BoilerCqlculotions
A.
Whntis equivalentevaporation?
Rateof heatsuppliedby fuel - Gf X (CV)lkcaVs
whereGr - rateof fuel buming,kgls
G' (H H*t)
.. n.
'rnoiler
Gf (Cv),
Ans It is thequantityof waterevaporated
from and
at 100'C to producedry saturatedsteamat 100"Cby
absorbingthe sameanount of heat as usedin the
boilerunderactualoperatingconditions.
= Mnr(H-Hnr)/(U)1
M"q= Mn (H - Hwt)/539
whereM"n - equivalentevaporation
Mact= actualmassof steamgeneratedper unit
massof fuel burnt
I/ - total specifrcenthalpyof steamunderoperating conditions,kcaUkg
H*, = specificenthalpyof feedwater,kcaVkg
Latentheatof dry, saturatedsteamat 100"c is 539
kcaVkg.
A.
What isfactor of evaporation?
whereG"/G1- actualevaporation- M*t
O.
Whatis ecornmizereficiency?
Ans. It is dcfinedas the ratio of the heatabsorbed
by theBFW in theeconomizer
to theheatsuppliedby
the flue gasesin the economizcr,the temperatureof
flue gasesbeingreckonedabovethe temperature
of
theair suppliedto the boiler
6rleon -=
MactA@
MrrCn(@1-@6)
Ans, It is the factorto be multipliedwith thequan- where AO - rise in BFW temperaturein the
tity of steamgeneratedunderworking conditionsto economizer
get the equivalentevaporation.
M1,= r[&SSof flue gasesper unit massof fuel
Equivalentevaporation- Actual evaporation
Cp- sPecificheatof flue gases
x (f)
@f- flue gastemperature
at inlet to economizer
or M"n=M*.(f )
O"ir
temperature
of
air
delivered
to theboiler
or M$(H - Hwt)/539= M*rf
Ptoblem 6.1 A boiler generates4.5 t of superf=(H_H*t)/539
heatedsteam(500'C,9}kgflcmz abs.)pertonof coal
feed.
a. Wtut is boiler eftcienq?
The BFW temperature- 45'C
Ans. It is lhe ratio of theheatload of he generated
What is the equivalentevaporationfrom and at
steamto the heatsuppliedby the fuel overthe same 100'Cpcr tonofcoal?
period.
Solution
Heatload of generatedsteam
= G"(H - H*,) kcaVs
whereG, = rateof steamgeneration,kg/s
Specific Enthalpy
= 809 kcaVkg
82 Boller Operolion Englneedng
Sensibleheatof feedwaterat 45"C- 45 kcal/kg ^
Heat requiredto produce4.5 t steam(90 kgflcm'
abs.,500"C)
= 4.5 x f03 x (809 - 45)
Solution The equivalentevaporationfrom and at
100"Cis
M"q=Mn (H-Hwt)/Lrco
Now
= 3 438 x 103kcal
Mrct- 8.5 kg steamPerkg of coal
Ijtent heatof dry, saturatedsteamat 100'C
H*r'163'4kl&;g
- 539 kcaVkg- 539 x ld kcaUt
H,r,- 830kJ/kg (at 14 bar)
Therefore, equivalent evaporation from and at
100'c
3 438x td tcat _ 6.379t per ton of coal
= __:_
539x 105 kcat/t
Ans.
7.5tonsof
Problem 5.2 A steamboilergenerates
equivalent
the
steamper ton of coal burned.Calculate
evaporationftom and at 100'C per ton ofcoal from
the following data
Steampdssure- 10kgflcm2.abs.
Drynessfraction- 0.95
Feedwatertemperature- 50"C
fulution Working formula
Mg(H - H*,)
M.s=-BMnr- 7'5Utof coal
- 50x 103kcaVt
kcaUkg
"*,'S
Hr"= 181.3x 103kcal/t
.r - 0.95
L-483x 103kcaVt
x 103
+ 0.95(483)l
H = H* + x L =t181.3
-r - 0.96
L - 1957.7kJ&g (at 14 bar)
H=Hw+xL
= 830+ 0.96(1957.7)
=27W.39kJ/kg
Itoo - 2257klkg
M"q = 8.5 Q7w.39-163'4)/2257
= 9.588kg stearn/kgof coal
= 9.59kg steam,zkg
of coal
Ans.
Prcblem 6.4 A boiler produces 220 t o^fdry
steampcr hourat apressure60kgflcm', abs.
saturated
of 120"c.
from feedwaterat a temperature
Coalconsumption- 1200VdaY
Calorific value of coal - 42O0kcaVkg
unbumt.
1% of coalescapes
Determine
(a) the equivalentevaporationper ton of coal
fred
(b) the eff,rciencY
of the boiler
(c) the overallefficiencyof the boiler
Solution
Step(I) Heat l-oad of Steam/Ton
= 640x lG kcal/t
steamat 60kgflcm2abs.
inrhalpy of dry, saturated
- 50)x rc3/fi9
:665.4 kcaVkg
M.n=1.5(6<O.tS
Enthalpyof waterat 120"C= 120kcaVkg
= 8.211t of steam/tof coal
Therefore,heatrequiredto raiseI ton of steam
Ans.
= td (66s.+- l2o) = 545.4x103kcal
Problem 6.3 A boileris workingat 14 bar and
EvaPoration
8.5kg of waterperkg of coalfired from Step(II) Equivalent evaporates
1200VdaY= 59 Y6
consumption
Coal
equivalent
the
BFW enteringat 39"C.Determine
per
ton ofcoal fred
gencrated
Steam
fromandat 100"Cif thesteamis0.96dry
evaporation
-22U50 - 4.4t
at thestopvalve.
Boller Colculoflont 83
Therefore,equivalentevaporation
(545.4x to3)
= 4.4
539x 103
- 4.452ton of steam/tof coal
Equivalentevaporation,
M"= M^(H - H)/L
= 8.321(2705- 425.036)x r03/Q257 x t}l)
- 8.405ton stearn/ton
of coal
Ans, Step(III) Boiler Thermal Efficiency
Step(IID Boiler Efficiency
Energyoutput=220 (545.4x 103)kcaVh
Coal chargedto the boiler = 50 t/h
Actualcoalbumt - 50 (l - l/lm) - 49.5tJh
Therefore,energyinput- 49.5x I 03 (4200)kcaUh
Energyto steam
= 8.32r (2705- 425.03Ox t03 U
= 5400x t03 kcavt ofcoal
= 5400x 4.1868x 103kJ/t of coal
49.5x lo3 x42oo
= 0.577 i.e. 57.7Vo
Ans.
Step(IV) Overall Efficiencyof the Boiler
220(545.4xt03)
r1
L'rboilerl- 5ox lo3 x42cn
-
WorkingFormula: Boiler ttrermalefficiency
- energyto stearn/energy
from fuel
Energyfrom fuel
(s4s.4xrd)
_'rboiler
- 220
r
Ans.
Boiler thermalefficiency
_8.32r (2705- 425.036)xrO3
5400x4.1868x103
= 0.8391
-
= 0.5713
=83.91%
= 57.137o
-84%
Ans.
Ans.
Problem 6.5 A boilerconsurrc,s224tons
of coal
to produce1864tonsofsteamperday.The steamis
dry, saturatedat 90 afrrr abs. Calculatethe boiler
thermalefficiency, and the equivalentevaporation
per ton of coal if the calorific value of coal is 5400
kcal/kg of coal, the specihcenthalpyof feedwater
beinga25.036U/kg of water.
Problem 6.6 A boilersenerates
7.5tonsof steam
per hour at 18 bar ( 1 b; - td ttlm2. The steam
temperature
is 598Kandthefeedwatertemperature
is
328K.
Whenfiredwith oil of calorificvalue47250kJkg,
theboilerplantachicvesanefficiencyof 85%.
Thegenerated
steamis fedto drivea turbinewhich
develops0.75 MW andexhaustsat 1.8bar,the drynessfractionof the steambeing0.97.
Determinethe rate of fuel consumptionand the
fractionof enthalpydrop,throughturbine,converted
to usefulwork.
If theturbineexhaustis directedfor processheating, estimatethe heat transferavailableper ton of
exhauststeamabove322.4K.
Solution
Step(I) Rateof Evaporation
Massof steamproduced- 1864ton
Massof coal consumed-224ton
Actualevaporationcapacity= 1864/224
= 8.321t /t of coal
Step(II) EquivalentEvaporation
Evaporation
capacity,M^= 8.321tltof coal
Sp.enthalpyof dry, satd.steam(90 atm.abs.),
H =2705kJ/kg
=2705 x t03kJtt
Sp. enthalpyof BFW, Hw = 425.036W1k1
= 425.036x l0r U/t
Solution
Step (I) Energy to RaiseSteam
Specificenthafy of generatedsteam
= 3106- 0.84(3105- 3083)
= 3 086.6 kJAg (by interpolation)
Specificenthalpyof BFW at 328K
&1 Boller Operotlon Englneedng
-230.TlakJkE
Specific energyto nrise steam
= 3086.68-230.274
=2 856.4kJ&g.
