Boiler Efficiency calculations

Efficiency is a very important criterion in Boiler selection and Design. Efficiency figure depends upon the type of boiler as well as on the type of fuel and it’s constituents. For example, efficiency of a Bagasse fired boiler is about 70% where as that of oil fired boilers is about 85 %. Higher moisture content in Bagasse reduces it’s efficiency. So better criterion is efficiency based on LCV or NCV. This is widely used in Europe and efficiency based on HHV or GCV is used in other parts of the world.

There are basically two methods to calculate efficiency of the boilers : Input-Output method and Heat Loss method. In Input-output method, boiler must be in steady running condition and the data of heat input in the form of  fuel and air and heat output in the form of steam and other losses is taken.

Here we are going to discuss the second and more popular method. In this method, first we calculate the heat input. Then all heat losses are calculated. Effective heat output is heat input less the heat losses. Output to Input ratio gives the efficiency.

 Heat losses in fired boiler are :

 a)     Dry gas losses

b)     Loss due to moisture in fuel

c)     Loss due to moisture formed during combustion

d)     Loss due to moisture in combustion air

e)     Unburnt fuel loss

f)       Loss due to radiation from Boiler to surroundings

g)     Manufacturers Margin OR unaccounted losses

 
Sample Case :

 Let us calculate Boiler efficiency of coal fired boiler. Ambient temp is 80 F and Back End Temperature (Exh gas temp) is 302 F. The percent composition of Coal is as under:

Carbon , C - 76.0 ; Hydrogen, H₂ - 4.1 ; Nitrogen , N₂ - 1.0 ;   Oxygen, O₂ - 7.6 ;   Suphur, S - 1.3 ;   Moisture, H₂O - 3.0 ;   Ash - 7.0 ;   

 
The Combustion calculations of the above fuel is already explained in detail in the other article.

From the above calculations, Unit Wet Gas, Kg / Kg of fuel = Unit Wet Air + (1-Ash)

                                      = 13.12 + (1-0.007)

                                      = 14.05

Unit Dry Gas, Kg / Kg of fuel = Unit Wet Gas – (Moisture in Air + Water produced  during combustion)

                                      = 13.484

Higher Heating Value, HHV or Gross Calorific Value, GCV in BTU/Lb

            =  14600.C  + 62000 (H2-O2/8) + 4050.S

 Lower Heating Value, LHV or Lower Calorific Value, LCV or Net Calorific Value, NCV, BTU/lb

           = HHV – 1030(9.H2 + Moisture)

 Let us use HHV and LHV notation.

            HHV =  (14600 x 76 +62000 (4.1-7.6/8) + 4050 x 1.3 )/100

                   = 13101.65 BTU/lb (7278.7 Kcal/kg )

          LHV =  13101.65 – 1030(9*4.1+3)/100

                  = 12690.6 BTU/lb  (7050 Kcal/kg)

a)     Dry gas losses:

Exhaust gases always leave the boiler at a higher temp than ambient. Heat thus carried away by hot exhaust gases is called Dry gas losses

Heat Losses, La = UnitDryGas x Cp x (Tg-Ta) x 100/HHV

                       = 13.478 x 0.24 x (302 -80) x 100 / 13101.65

                       = 5.48 %

  b)     Loss due to Moisture in fuel :

  The moisture present in the fuel absorbs heat to evaporate and get superheated to exit gas temperature.

  Lb =  MoistureInFuel x (1089-Ta+0.46xTg)x100/HHV

     = 0.03 x (1089 – 80 +0.46 x 302) x100 / 13101.6

     = 0.263 %

 

c)     Loss due to Moisture Produced during combustion :

Lc =  MoistureProduced x (1089-Ta+0.46xTg)x100/HHV

     = 0.369 x (1089 – 80 +0.46 x 302) x100 / 13101.6

     = 3.23 %

 
 

d)     Loss due to Moisture in air :

 
 

Ld =  MoistureInAir x Cp of Steam x (Tg-Ta) x 100/HHV

     = 0.0132 x 12.95 x  0.46 x (302 - 80) x100 / 13101.6

     = 0.133 %

 

     Here, Moisture in Air = 0.0132 lb/ lb of dry air at 60% Relative Humidity

              Cp of steam = 0.46

 

e)     Unburnt fuel loss :

 

This is purely based on experience. Unburnt fuel loss depends up on type of Boiler , grate, grate loading  and type of fuel. For Bio-Mass fuels, it ranges from 1.5 to 3 %, for oils from 0-0.5 and almost nil for gaseous fuels.

