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