solutionsManohar Tatwawadi Page37
e4 Extraction 4LP Heater 46.91136423838762.6918,181,004
Ed DeaeratorDeaerator12.964495984802.924,804,673
e5 Extraction 5HP Heater 512.9644416640802.9213,360,589
e6 Extraction 6HP Heater 628.0732740391732.9929,606,199
e7 Extraction 7HP Heater 742.1838131844755.9724,073,109
Total Extraction Flow / Enthalpy177088TPH129,746,812
Total Steam Extracted*100177088 * 100
Percentage of quantity = = 27.16%
of steam extractedTotal Steam supplied652000
Total Heat to Extractions *100129746812*100
Percentage of Heat Tapped == = 39.61%
Total Heat input to the cylinders327482444
Out of the above heat given to the extractions, total heat regained by feed water
regenerative heating cycle:-
?Upto LP Heater No 2480106 * (100.86-43.16)= 27702116 Kcal
?After LP Heater No 3555281 * (156.52 100.86)= 30906940 Kcal
?After Deaerator555281 * (166.71 156.52)= 5658313 Kcal
?In Feed water Pump652000 * (173.12 166.71)= 4179320 Kcal
?After HP Heater 7652000 * (253.94 173.12)= 52694640 Kcal
Total heat regained by Feed Water= 121141330Kcal
Heat Loss in Feed Water Regenerative Heating Cycle
= Total heat to extractions Heat regained
= 129746812 121141330 =8605482 Kcal
4.3Total Heat Lost to cooling water
= (Enthalpy of steam at exhaust Enthalpy of condensate in Hotwell)*Quantity of
Exhaust steam
= (589.18 43.16) * 460008 =251173568 Kcal
e4 Extraction 4LP Heater 46.91136423838762.6918,181,004
Ed DeaeratorDeaerator12.964495984802.924,804,673
e5 Extraction 5HP Heater 512.9644416640802.9213,360,589
e6 Extraction 6HP Heater 628.0732740391732.9929,606,199
e7 Extraction 7HP Heater 742.1838131844755.9724,073,109
Total Extraction Flow / Enthalpy177088TPH129,746,812
Total Steam Extracted*100177088 * 100
Percentage of quantity = = 27.16%
of steam extractedTotal Steam supplied652000
Total Heat to Extractions *100129746812*100
Percentage of Heat Tapped == = 39.61%
Total Heat input to the cylinders327482444
Out of the above heat given to the extractions, total heat regained by feed water
regenerative heating cycle:-
?Upto LP Heater No 2480106 * (100.86-43.16)= 27702116 Kcal
?After LP Heater No 3555281 * (156.52 100.86)= 30906940 Kcal
?After Deaerator555281 * (166.71 156.52)= 5658313 Kcal
?In Feed water Pump652000 * (173.12 166.71)= 4179320 Kcal
?After HP Heater 7652000 * (253.94 173.12)= 52694640 Kcal
Total heat regained by Feed Water= 121141330Kcal
Heat Loss in Feed Water Regenerative Heating Cycle
= Total heat to extractions Heat regained
= 129746812 121141330 =8605482 Kcal
4.3Total Heat Lost to cooling water
= (Enthalpy of steam at exhaust Enthalpy of condensate in Hotwell)*Quantity of
Exhaust steam
= (589.18 43.16) * 460008 =251173568 Kcal
total output power solutionsManohar Tatwawadi Page38
4.4Turbine gross heat rate
Turbine Heat Input-(Heat Extracted from all Ext.)+(Heat Lost in Regen. Heating)+(
Heat Lost to CW)
Power Generatedin kwh
327482444 129746812 + 8605482 + 251173568
211902
= 2159 Kcal/kwh
4.5Specific Steam Consumption
At normal specified steam pressure and temperature of Main Steam and Reheat Steam
Specific Steam Consumption of this turbine is as below:-
Load in kw211902202000176669151400101200
Specific Steam3.0042.9663.0063.0453.077
Consumption(in Kg/kwh)
******************
4.4Turbine gross heat rate
Turbine Heat Input-(Heat Extracted from all Ext.)+(Heat Lost in Regen. Heating)+(
Heat Lost to CW)
Power Generatedin kwh
327482444 129746812 + 8605482 + 251173568
211902
= 2159 Kcal/kwh
4.5Specific Steam Consumption
At normal specified steam pressure and temperature of Main Steam and Reheat Steam
Specific Steam Consumption of this turbine is as below:-
Load in kw211902202000176669151400101200
Specific Steam3.0042.9663.0063.0453.077
Consumption(in Kg/kwh)
******************
total output power solutionsManohar Tatwawadi Page39
210 MW Parameters at different regimes at CW Temp 30
0C
S.