Step (II) Rate of Fuel Oil Consumption
Raleof steamgeneration-7.5 tlh - 7500kg/h
sp. energyto raisesteilrl - 2856.akJ&g
Energy input to steanr/h- 7500 (2 856.4)kJ
Step (V) Heat Transfer Available in Exhaust
SteamAbove 322.4K
Sp.enthalpyof exhauststeam- 2635.37kJftg
Sp.enthalpyof water at322.4K = 207 kJ/kB
Heat transfer availablein exhauststeamabove
322.4K
-2635.37 -207
Ans.
-242E.37kl/Kg
Problem 6.7 The following observationswere
madein thecaseof a boiler fitted with aneconomizen
Rateof steamgeneration- 5 Vt of coal
=85*
Equivalentevaporationfrom andat 100'C
= 0.85
= 5.5 Vt of coal
Rateof firel consumption(Energyfrom oiVh)
Boiler feedwatertemp.inlet to economizer
= 100"C
- 7500 Q856.4Y(0.85)(47250)
of BFW inlet to boiler - 180"C
Temperature
- 533.408kg
Temlrrature of air suppliedto the boiler - 30'C
Ans.
of flue gasesenteringtheeconomizer
Temperature
Step (III) Rate of Sp. Enthatpy Drop in Turbine
- 4O0"C
Sp.enthalPYof exhauststeam'
Weightof flue gasesproducedper ton of dry coal
H 2 =H n + x ' L
=15t
"C
= 49O.7+ 0.97 Qzrl)
Meanspecificheatof flue glses- 0.20kcaVkg
=2635.37kJ/kg
Calorific valueof coal - 5400kcal/kg
Determine
Sp. enthalpyof inlet steam,Hr - 3086.68kJlkg
(a) the boilerefficiencY
Sp.enthalPYdroPin turbine,.
(b) theeconomizerefficiencY
A,H=Ht-Hz
(c) lhe combinedefficiencyof the wholeplant
= 3086.68-2635.37
Solution
Boilerefficien"y=:ry4
Energyfrom fuel./h
Step(I) Heat OutPut
100"C- 5.5Vtof coal
Steamgeneratedfromandat
burnt
Therefore,heatoutput= 5.5 x 10' (539)kcaVtof
coal burnt
Step(II) Heat InPut
-940.229kJ|s
Calorific valueof coal * 5400kcaVkg
Step (IV) Fraction of Enthalpy Convertedto Use'
Therefore,heatinput - 5400 x l0r kcaVtof coal
fulWork
burnt
Energyoutput ftom turbine
Step(lII) Boiler EfficiencY
-0.75 MW
HeatOtttotrt- 5.5x l0' (539)
llboiler- 0.75x 103kw
Heatllput
54O0x103
- 0.75x 103U.zs
= 0.5489i.e. 557o(approx)
Rateof enthalpydrop in turbine'94O'229 kJls
Ans.
Fractionof enthalpydropconvertedto usefulwork
Gases
Flue
Step0V) Heat of
- 0.75xro3/90.229
Heatof theflue gasesenteringtheeconomizer
- 0.7976- 0.8
= 451.31kJAg of steam
Steamfeed - 7.5 tlh - 7500/3600kds
Rateof sp.enthalpy drop in turbine
- 451.31(7 500/3600)kJ/s
I
.
.
.
=
Boller Cslculollonr 85
= 15x 103(0.20)(400- 30)kcayt
= 111x t04kcaVtofcoal
Step (V) Heat Absorbed by BFW in the
Economizer
Heat absorbedby BFW in the economizer
= 5 x ld (180- 100)kcaVtof coal
=40x t04fcaUtofcoal
Step(VI) EconomizerEfficiency
i.e.36%
I"-o = (e0x t04 )/(ttl x 104)-0.3603
Ans'
step (vII) combined Efficiency
Heatabsorbed
in theboiler- 5.5 x ld (539)kcaVt
ofcoal
Heatabsorbedin the@onomizer- 40 x 104kcaVt
ofcoal
Total heatabsorbedin boilerandeconomizercombined
- 5.5x 103(539)
+ 40 x loa
- 336.45x 104kcaVtof coal
Energy releasedby burning co3l - 5a00 x ld
kcaVtofcoal
= 336'45x lOa -0.623i.e.62.30%
'rcornD
n^^-.
5aoox ld
Ans.
steamat 90
Problem6.6 A boilerproduces
Boiler HorsePower is a very commonlyusedunit
for measuringthe capacity of a boiler. ASME
(American Society for Mechanical Engineers)
definesaunitboilerhorsepowerastheboilercapacity
to evaporate15.653kg of BFW per hourfrom andat
373 K into dry, satunted steam or equivalentin
heatingeffect
Boiler h.p.- Equivalentevaporationfrom andat
373"Kperhour/l5.653
Problem 6.9 Aboilergenerates
6.5tof steamper
ton of coal fired.
The steamis at l8 kgflcm2gauge
The boiler feedwatertemperature- 110"C
downstreamof deaerator
Boilerefficiency-75%
Factorof evaporation- l.l5
Coof steam- 0.55kcaVkg"C
Determine
(a) the tempcratxreof the steam
(b) thedegreeof superheat,
if any
(c) the equivalentevaporationper ton of coal
burnt
(d) thecalorific valueofcoal
Solution
StepP (I) SteamParameters
- 19kgf/cm2.abs.
Pressure
Sensibleheat,I/* - 213.1kcakg
l,atentheat,L-455.1 kcaVkg
Satrrration
temp.,@"- 20E.E'C
kgttcn? abs.at the rate 150t/h from the feedwaterat
Whatis theboiler
120"C.The steamis dry, saturated.
horsepower?
Step(II) DegreeofSuperheat
Total heatof the steam- Hw + L + CoAO
Solution
where,AO - degreeof superheat
Steanr,90kgflcm2
Totalheat- 655.7kcal/kg
Sp.enthalpyof feedwater- Hrw
abs.Dry,sanrrated
Therefore,the factorof evaporation
Sensibleheatof BFlv at 120'C- 120kcaVkg
Equivalentevaporationfrom andat 100'C
_H*+L+CoL/g.-Hr*
539
- 150x 103(655.7- t20)/539
213.1+ 455.1+ 0.55(AO)- ll0
- 149.08x td fgn
or 1.15=
539
Therefore,boilerhorsepower
AO= 112.09"C
= 149.08xrc3/$.653
Ans.
=9524.15
SteamTemperature
Step(III) Superheated
Ans.
AO- lrz.Ogrc
86 Boller Operollon Englneerlng
or, @-@"-112.09"C
or
-321"C
O=208.8+ 112.09=320.89'C
(c) thepercentage
of heatlossto the ash
(d) thepercentage
for
of heatlossunaccounted
Ans. Solution
Step (IV) Heat Output
Heat requiredto generatestearn
= 6.5x ld (213.1+ 455.r+ 0.55AO - 110)
kcaVtof coal
= 6.5x 103(558.2+ 0.55x 112.09)kcaVtof coal
= 4029.021x103kcavt of coal
Step(V) Heat Input
Calorific valueof coal - CV kcaVkg
Energyreleasedper ton of coal burnt
- 103x CV kcal
Step(I) SteamParameters
SensibleHeat,If* - ?frO.1kcaVkg
l---Latent
Heat, L - 466 kcakg
Step(II) Heat Output Rate
Rateof stearngeneration- 16 t/h
Heatoutputrate=
16x t03 (H, +xL-Hp)
= 16x 103(200.7+ 0.g x 466-30)
=9441.6x 103kcaUh
Step(III) Heat Input Rate
Coalconsumption- 2.5 Uh
Calorific valueofcoal = 6540kcaVkg
Heatinput rate=2.5 x 103x 6540kcaVh
Step(IY) Boiler Efficiency
Iboiler = Heatoutputrate,/Heatinput rate
Step(VI) Boiler Efficiency
-4029'o2lxlo'
11.
.. HeatoutDut
''borrer=E;ffi
ld x cv
or 0.75 4029.021ICY;
CY - 5372kcaVkgofcoal
Ans.
= 9441.6x103/Q.5x 103x 6540)
=0.5774 i.e. 57.74V0
Step(VII) EquivalentEvaporation
Step (V) Heat Load of Flue Gases
- 110)
- 15Ut ofcoal
Fluegasesgenerated
6.5(2r3.r+ 455.1+ 0.55x 112.09
,r"q =
Ans.
Heatloadof flue gases
= 15x 103x 0.25(350- 25)
Ans.
= 1218.75x ld kcaVtof coal
Prcblem 6.10 fbe following observationswere Step(VI) Heat GeneratedBy l Ton of Coal
madeduring the trial run of a boiler.
kcal
Heatproduced
by ltofcoal - 103x 65210
Steamgenerationrate= 16 Uh
Step (VID Percentageof Heat llss to Flue Gases
Feedwatertemperature= 30"C
= Ir2t8.75 x r03/ 651CIxl03l (100)
SteamqualitY= 0.9 dry
= 18.63%
Steampressure- 15kgflcm' abs.
