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
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
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
[email protected] + 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
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@)
[email protected]] = 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
[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 0.0 97 0.0 92 0.0 87 0.0 83 0.0 78 0.0 74 0 . 0 70 0.0166 0.0162 0 .0 5 8 0.0 54
4 70 . 8
224.6
2655.4
0.0
479.O
212.2 199.8 187.3
2 6 9 1. 2 2687.0 2682.7
0.0 47 0.0 44 0.0 4 1
256.7 2 6 1. 9 267.0
326.2 330.9 335.7 340.3 349.6
354.2 358.8 JOJ.J
367.8 376.8 385.7 394.5 403.2 411 I
420.5 429.0 437.s
487.2 495.4
sprrax ' tsarco
1.03
