Boiler Efficinecy Calculation
Thermal efficiency of a boiler is defined as “the percentage of heat energy input that is effectively useful in to generated steam.”
There are two methods of assessing boiler efficiency:
1. The Direct Method:
The energy gain of the working fluid (water and steam) is compared with the energy content of the boiler fuel .
This is also known as ‘input-output method’ due to the fact that it needs only the useful output (steam) and the heat input (i.e. fuel) for evaluating the efficiency. This efficiency can be evaluated using the formula:
Parameters to be monitored for the calculation of boiler efficiency by direct method are:
And where
– hg – Enthalpy of saturated steam in kcal/kg of steam
– hf – Enthalpy of feed water in kcal/kg of water
– Needs few instruments for monitoring
– Easy to compare evaporation ratios with benchmark figures
Disadvantages of direct method
– Does not give clues to the operator as to why efficiency of the system is lower
– Does not calculate various losses accountable for various efficiency levels
2. The Indirect Method:
The efficiency is the difference between the losses and the energy input.
The reference standards for Boiler Testing at Site using the indirect method are the British Standard, BS 845:1987 and the USA Standard ASME PTC-4-1 Power Test Code Steam Generating Units.
The indirect method is also called the heat loss method. The efficiency can be calculated by subtracting the heat loss fractions from 100 as follows:
Efficiency of boiler (n) = 100 – (i + ii + iii + iv + v + vi + vii)
Whereby the principle losses that occur in a boiler are loss of heat due to:
i. Dry flue gas
ii. Evaporation of water formed due to H2 in fuel
iii. Evaporation of moisture in fuel
iv. Moisture present in combustion air
v. Unburnt fuel in fly ash
vi. Unburnt fuel in bottom ash
vii. Radiation and other unaccounted losses
Losses due to moisture in fuel and due to combustion of hydrogen are dependent on the fuel, and cannot be controlled by design.
The data required for calculation of boiler efficiency using the indirect method are:
– Ultimate analysis of fuel (H2, O2, S, C, moisture content, ash content)
– Percentage of oxygen or CO2 in the flue gas
– Flue gas temperature in oC (Tf)
– Ambient temperature in oC (Ta) and humidity of air in kg/kg of dry air
– GCV of fuel in kcal/kg
– Percentage combustible in ash (in case of solid fuels)
– GCV of ash in kcal/kg (in case of solid fuels)
A detailed procedure for calculating boiler efficiency using the indirect method is given below. However, practicing energy managers in industry usually prefer simpler calculation procedures.
Step 1: Calculate the theoretical air requirement
= [(11.43 x C) + {34.5 x (H2 – O2/8)} + (4.32 x S)]/100 kg/kg of fuel
Step 2: Calculate the percent excess air supplied (EA)
O2 percent x 100
= ———————
(21 – O2 percent)
Step 3: Calculate actual mass of air supplied/ kg of fuel (AAS)
= {1 + EA/100} x theoretical air
Step 4: Estimate all heat losses
i. Percentage heat loss due to dry flue gas
m x Cp x (Tf-Ta) x 100
= —————————-
GCV of fuel
Where, m = mass of dry flue gas in kg/kg of fuel
m = (mass of dry products of combustion / kg of fuel) + (mass of N2 in fuel on 1 kg basis ) + (mass of N2 in actual mass of air we are supplying).
Cp = Specific heat of flue gas (0.23 kcal/kg )
ii. Percentage heat loss due to evaporation of water formed due to H2 in fuel
9 x H2 {584+Cp (Tf-Ta)} x 100
= ————————————–
GCV of fuel
Where, H2 = percentage of H2 in 1 kg of fuel
Cp = specific heat of superheated steam (0.45 kcal/kg)
iii. Percentage heat loss due to evaporation of moisture present in fuel
M{584+ Cp (Tf-Ta)} x 100
= ———————————-
GCV of fuel
Where, M – percent moisture in 1kg of fuel
Cp – Specific heat of superheated steam (0.45 kcal/kg)
iv. Percentage heat loss due to moisture present in air
AAS x humidity factor x Cp (Tf-Ta)} x 100
= —————————————————
GCV of fuel
Where, Cp – Specific heat of superheated steam (0.45 kcal/kg)
v. Percentage heat loss due to unburnt fuel in fly ash
Total ash collected/kg of fuel burnt x GCV of fly ash x 100
= ———————————————————————–
GCV of fuel
vi. Percentage heat loss due to unburnt fuel in bottom ash
Total ash collected per Kg of fuel burnt x G.C.V of bottom ash x 100
= ———————————————————————————–
GCV of fuel
vii. Percentage heat loss due to radiation and other unaccounted loss
The actual radiation and convection losses are difficult to assess because of particular emissivity of various surfaces, its inclination, airflow patterns etc. In a relatively small boiler, with a capacity of 10 MW, the radiation and unaccounted losses could amount to between 1 percent and 2 percent of the gross calorific value of the fuel, while in a 500 MW boiler, values between 0.2 percent to 1 percent are typical. The loss may be assumed appropriately depending on the surface condition.
