60 Ton Per HR Advance Empire Boiler Operation Manual

 

60 Ton/h Advance Ton/hrr @ 31 bar operation Advance Empire Boiler  1.  Setting Operation Pressure with UT UT 35 A FD Fan & Fuel Feed Controller ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

31 BAR Operation Parameters

FD FAN

--------------

--------------

SP

Fuel Feed Controller

Remarks

------------------------------------------------------------

31 bar

31 bar

---------------------------------------------------------Operation Pressure Set pt.

AL 1

-

29.5 bar

Cut off autogate 2

AL 2

-

30.0 bar

Cut off autogate 3

22 BAR Operation Parameters ---------------SP

FD Fan ---------------

Fuel Feeder Controller

Remarks

-------------------------------------------------------------------

22 bar

22 bar

------------------------------------------------------

Operation Pressure Set pt.

AL 1

-

21.5 bar

Cut off autogate

2

AL 2

-

22.0 bar

Cut off autogate

3

2.  Check Safety Systems before entering fuel to boiler ---------------------------------------------------------------------Blow water cooling chamber of level gage /water level transmitter =check safety alarm Hi,Normal,Low & LowLow sirens and alarms is working , cut off ID , FD fan ,Augur , Augur Conveyor 1,2,3,&4 upon lowlow lowlow water level alarm signal.

High alarm = 75% Yellow Spinning Light + Alarm Bell Normal

= In between 40.0 + to 75.0- %

Low

= 40.0% Yellow Spinning Light + Alarm Bell

LowLow

= 25.00% Cut Off ID.Fan,FD Fan , Fuel Feed Augur Conveyor 1,2,3 & 4 and

 

 

Cut Off Auto Fuel Gate 1 , 2 ,3 & 4 upon signal of 25% Low Low Water Level sign signal. LOW LOW Siren & RED Spinning Alarm Light.

3.  Fuel Feeding On Fuel Feeder Fan. On Fuel Feeder Augur Conveyor 1,2,3 & 4 . Open Auto Fuel Gate 1,2,3, & 4 in Manual Mode . On the other Fuel Feed Conveyor systems. Start fuel feed until you sufficient fuel in the furnace.Start the fire with a bit of diesel. Warming up the boiler for 1 or 2 hours before raising pressure.

4.  When ready to raise pressure,On the moving grate system. system. Timer can be adjusted : Fast = Timer 1 = 0 sec , Timer 2 = 0 sec Emptying Emptying Ash when stopping stopping boiler Med = Timer 1 = 1 sec , Timer 2 = 1 sec Slow = Timer 1 = 2 sec , Timer 2 = 2 sec Not Not much fuel coming Very Slow Timer 1 = 3 sec , Timer 2 = 3 sec etc

5.  Using one ID Fan & 50% of FD Fan Open Top Damper ID FAN no.1 =100% Open

Close Top Damper ID FAN no.2 =100% Close

Open 3-set Damper below ID FAN no.1 = 10%

Close 3-set Damper ID FAN no.2 = 100% Close

Open FD FAN Damper 50 % Open Boiler is ready for One ID Fan & 50% FD Fan Operation. On ID FAN no.1 at 35% minimum Frequency as set set in parameter. While adjusting –ve draft of Furnace Controller at –ve 10mm to  –ve 12ve mm H2O during Steam load & Fuel Feeding operation, you may m ay need to open more the I.D. 3 set-damper to 30% ,40% ,50% ,60% ,70% , 80% , 90% or up to 100% open to adjust the  –ve draft as the requirement of –ve 10 mm to – 12 mm H2O Draft. When switching the ID Fan & FD Fan to AUTOMATIC CONTROL MODE from MANUAL CONTROL MODE.

