86311502 effisiensi boiler
Post on 24-Sep-2015
216 Views
Preview:
DESCRIPTION
TRANSCRIPT
1.Diketahui : Sebuah PLTU berbahan bakar batu bara
Tekanan uap super heat = 98 bar
Temp uap superheat = 540 oC
Produksi uap = 150 ton/jam
Temp feed water = 215 oC
Fuel firing rate = 27200 kg/jam
Temp flue gas = 145 OC
Temp permukaan boiler = 75 OC
Teperatur udara ambient = 32 OC
Humidity ratio udara ambient = 0,0204 kg/kg dry air
Kecepatan udara sekeliling boiler = 3 m/s
Luas permukaan total boiler = 1600 m2 Ratio bottom ash to fly ash = 85 : 15
Kandungan CO2 dalam flue gas = 14 %
Kandungan CO dalam flue gas = 0,6 %
Kandungan abu = 3,3 %
Moisture dalam batu bara = 35 %
Carbon content = 45 %
Hydrogen content = 3 %
Nitrogen content = 0,7 %
Oxygen content = 12,9 %
GCV batu bara = 4100 kCal/kg
GCV bottom ash = 800 kCal/kg
GCV fly ash = 452,5 kCal/kg
Ditanya : 1. Efisiensi dengan direct method (apakah 91%)2. Efisiensi dengan indirect method
2. Diketahui :
Sebuah PLTU berbahan bakar minyak
Ultimate analysis :
Carbon = 84,8 %
Hydrogen = 11, 8 %
Nitrogen = 0,42 %
Oxygen = 1,3 %
Sulphur = 1,2 %
Moisture = 0,48 %
GCV = = 10500 kCal/kg
Flue gas analysis :
Temperatur flue gas keluar boiler = 192 OC
Kandungan CO2 dalam flue gas = 10,6 %
Kandungan O2 dalam flue gas = 7,5 %
Temp udara ambient = 32 OC
Humidity ratio udara ambient = 0,0204 kg/kg dry air
Kecepatan udara di sekeliling boiler = 3 m/s
Temp permukaan boiler = 80 oC
Luas permukaan total boiler = 1600 m2Ditanya :
1. Laju bahan bakar bila diinginkan eff boiler 91 %
2. Efisiensi boiler dengan menggunakan indirect method
Jawab :
1. Effisiensi boiler batu bara
Boiler efficiency = (Q x (H-h) x100) / (q x GCV)
Q = 150000 kg/jam
H = 830,594 kCal/kg
h = 220,008 kCal/kg
q = 27200 kg/jam
GCV = 4100 kCal/kg
Boiler efficiency = 82 %
Jadi Efisiensi boiler dengan direct method tidak sesuai dengan yang diklaim oleh pembuat boiler.
2. Effisiensi boiler berbahan bakar minyak :
Boiler efficiency = (Q x (H-h) x100) / (q x GCV)
Boliler eff = 91 %
Q = 150000 kg/jam
H = 830,594 kCal/kg
h = 220,008 kCal/kg
GCV = 10500 kCal/kg
q = 9585,34 kg/jam = 9,58534 ton/jam
Perhitungan menggunakan indirect method :
Perhitungan stoichiometric (berlaku juga untuk soal no.2)1. Theoretical air required for combustion
noC (%)H (%)O (%)S (%)Theoritical air
145312,95,70
284,811,81,31,213,94
C = Carbon content (%)
H = Hydrogen content (%)
O = Oxygen content (%)
S = Sulphur content (%)
Theoretical air = Theoretical air required for combustion (kg/kg of fuel)
2. Theoretical CO %
noWt N in theo airWt N in fuelMol wt of NMoles of NC Moles of CCO theo
14,3890,007280,1570,450,03750,192802
Wt N in theo air = Weight of Nitrogen in theoretical air
Wt N in fuel = weight of Nitrogen in fuel
Mol wt of N = Molecular weight of Nitrogen
Moles of N = Mol of Nitrogen
C (%) = Carbon content in fuel analysis
Moles of C = Mol of Carbon
% CO theo = Theoretical Carbondioxide
3. Excess air supplied
no% CO t% CO aEAno% O EA
119,281436,911
227,555,56
EA = Excess air supplied (%)
% CO t = Theoretical of CO (%)
% CO a = Percent of CO in flue gas
4. Actual mass of air supplied
noEATheoritical airAct mass supp
136,915,77,80
255,5613,9421,68
EA = Excess air supplied
Act mass supp = actual mass of air supplied
5. Mass of dry flue gas
noMass of COMass of N Mass of N in Mass of O Mass of dry flue gas
11,650,0076,0060,4838,146
noMass of COMass of N Mass of SO Mass of O Mass of N on air suppMass of dry flue gas
23,110,0240,00421,72516,6921,56
Perhitungan heat losses L1-L8
1. Heat loss due to dry flue gas
