Simplified Formulae for Boiler Efficiency

Simplified Formulae for Boiler Efficiency

Boiler efficiency mainly depends upon the excess air and exit gas temperature and refer- ence temperature. The following equations provide a quick solution based on 1% casing and other unaccounted losses.

For natural gas: η HHV = 89.4 – (0.002021 + 0.0351 × EA) × ΔT (1.16a) η LHV = 99 – (0.00202 + 0.0389 × EA) × ΔT

(1.16b) For fuel oils (no 2, diesel; may be used for heavy oil with lesser accuracy) η HHV = 92.9 – (0.002336 + 0.0351 × EA) × ΔT

(1.16c) η LHV = 99 – (0.00249 + 0.03654 × EA) × ΔT

(1.16d) ΔT is the difference between exit gas temperature and reference temperature in °C.

EA is the excess air factor. If excess air is 15%, then EA = 1.15.

Example 1.9

Determine the efficiency of a boiler firing the natural gas fuel as in Example 1.2. Assume exit gas temperature = 204°C and ambient temperature is 20°C. Relative humidity is 80%. Use a casing loss of 1%. HHV = 12,882 kcal/kg and LHV = 11,655 kcal/kg.

Solution

Combustion calculations have already been done in Example 1.1. Let us use the results. Dry flue gas w dg = 18 kg/kg fuel. Moisture in air = w wa –w da = 19.52 – 19.29 = 0.23 kg/ kg fuel.

24 Steam Generators and Waste Heat Boilers: For Process and Plant Engineers

Water vapor formed due to combustion of fuel = 20.4 – 0.23 – 18 = 2.17 kg/kg fuel 1. Dry gas loss = 100 × 18 × 0.24 × (204 – 20)/12,882 = 6.17%

2. Loss due to combustion of hydrogen and moisture in fuel = 100 × 2.17 × (584 +

0.46 × 204 – 20)/12,882 = 11.1% 3. Loss due to moisture in air = 100 × 0.23 × 0.46 × (204 – 20)/12,882 = 0.15% 4. If CO in flue gases = 100 ppm, then CO = 0.01% volume in flue gases. CO 2 = 8.5% from earlier combustion calculations. Carbon content in fuel was estimated as 0.7539 kg/kg fuel earlier.

L 4 = 100 × (0.01/8.5) × 0.7539 × 5,644/12,882 = 0.039% 5. Casing loss = 1.0% Total losses = 6.17 + 11.1 + 0.15 + 1.0 + 0.038 = 18.458%. Hence, efficiency on HHV basis = 100 – 18.458 = 81.54%. Efficiency on LHV basis = 81.54 × 12,882/11,655 = 90.12%.

Using the simplified formula, η HHV = 89.4 – (0.002021 + 0.0351 × 1.15)×184 = 81.6% η LHV = 99 – (0.00202 + 0.0389 × 1.15) × 184 = 90.39%

Example 1.10

Casing temperature of a boiler was measured as 82°C when the ambient temperature was 29°C. Wind velocity = 134 m/min. Determine the casing loss if the surface area of

the boiler is 231 m 2 . Casing emissivity may be taken as 0.1.

Solution

From Chapter 6, one may determine the casing loss q. q = 5.67 × 10 –11 × 0.1 × [(273 + 82) 4 – (273 + 29) 4 ] + 0.00195 × (82 – 29) 1.25 × [(134 + 21)/21] 0.5

= 0.043 + 0.755 = 0.798 kW/m 2 (252 Btu/ft 2 h)

The total heat loss = 231 × 0.798 = 184.3 kW (231 m 2 is the total surface area of the boiler casing). If boiler duty is 50 MW, the heat loss = 184/50,000 = 0.0037 or 0.37%. Note that as boiler load decreases, the heat loss in kW will not diminish (as it depends only on ambient conditions and wind velocity) while the duty decreases. Hence, cas- ing loss as a % will increase. At 50% load, the casing loss will be about 0.75% in our example. While the flue gas heat losses decrease as the load decreases, the casing loss increases in indirect proportion to the load. Thus, a parabolic trend is seen for the effi- ciency versus load curve. Table 1.11 shows how the boiler efficiency and various losses vary with load.

Example 1.11

If fuel oil is fired at 20% excess air in a boiler and ΔT = 150°C, then η HHV = 92.9 – (0.002336 + 0.0351 × 1.2) × 150 = 86.23%

η LHV = 99 – (0.00249 + 0.03654 × 1.2) × 150 = 92% Note that these equations assume a total heat loss of 1%. At part loads, the heat loss

will be higher as explained earlier and in Chapter 6. One has to adjust the efficiency to account for the higher heat loss at lower loads.

Combustion Calculations

TABLE 1.11

Typical Boiler Performance versus Load

Boiler Load, %

75 50 25 Boiler duty

72.81 54.61 36.41 18.20 MM kcal/h Ambient temp.

26.7 26.7 26.7 26.7 °C Relative humidity

60 60 60 60 % Excess air

17 17 17 17 % Flue gas recirculation

22 22 22 22 % Fuel input (HHV)

88.61 66.42 44.34 22.34 MM kcal/h Heat release rate (HHV)

372,038 187,434 kcal/m 3 h Heat release rate (HHV)

260,562 131,275 kcal/m 2 h Steam flow

37,432 kg/h Process steam

0 0 0 0 kg/h Steam pressure

76.5 76.5 76.5 76.5 kg/cm 2 g Steam temp.

293 ±5°C Feed water temp.

172 ±5°C Water temp. lvg eco

212 ±5°C Blowdown %

2 2 2 2 % Boiler exit gas temp.

307 ±5°C Eco exit gas temp.

176 ±5°C Air flow

34,394 kg/h Flue gas to stack

36,131 kg/h Flue gas through boiler

44,080 kg/h Flue gas analysis, losses, efficiency, %

Dry gas loss 5.42 5.29 5.19 5.13 % Air moisture

0.15 0.14 0.14 0.14 % Fuel moisture

11.04 11.01 10.99 10.98 % Casing loss

0.35 0.47 0.70 1.40 % Unacc/margin

0.50 0.50 0.50 0.50 % Efficiency, lhv

91.58 91.63 91.50 90.81 % Efficiency, hhv

82.55 82.59 82.48 81.85 % Furnace back pr.

91.18 23.20 mm wc % vol CO 2 8.13 8.13 8.13 8.13 H 2 O

18.02 18.02 18.02 18.02 N 2 71.11 71.11 71.11 71.11 O 2 2.74 2.74 2.74 2.74 SO 2 0.00 0.00 0.00 0.00

Fuel flow

1736 kg/h

Gas-% volume

Methane

98.0 Boiler surface areas—m 2

Ethane

0.5 Furnace volume

Propane

0.1 Furnace proj. area

Carbon dioxide 0.6 Nitrogen

0.8 LHV-kcal/kg

11,594.0 HHV-kcal/kg

26 Steam Generators and Waste Heat Boilers: For Process and Plant Engineers