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Boiler Efficiency Computation , Assessment & Factors

Affecting Efficiency

Dr. D. Banerjee

[email protected]

[email protected]

Boiler Performance Computations Factors affecting Performance Field Tests for Evaluation & Diagnosis Unit Heat Rate Specific coal consumption CONTENT

Boiler Efficiency

Indirect method or Loss method – The efficiency of a boiler equals 100% minus the losses. Thus, if the losses are known the efficiency can be derived easily. This method has several advantages: 1.Errors are not so significant; for example. if the losses total 10% then an error of 1.0% will affect the result by only 0.1 %,

2. The individual components losses are quantified. The losses method is now the usual one for boiler efficiency determination. Boiler Efficiency = 100 - Losses in % The unit of heat input is the higher heating value per kg of fuel. Heat losses from various sources are summed & expressed per kg of fuel fired.

Different Boiler Losses

Dry flue gas loss Calculation The only components of a fuel which burn to form dry products of combus- tion are the carbon and sulphur. Of these two carbon has the greater significance, so for the present ignore the sulphur. The carbon can burn to either carbon dioxide or carbon monoxide thus: (a) Carbon to carbon dioxide C + 02 = CO So masses = 12 + 32 = 44 44 kg of C02 contains 12 kg carbon. So 1 kg of CO2 contains12/44kg of carbon, i.e. 3/11 kg carbon. (b) Carbon to carbon monoxide 2C + 02 = 2C So masses = 24 + 32 = 56. 56 kg of CO contains 24 kg carbon. So 1 kg of CO contains 24/56 kg of carbon, i.e. 3/7 kg of carbon.

Dry flue gas loss

• (^) Total dry flue gas = kg carbon x (dry flue gas/kg carbon burned) Total Dry flue gas
• (^) So dry flue gas/kg carbon burned = ----------------------------- kg carbon in flue gas The total dry flue gas consists of the sum of all the dry constituents, i.e. CO2% + 02% + N2% +CO%, and they will add up to 100 kg mol. For example, suppose the percentage by volume of the various constituents is: CO2 15.0%; 02 4.4%; N2 80.5%; CO 0.l% Then the Relative Mass of each is: Contituent Vol% Moecular mass Relative Mass CO2 15.0 × 44 = 660 O2 4.4 × 32 = 140, N2 80,5 × 28 = 2254. CO 0.1 × 28 = 2.

100 3057. 

Therefore 3057.6kg of gas=100kgmole

Dry flue gas loss Carbon burned= C/100 - C in A Where, C = % carbon in the fuel. C in A = carbon in rough ash and dust, in kg/kg fuel 100 Dry flue gas = -----------------------( C/100 - C in A) kg moles/per kg fuel 12 (CO2 + CO) There is a further complication. S in fuel is almost all burned to SO2. Normally this effect can be ignored unless the sulphur content is very high, but if it is desired to allow for it the expression becomes: 100 Dry flue gas = -----------------------( C/100 + S/267 - C in A) kg moles/per kg fuel 12(CO2 + CO) Where S = % sulphur in fuel. The ratio of the atomic weights of carbon to sulphur = 1/2. Equation for mass of dry flue gas may be simplified where CO in flue gas is less than 50ppm by ignoring CO Mass of dry flue gas = (C+ S / 2.67 – 100* C in A ) / 12 CO 2 kg moles/per kg fuel The sensible heat loss per unit mass of fuel Sh = dry flue gas x kg mol Cp gas x (T - t) kJ/kg. Where kg mol Cp = kilogram molecular specific heat = 30.6 kJ/kg mol T = A/H gas outlet temperature in C. t = Temperature at F.D. duct inlet in C Dry Flue Gas Loss % = (Sh / (GCV of Fuel * 4.2)*100)

Wet Flue gas Loss Calculation The wet products of combustion are derived from the moisture and the hydrogen in the fuel. The combustion of hydrogen is represented by: 2H2 + 02 = 2H expressed as masses 4 + 32 = 36 So the combustion of 1 kg of hydrogen produces 9 kg of moisture. WFG (%) = (M+9H)SW/(GCV4.2) SW = [1.88(T-25)+2442+4.2(25-t)] kJ/kg** where SW is sensible of moisture per kg of fuel M = % moisture per kg fuel ( 12.7%) H = % hydrogen per kg fuel (3.2%) T = Air heater gas outlet temperature (C) (143C) t = Air temperature at F.D. intake (C) (34C) GCV= 4850 kcal/kg

13 Combustible in ash loss (kJ/kg of fuel) = cA

• 33820 100 c= % of carbon in dry ash A= Mass of ash kg/kg of fuel CV of carbon burnt to CO 2 = 33820 kJ/kg (8077.8 kcal/kg) or Loss due to Unburned carbon in ash (kcal/kg)= [cA8077.8] 100 Loss in (%) = [cA8077.8]100 = [cA8077.8] 20% of total ash constitute as bottom ash 80% of total ash constitute as fly 100*GCV GCV ash to be considered for computing (c) carbon percentage in dry ash

Compute Boiler efficiency loss % due to C in Ash Where c = 0.8 % , Ash

35% & GCV is 3500

U n b u r n e d g as (CO) l o s s ( kJ / k g fuel)=

23717 kJ/kg = CV of burning 1 kg of carbon in CO to CO 2 CO 2 , CO=% volume in dry gas C, S=% in fuel

S in _ A

28 3 ( CO

* [

7 CO  C )] * 23717

kj/kg

C

CO

Loss due to C in ash=0.65%

Other Losses…

3. Coal Mill Reject

Loss due to Mill Rejects = X / (Coal Flow * GCV * 1000)

X = [Rejects * (CVREJECT + CpREJECT (Tmillout – Trai))* 100 ]

Where

• (^) Coal Flow
• (^) Coal Mill Rejects
• (^) GCV of Coal
• (^) CV of Rejects
• Mill Outlet Temp Tmillout
• (^) Reference Temperature Trai
• (^) Specific Heat of Rejects CpREJECT

Factors affecting Boiler efficiency include

• (^) Design
• (^) Coal Quality
• (^) Operating parameters

➢ Mill Performance - PF Fineness

➢ Burner-to-burner PF balance

➢ Excess Air Level

➢ Water Chemistry

➢ Boiler loading

• (^) Component condition

➢ AH Performance

➢ Boiler Air Ingress

➢ Furnace / Convective section Cleanliness

➢ Insulation

➢ Quality of Overhauls

Efficiency Vs Moisture . in Coal Assumptions Exit Gas Temp - Constt. Fuel Moisture - 20.5 % Excess Air - 20 % GCV - 3700 kal/kg