Ans.
Coal consumption= 2.5 t/h
(VIID
Ash
Loss
to
of
Heat
Percentage
Step
Calorific valueofcoal = 6540kcaVkg
Ash + unbunttcoalcollected- 0.2 Vh
Ash + unburnt coal collectedfrom beneaththe
Heatlossdueto ash+ unburntcoal
grrtes= 0.2llh (Calorific value- 700 kcal/kg)
=0.2x ld x 700kcaVh
Weight of flue gts€s= 15Vt of coal fred
Fluegastemperature- 350"C
Heatgencratcdin the fumace
Average specific heat of flue gases= 0.25
=2.5x ld x 654OkcaVh
kcaVkg'C
of heatlossto ash
Therefore,percentage
Ambient air temperature- 25"C
0.2x 103_x7ffi ,
__________
Calculate
.
rl00)=0.85%
(a) the boilerefficiency
2.5x 10'x 6540
Ans.
(b) the percentage
of heatlossto the flue gases
= 7.474t/t of coal bumt
Boller Cdculctlonr 87
Step (IX) Percentageof Unrccounted Heat
Usefulbeat- 57.74%
Heatlost to flue gases- 18.63%
Heatlost to ash- 0.85%
Total rccountedheat= 57.74+18.63+0.85
--77.22%.
.'. Unaccountedheat- 100- 77.22- 22.77%
Ans.
=7.5x rG x2879.55/tfr
x 45x ld
rt =firlkelh
Step (V) Specific Enthalpy Drop in Turbine
Specificenthalpyofelhaust steam
- H* + x L - 490.7+ 0.95(2210.8)
- 2590.96U/kg
A boiler genentes75 t of steam
Ptoblem 6.ll
.'. Specificenthalpydrop in turbine
per hour at pressure 1.8 MN/h' and temperature
325'C ftom feedwaterat 49.4"C.Whenfircd with oil
= 3086.45-2590.96
of caiorific value 45 MJftg, the boiler attains an
= 495.45kl/lrg
efficiency of 78%. The steam (325'C, is fed to a
hubine that develops650 kW and exhaustsat 0.18 Step (VI) Rate of Enthalpy Drop in Turblne
Rateof steamfed to turbine
I\,N/m2, the drynessfraction of steambeing0.95.
Determine
-7.5llh
(a) the massof oil fired Perhour
- 7.5x 1d/3600kgs
(b) the fraction of the enthalpydrop through the
Specificenthalpydrop in h[bine - 495.45kJlkg
turbine which is convertedto useful work
.'. Rateof enthalpydropin turbine
Also determinethe heat transferavailableper kg
of exhauststeamabove49.4"C,if theturbineexhaust
= 495.45(7.5x 1dl3600) kJls
is usedfor processheating.
= 1032.187
U/s
blution
(YID
Fraction of Enthalpy Drop Converted
Step
Step (I) SpecificEnthalpy of GeneratedSteam
to Usefull{ork
H =3rM - 0.84(3106- 3083)
Energyoutprt from trbine
= 3086.45kJ/ke by interpolation)
Step (II) SpecificEnthalpy of BFW (49A"C)
Hr*=2A6.9tJ/lrg
Step (IID Heat OutPut
Energyrequiredto generatesteam
= 3086.45-206.9
=2879.55kJftg
The rate of steamgeneration- 7.5 Uh
.'. Heatouput =7.5 x 103x2 879.55U/h
Step (IV) Rate of Oil Burning
I.et the massof oil fired be fi vgn
Heatinput - Ifr x 45 x 103U/h
Boilerefficierc! =78%
0.78= Heatouput/Heat input
- 650kw
- 650kJ/s
Energyinputto turbine- 1032.187U/s
.'. Fraction of enthalpy drop convertedto useful
work
- 65Cl/1032.187
-0.629
Ans.
Step(VIID Heat Transfer from ExhaustSteam
The net heat available,for processheating,from
exhauststeamabove 49.4"C- ?59O.96-?n6.9 2384.06kI/lr9
Ans.
Pr<iblem 6.12 A steamgenerationplant supplies
8500 kg of steamper hour at pressure0.75 MN/m".
The steamis 0.95dry.
Feedwatertemlrnture = 41.5"C
Coal consumption- 900 kg/h
88 Bolbt Operollon Englneedng
Equivalentevaporation= 9.44Q488.125>/2?56.9
Calorific value of coal -32450kJilrg
Determine
= 10.40ke/ke of coal
(a) theboilerefficiency
(b) theequivalentevaporationfrom andat 100"C Step (YI) Energy Required to Generate Sm
(c) thesavingin fuel consurnption,if by installing Under New Conditions
an economizerit is estimatedthat the feedSpecificenthalpyof BFW at 100'C- 419.1U/kg
water ternperatue could be raisedto 100"C,
Energy required to generate steam when
assumingthat other coditions rcmainedun- econqnizer is incorporated
changedand the efficiercy of the boiler in-2662.U25- 419.1
creases
by 6%.
-2242.9?5Wfrg
blwion
Energyto steury'h
Step (I) SteamGeneration Fer Ton of Coal
=2242.925x 8 500kJ
generationkgh
Rateof steam
8500
kg/h
Coal consumption 900
Step (VII) Rate of Coal Consumption when
Therefore,steamgenerationper kg of coal
Economizer is Fitted
- 8500/900
Erergy output -2242.98 x 8500kJ/h
- 9.44kg
Energyinput- Ifr x32a50Hth
.'. St€amgeneration/tonof coal
- 9440kg
Boilerefficiercy-72.38 + 6 -78.38%
0.7838
-9.44t
2242.Y25x 8500
Nl x32450
kglh
rt =749.57
Step (II) Specific Enthalpy of Steam Rais€d
H -- H* + x L -7@.3 + 0.95(2055.5)
=2662.025kl/kg
Step(III) Energr Required to GenerateSteam
Specificenthalpyof steamraised
-262.0?5kr/ltg
Step(VIII) Savingin Fuel Consumption
Initial fuel consumptionrate - 900 kg/h
Modified fuel consumptionratewheneconomizer
is fiued -749.57 kglh
Savingin frcl consumption= 900- 749.57
= 150.43kg coal,zh
Ans.
Specificenfhalpyof BFW - l139kJ/lKE
Hearouput = 26tr2025- 173.9
=2488.125kJAg of steam
(IV)
Step
Boiler Efficiency
HeatoutDut/ksof coal 2488.125,^
_
PtoHem 6.13 The following obsenationswere
madeduring tbe trial run of a boilen
Rateof steamgeneration= 5 Uh
tl*r",=ffi#ff=ffi(g.u't
Steamquality: dry, saturated
Steampressure= l0 kgf/cm2gauge
=0.7238i.e. 72.38%
Ans.
Step(V) Equivalent Evaporatbn
Steamraisedper kg of coal - 9.4k9
Energyrequiredto gerrcmtethis steam
-9.44 (2488.125)
U&g coal
Specificenthalpyof evaporationfrorn andat
100"C- 2?56.9kJft:g.
"C
Averagespecificbeatof steam- 0.55kcaUkg.
Redwater ternperature- 85'C
Roomternperature- 25"C
Atnospheric pressure- 1 kgflcm2
Fuel consumption- 650 kg coaVh
Calorificvalueofcoal - 7500kcaVkgofcoal
Boiler Cqlculqlions 89
Moisturecontentof coal-2.5%
Fuelcontains: C - 86%:H - 5%: Ash - 9%
Flue gastemperature- 300"C
Mean sp. heatof flue gines- 0.25kcaVkg"C
Analysisof dry flue gases:
C O z -l 0 % ; O 2 - 8 % ; N 2 - 8 2 %
Producea completeheatbalancesheettaking I kg
dry coal asthe basis.
Solution
Step (I) Energr to SteamPer kg of Dry Coal
= l0 kgf/cn? garye
Steampressure
Step(III) Coal Analysis
Basis: lkg dry coal
Constitu{hemicalReaction
nt
during combustion
Remorks
C+Or-+CO,
(r2)
(44)
Wt. of dry flue gas
produced
ur+)or-+ Hro
Wt. of water vapour
produced
- (18/2)(5/loo)
-0.a5kglkg of dry coal
(2)
(18)
- (2992n04)(86trm)
- 24.741kglkgof coal
= l1 kgflcm2 abs.
WL of moisturefired
=0.025/0.975
= 0.02564kg/kg of dry coal
Latentbeatofevaporationat 1I kgflcm2abs.
Total wt. of watervapourin flue gases
- 478.4kcaVkg
- 0.45+ 0.02564
= 0.4756kg/kg of dry coal.