 

When the OPERATING PRESSURE &  –ve Draft of the Furnace can not be achieved with nd

nd

One ID FAN / 50% FD FAN , start the 2  ID Fan . Slowly open TOP damper of 2  ID FAN to check the –ve Draft of Furnace. IF cannot achieve achieve , slowly increase 2nd ID Fan 3-set Damper to 10 % , 20 % , 30 % , 40 % etc etc . While trying to achieve the  –ve Furnace Draft , you also need to increase the FD Fan Damper To 60 % , 70 % , 80% , 90% or 100 % to achieve the necessary Forced Draft to raise pressure and keeping in check the –ve Draft of Furnace around –ve 10 to 12 mm H2O. Photo 1 & 2 shows Main Damper Closed . Only the small damper 100% open. The control of draft controlled by Manual signal or auto signal from steam pressure and -ve draft of furnace for ID Fan.

6.  Stop Boiler Stop boiler ,just off the Auto Fuel Feeding Gates. Off FD Fan. OFF ID FAN 1 & 2. Off Fuel Feeder Fan & Sec. Air Air Fan. Keep MOVING GRATE on until the fire has gone down.Then Off Moving Grate and all the Deashing conveyors after 90% of the ashes have been extracted. extr acted. Rolling water = changing water in the boiler by blowdown and refilling for for 10 minutes . Then off the boiler. This is to prevent high concentration co ncentration of TDS crystallising onto the tube,s surface when cooling further. Do not blow off steam pressure through the Air Vent Valves at the steam Drum or SuperHeater SuperHeater Header when stopping the boiler.As the boiler cools down further , air & oxygen will be sucked into the Boiler through the air Vent valves.Any parts & Drum Plates above the water level mark will be under corrosion pitting by sucked in air through the air vents . Just keep the drain valve of superheater crack a little open for rreleasing eleasing steam slowly through the superheater tubes when stopping Boiler.

Note :- 1. Chemical Pumps and chemical dosage for the t he boiler. 2. Boiler Blow Blow down 5% to 10% per hour calculated from steam flow /hr .

 

 

Steam Load & Fuel Consumption of Boiler Estimation ========================================== 1000 Kw Back pressure at 3.5 bar Steam turbine at 30 bar Steam consumption : = 1250 Kva @ 0.8 pf x 14 kg/KVA = 17,500 kg of Steam/hr + Balance of 60,000 – 17,500 kg = 42,500 kg @ 30 bar goes to the 5 Megawatt turbine. 5000 Kw GE 3 stages 100% Condensing Turbine with input steam pressure at 30 bar@350 C. If Anticipated steam consumption consumption = 8 kg /KVA at 30 bar steam pressure 42,500 kg steam/hr ---------------------------------------------------- = 5312.5 KVA 8 kg/KVA 5312.5 KVA x 0.8 pf

= 4250 Kw without Capacitor Bank to raise power factor.

If the GE turbine consumption rate is =14 kg/kva at 30 bar steam pressure 42,500 kg steam /hr -------------------------------- = 3036 kva 14 kg/kva If power factor achieved is 0.8 pf , then consumed kw = 3036 kva x 0.8 pf = 2,428.8 kw If power factorachieved is 0.9 pf , then consumed kw = 3036 x 0.9 pf = 2,732.4 kw

Fuel Consumption of 60 ton/hr Boiler at 30 bar : 60,000 kg/hr x ( 669Kcal/kg steam + 2 Kcal/kg raisi raising ng from 98 C to 100 C ) -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------0.82 Eff of Heat Transfer x 3200 Kcal/kg Gross Heat Calory of Fuel Mixture

 

  40,260,000 kcal =

-----------------------------2624 Kcal/kg fuel mix

=

15,343 kg of fuel mix / hr + -

Fuel from Mill Process : 4 % Shell / ton FFB processed 13.5 % Fibre / ton FFB Processed 22% shredded EFB/ Ton FFB Processed

Fuel Mix : 4 % Shell + 13.5% Fibre + 10 % Shredded EFB fibre = 27.5 % (17.5 x157.2 %) / 100% = 27.51 % from a total 39.5 % fuel available / ton FFB processed.

Mill Process required / hr to produce enough fuel to run boiler at 60 Ton/hr capacity 15,343 kg/hr fuel steam flow at 60T/hr =

----------------------------------------------------------------0.2751 fuel factor useable

= 55,773 Kg FFB / hr Mill Mill Process to maintain the boiler .