N
PARAMETERS PRESSURE IN KG/SQCM ABSTEMPERATURE / DRYNESS
Regime IN KW211902202200176669151400101200211902202200176669151400101200
AMAIN STEAM
1MAIN STEAM130130130130130535535535535535
2
COLD REHEAT
(CRH) 28.0726.422.5218.9612.22327322311301286
3
HOT REHEAT
(HRH) 24.1923.220.1717.2211.46535535535535535
3.1
STEAM AFTER
IPT 1.3691.2991.1340.9710.6725183184184185190
3.2
STEAM BEFORE
LPT 1.3431.2741.1180.95190.6594183183183183183
4
LPT EXHAUST TO
COND 0.08890.0860.07980.07390.0637 0.95290.95490.90010.9663
B
EXTRACTION
STEAM
e7TO HEATER NO 742.1839.7734.3829.0619.68381375364354349
e6TO HEATER NO 628.0726.722.5818.9612.22327322311301286
e5TO HEATER NO 512.9612.2710.659.0686.2444444444444446
edTO DEAERATOR12.9612.2710.659.06812.22444444444444286
e4TO HEATER NO 46.9116.5455.6944.8583.334364364364365365
e3TO HEATER NO 32.9742.8082.4532.0971.44264265265265271
e2TO HEATER NO 21.3691.2991.1340.9710.6729183184184185190
e1TO HEATER NO 10.28950.27540.2410.20730.14490.9920.99350.996561 68
CSERVICE STEAM
d1Gland Sealing 1.03 143
d2Ejector of GC1 4.5 155
d3Leakage into LPT0.08890.0860.07980.07390.0637 139
d4Main Ejector 4.5 155
210 MW Parameters at different regimes at CW Temp 30
0C
S.
N
PARAMETERS PRESSURE IN KG/SQCM ABSTEMPERATURE / DRYNESS
Regime IN KW211902202200176669151400101200211902202200176669151400101200
AMAIN STEAM
1MAIN STEAM130130130130130535535535535535
2
COLD REHEAT
(CRH) 28.0726.422.5218.9612.22327322311301286
3
HOT REHEAT
(HRH) 24.1923.220.1717.2211.46535535535535535
3.1
STEAM AFTER
IPT 1.3691.2991.1340.9710.6725183184184185190
3.2
STEAM BEFORE
LPT 1.3431.2741.1180.95190.6594183183183183183
4
LPT EXHAUST TO
COND 0.08890.0860.07980.07390.0637 0.95290.95490.90010.9663
B
EXTRACTION
STEAM
e7TO HEATER NO 742.1839.7734.3829.0619.68381375364354349
e6TO HEATER NO 628.0726.722.5818.9612.22327322311301286
e5TO HEATER NO 512.9612.2710.659.0686.2444444444444446
edTO DEAERATOR12.9612.2710.659.06812.22444444444444286
e4TO HEATER NO 46.9116.5455.6944.8583.334364364364365365
e3TO HEATER NO 32.9742.8082.4532.0971.44264265265265271
e2TO HEATER NO 21.3691.2991.1340.9710.6729183184184185190
e1TO HEATER NO 10.28950.27540.2410.20730.14490.9920.99350.996561 68
CSERVICE STEAM
d1Gland Sealing 1.03 143
d2Ejector of GC1 4.5 155
d3Leakage into LPT0.08890.0860.07980.07390.0637 139
d4Main Ejector 4.5 155
total output power solutionsManohar Tatwawadi Page40
DLeak off Steam
g1Gland Cooler 1 0.97 286
g2Gland Cooler 20.3610.3430.3080.2650.197346337329318296
g3TO HEATER 4 6.9116.5455.6944.8533.334403400393386383
g4
FROM SPINDLES
TO DEAERATOR 7.5 479
g5TO HEATER 528.0726.422.9818.9612.22464457448438433
DLeak off Steam
g1Gland Cooler 1 0.97 286
g2Gland Cooler 20.3610.3430.3080.2650.197346337329318296
g3TO HEATER 4 6.9116.5455.6944.8533.334403400393386383
g4
FROM SPINDLES
TO DEAERATOR 7.5 479
g5TO HEATER 528.0726.422.9818.9612.22464457448438433
210 MW Parameters at different regimes at CW Temp 30
0C
Tag
No
PARAMETERS ENTHALPY KCAL/KG QUANTITY IN KG/HR
Regime IN KW211902202200176669151400101200211902202200176669151400101200
AMAIN STEAM
1MAIN STEAM819.94819.94819.94819.94819.94652000616000531000449000304000