nom (kg/kg)Cp (kCal/kgC)Tf(C)Ta(C)GCV of FuelL1
18,1460,261453241005,84
221,560,2619232105008,54
where:
L1 = % Heat loss due to dry flue gas
m = Mass of dry flue gas in kg/kg of fuel
Cp = Specific heat of flue gas in kCal/kgC
Tf = Flue gas temperature in C
Ta = Ambient temperature in C
2. Heat loss due to evaporation of water formed due to H in fuel (%)
noH (kg)Cp (kCal/kgC)Tf(C)Ta(C)GCV of FuelL2
10,030,61453241004,29
20,120,619232105006,99
H = kg of hydrogen present in fuel on 1 kg basis
Cp = Specific heat of superheated steam in kCal/kgC
Tf = Flue gas temperature in C
Ta = Ambient temperature in C
584 = Latent heat corresponding to partial pressure of water vapour
3. Heat loss due to moisture present in fuel
noM (kg)Cp (kCal/kgC)Tf(C)Ta(C)GCV of FuelL3
10,350,61453241005,56
20,00480,619232105000,03
M = kg of moisture present in fuel on 1 kg basis
Cp = Specific heat of superheated steam in kCal/kgC
Tf = Flue gas temperature in C
Ta = Ambient temperature in C
584 = Latent heat corresponding to partial pressure of water vapour
4. Heat loss due to moisture present in air
noAAS (kg)Humidity FactorCp (kCal/kgC)Tf(C)Ta(C)GCV of FuelL4
17,80,02040,61453241000,26
221,680,02040,619232105000,40
AAS = actual mass of air supplied per kg of fuel
Cp = Specific heat of superheated steam in kCal/kgC
Tf = Flue gas temperature
Ta = Ambient temperature
Humidity Factor = kg of water/kg of dry air
5. Heat loss due to incomplete combustion
no% CO% COCGCV of FuelL5
10,6140,4541002,59
CO = volume of CO in flue gas leaving economizer (%)
CO = actual volume of CO in flue gas(%)
C = carboncontentkg/kgoffuel
6. Heat loss due to radiation and covection
noTs (K)Ta (K)Vm (m/s)L6 (W/m)
1348,15305,1531013,161
2353,15305,1531161,056
L6 = Radiation loss in W/m
Vm = Wind velocity in m/s
Ts = Surface temperature (K)
Ta = Ambienttemperature(K)
Perhitungan L7 dan L8 hanya berlaku pada soal 1 (bahan bakar batu bara)
7. Heat loss due to unburnt in fly ash (%)
noTotal ash collectedfuel burntGCV of fly ashGCV of FuelL5
10,0330,15452,541000,05
Total ash = Total ash collected in kg
Fuel burnt = Fuel burnt in kg
GCV of fly ash = Gross calorific value of fly ash
GCV of fuel = Gross calorific value of fuel
8. Heat loss due to unburnt in bottom ash (%)
noTotal ash collectedfuel burntGCV of bottom ashGCV of FuelL5
10,0330,8580041000,55
Total ash = Total ash collected in kg
Fuel burnt = Fuel burnt in kg
GCV of bottom = Gross calorific value of bottom ash
GCV of fuel = Gross calorific value of fuel
Summary Heat Balance
1. Boiler berbahan bakar batu bara :
Input/Output ParameterkCal/kg of fuel% loss
Heat Input=4100100
Loses in boiler
1. Dry flue gas (L1)=23944,005,84
2. Loss due to hydrogen fuel (L2)=17589,004,29
3. Loss due to moisture in fuel (L3)=22796,005,56
4. Loss due to moisture in air (L4)=1066,000,26
5. Partial combustion of C to CO (L5)=10619,002,59
6. Surface heat losses (L6)=5125,001,25
7. Loss due to unburnt in fly ash (L7)=205,000,05
8. Loss due to unburnt in bottom ash (L8)=2255,000,55
Boiler Efficiency79,61
2. Boiler berbahan bakar minyak :
Input/Output ParameterkCal/kg of fuel% loss
Heat Input=10500100,000
Loses in boiler
1. Dry flue gas (L1)=89689,608,542
2. Loss due to hydrogen fuel (L2)=73395,006,990
3. Loss due to moisture in fuel (L3)=315,000,030
4. Loss due to moisture in air (L4)=4200,000,400
5. Partial combustion of C to CO (L5)=0,000,000
6. Surface heat losses (L6)=16695,001,590
7. Loss due to unburnt in fly ash (L7)=0,000,000
8. Loss due to unburnt in bottom ash (L8)=0,000,000
Boiler Efficiency82,448
top related