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 m2

Ditanya :

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

no C (%) H₂ (%) O₂ (%) S (%) Theoritical air

1 45 3 12,9 5,702 84,8 11,8 1,3 1,2 13,94

C = Carbon content (%)H₂ = Hydrogen content (%)O₂ = Oxygen content (%)S = Sulphur content (%)Theoretical air = Theoretical air required for combustion (kg/kg of fuel)

2. Theoretical CO₂ %

no Wt N₂ in theo air

Wt N₂ in fuel

Mol wt of N₂

Moles of N₂ C Moles of C CO₂ theo

1 4,389 0,007 28 0,157 0,45 0,0375 0,192802

Wt N₂ in theo air = Weight of Nitrogen in theoretical airWt N₂ in fuel = weight of Nitrogen in fuelMol wt of N₂ = Molecular weight of NitrogenMoles of N₂ = Mol of NitrogenC (%) = Carbon content in fuel analysisMoles of C = Mol of Carbon% CO₂ theo = Theoretical Carbondioxide

3. Excess air supplied

no % CO₂ t % CO₂ a EA no % O₂ EA

1 19,28 14 36,91 1 2 2 7,5 55,56

EA = Excess air supplied (%)% CO₂ t = Theoretical of CO₂ (%)% CO₂ a = Percent of CO₂ in flue gas

4. Actual mass of air supplied

no EA Theoritical air

Act mass supp

1 36,91 5,7 7,802 55,56 13,94 21,68

EA = Excess air suppliedAct mass supp = actual mass of air supplied

5. Mass of dry flue gas

no Mass of CO₂

Mass of N₂

Mass of N₂ in

Mass of O₂

Mass of dry flue gas

1 1,65 0,007 6,006 0,483 8,146

no Mass of CO₂

Mass of N₂

Mass of SO₂

Mass of O₂

Mass of N₂ on air supp

Mass of dry flue gas

2 3,11 0,024 0,0042 1,725 16,69 21,56

Perhitungan heat losses L1-L8

1. Heat loss due to dry flue gas

no m (kg/kg)

Cp (kCal/kg⁰C)

Tf(⁰C) Ta(⁰C) GCV of Fuel

L1

1 8,146 0,26 145 32 4100 5,842 21,56 0,26 192 32 10500 8,54

where:L1 = % Heat loss due to dry flue gasm = Mass of dry flue gas in kg/kg of fuelCp = Specific heat of flue gas in kCal/kg⁰CTf = Flue gas temperature in ⁰CTa = Ambient temperature in ⁰C

2. Heat loss due to evaporation of water formed due to H₂ in fuel (%)

no H₂ (kg)Cp (kCal/kg⁰C)

Tf(⁰C) Ta(⁰C) GCV of Fuel

L2

1 0,03 0,6 145 32 4100 4,292 0,12 0,6 192 32 10500 6,99

H₂ = kg of hydrogen present in fuel on 1 kg basisCp = Specific heat of superheated steam in kCal/kg⁰CTf = Flue gas temperature in ⁰CTa = Ambient temperature in ⁰C584 = Latent heat corresponding to partial pressure of water vapour

3. Heat loss due to moisture present in fuel

no M (kg)Cp (kCal/kg⁰C)

Tf(⁰C) Ta(⁰C) GCV of Fuel

L3

1 0,35 0,6 145 32 4100 5,562 0,0048 0,6 192 32 10500 0,03

M = kg of moisture present in fuel on 1 kg basisCp = Specific heat of superheated steam in kCal/kg⁰CTf = Flue gas temperature in ⁰CTa = Ambient temperature in ⁰C584 = Latent heat corresponding to partial pressure of water vapour

4. Heat loss due to moisture present in air

no AAS (kg) Humidity Factor

Cp (kCal/kg⁰C)

Tf(⁰C) Ta(⁰C) GCV of Fuel

L4

1 7,8 0,0204 0,6 145 32 4100 0,262 21,68 0,0204 0,6 192 32 10500 0,40

AAS = actual mass of air supplied per kg of fuelCp = Specific heat of superheated steam in kCal/kg⁰CTf = Flue gas temperatureTa = Ambient temperatureHumidity Factor = kg of water/kg of dry air

5. Heat loss due to incomplete combustion

no % CO % CO₂ C GCV of Fuel

L5

1 0,6 14 0,45 4100 2,59

CO = volume of CO in flue gas leaving economizer (%)CO₂ = actual volume of CO₂ in flue gas(%)C = carboncontentkg/kgoffuel

6. Heat loss due to radiation and covection

no Ts (K) Ta (K) Vm (m/s)

L6 (W/m²)

1 348,15 305,15 3 1013,1612 353,15 305,15 3 1161,056

L6 = Radiation loss in W/m²Vm = Wind velocity in m/sTs = 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 (%)

no Total ash collected

fuel burnt

GCV of fly ash

GCV of Fuel

L5

1 0,033 0,15 452,5 4100 0,05

Total ash = Total ash collected in kgFuel burnt = Fuel burnt in kgGCV of fly ash = Gross calorific value of fly ashGCV of fuel = Gross calorific value of fuel

8. Heat loss due to unburnt in bottom ash (%)

noTotal ash collected

fuel burnt

GCV of bottom

ashGCV of

FuelL5

1 0,033 0,85 800 4100 0,55

Total ash = Total ash collected in kgFuel burnt = Fuel burnt in kgGCV of bottom = Gross calorific value of bottom ashGCV of fuel = Gross calorific value of fuel

Summary Heat Balance

1. Boiler berbahan bakar batu bara :

Input/Output Parameter

kCal/kg of fuel

% loss

Heat Input = 4100 100Loses in boiler 1. Dry flue gas (L1) = 23944,00 5,842. Loss due to hydrogen fuel (L2) = 17589,00 4,293. Loss due to moisture in fuel (L3) = 22796,00 5,564. Loss due to moisture in air (L4) = 1066,00 0,265. Partial combustion of C to CO (L5) = 10619,00 2,596. Surface heat losses (L6) = 5125,00 1,257. Loss due to unburnt in fly ash (L7) = 205,00 0,058. Loss due to unburnt in bottom ash (L8) = 2255,00 0,55

Boiler Efficiency 79,61

2. Boiler berbahan bakar minyak :

Input/Output Parameter

kCal/kg of fuel

% loss

Heat Input = 10500 100,000Loses in boiler 1. Dry flue gas (L1) = 89689,60 8,5422. Loss due to hydrogen fuel (L2) = 73395,00 6,9903. Loss due to moisture in fuel (L3) = 315,00 0,0304. Loss due to moisture in air (L4) = 4200,00 0,4005. Partial combustion of C to CO (L5) = 0,00 0,0006. Surface heat losses (L6) = 16695,00 1,5907. Loss due to unburnt in fly ash (L7) = 0,00 0,0008. Loss due to unburnt in bottom ash (L8) = 0,00 0,000

Boiler Efficiency 82,448