 
Let us consider Unburnt fuel loss, Le = 2.5 % for Coal.

  
f)       Radiation Loss:

 

Radiation Loss is because of hot boiler casing loosing heat to atmosphere. ABMA chart gives approximate radiation losses for fired boilers.

 

Let us take a radiation Loss , Lf = 0.4 % in this case.

 

      g) Manufacturer’s margin :

 

This is for all unaccounted losses and for margin. Unaccounted losses are because of incomplete combustion carbon to CO, heat loss in ash ..etc. This can be 0.5 to 1.5 % depending up on fuel and type of boiler.

 

In this case, let us take, Manufacturer’s margin Lg = 1.5%.

 

Total Losses =  La + Lb + Lc + Ld + Le + Lf + Lg

 

                   = 5.48 + 0.263 + 3.23 + 0.4 +0.133 +2.5 + 1.5

 

                    = 13.506 %

 

Therefore, Efficiency of the boiler on HHV basis = 100 – Total Losses

                                      = 100 – 13.506

                                      = 86.494 %

Efficiency based on LHV:

Efficiency based on LHV =  EfficiencyOnHHV x HHV/LHV

                                 =  86.494 x 13101.6/12690.6

                                 =  89.29 %

1.Convert actual steam rating into From and At 100 C

Steam capacity from and at 100 C (212 F) is equivalent steam capacity if operating conditions are reduced to atmospheric pressure.

          

Steam capacity = 8000 kg/hr at 10.5 kg/cm2 saturated
Feed water inlet = 30 C
Heat load = 8000 (664-30)  Kcal/hr
        = 5.072e06 Kcal/hr  = 20.127e06 btu/hr = 5.8976 MW

          where Sat steam enthalpy = 664 Kcal/kg
        Inlet water enthalpy  = 30 Kcal/kg
Steam enthalpy at 100C and 1 atm pressure = 540 Kcal/kg

Therefore, steam capacity F&A 100 C = 5.072e06/540
                               =  9392 Kg/hr


2. Heat Duty Calculations :

Let us calculate heat duty of a boiler generating 50,000 kg/hr at 65 bar and 485 C
Water inlet temperature  = 105 C

Steam & water properties:

Superheated steam enthalpy at 65 bar & 485 C = 808 Kcal/kg
Saturated water enthalpy =  295 Kcal/kg

Heat Duty     = 50000 x (808 - 105)
= 35.15e06 Kcal/hr (139.48e06 Btu/hr or 40.87 MW)

Usually 1 – 3% of the water flow is used for blowdown.

Considering 2% blow down , heat in blowdown water = 50000 x 0.02 x (295 – 105)
                                  = 0.19e06 kcal/hr

Total heat duty  = (35.15 + 0.19) e06 = 35.34e06 kcal/hr
                         =  140.24e06 Btu/hr = 41.09 MW

In case of Hot water generator or hot water boiler,

Heat duty = Water flow x Cp of water x Temp gain

For example, 200,000 kg/hr of water is heated from 70 to 90 degC,

Heat Load = 200,000 x 1 x (90-70)
         =  4.0e06 Kcal/hr
         =  15.873e06 Btu/hr or 4.651 MW

3. Heat Transfer calculations:

Over all heat transfer coefficient,

Uo  = 1/(1/Ho+Rm+1/Hi*(TubeOD/TubeID)+Ro+Ri*(TubeOD/TubeID))

Where Ho = Outside heat transfer coefficient
    Hi =  Inside heat transfer coefficient
          Rm = tube metal resistance
          Ro  =  Fouling resistance on outside tubes
          Ri  =  Fouling resistance on inside tubes

Inside Heat Transfer coefficient can be calculated using the following correlation :

NuInside=0.023* (ReInside^0.8)*(PrInside^0.4)

        Where NuInside = Hi x TubeID / Gas Cond

Outside heat transfer coefficient during boiling is very high and so resistance offered is negligibly small. There are many correlations available to predict Ho, but Ho can be safely assumed to be about 10000 Kcal/hr/m2/C.

Sample sizing calculations for BFW pumps and Fans for a typical Coal fired Boiler generating steam of 50,000 Kg/hr at 67 kg/cm2 and 485 degC. (110,000 lb/hr at 950 PSI & 905 F). Feed Water inlet at 105 C and Exhaust gas temp at 150 C.

Let us first calculate heat load and fuel consumption of the above boiler.