2
COLD REHEAT
(CRH) 732.99730.57727.11723.41720.69566369534409484930395441259703
3
HOT REHEAT
(HRH) 845.83846.22846.86847.54848.82566369534409484930395441259703
3.1
STEAM AFTER
IPT 678.41678.84679.34680682.82470857447226390074333899230723
3.2
STEAM BEFORE
LPT 678.41678.84679.34680682.82470857447226390074333899230723
4
EXHAUST STEAM
TO COND 589.18590.08592.41595.31606.03230024218790191648134954115604
B
EXTRACTION
STEAM
e7TO HEATER NO 7755.97753.53750.42746.79746.643184429654249252026313294
e6TO HEATER NO 6732.99730.57727.11723.41720.65403913713429752232614638
e5TO HEATER NO 5802.92803.23803.88804.52808.216440131296605545
edTO DEAERATOR802.92803.23803.88804.52720.6959847937116101430618728
e4TO HEATER NO 4762.69763.02763.73764.56767.84238382231118460149453108
e3TO HEATER NO 3715.68716.07716.63717.2720.32210331954716312135848488
e2TO HEATER NO 2678.41678.84679.36680682.322536913756196791627610038
e1TO HEATER NO 1622.05622.43623.07623.78626.45119891078779195132634
CSERVICE STEAM
d1
TO GLAND
SEALING 659.83 2930
d2
TO EJECTOR OF
GC1 659.8 400
d3
LEAKAGE INTO
LPT 659.83 570
d4
TP MAIN
EJECTOR 659.83 1500
D
LEAK OFF
STEAM
g1
TO GLAND
COOLER 1 728.2728.22728.29728.36728.48 1260
g2
TO GLAND
COOLER 2 756.8752.66748.77743.1732.9249494733422537352863
g3TO HEATER 4782.31780.58777.65774.47773.9749354663402033992301
g4
FROM SPINDLES
TO DEAERATOR820.28820.29820.32820.35820.39 2300
g5TO HEATER 5806.5803.66799.78795.63795.443814140356830172043
0C
Tag
No
PARAMETERS ENTHALPY KCAL/KG QUANTITY IN KG/HR
Regime IN KW211902202200176669151400101200211902202200176669151400101200
AMAIN STEAM
1MAIN STEAM819.94819.94819.94819.94819.94652000616000531000449000304000
2
COLD REHEAT
(CRH) 732.99730.57727.11723.41720.69566369534409484930395441259703
3
HOT REHEAT
(HRH) 845.83846.22846.86847.54848.82566369534409484930395441259703
3.1
STEAM AFTER
IPT 678.41678.84679.34680682.82470857447226390074333899230723
3.2
STEAM BEFORE
LPT 678.41678.84679.34680682.82470857447226390074333899230723
4
EXHAUST STEAM
TO COND 589.18590.08592.41595.31606.03230024218790191648134954115604
B
EXTRACTION
STEAM
e7TO HEATER NO 7755.97753.53750.42746.79746.643184429654249252026313294
e6TO HEATER NO 6732.99730.57727.11723.41720.65403913713429752232614638
e5TO HEATER NO 5802.92803.23803.88804.52808.216440131296605545
edTO DEAERATOR802.92803.23803.88804.52720.6959847937116101430618728
e4TO HEATER NO 4762.69763.02763.73764.56767.84238382231118460149453108
e3TO HEATER NO 3715.68716.07716.63717.2720.32210331954716312135848488
e2TO HEATER NO 2678.41678.84679.36680682.322536913756196791627610038
e1TO HEATER NO 1622.05622.43623.07623.78626.45119891078779195132634
CSERVICE STEAM
d1
TO GLAND
SEALING 659.83 2930
d2
TO EJECTOR OF
GC1 659.8 400
d3
LEAKAGE INTO
LPT 659.83 570
d4
TP MAIN
EJECTOR 659.83 1500
D
LEAK OFF
STEAM
g1
TO GLAND
COOLER 1 728.2728.22728.29728.36728.48 1260
g2
TO GLAND
COOLER 2 756.8752.66748.77743.1732.9249494733422537352863
g3TO HEATER 4782.31780.58777.65774.47773.9749354663402033992301
g4
FROM SPINDLES
TO DEAERATOR820.28820.29820.32820.35820.39 2300