Specificenthalpyof dry, saturatedsteamgeneraEd
Step(IV) Heat L,oadof Water Vapour
- 185.7+ 478.4
- 664.1kcal/kg
= 0.4756[638.8+ 0.55(300- 90) - 251
=344.5kcaVkgof dry coal
Specificenthalpyof feedwater= 85 kcal./kg
where638.8kcaVkg= totalheatof watervapour
Coalconsumption
- 650kdh
at I kgflcm2abs.to which flue gasesarc discharged
= 650 (100-2.r/lm
Dry, coalconsumption
Step(V) Heat Load of Dry Flue Gases
= 650x 0.975kg/h
Heatlossto flue gases
Energy to steam/kgcoal
=24.741(0.25)
(300- 25)
- (664.r _ 85) (5000)/(650x 0.975)
- 4568.836kcal
= 17N.944 kcaVkgof dry coal
Step (II) Flue Gas Analysis
Step(VI) HeatBalance
Basis: I kg of dry coal
Basis: 100m3 of dry flue gas
Sensibleheatof steamat 1l kgflcm2abs.
- 185.7kcaUkg
Constituent
coz
Volume Mol.v,t. Proportional
-3
Mass
10
44
Heat Input
%
Heat Expenditure
Total hcat
supplied
- 7500kcal
t00
Ijcat consumedin stearn @.92
fonnation
- 4568.836kcal
Ileat lost to flue gas
22.70
- 1700.944kcal
Heat lost to vapour
4.ffi
- 344kcal
Heat unaccounted for
I1.80
- 886.220kcal
7 500 kcal
100 7 500 kcal
Remarks
44(lO)= 440 Carbon
content
= 410(12/44)
= l2O
o2
8
32
82
28
100
t04
32(8\-2s6
28(82)-229<
2992
7o
100.00
90 Bolter OPerollon Englneedng
Prcbtem 6.t4 During the tial run of aboilerthe
following datawere recorded
8 3 . 1t
606 t
CoalconsumPtion
Steamproduced
Boiler:
SteamPressure
SteamtemPeraturc
Superheater:
steamtempcrature
Superheated
Economizer:
Water inlet temPerature
Watelroutlct tcmPerature
Air heater:
Air inlet temPerature
Air outlet temPerature
FIuegasinlet temPerature
Flue gasoutlet temPerature
1.461MN/m2
-14.42^tn
470 K
6 1 0K
353K
400 K
320K
380 K
503 K
405 K
Coalanalysis(bYweight)
Fluc gasanalYsis(drYbasis)
c
coz
o2
62.5%
4.25%
H
5.tt%
o
t2%
N
o.85%
s
9.85%
Ash
16.24%
Moisture
Total
N2
13.2%(by volume)
4.85%( -do- )
81.95%( do- )
(ii) suPerbeater
(iii) air heater
(iv) economizer
(e) heatlost in the flue gas
Summarizethe overall result on the basisof I kg
coal burnt.
Sohttion
Step (I) Theoretical Air Requirements
Basis: 100kgcoal
Molecular
Constituenl % W
Weight Weight
Element
c
62.5
4.25
H
o
N
S
Ash
Moisture
t2
2
5 . 1 I 32
28
t.2
0.85 32
9.85
16.24 IE
klmol
kmol of Ot
requircd for
complete
combustion
5.208
2.12512
- 1.0625
0 . 1 5 9 (-) 0.lse
5.208
2.r25
o.o42
o.o27
o.027
E - 6.1385
There,theoreticalair requirement
= 6.1385(100/21)
49.23 kmoV100kgcoal
=29.23 (28.9)k9100 kg coal
rco%
=844.744kglm kg coal
Grosscalorific value- 30550kJftg (dry coal)
= 298 K
Boiler housetemPerature
Enthalpyof dry, saturatedsteamat 1.451MN/m2
=2791L<IkE.
Substances
Sp ecif c H e ats(kJltg'K)
Dry flue gas
Watervapor in flue gas
Water
1.005
2.095
4.t81
Determine
(a) theoreticalair requirementsper kg of coal
(b) actualair suppliedperkg of coal fred
(c) weightof flue gasperkg of coal bumed
(d) thermalefficienciesof
(i) boiler
=8.447 kgftg of coal
AnS.
28'9)
is
air
of
weight
(c/ The averagemolecular
Step (II) Actual Air SirPPlied
100kg coal contains5.208kmol of C
100kmol of dry flue gascontains13.2kmol of C
Therefore,the amountof flue gasproduced
- 5.208(100/13.2)
=39.45kmoV100kg coal
Let r mol of air be suppliedper 100 kg of coal
burnL
thereforeby nitogen balancewe get'
79x 4ffi+ 0.042-H
(39.45)
Boller Colculollonc 9l
.'. .r - 4O.87kmoU100kg coal
-2E9.25lU.kg
Therefure,the weightof air supplied
- 40.87(28.9)
- l l8l.l4kg/l00kgcoal
Rateof steamgeneration/tof coal
=66/83.1r/t
-7 .292Vt of coal (or kJlkg coal)
- 11.81kgfu coal
Therefore,the heattransferredto steany'kgof coal
burned
=7.2V) (2259.251)
Nore % excessxi1= (ll.8l -8.47)(tU.,/8.#
=39.81%
= 16474.458
kJ./kgcoal
Step(III) Weight of Flue Gas
Basis: 100kmol
Fluc Gas koiol
Conslitaen
t
coz
o2
N2
Mohca- Wcight
ls
Weight
r3.2 4
4.85 32
81.95 28
Grosscalorific valueof coal asfired
= 30550(100- 16.24)/100
Wcightin39.45
kaolof Fluc Gas
13.2(44) 13.2(,14X39.45/100)
-229.t25
4.85(32)4.8s(32)(3e.4sll
m)
- 61.226
8 1.95(2E)
8 l.9s(28x39.4sl
IOO)
= 905.219
E= il95.57k9
Waterproduceddue to combustionof hydrogen
contentofcoal
-2.18 kmol- 2.125(18)- 3E.25kg
Freemoisture- l6.24kg
Therefore,the total weight of wet flue gas
- 1195.57+ 38.25+ 16.24
Ans'
Total heatcontentof waterchargedto boiler
=4.187(4N-273)
= 531.749U/kg
Therefore,the netheattansfened to steam
=2791- 531.749
Therefore,thermalefficiencyof the boiler
= 16474.458/25588.68
= 0.6438 i.e. 64.38%
Ans,
2. Superheater
Net heattansferredto steamin the superheater
=7.2T2(2.095)(610_ 470)
=2138.74kJ/kgcoal
Therefore,thermalefficiencyof lhe superheater
= 2138.7
4/25588.68
= 0.0835 i.e. 8.35%
Ans.
3. Air Heater
Weightof air chargedto the boiler
- 11.81kJ/kgcoal
- I 250.06kglm kg coal
- 12.50kgltg coal
Step(IV) Thermal Efliciencies
l. Boiler
Total heatcontentof steamat 1.461MN/m2
-279lkItkg
= 25588.68kI/kg coal
Heatabsorbedby air in the air heater
- 320)
= 11.81(1.005X380
=712.143kJ/kgcoal
Therefore,the efficiencyof the air heater
=712.143/25588.68
= 0.0278 i.e. 2.78%
Ans.
4. Economizer
Heattransfcncd
to BFW
=7.292(4.187X400
- 353)
: 1434.985
kJ/kgcoal
92 Boller Operotlon Englneedng
Therefore,thermal efficiency of the economizpr
- r434.985t25
588.68
:0.0560 i.e. 5.6Vo
- (2950.29
100)
t?5588.68)(
= 1L.52%
Tabulation
of Result(Basis: I kg coalbumed)
Ans.
Heatrecovered %GCV % Elficiency
(kJ)
t^,ctcoal
Step (V) Heat Lost to the Flue Gas
Weight of the dry flue gas
= 11.95kJ/kgcoal
Enthalpyof the dry flue gas
= 11.95(1.005)(405- 298)
= 1285U/kg coal
Watercontentin flue gaseswhen100kg coalburnt
-38.25 + 16.24
Boiler
Air Heater
Superheater
Economizer
Heatto flue gas
Heat
unaccounted
r5762
712
2t39
r435
29fl
259r
61.591
2J8
|
78.34
8.36
r
s.6lJ
I l.53\
21.66
1o.t3i
>- 100.00
= 54.49k9
ProUem 5.15 A boiler generates6000kg steam
per
hourat 10kgf/cnt2from BFW at 4O"C.The steam
Therefore,the weight of watervapour/kgof coal
is 0.97dry. The boiler is fired with coal at the rateof
burnt
700ke/husing16kg ofair (at 15'C) perkg ofcoal
- 0.5,149kg
fired.
Therefore,the enthalpyof waterin the flue gases
Assumingtlre boiler efficiencyto be 70%, determine
=0.5M9 t2.095(405- 311*)+24t1.2*
(a) excessair coefficient
+ 4.187(405- 298)
(b) flue gasternperaureleavingthe boiler
= 1665.29kJ/kgcoal
Given
* fDewPoittt ofWet Flue Gasl
The coal is composedof carbon and hydrogen
besidesits ashcontcnt12%.