Pressure and temp at 1. Superheater Outlet :  67 Kg/cm2 & 485 C
             2. Steam Drum        :  73 Kg/cm2 & Saturated
             3.  Economizer inlet    :   Water inlet at 105 C

From Steam tables,

Enthalpy of Superheated steam , Hsh = 809 Kcal/ kg = 1456 BTU/lb
Enthalpy of Drum water , Hdwat         =  305 Kcal/kg = 549 BTU/lb
Enthalpy of inlet water   , Hwat          =  105 Kcal/kg  =    189 BTU/lb

Assume 3% Blowdown from Boiler.

Total Heat Load of the Boiler = Total heat absorbed by water to convert to steam + heat absorbed to get superheated + Blow down losses
            = 50000(809-305) + 50000 x 1.03 x (305-105)
            = 35.5e06 Kcal/hr = 140.87e06 BTU/hr

Fuel consumption = Heat Load/ (HHV x Efficiency)
              =  35.5e06/ (7278 x 0.8649)
              =  5639 Kg/hr = 12428 Lb/hr of coal

From previous article on Combustion and efficiency,
        Wet gases = 14.05 and Air = 13.12 kg / kg of coal
        

Therefore, Exhaust gases produced = Fuel consumption x UnitWetGas
       = 5639 x 14.05
                  =  79,228  Kg/hr of wet gases
Combustion Air required   =  5639 x 13.12
                  =  73,984 Kg/hr of combustion air

Feed Water required           =  50,000 x 1.03    : 3% Blowdown
                  =   51,00 Kg/hr

Sizing Calculations :

a)    Boiler feed Water Pumps :

Two pumps of 100 % capacity are required one for working and one for standby.

Each pump discharge capacity minimum= 51500 Kg/hr
                    = 51500/950        : 950 kg/m3 water density
                    = 53.8 m3/hr
Margin on discharge capacity : 15- 25 %.
Take 20% margin in this case.

So discharge capacity of each pump : 53.8 x 1.2
                    = 64.6 m3/hr =say 65 m3/hr

If Recirculation valves are not provided, you need to add min recirculation flow to the above figure, which may be about 6-10 m3/hr depending up on pump type and make.

Pump head required = Drum Pressure + Drum elevation + Piping Losses + Control Valve Loss + Other valve losses

            = 75 Kg/cm2 + 2.0 + 2.0 +5.0 +2.0
            = 86 Kg/cm2
            = 86 x 10/0.95 mts of water head at 105C
            = 905 mts of WC

Provide up to 5% margin on head. So final Pump head is 905 x 1.05 = 950 m of WC

So BFW pumps (2 nos) rating is 65 m3/hr at 950 m of WC with feed water at 105 C.

b) Sizing calculations of FD Fan :

Forced Draft Fan is required to pump in primary combustion Air into the Boiler furnace. Air from FD fan passes through Air Heater before entering furnace through Grate. Secondary Air Fan (SA fan) supplies secondary combustion air in to the furnace. Usually primary air is 70 -80 % of the total air and balance is supplied as secondary air through SA fan. Secondary air is supplied at a higher pressure to help fuel spreading on the grate called as pneumatic spreading.

Total combustion Air, Kg/hr = 73,984
= 73994/(1.17 x 3600) m3/s     :Air density-1.17kg/m3
= 17.56 m3/s

Primary Air , 70% of total , m3/s    = 0.7 x 17.56
                     =  12.3 m3/s

Take 20% margin on discharge capacity. So FD Fan flow is 1.2 x 12.3 = 14.76 m3/s

Head required = Draft loss across Air Heater + Grate + Ducting & others
         = 75 mmWC + 75 + 50 mm    : Approximate
         = 200 mm WC  (approximate)
Take 15-20 % margin on head.  So FD fan head should be about 230 mm of WC.

Therefore, FD fan rating is 15 m3/s of air at 230 mm WC static head.

Power requirements of FD Fan :

Let us assume Fan efficiency as 75% and Motor Efficiency as 90%.

Power required for FD Fan, BHP = Flow x Head / (Efficiency x 75.8)
                      =  15 x 230 / (0.75 x 75.8)
                = 60.7 HP

Motor HP required = 60.7 / 0.9  = 68 HP


Annual cost of operation assuming 7 cents per KWH and 7200 hrs of operation per annum. 0.74 is factor for converting HP to KW. Pl note that unit Electricity charges vary widely across different countries.

= 68 x 0.74 x 0.07 x 7200
= $ 25, 362 /-



c) Sizing calculations of SA Fan :

Secondary Air Fan (SA fan) supplies secondary combustion air in to the furnace.

Secondary Air , 30% of total , m3/s    = 0.3 x 17.56
                     =  5.27 m3/s

Take 20% margin on discharge capacity. So SA Fan flow is 1.2 x 5.27 = 6.3 m3/s

SA fan static head is about 630 mm WC.