g5TO HEATER 5806.5803.66799.78795.63795.443814140356830172043
SYNERGEN HEAT RATE Page42
210 MW Parameters at different regimes at CW Temp 30
0C
Tag
No
PARAMETERS ENTHALPY KCAL/KG QUANTITY IN KG/HR
Regime IN KW211902202200176669151400101200211902202200176669151400101200
E CONDENSATE
5.0HOTWELL 43.1642.5341.1139.6636.92460008437577383296329907231289
5.WCEP SUCTION43.1642.5341.1139.6636.92480106456529398599341934237891
5.DBEFORE EJECTOR43.1642.5341.1139.6636.92480106456529398599341934237891
5.AAFTER EJECTOR45.0744.5443.4143.3540.79480106456529398599341934237891
5.1AFTER GC147.1846.7645.9645.3244.86480106456529398599341934237891
5.2AFTER HEATER 161.0559.9257.0853.7746.35480106456529398599341934237891
5.3AFTER GC2 68.166.9964.2961.1754.47480106456529398599341934237891
5.4AFTER HEATER 2100.8699.3195.991.0781.32480106456529398599341934237891
5.5
BEFORE HEATER
3 101.5299.9796.1291.6981.87555281526534463675390683376826
5.6AFTER HEATER 3125.27123.83118.85113.82102.33555281526534463675390683376826
5.7AFTER HEATER 4156.52154.33148.83143.18129.23555281526534463675390683376826
F FEED WATER
6.0BFP SUCTION 166.71 652000616000531000449000304000
7.0
AFTER FEED
PUMP 173.12 652000616000531000449000304000
7.1
AFTER HP
HEATER 5 189.61187.14180.76173.75173.12652000616000531000449000304000
7.2
AFTER HP
HEATER 6 228.8225.28216.632207.35185.82652000616000531000449000304000
9.0
AFTER HP
HEATER 7 253.94250.11241.1231.11209.73652000616000531000449000304000
G DRAINS
gd1FROM GC1 100 1660
hd1
FROM HEATER
NO1 65.9364.861.8658.654.14119891078779195132634
gd2FROM GC2 72.9471.7869.2665.9759.3849494733422537352868
hd2
FROM HEATER
NO 2 105.78104.24100.3496.0186.237517570275650764874929935
hd3
FROM HEATER
NO 3 130.26128.32123.83118.77107.254980646519453973247319897
hd4
FROM HEATER
NO 4 161.73159.51153.97148.72134.22877326974290851863911409
hd5
FROM HEATER
NO 5 189.51186.91180.38173.22 93265840596605545
hd6
FROM HEATER
NO 6 198.05195.43188.86181.64180.997661670930532454654119976
hd7
FROM HEATER
NO 7 238.65234.98225.92216.29194.083134429654249252026313294
210 MW Parameters at different regimes at CW Temp 30
0C
Tag
No
PARAMETERS ENTHALPY KCAL/KG QUANTITY IN KG/HR
Regime IN KW211902202200176669151400101200211902202200176669151400101200
E CONDENSATE
5.0HOTWELL 43.1642.5341.1139.6636.92460008437577383296329907231289
5.WCEP SUCTION43.1642.5341.1139.6636.92480106456529398599341934237891
5.DBEFORE EJECTOR43.1642.5341.1139.6636.92480106456529398599341934237891
5.AAFTER EJECTOR45.0744.5443.4143.3540.79480106456529398599341934237891
5.1AFTER GC147.1846.7645.9645.3244.86480106456529398599341934237891
5.2AFTER HEATER 161.0559.9257.0853.7746.35480106456529398599341934237891
5.3AFTER GC2 68.166.9964.2961.1754.47480106456529398599341934237891
5.4AFTER HEATER 2100.8699.3195.991.0781.32480106456529398599341934237891
5.5
BEFORE HEATER
3 101.5299.9796.1291.6981.87555281526534463675390683376826
5.6AFTER HEATER 3125.27123.83118.85113.82102.33555281526534463675390683376826
5.7AFTER HEATER 4156.52154.33148.83143.18129.23555281526534463675390683376826
F FEED WATER
6.0BFP SUCTION 166.71 652000616000531000449000304000
7.0
AFTER FEED