FUEL MIX ---------------Shell = 55773 kg FFB/hr x 0.04 =

2230 kg /hr

Fibre= 55773 kg FFB/hr x 0.135 = 7529 kg/hr Shredded EFB = 55773 kg FFB/hr x 0.10 = 5579 kg/hr -----------------------------------------------------------------------Total Fuel Feed

= 15,338 kg /hr + -

Excess EFB shredded fibre = 6691 kg/hr accumulated for n next ext day start up from 60 T/hr Process.

 

 

60 Ton/hr DEAERATOR ===================

60 ton /hr of water raised from 70 C to 103 C requires 33 Kcal/kg of water Total Kcal/hr = =

60,000 kg of water/hr x 33 Kcal/kg of water 1,980,000 Kcal/hr

Total Steam required required to Heat Heat up the water = 1,980,000 kcal / 535.74 kcal/kg steam side at 0.23 bar =

3695.8 kg/hr

=

3.696 Ton/hr

Set Pressure of PRV or Modulating Valve Steam supply to Deaerator is 0.23 minimum to 0.35 Bar Max. The higher the set set pressure , more steam is required required but the rate of superheating the water droplets is faster due to higher temperature of the steam.

Steam Pressure

Steam Temperature

Kcal in Water/kg

Kcal in Steam/kg

--------------------

--------------------------------------------------

-------------------------------------------

----------------------

0.12 bar

103.148 C

103.32

537.48

0.23 bar

105.7

C

106

535.74

0.24 bar

106.0

C

106.27

535.59

0.25 bar

106.245 C

106.5

535.44

0.26 bar

106.474 C

106.73

535.29

0.27 bar 0.28 bar

106.703 C 106.932 C

106.96 107.19

535.14 535.0

0.29 bar

107.161 C

107.41

534.85

0.30 bar

107.39 C

107.64

534.70

0.35 bar

108.47 C

108.76

534.0

At 0.23 bar the Heat of 535.74 Kcal/kg of steam side is given out to deaerate the soft water in the small capsule by superheating the sprinkled tiny droplets of water ,thereby ,t hereby releasing oxygen at a critical temperature of 103 C.

As the pressure setting goes higher , less heat energy is present in the steam side for heating

 

release.

Palm Fibre/shell Fuel Heating Calorific Values at 3% to 5% moisture ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ A. 1 kg of fibre @ 5% moisture moist ure = Heat CV 3600 kcal C. Heat CV 3200 kcal@30% moisture B. 1 kg of shell @ 5% moisture

=Heat CV

4600 kcal D. Heat CV

4500 kcal@30% moisture

Fuel mixtures of 97/3 at 5 % moisture ------------------------------------------------------- :  A 1) HCV fibre 0.97 x 3600 kcal/kg = 3492 kcal 5% moisture HCV shell 0.03 x 4600 kcal/kg = 138 kcal 5 % moisture -------------gross H.C.V. = 3630 kcal/kg of fuel mixture  A2) Fuel mixture of fibre f ibre 100 % at 30% moisture ---------------------------------------------------------------------------------------------------------------------------------HCV fibre 100% x 3200 kcal/kg = 3200 kcal HCV shell 0% x 4500 kcal/kg = 0 kcal -------------------Gross HCV = 3200 kcal/kg of fuel mixture B1) Fuel mixture of 90/10 at 30% moisture ----------------------------------------------------------------------------------------------------------------------------HCV fibre 90% x 3200 kcal/kg = 2880 2 880 kcal HCV shell 10% x 4500 kcal/kg = 450 kcal -----------------------Gross HCV = 3330 kcal/kg fuel mixtures B2) Fuel mixture of 98/2 at 30% moisture ------------------------------------------------------------------------------------------------------------------HCV fibre 98% x 3200 kcal/kg = 3136 kcal HCV shell 2% x 4500 kcal/kg = 90 kcal ---------------------Gross HCV = 3226 kcal/kg of fuel mixture.