Flue Gas
Conslituent
coz
o2
N2
Hzo
kmol
%
Composition
- s.2 12.24
(13.2/100x39.4s)
- l.e
(4.82/lm)(3e.4s)
4.47
(81.95/
.4s)- 32.33 7 6 . 1 3
100)(3e
-3.02
1.tl
54.49t18
=242.46
> - 99.95
Combustion
Heat Of Combustion
C+Or-rCO,
Hr+ Or-+ H, O
8075kcaVkgofcarbon
34500kcaVkgof hydrogen
'C
Specificheatof flue gas- 0.25kcaVkg
18% of total heat generatedby coal is lost to
substances
otherthancoal.
Therefore,
of watervapour
thevapourpressure
- (7.ru99.9s)
(101.3)
=7.z}kN/m2
Solution
Step(I) Heat Contentof GeneratedSteam
H=Hw+xL-Hrn
whichcorresponds
to thedewpoint311'Kand
= 181+ 0.97(482)- { = 608.54kcal/kg
Step(II) Calorific Value of Coal
It can be determinedfrom the boiler efficiencv
relationship.
Iboil", = HeatoutpuvHeatinput
latentbeatof evaporation2411.2kJkg
Therefore,the total heatlost to flue gas
- 1285+ 1665.29
=2950.29kJ/kgcoa.
Hence,the percentage
of heatlost to the flue gas
Now HeatOutput= 6000(608.54)kcaVh
Boller Colculqllonr 93
Heatinput- 700 x CV whereCV - calorif,rcvalue
ofcoal
or 0.70- 6000 (608.54)/(700x CV)
CV - 7 4l?kcallkg
Step(IID Carbon and HydrogenContentsof'Coal
Basis: I kg coal
Therefore,the coalcontains(l - 0.12),i.e.
0.88kg ofC + H per kg ofcoal fired.
Heat Generated
(kcaVh)
by coal: 70{l,Q452)
- 521640,0
Heat Rcceivcd by
(kcaVh)
Fluegas: 16.88(700X0.25XA@)
- 29s4(A@)
Steam: 6000(608.54)- 3651240
Sirbstances
otherthanflue gas:
7 OO(7
4s2)(O.r8)- 9389s2
By heatbalancing
2954(A@)+ 365Qa0 + 938952- 52164A0
*212C
.'. AO=211.98"C
If .r be thepart of CYkgof coal,then
=7452
.t (8075)+ (0.88- -rX3a500)
.r = 0.8669
lc-"'b*l*lHyd'os;l
I kg Coal 0.8669kg
0.0131kg
Step (IV) Theoretical Air Requirement for Complete Combustion
Basis: I kg coal
Hencetheternperature
of the flue gasatthe biler
outlet
-212 + L5-227'C
Ans.
PtoUem 6.16 A boiler is fired with coal having
following percentage
compositionby mass:
C45Vo; H-54o; S-lVo; O-2.59o: IncomElzme
Combustion
Weight
OrRequirement
bustible-6.5%.
nt
Reaction
Determinethe boiler efficiency from the given
C + O, -+ CO, 0.8669kg
o.E669QAn)
data:
(r2) (32)
-2.3rr7 kg
Excessair supplied= 407o
Fluegastemperatureat boiler exit = 170"C
2H, + O, + 2HrO 0.0131kg o.or31(32t4)
Ambient air temperature- 25"C
(4)
- 0.1048
(32')
kg
Specificheatof flue gas- 0.25kcaVkg'C
Specificheatof steam= 0.48kcaVkg"C
E - 2.4165
kg
Since air contains 23% Oz by mass, the
stoichiometric(theoretical)air requirenrnt for completecombustion
-2.4165 (100/23)
= t0.506kg
Step (V) ExcessAir Coefficient
Excessair coefficient= Actual air,/minimumair
= 16/10.506
= I.523
Combustion
Heat of Combustion
C + Or--+
CO,
8075kcaVkg
S + Or---r
SO,
222Okcdlkg
H, + Or----+ HrO
34500kcaVkg
Unaccounted
heatloss- lSVo
Solution
Step (I) Calorific Value of Coal
Cv - 8075(c)+ 2220(5)+ 3a500(H- O/8)
AIlr.
Step(VI) Enthalpyof Flue Gas
where, C, S, H & O standfor carbon,sulphur,
hydrogenandoxygenpercentage.
Massof coal + massof air = Massof flue gas
(1 - 0.12)kg + 16kg = Massof flue gas
- 8075(0.85)+ 2220(0.0r)
+ 3a500(0.05- 0.025/8)
or Massof flue gas- 16.88kg.
- 8503 kcaUkg
94 Boiler Operolion Englneerlng
Step(II) StoichiometricOxygen
Basis: I kg fuel
Element
Oxygen Rcquircd Per
Kg of Fuel
Combustion
Reoction
C + Or --r
s) - 2.2666ks
COt (32t12)(0.8
(r2) (32)
- 0.01kg
S + O, ----+ SOt (3u32)(o.ot)
(32)
(32)
ZHr+ Or-+ZHrO
(324)(0.05)- 0.4kg
2-2.6766k9
= 2 (18\14
-9kg
Mass of waterproduced/kgof coal (H-content:
0.5%)burned
= 9 (0.05)kg
= 0.45kg
Massofdry flue gasproduced/kgofcoal burnt
= 17.075- 0.45
= 16.625k9
Step(VI) Heat Balance
Basis: I kg coal
Sincethefuelcontains0.025kg oxygen/kgof fuel'
Heal Lost To
of 02 requirementperkgof coalburnt Heat Evolved
theactualmass
-2.6766- 0.025
Fluegas(dry)
8 503kcal
- r6.62s(0.25)(l7O-2s)
=2.6516k9
602.65 kcal
Step(III) Air Supplied
Theoreticalmassof air requtement
=2.5516(1m/23)
- 11.5289
kg
40%excessair suPPlied.
Hencethe actualair suPPlied
= 1.4(11.5289)
= L6.l4kg/kg of coal
Step (IV) Massof Flue Gas
Massof combustiblesPerkg of coal
- 1 - 0.065
= 0.935kg
Air = FlueGas
0.935kg 16.14kg = FlueGas
Hencethetotalmassof flue gas(inclusiveof waler
vapour)producedperkg of coalburnt
= 0.935+ 16.14
- t7.075kg
Fuel +
Step (V) Massof Dry Flue Gas
2H2 + O, ------+2HrO
(2 x l8)
(4\
Massof waterproduced/kgof H2 burned
Stcam (l atm pr€s$.)generatedfrom
fuel burning
= 0.45 [/J + Cp (A@) -@.ir]
= 0.45 [639 + 0.48 (170 - 100) - 25]
=297.42kcal
Unaccountcd sourceg
- (18/100)(E 503)
- I 530.54 kcal
Total - 2 424.61kca|
Heatutilized= 8503 -2424.61
= 6078.39kcalAg coal
(VID
Efficiency
Boiler
Step
Heatutilized
TlboiLr=
H*t ga"*aLd
= 6078.39
8503
= 0.7I 48 i.e. 7 1.487o
Ans.
fuoblem 6.17 A water tube boiler operates
8400 h/year al 80o/oefficiency. The unit rated at
at7.82atm.
272L5kgh operates
It burnsnaturalgasfor six monthsof the yearand
No.2 fuel oil for the rest.
Averageaurual boiler loadingis 6O%with an
kcaVh.
inputof 11347303
Boiler Colculotionr
lYlthout Economlzer
^ \
Naturalg:rsconsumption = 1274,25Nmr/h latwo
Fueloil (No.2)consumption= t.+S9m3n,J bd
Afier Addlng An Economl,zer
BFW flowate (includingblowdown)at60%lofi
- 17145kg/h
F l u og o ! o x h o u r t
Feedwatertenperatureat the economizerinlet
= 105"C
Feedwatertemperatureat the economizeroutlet
- 136'C
Fluegast€mperature
ateconomizerinlet = 260'C
Fluegastemperatur€at economizeroutlet- 149"C
Determine
(a) the fuel savingusing the economizer
(b) totat annualfuel cost without insrallingthe
economizer
(c) total annualsavingof fuel alter installingthe
economizer
(d) thepaybackmonths,if theeconomizercostis
Rs. 500,000installed
Given: Nanral gascost = Rs. 1.06per Nm3 of
gas
Flg. 6.1 Figureto the Problem6.17
Fueloil (No.2)cost- Rs I 255 perm3 of F.O.
x 100:5.85%
s - 664368.78
Solwion The additionof an economizqrto awaterrr347303
tubeboiler systemreducesfuel cosl
Ans'
The fuel savingusing the economizeris
(b) Total operatingperid = 8400hlyear
Ilx 100
Naturalgasburnedfor 4200h andF.O.bumedfor
"- = _ _ _
rest4200h overtheyear.
Annualcostof naturalgas
whereS- fuel savingin perceht
/^- r\
=ry
H= heatrecovere4
kcavh
F - BFW flowrate, kg/h
AO = 6z - 8l = temperaturcdifference of BFW
beforeand after the economizer
Or - BFW temperatueat economizerinlet, 'C
Oz = BFW temperaturcat the economizeroutlet,
'c
B = boiler efTiciency
(a) H= (17t45X136-105)/0.8
= 664368.78kca1/h
=,zzo.zs[*l'
lx+200
rj-), r.06
r-R'-)
\n/
1v.n/
= Rs.5672961.