PUMP 173.12 652000616000531000449000304000
7.1
AFTER HP
HEATER 5 189.61187.14180.76173.75173.12652000616000531000449000304000
7.2
AFTER HP
HEATER 6 228.8225.28216.632207.35185.82652000616000531000449000304000
9.0
AFTER HP
HEATER 7 253.94250.11241.1231.11209.73652000616000531000449000304000
G DRAINS
gd1FROM GC1 100 1660
hd1
FROM HEATER
NO1 65.9364.861.8658.654.14119891078779195132634
gd2FROM GC2 72.9471.7869.2665.9759.3849494733422537352868
hd2
FROM HEATER
NO 2 105.78104.24100.3496.0186.237517570275650764874929935
hd3
FROM HEATER
NO 3 130.26128.32123.83118.77107.254980646519453973247319897
hd4
FROM HEATER
NO 4 161.73159.51153.97148.72134.22877326974290851863911409
hd5
FROM HEATER
NO 5 189.51186.91180.38173.22 93265840596605545
hd6
FROM HEATER
NO 6 198.05195.43188.86181.64180.997661670930532454654119976
hd7
FROM HEATER
NO 7 238.65234.98225.92216.29194.083134429654249252026313294
SYNERGEN HEAT RATE Page43
H PRESSURE IN HEATERS
HEATERS DATA
AT LOADS211902202200176669151400101200TTD
HP7HEATER NO 738.836.5931.6326.7418.12.5
HP6HEATER NO 626.6725.0921.4418.0111.64.5
HP5HEATER NO 511.9211.299.88.343 2
LP4HEATER NO 46.3586.0215.2384.4693.0675
LP3HEATER NO 32.7362.5842.2671.9291.3255
LP2HEATER NO2 1.261.1951.0430.89330.4675
LP1HEATER NO 10.26640.25340.22170.19070.13335
I HEAT RATE IN KCAL/KWH
Regime IN KW211902202200176669151400101200
HEAT RATE
Kcal/KWH 20402039205120662154
*********************
H PRESSURE IN HEATERS
HEATERS DATA
AT LOADS211902202200176669151400101200TTD
HP7HEATER NO 738.836.5931.6326.7418.12.5
HP6HEATER NO 626.6725.0921.4418.0111.64.5
HP5HEATER NO 511.9211.299.88.343 2
LP4HEATER NO 46.3586.0215.2384.4693.0675
LP3HEATER NO 32.7362.5842.2671.9291.3255
LP2HEATER NO2 1.261.1951.0430.89330.4675
LP1HEATER NO 10.26640.25340.22170.19070.13335
I HEAT RATE IN KCAL/KWH
Regime IN KW211902202200176669151400101200
HEAT RATE
Kcal/KWH 20402039205120662154
*********************
SYNERGEN HEAT RATE Page44
SYNERGEN HEAT RATE Page45
SYNERGEN HEAT RATE Page46
Boiler Efficiency calculation by
HEAT-LOSS METHOD
Proximate Analysisof fuel / Coal
??Following parameters are to be obtained from the lab, on day-to-day basis, for
determining the Proximate Analysis of coal.Those usually are onAir-dried
basis.
#Parameters(Air-dried basis)Unit
1Fixed Carbon %
2Inherent Moisture%
3Volatile Matter%
4Ash %
5Gross Calorific Valuekcal/kg
6Total Moisture %
??A constant (K) is derived by the given formula:-K = (100 TM) / (100 M)
??By multiplying this constant K with the above parameters, Proximate Analysis
on As-Received basiscan be calculated & these parameters are used for the
calculation of Ultimate Analysis.
#Parameters(As-Received Basis)UnitAbbreviation
1Fixed Carbon % FC As perLab
2Inherent Moisture% M As per Lab
3Volatile Matter % VM As per Lab
4Ash % A As per Lab
5Gross Calorific Valuekcal/kgQ As per Lab
6Total Moisture % TM As per Lab
Boiler Efficiency calculation by
HEAT-LOSS METHOD
Proximate Analysisof fuel / Coal
??Following parameters are to be obtained from the lab, on day-to-day basis, for
determining the Proximate Analysis of coal.Those usually are onAir-dried
basis.
