ENERGY EQUIPMENT


Assessment of boilers and thermic fluid heaters


Indirect method of determining boiler efficiency

Methodology

The reference standards for Boiler Testing at Site using the indirect method are the British Standard, BS 845:1987 and the USA StandardASME 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 industries 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 % excess air supplied (EA)

= O 2% x 100
----------------
   (21 - O 2%)

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 C p x (T f-T a) 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 H 2 {584+C p (T f-T a)} 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+ C p (T f-T a)} x 100
   ---------------------------------
            GCV of fuel

Where, M – % 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 C p (T f-T a)} 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% and 2% of the gross calorific value of the fuel, while in a 500 MW boiler, values between 0.2% to 1% 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 utilised 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

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 % excess air supplied (EA)

Excess air supplied (EA)
     = (O2 x 100)/(21-O2)
     = (7 x 100)/(21-7)
     = 50%

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     0.03x64      20.74x77
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%

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%

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%

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%

iv. Heat loss due to radiation and other unaccounted losses

For a small boiler it is estimated to be 2%

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%
ii. Heat loss due to evaporation of water formed due to H2 in fuel: 7.10 %
iii. Heat loss due to moisture present in air : 0.317 %
iv. Heat loss due to radiation and other unaccounted losses : 2%

= 100- [9.29+7.10+0.317+2]
= 100 – 17.024 = 83% (approximate)

Evaporation Ratio = Heat utilised 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, which makes it easier to identify options to improve boiler efficiency

Disadvantages of indirect method

  • Time consuming
  • Requires lab facilities for analysis

Copyright© United Nations Environment Programme 2006