Indirect Method of finding Boiler Efficiency

Indirect method is also called as heat loss method. The efficiency is arrived at, by subtracting the heat loss fractions from 100. The standards do not include blow down loss in the efficiency determination process. 

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Principal heat losses that take place in a boiler are listed below:

•       Heat loss in dry flue gas

•       Heat loss due to evaporation of water formed due to hydrogen present in the fuel

•       Heat loss due to evaporation of moisture present in the fuel

•       Heat loss due to moisture present in the combustion air

•       Loss due to unburnt fuel in ash

•       Heat loss due to radiation and other unaccounted losses

Also, some heat is added to the boiler other than the fuel. These are called credits. Some of the credits are as follows:

•       Heat in feedwater

•       Heat in combustion air

•       Heat from auxiliary equipment

•       Sensible heat on fuel

Following steps are followed for the calculation of efficiency:

Theoretical air requirement for the combustion of fuel

4.35[(8/3C + 8H₂ + S) – O₂]/100 kg/kg of fuel.

Percentage of excess air

Excess air (EA) = O₂ x 100/(21 – O₂)

Actual mass of air supplied per kg of fuel

(1 + Excess air/100) x Theoretical air

Percentage heat loss due to dry flue gas

m x C_p(T_f – T_a) x 100/GCV of fuel

where, m is the mass of dry flue gas in kg per kg of fuel and C_p is the specific heat of flue gas (0.23 kcal/kg).

Here, T_f is flue gas temperature at boiler exit (°C) and T_a is ambient air temperature (°C).

Mass of dry flue gas can be calculated as

Mass of actual air supplied + 1(Mass of fuel supplied) – (M + 9H₂)

here, M is the percentage of moisture in 1 kg of fuel.

Heat loss due to evaporation of water formed due to hydrogen present in the fuel

9 x H₂[584 + C_p(T_f – T_a)]/GCV of fuel

where, H₂ is the percentage of hydrogen in 1 kg of fuel and C_p is the specific heat of superheated steam (0.45 kcal/kg).

Heat loss due to evaporation of moisture present in the fuel

M[584 + C_p(T_f – T_a)]/GCV of fuel

where, M is the percentage of moisture in 1 kg of fuel and C_p is the specific heat of superheated steam (0.45 kcal/kg).

Heat loss due to moisture present in the combustion air

Actual mass of air supplied x Humidity ratio of air x C_p(T_f – T_a) x 100/GCV of fuel

Loss due to unburnt fuel in ash

Ash collected per kg of fuel x GCV of Ash x 100/GCV of fuel

Heat loss due to radiation and other unaccounted losses

For a smaller boiler, this loss may be assumed as 1% to 2% and for a larger boiler, this can be assumed between 0.2% to 1%.

Calculate the boiler efficiency

Boiler efficiency = 100 – Sum of all above losses

Example

Find the efficiency of a coal-fired boiler by heat loss or indirect method.

Ultimate Analysis of Coal

H₂ = 2%

O₂ = 5%

C  = 38%

S  = 1%

Moisture = 5%

 Ash = 47%

Flue Gas Analysis (O2) = 3%

Flue Gas Temperature at Boiler Exit = 140 °C

Ambient Air Temperature  = 40 °C

Humidity Ratio = 0.04 of fuel kg/kg

Relative Humidity = 80%

GCV of Fuel = 3400 kcal/kg

Ash Generation = 0.47 kg/kg of fuel

GCV of Ash = 200 kcal/kg

Specific Heat of Flue Gas = 0.23 kcal/kg

Superheated steam = 0.45 kcal/kg

Solution

Theoretical air requirement for combustion

=(4.35/100)[{(8/3)C + 8H₂ + S} - O₂]

= (4.35/100)[{(8/3)(38) + 8 x 2 + 1} - 5]

= 4.93 kg/kg of coal

Percentage of excess air

= O₂ x [100/(21 - O₂)]

= 3 x (100/(21 - 3)

= 16.67%

Actual mass of air supplied

= (1 + Excess air %/100) x theoretical a

= (1 + 0.1667 ) x 4.93 = 5.75 kg/kg of coal

Dry flue gas loss

Mass of dry flue gas

= Mass of actual air supplied + 1 – (M + 9H₂)

= 5.75 + 1 – (0.05 + 9 x 0.02) = 6.52

Dry flue gas loss

= m x C_p(T_f  – T_a) x 100/GCV of fuel

= 6.52 x 0.23 (140 – 40) x 100/3400 = 4.41%

Heat loss due to evaporation of water

= 9 x H₂[584 + C_p(T_f – T_a)}/GCV of fuel

= 9 x 2[584 + 0.45(140 – 40)]/3400 = 3.33%

Heat loss due to evaporation of moisture present in fuel

= M[584 + C_p(T_f – T_a)]/GCV of fuel

= 5[584 + 0.45 (140 - 40)]/3400 

= 0.925%

Heat loss due to moisture present in combustion air

 = Actual mass of air supplied x Humidity ratio of air x C_p(T_f – T_a) x 100/GCV of fuel

= 5.75 x 0.04[0.45(140 – 40)] x 100/3400 

= 0.3%

Loss due to unburnt fuel in ash

= Ash collected per kg of fuel x GCV of ash x 100/GCV of fuel

= 0.47 x 200 x 100/3400 

= 2.76%

Heat loss due to radiation and other unaccounted losses

Let these be 1%.

Boiler efficiency

= 100 – Sum of all above losses

= 100 – (4.41 + 3.33 + 0.925 + 0.3 + 2.76 + 1)

= 100 – 12.88 = 87%

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