1N*,.,;
Annualcostof fuel oil
/ r\
=r.45e
f+ l,<4200
r-!-)x tzss[4)
\o/
\Year;
l-'J
- Rs.7690389
Total annualcost of fuel prior to installation
of economizer
= Rs.5672961+ Rs.7690389
96 BollEr OPerollon Englneedng
- Rs.13,363,350/-
Ans.
a
(c) After the installation of the economizer,
5.85%savingin fuel results'
Annualsavingin naturalgas
= Rs.5572961x 5'85/100
= Rs.331868/Annualsavingin fuel oil cost
= Rs.76903g9x 5.95.2100
- Rs. 449888/instalation
.'. Total annualsavingof fuel costafter
of the economizer
= Rs' 781756/= Rs.331868+ Rs' ut49888
Ans.
is
(d) The PaYback
= Rs 3,000000/Maintenanceatrdoverheadexpenses
- 12%of the cAstof WHB
the principal
Rate of interest payable 20% on
arnount
StcanGenerationRate
DG SetLoad
4OVdaY
80 Uday
@%
too%
steamat
1 ton ofcoal generates4'5 t of saturated
10kg/cm2g
Costof coal - Rs 750 Perton
blution
y-HB,.
ia)
'-'rn"Bco"o*ics of lncorporating
from the grid
puphased
"ott of ebcricity
ro -- E x 1 2
A
-Rs l.l?KWH
the installaThe cost of generatedelectricity after
'0.95/KWH
tion
-- of WHB
generated
frfonetarysavingsper unit electricity
= Rs (1.12- 0.95)/KWH= Rs 0'17/KWH
whereP-paybackmonths
E - installedeconomizercost' Rs
Rs
A - annualfuel savingswith economizer'
. 'p-!''9SS
,.rz
Rs.781756
= 9.21month,
year - 270
Number of operatingdays per
Costof wasteheatboiler installed
on .
Averageelectricitygeneratior/day
=g#*MwH =67.5MwH
is hookedup
Problem 6-t8 A wasteheatboiler
steam from
-270
produce
to
set
with a diesel generator
Numberof working daysper lear
power
respectto
with
wasteheat.
Therefore,monetarysavings
l0kg/c#
steamof
saturated
ATQVoDG'setload'
purchasedfrorn the grid
rate of 40
g is producedin the wasteheatboiler at the
tor/daY.
--per day varies
nn.tug" ebcric energygenerated
from65 to 70MWH.
Rs3,098250/-peryear
, zzo(,days')=
Esttnate
[vear''1
thewasteheat
tui-,n" oonomics of incorporating
(b)
PaybackPeriod Of WHB
boiler
boiler
heat
Costof wasteheatboiler
(b) th; paybackperiodof the waste
= Rs 3,000'000/Given
from the
energy
electrical
purchased
of
"o.t
ih"
Maintenarpeard overheads
prid
- - Rs l.IZKWH
'= l2Voof caPitalcostof WHB
cor, of generatedelectricityafter the installation
= Rs 360'000/of
'- WHB - Rs0.95/KWH
(t* includesoverheadsand depreciationcharInterestol PrinciPalarnount
ges)
- o/tl
[nffr)'
x'.*
67.5
[#, )
Boller Colculollons
= 20%of Rs.3,000,000/= Rs 600,000./Total steamgeneratedon 100%load- 80 Vday
4.5tofcoal generat€I ton of steam
Amountof coal saved- 8C/4.5- 17.777tlday
Monetarysavings,on the basisof coal,per year
- 750f-n,-)xn.t77(tro')"zzo[-@-)
(ton/
- Rs359984?-
(oayJ
Averagefuel consumptionper day
-6750014.025
- 16770lL
Massrateof fuel consumption
(. r<e) I
---^ / rt )
I
[vear.1
WHB is an energy savingequipment.So it
qualifiesfor 100%depreciationin the lst year.
Approximatesavingsin corporatetaxes(@ 55%)
perye:u
- Rs 3000000x 0.55
- Rs 16500004
Net Savingsperyear
- Rs (3599842- 360000- 600000+ 1650000)
- Rs 4289842Ps
= 3000000
Payback
period
ffiffi
(a) Rateof Fuel Consumption
Averageelcclricenergygenerated
per day
:67500 KWH
x 12months
= 8.39months
Ans,
=16770 o.e
[dj,.
[sf,j' ;i $6
- 628.875ke/h
Ans.
(b) Flowrate of Flue Gas at 60% l-oad
-7.55 x 3600- 27180kgh
(c) UsefulHeatof Flue Gas
Total heatrejectedby hot flue gasin the WHB
=27180
("c)
f+l'0.26
f'9)x ezl -r70)
(n
J
[K8-u'
= 1060020
kcaVh
Heatlostto radiation
- 106(n20x 5/100kcaVh
= 53001kcaVlt
Usefulheatavailablefor steamgeneration
Ptoilem d.l9 lreterminethe
(a) rateof fuel consumptionin kg/h
= 1060020-53001
(b) efficiencyof WHB of problem6.18
= 1007019kcaVh
Given
- Heatinput rate
I lt. of fuel generates4.025 KWH of electrical
energy
(d) Heat Output
Specificgravityof liquid fuel - 0.90
Averagesteanr(10 kg/cm2g and saturated)
Exhaustgasflowrateandtemperature
at 68%load generationrate
are7.55kg/sand325'C respectively.
- 40ttday
Fluegastemperature
at WHB inlet - 320'C
= 40x l0ffil24kg/b
Fluegastempentureat WHB outlet= 170'C
Averagefeedwatertemperatureto the boiler
= 1666.66kg/h
- 75'C
Averagefeedwatertemperature
= 75'C
Heatrequiredto generate1666.66kg steam
Specificheatof flue gas- 0.26kcaVkg'C
(10kg/cmzg andsaturated)
Assume59oradiationlosssufferedby theflue gas
in theWHB.
- 1666.66x (183- 75) + 1666.66x 478.4kcal
=977329kcal
Solution The determinationof efficiency of the
wasteheatboileris to be madeon the basisof heat
Heatoutputrate= 917329kcat/h
balance.
Heatinputrate= 1007019kcaVh
98 Boller OPerotlon
Englneerlng
-mxLoo%
ni - numberof molesof i-th componentpresentin
theflue gasproduceddueto combustionof 1 kg fuel.
En,- the meanspecificheatof i-th componentat
-97'05%
ofc
.'. Efficiency of waste heat boiler
Ans.
BOILER HEAT BALANCE CALCULATIONS
Basis: lkg fuel
HeatInput
(A) Ilr - Grosscalorific valueof fuel' kcal
(B) Hz= Heatinput of fuel
= c' (@r- @r)'kcal
"C
c specificheatof fuel, kcaVkg
'C
of fuel,
Ot - temperature
"C
@r- r€ferencetemperature,
(c) tt = Heatinput of air
= M"cr(@" - @r),kcal
M" massof input dry airlkg fuel, kg airlkg fuel
ca- sP€ciltcheatof humid air
-0.24+ 0.46H,kcaVkgdry air "C
I/ - humidig of air, kg moisture&g dry air
'C
@a- air temperature,
Totalheatinput,I{ = Hr* Hr+ H3,kcal
HeatOutput
(A) Heat consumedin generatingsteam
l. EconomizerHo= Mw (he,,- h6r),kcal
Mw - fllass of feedwaterper unit massof fuel,
"C
@1,= flue gastcmPerature'
(C) Heat lossdue to evaPoration
l. Moisture is formed due to combustionof
hydrogenin the fuel. Loss of heatto evaporatethis
moisture
H8=M^L' kcal
Mm = tnilss of moisture formed by burning of
hydrogenperkg of t'uel,kg H2Olkgfuel
of themoistureat the
L = latcnthcatof evaporalion
dew point of the flue gases,kcaVkg
2. Heat loss due to evaporationof moisture
presentin thc luel
Hs= M*f L,kcal
Mnrf= massof moisturepresentin the fuel' kg/kg
fuel
(D) Heat Lossdueto incompletecombustionof
carbon as carbon monoxide.
kcal
a,o=I c o l .x c x 5636.7
fEofrE.]
