Sample Boiler Calculations

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

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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.

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)

Calculations of the Losses based on Higher Heating Value:

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 %

article from : http://www.firecad.net/

Boiler Economizers

A boiler economizer is a device that reduces the overall fuel requirements a boiler requires which results in reduced fuel costs as well as fewer emissions - since the boiler now operates at a much higher efficiency. Boiler economizers recover the "waste heat" from the boiler's hot stack gas from transfers this waste heat to the boiler's feed-water. Because the boiler feed-water is now at a higher temperature that it would have been without a boiler economizer, the boiler does not need to provide as much additional heating to produce the steam requirements of a facility or process, thereby using less fuel and reducing the fuel expenses. Boiler economizers also help improve a boiler's efficiency by extracting heat from the flue gases discharged from the final super-heater section of a radiant/reheat unit or the evaporative bank of a non-reheat boiler. Heat is transferred, again, back to the boiler feed-water, which enters at a much lower temperature than saturated steam.

Boiler Economizers are a series of horizontal tubular elements and can be characterized as bare tube and extended surface types. The bare tube includes varying sizes which can be arranged to form hairpin or multi-loop elements. Tubing forming the heating surface is generally made from low-carbon steel. Because steel is subject to corrosion in the presence of even low concentrations of oxygen, water must be practically 100 percent oxygen free. In central stations and other large plants it is common to use deaerators for oxygen removal.