The performance parameters of boiler, like efficiency and evaporation ratio reduces with time due to poor combustion, heat transfer surface fouling and poor operation and maintenance. Even for a new boiler, reasons such as deteriorating fuel quality, water quality etc. can result in poor boiler performance. Boiler efficiency tests help us to find out the deviation of boiler efficiency from the best efficiency and target problem area for corrective action.

## Boiler Efficiency

Thermal efficiency of boiler is defined as the percentage of heat input that is effectively utilised to generate steam. There are two methods of assessing boiler efficiency.

**The Direct Method:**Where the energy gain of the working fluid (water and steam) is compared with the energy content of the boiler fuel.**The Indirect Method:**Where the efficiency is the difference between the losses and the energy input.

## Direct Method

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

*Boiler Efficiency *= *Heat Output* * 100/Heat Input

Parameters to be monitored for the calculation of boiler efficiency by direct method are :

- Quantity of steam generated per hour
(Q) in kg/hr.
- Quantity of fuel used per hour (q) in kg/hr.

- The working pressure (in kg/cm2(g)) and superheat temperature (°C), if any

- The temperature of feed water (°C)

- Type of fuel and gross calorific value of the fuel (GCV) in kCal/kg of fuel

*Boiler Efficiency *(h) = *Qx *(*h g *– *h f *) * 100 / (q * GCV)

Where, h_{g} – Enthalpy of
saturated steam in kCal/kg of steam h_{f} – Enthalpy of feed water in
kCal/kg of water

### Example

Find out the efficiency of the boiler by direct method with the data given below:

Type of boiler = Coal fired

Quantity of steam (dry) generated = 8 TPH

Steam pressure (gauge) / temp = 10 kg/cm2(g)/180°C

Quantity of coal consumed = 1.8 TPH

Feed water temperature = 85°C

GCV of coal = 3200 kCal/kg

Enthalpy of steam at 10 kg/cm2 pressure = 665 kCal/kg

Enthalpy of feed water = 85 kCal/kg

*Boiler Efficiency *(h) = 8 * (665 – 85) *1000 / (1.8 * 3200 * 1000) =.80=80

It should be noted that boiler may not generate 100% saturated dry steam, and there may be some amount of wetness in the steam.

### Advantages of direct method:

- Plant people can evaluate quickly the efficiency of boilers
- Requires few parameters for computation
- Needs few instruments for monitoring

### Disadvantages of direct method:

- Does not give clues to the operator as to why efficiency of system is lower
- Does not calculate various losses accountable for various efficiency levels

## Indirect Method

There are reference standards for Boiler Testing at Site using indirect method namely British Standard, BS 845: 1987 and USA Standard is ASME PTC-4-1 Power Test Code Steam Generating Units’.

Indirect method is also called as heat loss method. The efficiency can be arrived at, by subtracting the heat loss fractions from 100. The standards do not include blow down loss in the efficiency determination process. A detailed procedure for calculating boiler efficiency by indirect method is given below. However, it may be noted that the practicing energy mangers in industries prefer simpler calculation procedures.

The principle losses that occur in a boiler are:

- Loss of heat due to dry fluegas
- Loss of heat due to moisture in fuel and combustion air

- Loss of heat due to combustion of hydrogen

- Loss of heat due to radiation

- Loss of heat due to unburnt

In the above, loss due to moisture in fuel and the loss 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 indirect method are:

- Ultimate analysis of fuel (H
_{2}, O_{2}, S, C, moisture content, ash content)- Percentage of Oxygen or CO
_{2}in the flue gas

- Flue gas temperature in °C (T
_{f})

- Ambient temperature in °C (T
_{a}) & 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)

- Percentage of Oxygen or CO

### Solution :

Theoretical air requirement

=[(11.6 × C) + {34.8 × (H_{2} – O_{2}/8)} + (4.35 × S)]/100 kg/kg of fuel

Excess Air supplied (EA) =* O *2 % ´100 21 – *O *2 %

Actual mass of air supplied/ kg of fuel (AAS) = {1 + EA/100} *× *theoretical air

*m × C _{p} × (T_{f} – T_{a} ) × *100

Percentage heat loss due to dry flue gas =

m = mass of dry flue gas in kg/kg of fuel

*GCV of fuel*

m = Combustion products from
fuel: CO_{2} + SO_{2}
+ Nitrogen in fuel
+ Nitrogen in the actual mass
of air supplied + O_{2} in flue gas. (H_{2}O/Water
vapour in the
flue gas should not be considered)

C_{p} = Specific heat of flue gas (0.23 kCal/kg °C

Percentage heat loss due to evaporation of water formed due to H_{2} in fuel= 9 ´ H _{2} ´ {584 + C _{p} (T _{f}-T _{a} )} ´ 100