CO - Vo(by volume)of carbonmonoxidein the
flue gas
COt- qo(by volume)of carbondioxidein theflue
gas
kg/kg tuel
C = c:ubonbumt per kg fuel burnt,kglkg fuel
iew - eothalpy of water at economizeroutlet'
(E) Heat lossdue to unburnt carbon
kcaVkg
Htr= M"(7 837'5)'kcal
ftfw= enthalpYof feedwater,kcaVkg
carbonin refuse,kg/kg
M" = 62ss of unconsunred
2. Evaporator(Boiler) Hs= M"(/t, - i"*),kcal
fuel
M. - massof steamgenerationper unit massof
(F) Heatlossdueto blowdorvn
fuel, kg steam/kgfuel
kcal
H tz= Mt t (hu* - /t1.."),
=
generated,
kcaVkg
steam
of
h, enthalpy
Mbl = massof blowdownwater,kg/kg fuel
3. SuPerheaterHu= M"(Hu- hr)' kcal
ftuw= enthalpyof boilerwater,kcaVkg
steam,kcal/kg
/rr. = enthalpyof superheated
(G) Unaccountedheatloss
(B) Heat lost in flue gases
Hrt= Hi- (Hc+ H5+ Hu+ H, + Hg
+ H n + H r c + H 1 . +H p )
Ht =2 n; ?0,(@s,- 25)' kcal
Boller Colculotlons 99
Therefore,theoxygenutilizedfor hydrogenburnProdem 6.fr A stoker-fired waterhrbeboiler
burnscoal-atthe rate of 4 Uh to generatesteamof 30 ing of fuel
kglcfl:Pabs and 430"C at the rate of 30 Vhour.
=21.438-(12.85+6.5)
Evaluatethe boiler performancefrom the following
- 2.088kmol
data
2Hz + 02 -----s 2H2O
(a) Component Proximateanalysisof coal
2
kmols
I kmol
2 kmols
Ash
12.7%(by weight)
Hydrogenbumt= 2 (2.088)= 4.176kmol
Moisture 7.9%(by weight)
Waterproduced- 4.176kmol
(b) Grosscalorific valueof coal - 6 250kcaUkg
(c) Component
coz
(d)
(e)
(f)
(g)
Flue Gas Arnlysis
t2.85%
02
6.580
N2
rest
Carbonpresentin the cinder asunburntcombustible-2.75%
The feedwatertempenture - 90"C
Flue gastemperaturcat economizeroutlet
- 150\C
Fluegaspressureat economizeroutlet
- 755mmHg
Air tempqratures
at burnerinlet 30"CDB and
22"CWB. Ignorethepresence
of sulphurand
oxygenin coal.
blwion The boiler performance,i.e. the overall
thermalefficiency of the boiler is to be evaluatedon
thebasisof heatbalance.
Basis: 100kmol of dry flue gas.
1. OxygenSuppliedwith CombustionAir
N, in the flue gases
- 100- (12.85+ 6.5)
- 80.65kmol
l0Okmol aircontains79kmol N2 and21 kmol 02
.'. 02 suppliedfor combustion
-(21179)x 80.65- 21.438kmol
2. Water Vapour ProducedDuring
Combustion
C + 02 -----+ CO2
I kmol I kmol
I kmol
I kmol of CO2requiresI kmol of O, forcombustion.
12.85kmol of CO2require12.85kmol of 02
for combustion.
3. Unburnt Carbon for 100kmol of
Dry Flue Gas
Carbonretainedin thecinder
)1\
-ffix
4x 1000kg/h
- ll0k9h
(FC+ VM) in coal
= 100- (asb%+ moistureTo)
in coal
= 100- (12.7+7.9)=79.4Vo
unbumtcarbon
".
(FC + VM - unburntcarbon)
)7\
===-?
't9.4-2.75 =0.0358
Carbonin the flue gas
- carbonin the CO2in the flue gas
- 12.85kmol
= 12.85x 12
= 154.2kg
Hydrogenin the flue gas
- 4.176kmol
= 4.176x2
= 8.352kg
Total burntcombustible
- 154.2+ 8.352
- r62.552kg
Carbonunburntfor 100kmol ofdry flue gas
= 0.0358x 162552
= 5.8193
kg
= 0.4849kmol
100 Boller Operotlon Englneedng
- (4.176+ 0.8985+ 2.203)kmol
=7.277Sklrrol
4. ExcessAir
C +
O2+CO2
l kmol
l kmol l kmol
6. Compositionof Flue Gas
I kmol of C requiresI kmol . 3r for combustion
.'. 0.4899kmol of C requires0.4899kmol of 02
for combustion
.'. Oxygenrequiredto bum that unbumtcarbon0.4899kmol
.'. Excess02 supplied
= 6.5- 0.4899= 6.0151kmol
Component
kmol
mol%
coz
o2
12.85
6.5
10
N2
*x21.$8-80.&'7
zl
Hzo
Total stoichiometric02 required
= 12.85+ 2.088+ 0.4899
rr.97
6.05
75.179
7.275
6.781
lm.272
D.980
7. Heat Input Rate
Rateof fuel buming- 4 ton coaVh
= 15.4229lnnol
Grosscloritic valueof fuel - 6250kcaVkg
Therefore,excessair supplied
Heatinput rate= 4 x 1000x 6250
= 25000000kcaVh
_ 6.0151"Y rfl)
'""
15.4229
Ignoringtheheatinputof air (at 30'C), thenetheat
=39Vo.
input rate - 25000000kcavh
8. Heat Output Rate
5. Moisture Content of the Flue Gas
[A] Heat absorbedin generating superheat,ed
100 kg coal contains7.9 kg free moisture and
steam
(12.7
+ 7.9)l i.e.T9.4kgcombustibles
ll00
Hn+ Hr+ Hu
Freemoistureappearingwith the combustionof
- 30,000x (787.8- 90.04)kcaUh
162.552kg combustibles
= 20932800kcaI/h
? o
t9.4
where 787.8 kcal/kg - enthalpy of steam at
30 kgf/cm2absand at 4t0"C = h*
- 16.173kg= t6.tl3t18 kmol - 0.8985kmol
90.04kcaVkg= enthalPYof waterat 90'C
Frompsychrometricchart,humidityof air at 30'C
lBl Combustiblesleft in thecinder(asC)
DB and22'C WB
- I l0 kgih
- 13.4grr/kg dry air
Calorihcvalueof carbon(GCV - NCV)
1 ? 4
I
=
kmol
HrO$
kmol
dry
air
= 94.05kcaVmol
lg . 1000
q4 ()5
=#
0.02158kmol water
= -ktnol dry uit
Therefore,thetotalmoistureenteringthecombustion zone
= 0.02158x kmol of air containing21.438kmol
o2
= (0.02158)
r [+ x 2r.+38-]
lLt
I
- 2.203kmol
Therefore,total moisturein the flue gases
x 1000kcal/kg
tz
Heatlost in the combustibles
Hrr=Wx
1 0 0 0 xl l 0 k c a V h
=862125kcal,/h
lCl Total freemoistureevaporatedfrom coal
=l#x4ooo"%E
= 305.055kg/h
Boiler Colculoliom
Now, partial pressureof water vapour in the flue
gas
= 755 x rnol frrction of water vap. in the flue gas
- 755 (7.I7 5l W7.2:12)mmltg
-51.20mmH9
Dew point of flue gases- 38"C
Latentheatof waterat 38"C- 575.83kcavkg
Heat lost due to evaporationof free mois[re
- 305.055x 575.83- 175660kcaYh
[D] Heat loss due to evaporation of moisture
fonned due to cornbustion of hydrogen in the coal
burnt
,, =l#r)G.*)
[4#]
I
lI 1g coarlf
' ' -ks
- - cornbustiblesl
-------------'
kgcoal I
h
fkgcorrbustibleslf
lL
I| _--tgHro lfll -:-r""rtlI kmolH2oll tg J
815408kcarl/h
lEl Heat lost in flue gases(I/r) is evaluatedm the
basis of rnean specific heat data of flue gas components:
Componcfi
Mcan Sp.Hcat intlu Rangc25"-l50"C
coz
9.5 kcal/tmol'C
"C
8.12kcalrtmol
"C
7.l2kcal/kmol
7.00kcal,/kmol'C
Hzo
o2
N2
Thercfore,for 1f/.272krnol of flue gas
= 12.E5
(95)+7.275(8.12)+ 6.5(7.12\
r\c^
' r t
+8O.&7(7)kcaV'C
=79l.957kcal/"C
='lrt:lrT"i!;"'
innue
Heatrost
83s€s
Therefore,the rateof heatlossof flue gases
- ee,ee
4,00cy
rd2.
