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 its 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 its 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:

Dry gas losses

Loss due to moisture in fuel

Loss due to moisture formed during combustion

Loss due to moisture in combustion air

Unburnt fuel loss

Loss due to radiation from Boiler to surroundings

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 (Exhaust 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;

Sulphur, 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:

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 = Unit Dry Gas x C_{p} x (T_{g}-T_{a}) x 100/HHV

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

= 5.48 %

Loss due to Moisture in fuel:

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

Lb. = Moisture in Fuel x (1089-T_{a}+0.46xT_{g}) x100/HHV

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

= 0.263 %

Loss due to Moisture Produced during combustion:

Lc = Moisture Produced x (1089-Ta+0.46xT_{g}) x100/HHV

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

= 3.23 %

Loss due to Moisture in air:

Ld. = Moisture in Air x C_{p} of Steam x (T_{g}-T_{a}) 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

C_{p} of steam = 0.46

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.

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 L_{g} = 1.5%.

Total Losses = L_{a} + L_{b} + L_{c} + L_{d} + L_{e} + L_{f} + L_{g}

= 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 = Efficiency On HHV x HHV/LHV

= 86.494 x 13101.6/12690.6

= 89.29 %

Sample sizing calculations for BFW pumps and Fans for a typical Coal fired Boiler generating steam of 50,000 Kg/hr. at 67 kg/cm^{2} and 485 deg C. (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

- Superheater Outlet :67 Kg/cm
^{2}& 485 C - Steam Drum :73 Kg/cm
^{2}& Saturated - Economizer inlet : Water inlet at 105 C

From Steam tables,

Enthalpy of Superheated steam, H_{sh} = 809 Kcal/ kg = 1456 BTU/lb

Enthalpy of Drum water, H_{dwat} = 305 Kcal/kg = 549 BTU/lb

Enthalpy of inlet water, H_{wat} = 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 Unit Wet Gas

= 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 m^{3}/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 m^{3}/hr =say 65 m^{3}/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/cm^{2} + 2.0 + 2.0 +5.0 +2.0

= 86 Kg/cm^{2}

= 86 x 10/0.95 mts of water head at 105 C

= 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 m^{3}/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 m^{3}/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 m^{3}/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, m^{3}/s = 0.3 x 17.56

= 5.27 m^{3}/s

Take 20% margin on discharge capacity. So SA Fan flow is 1.2 x 5.27 = 6.3 m^{3}/s

SA fan static head is about 630 mm WC.

Therefore, SA fan rating is 6.3 m^{3}/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 Nm^{3}/hr

= 16.6 Nm^{3}/s

Gas flow at 150 C in m^{3}/s = 16.6 x (273+150)/273 = 25.7

ID Fan capacity taking 20% margin on flow = 25.7 x 1.2

= 30 m^{3}/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 /-

Combustion calculations are starting point for Boiler or any Thermal equipment design. Here is a sample calculation 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;

Sulphur, S – 1.3;

Moisture, H2O – 3.0;

Ash – 7.0;

Excess Air: 30 %

Molecular Wt of C – 12,

O_{2} -32,

H_{2} – 2,

S – 32

The Chemical equations are as follows: C + O2 –> CO^{2}; S + O2 –> SO^{2}; H2 + O –> H2O

———————————————————————

Comp Wt. O_{2} reqd CO_{2} H_{2}O N_{2} SO_{2}

——————————————————————————————————

C 0.76 2.0267 2.7867 – – –

H_{2} 0.04 0.328 – 0.369 – –

O_{2} 0.076 (-0.076)

N_{2} 0.01 – – – 0.01 –

S 0.013 0.013 – – – 0.026

H_{2}O 0.03 – – 0.03 – –

ASH 0.07

———————————————————————-

100.0 2.2917 2.7867 0.399 0.01 0.026

From the above table, Theoretical O_{2} required = 2.2917 kg/kg of coal

Therefore, Theoretical dry Air required = 2.2917/0.23 = 9.964 kg of dry air/ kg of fuel

considering 30% Excess Air, T_{h} 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:

————————————————————————————————

Gas Weight %Wt %WT/MW %Volume

————————————————————————————————

CO_{2} 2.7867 9.832 0.4507 13.381

H_{2}O 0.567 4.035 0.2242 6.655

SO_{2} 0.026 0.185 0.0029 0.086

N_{2} 9.9837 71.055 2.5376 75.339

O_{2} 0.6875 4.893 0.1529 4.539

————————————————————————————————

14.0509 100.0 3.3654 100

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