GCV of fuel Where, H_{2} – kg of H_{2} in 1 kg of fuel

C_{p} – Specific heat of superheated steam (0.45 kCal/kg °

Percentage heat loss due to evaporation of moisture present in fuel

= M x {584 + C _{p} (T_{f} -T _{a}
)} *x *100 GCV of fuel

Where, M – kg of moisture in 1kg of fuel

C_{p} – Specific heat of superheated steam (0.45
kCal/kg)°C

584 is the latent heat corresponding to the partial pressure of water vapour

Percentage heat loss due to moisture present in air

= *AAS *´ *humidity factor *´ C_{p} x (T_{f} -T_{a} ) ´ 100

C_{p} – Specific heat of superheated steam (0.45
kCal/kg °C)

Percentage heat loss due to unburnt in fly ash

= Total ash collected / kg of fuel burnt ´ G.C.V of fly ash

Percentage heat loss due to unburnt in bottom ash

= Total ash collected / kg of fuel burnt ´ G.C.V of bottom ash ´ 100

GCV of fuel

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, air flow pattern 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.

#### Efficiency of boiler (h) = 100 – (i + ii + iii + iv + v + vi + vii)

**Example: **The following are the data collected for a typical oil fired boiler. Find out the efficiency of the boiler by indirect method and Boiler Evaporation ratio.

- Type of boiler = Oil fired

- Ultimate analysis of Oil

C : 84.0 % H_{2} : 12.0 %

S : 3.0 % O_{2} : 1.0 %

- GCV of Oil : 10200 kCal/kg
- Steam Generation Pressure : 7kg/cm2(g)-saturated

- Enthalpy of steam : 660 kCal/kg

- Feed water temperature : 60 °C

- Percentage of Oxygen in flue gas : 7

- Percentage of CO
_{2}in flue gas : 11

- Flue gas temperature (T
_{f}) : 220 °C

- Ambient temperature (T
_{a}) : 27 °C

- Humidity of air : 0.018 kg/kg of dry air

### Solution

**Step-1:
Find the theoretical air requirement**

= [(11.6 ´ *C*) +{34.8 ´( *H *2 – *O*2 / 8)} +(4.35 ´ *S*)] /100

=[(11.6 × 84) + [{34.8 × (12 – 1/8)} + (4.35 × 3)]/100 kg/kg of oil

=14 kg of air/kg of oil

### Step-2: Find the %Excess air supplied

Excess air supplied (EA)

= (O_{2} × 100)/(21-O_{2})

= (7 × 100)/(21-7)

= 50%

### Step-3: Find the Actual mass of air supplied

Actual mass of air supplied /kg of fuel = [ 1 + EA/100] x Theoretical Air

### Step-4: Estimation of all losses

Dry flue gas loss

= [1 + 50/100] x 14

= 1.5 x 14

= 21 kg of air/kg of oil

*m × C _{p} × (T_{f} – T_{a} ) × *100

Percentage heat loss due to dry flue gas =

*GCV of fuel*

m = mass of CO_{2} + mass of SO_{2} + mass of N_{2} + mass of O_{2}

### Alternatively a simple method can be used for determining the dry flue gas loss as given below.

a) Percentage heat loss due to dry flue gas = *m *× *C _{p} *×

*(T*× 100

_{f}– T_{a})*GCV of fuel*

Total mass of flue gas (m) = mass of actual air supplied + mass of fuel supplied = 21 + 1 = 2

% Dry flue gas loss = 22 *x *0.23 *x*(220 – 27)*100/10200 = 9.57%

### Heat loss due to evaporation of water formed due to H_{2} in fuel

9 × H_{2} {584 + C_{p} (T_{f} – T_{a})}× 100 = GCV of fuel

Where, H_{2} – percentage of H_{2} in fuel 9 × 12 {584 + 0.45 (220 – 27)}× 100/10200

= 7.10%

### Heat loss due to moisture present in air

*AAS *× *humidity *× *C _{p} *×

*(T*× 100 =

_{f}– T_{a})*GCV of fuel*

= 21

*x*0.018

*x*0.45

*x*(220 -27)*100/10200

= 0.322%

### Heat loss due to radiation and other unaccounted losses

For a small boiler it is estimated to be 2%

## Boiler Efficiency

Heat loss due to dry flue gas | 9.14% |

Heat loss due to evaporation of water formed due to H_{2} in fuel | 7.10 % |

Heat loss due to moisture present in air | 0.322% |

Heat loss due to radiation and other unaccounted loss 2%

**Boiler Efficiency **= 100- [9.14 + 7.10 + 0.322 + 2]

= 100 – 18.56 = 81 %(app)

Evaporation Ratio = Heat utilized for steam generation/Heat addition

= 10200 × 0.83/ (660-60)

= 14.11

### Boiler Evaporation Ratio

Evaporation ratio means kilogram of steam generated per kilogram of fuel consumed. Typical Examples:

Coal fired boiler: 6

Oil fired boiler: 13

i.e 1 kg of coal can generate 6 kg of steam 1 kg of oil can generate 13 kg of steam. However, this figure will depend upon type of boiler, calorific value of the fuel and associated efficiencies.