Therefore, SA fan rating is 6.3 m3/s of air at 650 mm WC static head.

Power requirements of SA Fan :

Let us assume Fan efficiency as 70% and Motor Efficiency as 90%.

Power required for FD Fan, BHP = Flow x Head / (Efficiency x 75.8)
                      =  6.3 x 650 / (0.7 x 75.8)
                = 77.1  HP

Motor HP required = 77.1 / 0.9  = 86 HP


Annual cost of operation assuming 7 cents per KWH and 7200 hrs of operation per annum. 0.74 is factor for converting HP to KW. Pl note that unit Electricity charges vary widely across different countries.

= 86 x 0.74 x 0.07 x 7200
= $ 32,075 /-



d) Sizing calculations of ID Fan :

Induced draft fan or ID Fan is required to evacuate the exhaust gases from Boiler to atmosphere through Duct collectors and chimney. Usually ID should take care of draft loss across the Boiler from furnace to Air heater and then draft loss across Duct Collectors like ESP, Wet Scrubber or mechanical type Cyclone dust collectors .etc.
Total wet gases, Kg/hr = 79,228

Gas Density          =  1.3265 Kg/Nm3

Therefore, gas flow in Nm3/hr = 79,228 / 1.3265
                    =  59227 Nm3/hr
                    =  16.6 Nm3/s

Gas flow at 150C in m3/s = 16.6 x (273+150)/273  = 25.7

ID Fan capacity taking 20% margin on flow = 25.7 x 1.2
                          =  30 m3/s


ID Fan static Head = Draft Loss in (Boiler + Duct + Dust collector)
               =  150 + 50 + 50 mm WC    :  Approximate
               =  250 mmWC

Taking 20% margin on head, ID Fan head = 250 * 1.2 = 300 mm WC


Power requirements of ID Fan :

Let us assume Fan efficiency as 75% and Motor Efficiency as 90%.

Power required for ID Fan, BHP = Flow x Head / (Efficiency x 75.8)
                      =  30 x 300 / (0.75 x 75.8)
                = 158 HP

Motor HP required = 158 / 0.9  = 175 HP


Annual cost of operation assuming 7 cents per KWH and 7200 hrs of operation per annum. 0.74 is factor for converting HP to KW. Pl note that unit Electricity charges vary widely across different countries.

= 175 x 0.74 x 0.07 x 7200 = $ 65,268 /-

Sample Combustion Calculations by weight method

 

Combustion calculations are starting point for Boiler or any Thermal equipment design. Here is a sample calcualtion done for Coal with a calorific value of approximately 7200 Kcal/kg (12960 BTU/Lb). 

Coal composition in % is as follows :

Carbon , C - 76.0 ; Hydrogen, H₂ - 4.1 ; Nitrogen , N₂ - 1.0  ;   Oxygen, O₂ - 7.6 ;   Suphur, S - 1.3 ;   Moisture, H₂O - 3.0 ;   Ash - 7.0 ;   

Excess Air : 30 %

Molecular Wt of C - 12 ,  O2 -32 , H2 - 2, S - 32

The Chemicl equations are as follows: C + O2 --> CO2 ; S + O2 --> SO2 ; H2 + O --> H2O

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Comp    Wt.                    O2 reqd     CO2      H2O      N2        SO2      

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C        0.76                  2.0267      2.7867      -         -         -         

H2      0.041                  0.328         -         0.369      -         -         

O2       0.076                 (-0.076)

N2       0.01                     -               -            -      0.01      -

S       0.013                  0.013            -            -      -        0.026      

H2O    0.03                    -               -         0.03         -       -       

ASH    0.07

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From the above table, Theoritical O2 required  = 2.2917 kg/kg of coal

There fore, Theoritical dry Air required = 2.2917/0.23 = 9.964 kg of dry air/ kg of fuel

considering 30% Excess Air , Th air reqd = 9.964 x 1.3 = 12.9532 kg/kg

Wet air required (assuming 60% RH)     = 12.9532 x 1.013 = 13.12 kg/kg of fuel

Gas composition :

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Gas     Weight               %Wt                  %WT/MW           %Volume

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CO2    2.7867                  19.832               0.4507               13.381

H2O    0.567                    4.035                 0.2242               6.655

SO2    0.026                    0.185                 0.0029               0.086

N2     9.9837                  71.055                2.5376               75.339

O2    0.6875                  4.893                 0.1529              4.539

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      14.0509                  100.0                  3.3654                  100

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