4.62(
s rr>(#EJ
*"ro
= t,Ag6,06aY
9. Heat Balarrce
Hcd Input
Rate
(kcd/h)
Heat OutputRate
(kcaYh)
-2W3280083.73%
25,000,000 $1eamgeneration
(,sx57s.s3)rca,/h
tmot urO
l0l
-862125
Heatlossdueto
unburntcombustibles
Heatlossdueto
evaporationoffree
moisture
- 175660
- El64$
Heatlossdueto
evapcation of moishre
formed due to
combustionof
hydrogenh thc fuel
3.4%
0.70%
3.26%
Heatlost to flue gases - 1886064 7.54%
Unaccountedheatloss-326943
(by differerce)
Ln%
Total - 25 000 000
(10) Overall ThermalEfficiency of the Boiler
_ Heat oubut rate (steamgeneration)
Heat input rate (fuel combustion)
=#ffffix roo=
83.73%
Steam tables
Temperature
Pressure
bar
kPa
I Dpecrrrc
SpecificEnthalPY
(hfs) lSteam(hn)l Steam
Water(hf) | Evaporation
kJ/kq I
kJ/ks
I kJ/kg lmJ/kg
absolute
0.30
30.0
69.10
289.23
2336.1
2625.3
5.229
0.50
50.0
81.33
340.49
2305.4
2645.9
3.24Q
o.7s
0.95
75.O
95.0
91.78
98.20
384.39
411.43
2278.6
2261.8
2663.0
2673.2
2.217
1.777
gauge
100.00
419.04
2257.0
2676.0
1.673
0.10
10.0
102.66
430.2
2250.2
2680.2
1.533
0.20
0.30
0.40
0.50
0.60
20.0
30.0
40.0
50.0
60.0
105.10
107.39
109.55
1 11 . 6 1
113.56
440.8
450.4
459.7
468.3
476.4
2243.4
2237.2
2231.3
2225.6
2220.4
2644.2
2647.6
2691.0
2693.9
2696.8
1.414
't.312
0.70
70.o
115.40
444.1
2215.4
2699.s
1.024
0.80
0.90
1.00
1.10
80.0
90.0
100.0
110.0
1.20
1.30
491.6
498.9
505.6
512.2
518.7
524.6
2210.5
2205.6
2201 .1
2197.0
92.8
88.7
z 84.8
z 8 1. 0
77.3
2702.1
2704.5
2706.7
2709.2
2711.5
2713.3
2715.3
2717.1
2718.9
0.971
0.923
0.881
0.841
0.806
0.773
0.743
0.714
0.689
73.7
2 7 2 0. 8
0.665
1.90
2.00
2.20
2.40
2.60
2.80
3.00
3.20
3.40
3.60
3.80
4.00
4.50
5.00
5.50
6.00
120.0
130.0
140.0
150.0
160.0
170.0
180.0
190.0
200.0
220.0
240.O
260.0
280.0
300.0
320.0
340.0
360.0
380.0
400.0
450.0
500.0
550.0
600.0
117.14
118.80
120.42
1 2 1. 9 6
123.46
124.90
70.1
z 66.7
63.3
56.9
2 50.7
2 44.7
2722.4
2724.0
2725.5
2724.6
2731.4
2733.9
0.643
0.622
0.603
0.568
0.536
0.509
39.0
2736.4
0.483
33.4
24.1
2738.7
2741 .O
0.461
0.440
22.9
6.50
7.00
0
'1
.40
1.50
1.60
1.70
1.80
0
1.225
1.149
1.083
126.28
530.5
127.62
128.89
130.13
131 .37
132.54
'1
33.69
135.88
536.1
5 4 1. 6
547.1
552.3
557.3
562.2
5 7 1. 7
138.01
140.00
1 4 1. 9 2
143.75
145.46
1 4 7. 2 0
580.7
589.2
597.4
605.3
612.9
620.0
2742.9
0.422
148.84
150.44
1 5 1. 9 6
627.1
634.0
640.7
I
17.8
z 12.9
08.1
2744.9
27 46.9
2748.4
0.405
0.389
0.374
155.55
158.92
162.08
|65.04
656.3
670.9
684.6
697.5
2096.7
2086.0
2075.7
2066.0
2753.0
2756.9
2760.3
2763.5
0.342
0.315
0.292
0.272
650.0
700.0
167.83
170.50
709.7
721.4
2056.8
2047.7
2766.5
2769.1
0.255
0.240
z
7.50
750.0
173.02
732.5
2039.2
2771.7
0.227
8.00
800.0
175.43
743.1
2030.9
2774.0
0.215
8.50
9.00
9.50
850.0
900.0
950.0
177.75
179.97
182.10
753.3
763.0
772.5
2022.9
2015.1
2 0 0 7. 5
2776.2
2778.1
2780.0
0.204
0.194
0.185
10.00
1000.0
''|
84.13
7 8 1. 6
2000.1
2 7 8 1. 7
0.177
10.50
1050.0
186.05
790.1
1993.0
2783.3
0.171
1 10 0 . 0
18 8 . 0 2
798.8
1986.0
2744.4
0.163
1. 0 0
EDB/1
splrax
' tsarco
1.01
tables
Pressure
bar
I t.5U
gauge
kPa
1150.0
Temperature
Water(h1)
kJ/kg
189.82
S p e c i f i cE n t h a l p y
Evaporation (h1n) S t e a m ( h n)
kJ/kg
kJ/kg
807.1
12 . o 0
1250
Specific
Volume
Steam
ms/kg
ffi
ft-oo
-
IJ.3U
- 1t 4
4 5- o
0o
.-I C . U U
io_
I3.JU
16-00
17 . o 0
tsJo
19-OO
20^oo
ias
E.oo
22nO
23^00
24l 0
25-00
26-OO
2 7S O
T
Z8-oo
29.o0
30.00
31.00
32'00
33.00
34^OO
35-OO
36-OO
37-OO
38-00
39-OO
40-oo
iei
#r
irffi
?loo
43-OO
44nO
45-OO
46-OO
ffi
ffi
nno
Zs.oo
49nO
50s0
51i0
52-00
53.0O
54.oO
ffi
ffi
CC.UU
56-OO
1.02
splrax
' .lsarco
EDB/1
Steam tables
Pressure
Temperature
"C
Specific Enthalpy
@kJ/ks
qauoe
kPa
bar
s700.0
57.00
58.00
5800.0
59.00
5900.0
60.00
6000.0
61.00
6100.0
62.00
6200.0
63.00
6300.0
64.00
6400.0
273.45
274.55
275.65
276.73
277.80
278.85
279.89
280.92
202.1
207.8
213.4
218.9
224.5
230.0
235.4
240.8
65.00
6500.0
2 8 1. 9 5
66.00
67.00
68.00
69.00
6600.0
6700.0
6800.0
6900.0
282.95
283.95
284.93
285.90
246.1
251.4
70.00
7 1. 0 0
72.OO
73.00
74.OO
75.00
76.00
77.OO
78.00
29.00
80.00
8 1. 0 0
82.00
83.00
84.00
7000.0
7100.u
7200.0
7300.0
7400.0
7500.0
7600.0
7700.0
7800.0
7900.0
8000.0
8100.0
8200.0
8300.0
8400.0
28ti.85
247.4o
244.7 5
289.ti9
290.tt0
291.51
292.41
293.91
294.20
2 9 5 . 10
295.96
296.81
297.66
298.50
299.35
272.'l
277 3
282.3
247.3
292.3
297.2
302.3
307.0
311.9
J t o 7
321 .5
300.20
301.00
301.81
302.61
303.41
304.20
305.77
307.24
308.83
310.32
311.79
313.24
314.67
316.08
317.46
318.83
3 2 0. 1 7
3 2 1. 5 0
322.81
324.10
32s.38
345.0
85.00
86.00
87.00
88.00
89.00
90.00
92.00
94.00
96.00
98.00
100.00
102.00
104.00
06.00
08.00
10.00
12.00
14.00
16.00
18.00
20.00
EDB/1
8500.0
8600.0
8700.0
8800.0
8900.0
9000.0
9200.0
9400.0
9600.0
9800.0
0000.0
0200.0
0400.0
10600.0
10800.0
1000.0
12 0 0. 0
1400.0
1600.0
' 18 0 0 . 0
2000.0
I
kJ/ks
584.5
577.7
571.0
564.4
557.6
550.9
544.3
537.3
5 3 1. 2
524.7
518.1
511.6
lDpeolilc;
Volume
S t e a m ( h-o ) | S t e a m
kJ/ko
| m3/ko
2786.6
2785.5
2784.4
2783.3
2782.1
2780.9
2779.7
2778.5
2777.3
2 7 76 . 1
2774.8
2773.5
0.0337
0.0331
0.0325
0.0319
0.0314
0.0308
0.0303
0.0298
0.0293
0.0288
0.0283
0.0278
501.1
498.7
492.2
485.8
479.4
473.0
4tt6.6
460.2
453.9
4 4 7. 6
44't.3
435.0
424.7
422.5
416.2
410.0
403.8
397.6
2772.1
2770.4
27b9.5
2768.'t
276b.7
2765.3
2763.A
2762.5
2760.9
2759.5
2758.0
2756.5
2754.9
2753.4
2751.9
2750.3
2748.8
2 74 7 . 2
0.0274
O.A'274
0.0266
0.0262
0.0258
0.0254
0.0250
0.0246
Q.O242
0.0239
0.0236
0.0233
0"0229
0.0226
0.0223
0.0220
0.0217
0.0214
445.9
454.3
462.6
391.3
385.2
379.0
372.7
360.3
348.0
335.7
323.3
310.9
2 9 8. 7
286.3
274.0
2 6 1. 7
249.3
237.0
2745.5
2 74 4 . O
2 74 2 . 3
2 74 0 . 5
2737.1
2733.7
2730.2
2726.5
2722.8
2 71 9. 2
2715.3
2711.5
2707.6
2703.6
2699.6
0.0211
0.0208
0.0205
0.0202
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