Boiler Efficiency

There are basically two methods of determining boiler efficiency.

1. Direct method in which the energy output and fuel input are measured and efficiency = output/input Fuel input estimation requires accurate information on flow of fuel and its heating value. While with gaseous fuels a fairly accurate measurement of flow may

be obtained, with oil fuels due to its viscosity and variations with temperature,

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

accurate fuel consumption is difficult to measure. With solid fuels, it is more so difficult. With modern instruments, steam flow and temperatures can be measured to determine the energy output. Boiler blowdown is avoided during the testing as the amount of blowdown cannot be determined accurately due to the two-phase critical flow in the blowdown line. See Chapter 6 on miscellaneous calculations for the estimation of blowdown flow.

2. The indirect method simply estimates the various losses in the boiler as described here. An advantage of this method is that errors in measurements do not impact the efficiency much. For example, if the fuel input measurement to a boiler has an error of 1% in the direct method, the error will be 92 ± 0.9% and the efficiency can vary from 91.1 to 92.9%.

A 1% error in, say, exit gas temperature measurement (which is the major heat loss) will result in a variation of 8 ± 0.08 or 8.08–7.92%. (92% is typical boiler efficiency for gaseous fuels with about 15% excess air and about 150°C exit gas temperature); hence efficiency will be 100 – 8.08 = 91.92 to 100 – 7.92 = 92.08%. One may feel that the indirect method favors the boiler supplier rather than the boiler owner; however, if the exit gas temperature (for the same feed water inlet temperature in the economizer, steam parameters, and ambient conditions) is higher by, say, 10°C, the efficiency decreases by 0.45%, a significant deviation. Hence this method that requires lesser manpower and time for testing is widely accepted in industrial practice.

There are two ways of stating efficiency, one based on HHV and the other on LHV. As discussed earlier, η HHV × HHV = η LHV × LHV.

The European and Asian countries generally follow the LHV method while in the United States, efficiency is often stated based on HHV. Hence one should be aware of this practice and not be misled by the values of efficiency and investigate further. With oil firing, the ratio of the heating values is about 6.5%, while for natural gas, it is about 9.5%. So one should question the heating value basis used whether it is on HHV or LHV before comparing different boiler proposals or performance. The exit gas temperature from the boiler also is an indicator of efficiency.

Heat Loss Method

The various heat losses are [4]

1. Dry gas loss L 1 ,% L 1 = 24w dg (t g –t a )/HHV (1.15a)

2. Loss due to combustion of hydrogen and moisture in fuel, L 2 . L 2 = 100 × (9H 2 + W) × (584 + 0.46t g –t a )/HHV (1.15b) where 584 is the latent heat of water vapor in kcal/kg and 0.46 the specific heat of

water vapor in kcal/kg °C

3. Loss due to moisture in air L 3 L 3 = 46Mw da (t g –t a )/HHV (1.15c)

Combustion Calculations

4. Loss due to CO formation

L 4 = 100 × [CO/(CO + CO 2 )] × C × 5644/HHV

(1.15d)

5. Casing or radiation loss L 5

A more accurate way to compute this loss is to compute the heat losses at various sections based on the calculated or measured casing temperature, which depends on the insulation thickness used and ambient wind and temperature conditions as illustrated in Chapter 6 on miscellaneous calculations. Then the heat loss in

kW/m 2 may be estimated for the entire casing and then obtain the % heat loss as shown in the following. It is generally very small in oil- and gas-fired package boilers, in the range of 0.2%–0.5%. For smaller units (up to 20 t/h of steam), it may

be high, and for larger units it is smaller. We are not dealing with solid fuel-fired boilers, which can have refractory-lined casing with higher heat losses.

A quick estimate of L 5 = 10 0.62–0.42 log(Q) (1.15e) where Q is the boiler duty in MM Btu/h. If duty of a boiler is 200 MM kJ/h

(189.7 MM Btu/h), the heat loss = L 5 = 10 [(0.62–0.42 × log(189.7)] = 0.46%.

Boiler Efficiency