2. Theoretical Air Required & Combustion Flue Gases ========================================== a) Theoretical Air Required = Lt = 1 / 0.21 x ( 1.867 x C + 5.6 x H + 0.7 x S - 0.7 x O )

( i ) Low Low Calory Fuel = Lt low = 4.761904762 x ( 1.867 x 0.3334 + 5.6 x 0.0396 + 0.7 x 0 - 0.7 x 0.2904 ) = 4.761904762 x ( 0.6224578 + 0.22176 + 0 - 0.20328 ) = 4.761904762 x 0.6409378

 

  = 3.052084762 = 3.05 Nm3 of air / kg of fuel stoichiometrically

( ii ) Medium Calory Fuel = Lt Med = 1 / 0.21 x{ ( 1.867 x 0.35) + ( 5.6 x 0.05 ) + ( 0.7 x 0.02 ) - ( 0.7 x 0.21 ) }

= 1 / 0.21 x ( 0.65345 + 0.28 + 0.014 - 0.147 ) =

1 / 0.21 x 0.80045

=

3.8116666667

= 3.81 Nm3 of air / kg of fuel stoichiometrically

( iii ) High Calory Fuel = Lt high = 1 / 0.21 x ( 1.867 x 0.355 + 5.6 x 0.055 + 0.7 x 0.02 0.7 x 0.21 ) = 1 / 0.21 x ( 0.662785 + 0.308 + 0.014 - 0.147 ) = 1 / 0.21 x 0.823785 = 3.922786 Nm3 = 3.92 Nm3 of air / kg of fuel stoichiometrically

3. Stoichiometric air required per main constituents of molecular fuel in Chemical Combustion Process. ============================================================================= a) During combustion the fuel a always lways react with Oxygen and liberates heat energy.A knowlegde of the constituents of air therefore is required to find the amount of Oxygen in a given air bulk. Natural air contains many gases such as oxygen,nitrogen,argon , helium , neon , krypton , xenon , carbon dioxide , and traces contamination by carbon monoxide and sulphur dioxide together with some vapour moistures. For combustion calculation purposes it is assumed that the air consist entirely of 21 % Oxygen and 79 % Nitrogen by Volume or 23.2 % Oxygen and 76.8 % Nitrogen by Mass . Nitrogen is an inert gas and does not take part directly in the combustion but it will slow down the combustion and lower the temperature of the combustion by Oxygen.Whatever amount of Nitrogen introduced in the no react combustion , the exhaust flue gases will contain the same amount of Nitrogen as introduced before.

 

  Table of Molecular Masses of fuel constituents -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Elements

Unit Relative atomic mass

Relative molecular mass

carbon

=C=

12

-

Hydrogen

= H2 =

1

2

Oxygen

= O2 =

16

32

Nitrogen

=N2=

14

28

Sulphur

=S=

32

-

Compounds Carbon monoxide

= CO =

-

28

Carbon dioxide

= CO2 =

-

44

Water

= H2O

-

18

---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- ---( i ) C + O2 = CO2 ( ii ) 2 H2 + O2 = 2 H2O ( iii ) CH4 + 2 O2 = CO2 + 2 H2O ( Methane natural gas ) ( iv ) C6 H14 + 9.5 O2 = 6 C O2 + 7 H2O ( Hexane , parafin )

Combustion Analysis by weight/mass/gravimetric with relative atomic weight of Hydrogen = 1 ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------( i ) C + O2 = CO2 1x12 + ( 2 x 16 ) O2 = =

CO2

12 + 32 = 44

divide through by 12 =

=

1

=

1

that is

12 / 12 + 32 / 12 = 44 /12

8 8 + 2 ----- = 3 -------12 12 + 2 & 2/3 = 3 & 2/3 >>

1 kg C + 2 & 2/3 kg Oxygen = 3 & 2/3 kg of CO2

 

  thus : 2 & 2/3 kg O2 = stoichiometric mass of O2 2 & 2/3 kg O2 are contained in natural air air of weight = 2 & 2 2/3 /3 div. by .232 by mass stoichiometric mass of natural air But :