#Parameters(Air-dried basis)Unit
1Fixed Carbon %
2Inherent Moisture%
3Volatile Matter%
4Ash %
5Gross Calorific Valuekcal/kg
6Total Moisture %
??A constant (K) is derived by the given formula:-K = (100 TM) / (100 M)
??By multiplying this constant K with the above parameters, Proximate Analysis
on As-Received basiscan be calculated & these parameters are used for the
calculation of Ultimate Analysis.
#Parameters(As-Received Basis)UnitAbbreviation
1Fixed Carbon % FC As perLab
2Inherent Moisture% M As per Lab
3Volatile Matter % VM As per Lab
4Ash % A As per Lab
5Gross Calorific Valuekcal/kgQ As per Lab
6Total Moisture % TM As per Lab
SYNERGEN HEAT RATE Page47
Ultimate Analysis
??Using those above parameters, the Ultimate Analysis can be calculated as below:-
# Parameters UnitAbbreFormula
ASulphur Content in Coal S %Fixedor
Obtained
from Lab
BCarbon Content in Coal
= (1-0.01*Z)*Cp + (0.05*A) (0.5*S)
where,Z = Mineral matter content of the fuel burnt
= (TM + 1.1*A +0.1*S)
AndCp= Carbon content on Mineral matter free
basis =(0.0015782*Qp) (0.2226*Vp) + 37.69
WhereQp = Calorific valuein kCal/kg
= (100*Q*4.186)/(100-Z)
AndVp = Volatile matter content
= (100*(V-0.1*A-0.1*S))/(100 Z)
C %
%
%
%
Parr s
method
Parr s
method
Parr s
method
Parr s
method
Parr s
method
CH2Content in Coal
= (1-0.01*Z )*Hp +(0.01*A) (0.015 * S)
where,Hp = Hydrogen content on Mineral matter
free basis = (0.0001707*Qp) + (0.0653*Vp) 2.92
H2 %Parr s
method
DN2Content in Coal
= [2.1-(0.012*VM)]
N2 %Gebhardt s
formula
EAndO2Content in Coal
= 100-( C + H + S + N + A +TM )
O2 %By diff.
Ultimate Analysis
??Using those above parameters, the Ultimate Analysis can be calculated as below:-
# Parameters UnitAbbreFormula
ASulphur Content in Coal S %Fixedor
Obtained
from Lab
BCarbon Content in Coal
= (1-0.01*Z)*Cp + (0.05*A) (0.5*S)
where,Z = Mineral matter content of the fuel burnt
= (TM + 1.1*A +0.1*S)
AndCp= Carbon content on Mineral matter free
basis =(0.0015782*Qp) (0.2226*Vp) + 37.69
WhereQp = Calorific valuein kCal/kg
= (100*Q*4.186)/(100-Z)
AndVp = Volatile matter content
= (100*(V-0.1*A-0.1*S))/(100 Z)
C %
%
%
%
Parr s
method
Parr s
method
Parr s
method
Parr s
method
Parr s
method
CH2Content in Coal
= (1-0.01*Z )*Hp +(0.01*A) (0.015 * S)
where,Hp = Hydrogen content on Mineral matter
free basis = (0.0001707*Qp) + (0.0653*Vp) 2.92
H2 %Parr s
method
DN2Content in Coal
= [2.1-(0.012*VM)]
N2 %Gebhardt s
formula
EAndO2Content in Coal
= 100-( C + H + S + N + A +TM )
O2 %By diff.
SYNERGEN HEAT RATE Page48
??In addition to this, following parameters arerequired to benotedduring the test.
# ParametersUnitAbbreviationsValues
1Avg Flue Gas o/l Temp
0
C T
As per Control
Room
2Ambient Temp-Drybulb
0
C t
As per Control
Room
3Ambient Temp-Wet bulb
0
C
As per Control
Room
4 O2at APHInlet% O2
As per Control
Room
5 CO2at APHinlet% CO2
As per Control
Room
6 N2at APHinlet% N2
As per Control
Room
7Unburnts in Bottom ash% UBA As per Lab
8Unburnts in Fly ash% UFA As per Lab
9 CO at APH Outppm CO
As per
measurement
??In addition to this, following parameters arerequired to benotedduring the test.