Step 5: Calculate boiler efficiency and boiler evaporation ratio
Efficiency of boiler (n) = 100 – (i + ii + iii + iv + v + vi + vii)
Evaporation Ratio = Heat utilized for steam generation/Heat addition to the steam
Evaporation ratio means kilogram of steam generated per kilogram of fuel consumed. Typical Examples are:
– Coal fired boiler: 6 (i.e. 1 kg of coal can generate 6 kg of steam)
– Oil fired boiler: 13 (i.e. 1 kg of oil can generate 13 kg of steam)
However, the evaporation ratio will depend upon type of boiler, calorific value of the fuel and associated efficiencies.
Example
– Type of boiler: Oil fired
– Ultimate analysis of Oil :
C: 84 percent
H2: 12.0 percent
S: 3.0 percent
O2: 1 percent
– GCV of Oil: 10200 kcal/kg
– Percentage of Oxygen: 7 percent
– Percentage of CO2: 11 percent
– Flue gas temperature (Tf): 220 oC
– Ambient temperature (Ta): 27 oC
– Humidity of air : 0.018 kg/kg of dry air
Step-1: Calculate the theoretical air requirement
= [(11.43 x C) + [{34.5 x (H2 – O2/8)} + (4.32 x S)]/100 kg/kg of oil
= [(11.43 x 84) + [{34.5 x (12 – 1/8)} + (4.32 x 3)]/100 kg/kg of oil
= 13.82 kg of air/kg of oil
Step-2: Calculate the percent excess air supplied (EA)
Excess air supplied (EA) = (O2 x 100)/(21-O2)
= (7 x 100)/(21-7)
= 50 percent
Step 3: Calculate actual mass of air supplied/ kg of fuel (AAS)
AAS/kg fuel = [1 + EA/100] x Theo. Air (AAS)
= [1 + 50/100] x 13.82
= 1.5 x 13.82
= 20.74 kg of air/kg of oil
Step 4: Estimate all heat losses
i. Percentage heat loss due to dry flue gas
m x Cp x (Tf – Ta ) x 100
= —————————–
GCV of fuel
m = mass of CO2 + mass of SO2 + mass of N2 + mass of O2
0.84 x 44 0.03 x 64 20.74 x 77
m = ———— + ———- + ———– (0.07 x 32)
12 32 100
m = 21.35 kg / kg of oil
21.35 x 0.23 x (220 – 27)
= ——————————- x 100
10200
= 9.29 percent
A simpler method can also be used: Percentage heat loss due to dry flue gas
m x Cp x (Tf – Ta ) x 100
= ——————————–
GCV of fuel
m (total mass of flue gas)
= mass of actual air supplied + mass of fuel supplied
= 20.19 + 1 = 21.19
21.19 x 0.23 x (220-27)
= ——————————- x 100
10200
= 9.22 percent
ii. Heat loss due to evaporation of water formed due to H2 in fuel
9 x H2 {584+0.45 (Tf – Ta )}
= ———————————
GCV of fuel
where H2 = percentage of H2 in fuel
9 x 12 {584+0.45(220-27)}
= ——————————–
10200
= 7.10 percent
iii. Heat loss due to moisture present in air
AAS x humidity x 0.45 x ((Tf – Ta ) x 100
= ——————————————————
GCV of fuel
= [20.74 x 0.018 x 0.45 x (220-27) x 100]/10200
= 0.317 percent
iv. Heat loss due to radiation and other unaccounted losses
For a small boiler it is estimated to be 2 percent
Step 5: Calculate boiler efficiency and boiler evaporation ratio
Efficiency of boiler (n) = 100 – (i + ii + iii + iv + v + vi + vii)
i. Heat loss due to dry flue gas : 9.29 percent
ii. Heat loss due to evaporation of water formed due to H2 in fuel : 7.10 percent
iii. Heat loss due to moisture present in air : 0.317 percent
iv. Heat loss due to radiation and other unaccounted losses : 2 percent
= 100- [9.29+7.10+0.317+2]
= 100 – 17.024 = 83 percent (approximate)
Evaporation Ratio = Heat utilized for steam generation/Heat addition to the steam
= 10200 x 0.83 / (660-60)
= 14.11 (compared to 13 for a typical oil fired boiler)
Advantages of indirect method
A complete mass and energy balance can be obtained for each individual stream, making it easier to identify options to improve boiler efficiency
Disadvantages of indirect method
– Time consuming
– Requires lab facilities for analysis