Natural air will contain

=

11.5 kg. of air

11.5 kg - 2.66 kg of Oxygen

= 8.84 kg of Nitrogen N2. then : 1 kg pure Carbon + 11.5 kg of air air ( 2.66 kg O2 + 8.84 kg of N2 ) = 3 & 2/3 CO2 + 8.84 Kg N2 Flue = 3.6667 CO2 x 44 kg + 8.84 kg of N2 = 170.1748 kg of flue gas Flue gas analysis by Volume ---------------------------------------------------------------------a) 100 % combustion 1 m3

of O2 when 100 % combusted with C will produce 1 m3 of CO2

1 m3 of CO2 is contained in natural air of volume = Out of 4.76 M3 - 1 m3

Oxygen = 3.76 m3 of

1 m3/ 0.21 m3 = 4.76 M3 of air Nitogen N2

Conclusion is that : 4.76 m3 of of air 100 % combusted with C will produce 1 m3 of CO2 and 3.76 m3 of N2

b) Incomplete combustion ( Carbon monoxide ) Over feeding the boiler. 2 C + O2 = 2 CO

 Analysis by mass ============= (2 x 12) + ( 2 x 16 ) = 2 ( 12 + 16 ) 24

+

32

=

56

divde through by 24 : in weight Hence :

:

1 + 1 & 1/3 =

2 & 1/3

1 kg C + 1 & 1/3 kg O2 = 2 & 1/3 kg of CO

1 & 1/3 kg O2 = stoichiometric Mass of O2

Now 1 & 1/3 kg of O2 are contained in air of weight = 1 & 1/3 div. by 0.232 = 5.75 kg of air = stoichiometric mass of air This amount of air will contain 5.75 kg - 1.33 kg O2 = 4.42 kg of N2 Hence 1 kg C + 5.75 kg of air air = 2 & 1/3 kg of CO + 4.42 kg of N2

 

 

 Analysis by volume of air =================== 1 m3 of O 2 reacted with 1 C during incomplete combustion will produce 2 m3 of of CO . 1 m3 O2 is contained in air of v volume olume 1 / 0.21 = 4.76 m3 of air and will contain 3.76 m3 of N2 of also. Hence : 4.76 m3 of air reacted incomplete combustion with C will produce 2 m3 of CO together with 3.76 m3 N2.  A total volume increase of 2 + 3.76 = 5.76 m3 - 4.76 m3 = 1 m3 ******* sudden increase in volume = puffing and cooling down of furnace.

4) Complete combustion of CO carbon monoxide to CO2 =========================================

 Analysis by mass ============= 2 CO + O2

= 2 CO2

2 ( 12 + 16 ) + ( 2 x 16 ) = 2 { 12 + ( 2 x 16 ) } 56

+

32

=

divide . through by 56 : In weight

:

88 1

+

4/7 =

1 &4/7

1 kg of CO + 4 / 7 kg of O2 =

1 & 4/7 kg of CO2

Hence 4/7 kg of Oxygen O2 = stoichiometric mass of O2 in the combustion.

4 Now 4/7 kg of O2 is contained in natural air of weight = -------------------------- ------------------------------ = 2.46 kg of air 7 x 0.232 = stoichiometric mass of air this air will contain also (2.46 kg Air - 4/7 kg O2) = 1.89 kg of Nitogen N2

 

  lastly therefore : 1 kg of CO + 2.46 kg of air ( 0.57 kg O2 + 1.89 kg N2 ) = 1.57 kg of CO2 + 1.89 kg of N2

 Analysis by Volume of air ====================

2 CO + O2 = 2 CO2 Proportion by Volume : 2 + 1 = 2 divide. through by 2

: 1 + 0.5 = 1

Then :

1 m3 of CO + 0.5 m3 of O2 = 1 m3 of CO2

******

A sudden contraction of 0.5 m3 in Volume of combustible gases = implosion

Hence :

0.5 m3 O2 = stoichiometric volume of O2 in the combustion

Now : 0.5 m3 of O2 is contained in air of volume = 0.5 m3 / 0.21 = 2.38 m3 of natural air = stoichiometric volume of air in the combustion This air will will contain also N2 of ( 2.38 m3 - 0.5 m3 of O2 ) = 1.88 m3 of of N2 Lastly therefore : 1 m3 of CO + 2 2.38 .38 m3 of air = 1 m3 of CO2 + 1.88 m3 of N2 with a sudden contraction in volume of 1 m3. = implosion !