# ParametersUnitAbbreviationsValues
1Avg Flue Gas o/l Temp
0
C T
As per Control
Room
2Ambient Temp-Drybulb
0
C t
As per Control
Room
3Ambient Temp-Wet bulb
0
C
As per Control
Room
4 O2at APHInlet% O2
As per Control
Room
5 CO2at APHinlet% CO2
As per Control
Room
6 N2at APHinlet% N2
As per Control
Room
7Unburnts in Bottom ash% UBA As per Lab
8Unburnts in Fly ash% UFA As per Lab
9 CO at APH Outppm CO
As per
measurement
SYNERGEN HEAT RATE Page49
BOILERLOSSESCALCULATION
Loss Due to Unburnt Carbon
= (33820*Unburnt Carbon in ash kg/kg)*100/(4.186*Q)
Where, Unburnt Carbon in ash per kg of coal
= (0.00008*A%*% UFA) + (0.00002*A%*% UBA)
UCLOSS
Loss due to Dry Flue Gas
= (100/(12*(CO2+CO)))*(( C/100 + S/267-carbon in ash))*
(30.6*(T-t))*100/(4.186*Q)
DFLOSS
Loss due to Wet Flue Gas
= ((TM + 9*H )/100) * (1.88*(T-25) + 2442 + 4.2*(25-t )) *
100 /(4.186*Q)
WFLOSS
Loss due to Moisture in Air
= (Ma*h*1.88*(T-t)*100)/(4.186*Q )
Where, Ma = Dry air for combustion in kg / kg fuel
= (3.034*% of N2 in flue gas/(% of CO2 + CO ))*
(C/100 + S / 267-C in ash)
Where h = Mass of moisture per kg of dry air
= From Psychometric chart
MALOSS
Loss due to CO = CO in ppm*10
-6*Fuel Rate per Hour*28*5744/QCOLOSS
Unmeasured Losses (Including Radiation, Blowdown)ULOSS
Total Losses = (UC + DF + WF + MA + CO + U) LOSS
Boiler Efficiency= 100 Total Losses
BOILERLOSSESCALCULATION
Loss Due to Unburnt Carbon
= (33820*Unburnt Carbon in ash kg/kg)*100/(4.186*Q)
Where, Unburnt Carbon in ash per kg of coal
= (0.00008*A%*% UFA) + (0.00002*A%*% UBA)
UCLOSS
Loss due to Dry Flue Gas
= (100/(12*(CO2+CO)))*(( C/100 + S/267-carbon in ash))*
(30.6*(T-t))*100/(4.186*Q)
DFLOSS
Loss due to Wet Flue Gas
= ((TM + 9*H )/100) * (1.88*(T-25) + 2442 + 4.2*(25-t )) *
100 /(4.186*Q)
WFLOSS
Loss due to Moisture in Air
= (Ma*h*1.88*(T-t)*100)/(4.186*Q )
Where, Ma = Dry air for combustion in kg / kg fuel
= (3.034*% of N2 in flue gas/(% of CO2 + CO ))*
(C/100 + S / 267-C in ash)
Where h = Mass of moisture per kg of dry air
= From Psychometric chart
MALOSS
Loss due to CO = CO in ppm*10
-6*Fuel Rate per Hour*28*5744/QCOLOSS
Unmeasured Losses (Including Radiation, Blowdown)ULOSS
Total Losses = (UC + DF + WF + MA + CO + U) LOSS
Boiler Efficiency= 100 Total Losses
SYNERGEN HEAT RATE Page50
Reduction in fuel consumptionand saving
1.Boiler Efficiency beforetuning of boiler=Effy(Pre)
2.Boiler Efficiencyafter tuning of boiler=Effy(Post)
Effy(Post) Effy(Pre)
3.Reduction in fuel consumption = 100*
Effy(Pre)
4.Saving in fuel per day = Fuel consumption per day * Reduction in fuel
consumption
5.Cost of fuel saved per day = Saving in fuel per day* Cost of fuel
****************************
Reduction in fuel consumptionand saving
1.Boiler Efficiency beforetuning of boiler=Effy(Pre)
2.Boiler Efficiencyafter tuning of boiler=Effy(Post)
Effy(Post) Effy(Pre)
3.Reduction in fuel consumption = 100*
Effy(Pre)
4.Saving in fuel per day = Fuel consumption per day * Reduction in fuel
consumption
5.Cost of fuel saved per day = Saving in fuel per day* Cost of fuel
****************************
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