In between incomplete combustion and complete combustion of CO gases there will be profound sudden implosion and expansion of of furnace combustion gases causing movements of boiler furnace walls. Thereby causing ocasional puffing of flames at manual feed doors and also cooling down combustion in the furnace.

Black smoke and unburnt particles will be noticed at the chimney emmisions if the multi-cyclones are overloaded by their capacities.

Therefore in Tuning of Solid Fuel boilers , it is always advised to have excess combustion air of 10% to 30 % more than the actual required to prevent incomplete combustion due to lack of air and furnace being too cold if too much air is supplied especially the Under Fire Forced Draft air. The trick of balancing act is in keeping the variables in operation more constant and use the secondary air draft damper adjustments to make up for excess air although most % comes from fuel feeder air.When the secondary air fans are under size , then the question of difficulties adjustments for sufficient 10 % to 20 % excess air will arise. Correct it by pulley sizing.  A high capacity I.D. can be solved at site very easily to run at the corr correct ect capacity.A high capacity F.D . Fan can also be solved easily at site to run at the correct capacity by pulley sizing.

 

The capacity of over fire fan with higher capacity can be corrected with an additional damper.As for a low capacity over fire fan,the whole unit have to be changed to get the right capacity for excess combustion air.The correct maximum static pressure is 8” to 12 inch W.C. for all secondary air fan static pressure at positive side when the damper is adjusted.The Secondary air is for reducing combustion black smoke , while the Forced Draft air is to accelerates the combustion for more heat when pressure drop.

 As the analysis by weight will not show the increase or decrease in air volume amount , but the densities and volume analysis will show very apparently the implosion/expansion of gases under conditions of incomplete/complete carbon monoxide combustion and cooling down of furnace creating black smoke and particles emmission at the chimney. These situation will occur more frequently on manual feeding and during pressure drops because of increased steam loading on the steam turbine affecting sudden surge of steam flow on the boiler ,thereby requiring more feed water at minimum 70 to maximum 98 deg .C to be heated up to 100 deg. C, From and At 100 deg C & at whatever pressure in the boiler ,where it can maintain during irregular manual fuel feeding to boost the dropping boiler pressure.The higher the feed water temperature the better up to a maximum 98 C.

Don`t forget this is done on calculation basis with a lot of assumptions not taking into account of ashes , clinker waste ,intermittent blow down exercise, soot blowing exercise , drawing cleaning the furnace per 4 hourly , irregular manual feeding for pressure drop , fluctuating feed water temperatures , controlling the feed water modulating valve for optimum minute constant trickle feed as maintaining the maximising of the fuel ,steam flow and to withstand pressure drops. Setting the safety valves to blow at higher ranges within design also helps in the maximisation of fuel over steam flow as a higher steam pressure will reduce the steam consumption rate of Turbine . The actual perfomance still depends a lot on site datas , site work , solid hands on boiler operation experiences and technics by the boiler chargeman and the settings and advices from the boiler manufacturer.

5.Analysis by mass of air and fuel o off 80 % fibre/ 20 % shell from FFB process of 33 t/hr ================================================================= Total fuel available is 6150 kg/hr just nice for a 25 ton per hour boiler. F.D.Fan in take air pressure are are 101.5KN/m2 @ Absolute and at 30 deg C. and respectively, (R for air = 0.287 KJ/kg K ) 100 kN = 1 kg /cm2 = 14.22 psi 1 psi = 27.91 inches W.C. . 101.5 kN /m2 = (14.43psi -14.22psi ) x 27.91 in. W.C. = 0.21 x 27.91 inch. W.C. = 5.86 inches W.C. at positive side Palm fibre and shell mixture 80/20 fuel Carbon C = 35 % Hydrogen = 5 % Oxygen = 21 %

 

moisture  Ash

= 30 % = 9%

1 kg of C + 2 & 2/3 kg of O2 = 3 & 2/3 kg of CO2 stoichiometric of Oxygen required. For 1 kg of Fuel 0.35 kg of carbon require 0.35 x 2 & &2/3 2/3 kg of O2 = 0.93 kg of O2 0.05 kg of H2

require 0.05 x 8 kg of O2

= 0 0.4 .4 kg of O2 ------------------------= 1.33 kg of O2

total But there are 21 % Oxygen in the fuel.  Actual O2 required = 1.33 kg - 0.21 kg = Theoretical combustion air

1.12 kg of O2 per kg of Fuel

= 1.12 kg / 0.232 = 4.82 kg of air per kg of fuel

Based on 30 % Excess Air = 1.3 x 4.82 kg of air = 6.26 kg of Air/kg of fuel = the actual amount of natural air feed to the boiler/kg of fuel. The boiler use 6150 kg of fuel / hr to raise 25 tph of steam. Fuel reflected in kg / sec = 6150 kg / 3600 secs = 1.70 kg / sec  Actual Air required for 1.7 1. 7 kg of fuel / sec = 1.7 x 6.26 kg of air / sec m = 10.64 kg of air / sec for 1.7 kg/sec of fuel on mechanical feed.

Now PV = mRT /kgK

and T =

30 K + 273K = 303 K and P = 101.5 kN/m2 ; R=0.287 Kj

mRT 10.64 x 0.287 x 303 Volume of Air intake = V = ----------------------------------------------- = --------------------------------- ------------------------------for 1.7 kg of fuel /sec P 101.5 = 9.11 m3/sec at an average pressure of 0.3 in. W.C. after deduct the friction loss etc across the pin hole fire grates, fuel bed and with help of I.D. Fan suction neutralising the under draft. F.D. Air in take reflected in m3 / min = 9.11 m3 / sec x 60 sec = 546.60 m3 //min min F.D. Fan =

say 550 m3/min rounding it up.

 Analysis of Flue Exhaust =================== 0.35 kg Carbon of the fuel will produce = 0.35 x 3 & 2/3 kg of O2 = 1.28 kg of CO2 stoichiometric The 30 % excess of the 1.12 kg of O2 = 0.33 kg of O2 Theoretical air supplied based on 30 % excess 6.26 kg of air contains 76.8 % of N2 = 4.80 kg of N2

 

  So the dry flue gas per kg of fuel @ 295 deg C will be : CO2 = 1.28 kg = 20 % O2 = 0.33 kg = 5% N2 = 4.80 kg = 75 % -------------------------------------------------------------------------Total = 6.41 kg = 100 % VP =F/hr x mRT where T = 295 K + 273K = 568 568 K ; F/hr = 6150 kg /hr ; R=0.287 KJ/kg K m = 6.41 kg of Exhaust gases / kg of fuel/hr ; P = 100 kN/m2 = 1 kg/cm2

V = 6150 kg/hr ( 6.41 x 0.287 x 568 ) / 100 kN/m2 ( @ 1 atmosphere) = 64263.35244 m3/hr ( actual amount of flue gases / hr ) Reflected in m3 / min = 64263.35 m3 /60min = 1071 m3 / min = 17.85 m3/sec Rounding up

= 1100 m3/min = 18.33 m3/sec

Plus another another 10 % Allowance for moisture load and mechanical mechanical efficiency = 1100 + 100 m3/min  Actual sizing of I.D. Fan = 1200 m3 / min dry flue gas or equivalent to = 20 m3 / sec of dry flue gas The density of this flue gas per m3 = 6150 kg of fuel x 6.41 kg of flue /kg of fuel -----------------------------------------------------------------------------------------------------------64263.35244 m3 of flue = 0.61346717 kg/m3 Density,

rounding up

= 0.6135 kg/m3 at 295 deg C