Furnace Exit Gas Temperature Evaluation

Furnace has been often described as the heart of the boiler. Hence, its sizing assumes paramount importance. Too large a furnace would be expensive. The furnace exit gas temperature will also be low resulting in larger surface area requirements for superheaters and evaporators downstream of the furnace. Too small a furnace can cause overheating of superheater tubes, and if ash-containing oil fuels are fired, slagging or melting of ash can occur causing high-temperature corrosion problems in the superheaters. Hence, the furnace duty or energy absorbed by steam in the furnace, which also fixes the furnace exit gas temperature, should be evaluated with care. Note that the determination of furnace exit gas temperature is not an accurate science and has to be linked with the type of furnace, excess air, flue gas recirculation (FGR) rate used, location of burners, and boiler load. Hence, there is no universally accepted procedure for this calculation.

For the process and plant engineers, a simplified procedure for estimating the furnace performance is given here. Energy balance yields the following equation:

Q f =A p ∈ w ∈ f σ (T g 4 −T o 4 )=W f × LHV × (1 – casing loss-unaccounted loss) − W g ×h e (2.1) where

Q f is the energy transferred to the boiler furnace, kW W f , W g is the fuel, flue gas flow, kg/s LHV is the lower heating value of fuel, kJ/kg

h e is the enthalpy of flue gas at furnace exit, kJ/kg

A p is the effective projected area of the furnace cooling surface, m 2 ∈ w, ∈f is the emissivity of wall and flame

σ is the Steffan–Boltzman constant = 5.67 × 10 −8 W/m 2 K 4

T g ,T o is the furnace gas temperature, and furnace tube wall temperature, K T g is defined in a few ways. Some use the furnace exit gas temperature itself for T g , while

some others use the adiabatic combustion temperature. A reasonable agreement between measured and predicted values of the furnace exit gas temperature prevails when T g is taken as the furnace exit gas temperature plus 160°C–175°C. Using the procedure and example discussed in Appendix A, one may arrive at the furnace exit gas temperature from economizer exit gas temperature. A correlation may be developed based on heat input and the furnace exit gas temperature for a given type of furnace.

The emissivity of wall ∈ w varies depending on whether the furnace surfaces are clean or covered with slag. It can vary from 0.6 to 0.9. Soot blowing also changes this value consid-

erably. Flame emissivity ∈ f is given by

∈ f = β(1 – e –KPL ) (2.2) where

P is the gas pressure in atmospheres L is the beam length of the furnace, m K is the attenuation factor that depends on the fuel type and presence of ash and its con-

centration. For a nonluminous flame,

( 0 . 8 1 6P . w ) ( 1 . 0 000 38T g ) ×

0 . 5 (2.3a)

 ( P c + PL w ) 

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

P c and P w are partial pressures of CO 2 and H 2 O in atm. For a semiluminous flame, the ash particle size and concentration enter into the calculations.

K = ( 0 . 8 + . 1 6P w ) ×− ( 1 . 0 000 38T

(2.3b)

m  T ( ) ( e )

d m is the mean effective diameter of ash particles in microns µ m = 13 for coals ground in ball mills, 16 for coals ground in medium- and high-speed mills, and 20 for coals ground in hammer mills µ is the ash concentration in g/Nm 3

β = flame filling factor = 1.0 for nonluminous flames, 0.75 for luminous sooty flames of liquid fuels, and 0.65 for luminous and semiluminous flames of solid fuels. For a

luminous oil or gas flame, K = 0.0016T e − 0.5, where T e , the furnace exit gas temperature, is in K. These equations give only a trend or an approximation for the evaluation of furnace exit gas temperature. A wide variation between measured and estimated furnace exit gas temperature can exist due to the combustion process, location of the burners, type of firing (corner or front wall), how large the furnace is, and how much of it is filled with the flame whether there are soot particles in the fuel to mention a few. The unfilled portions are sub- ject to gas radiation only, and the emissivity (0.15–0.3) can be far less than that estimated for a luminous flame.

In practice, many flames are a combination of luminous and nonluminous portions. The emissivity then is obtained by combining the two. ∈ f = m∈ l + (1 − m)∈ n where ∈ f ,∈ l ,∈ n are emissivity of flame, luminous, and nonluminous portions and m is a coefficient. For natural gas, m = 0.1, while for fuel oils, it is 0.55. For solid fuels, since the flame is bright all over the furnace, m = 1.

Charts are available to estimate the furnace exit gas temperature directly based on furnace net heat input and its effective projected cooling area as shown later. However, the charts are developed based on furnace type and burner arrangement and evaluation of field data as discussed in Appendix A. Hence, each boiler manufacturer may have his chart or correlations for furnace performance evaluation, and no universal procedure exists as in the case of estimation of convective heat transfer coefficients. Furnace performance evaluation is described by a few as black magic and an art. The following chart shows how the furnace exit gas temperature varies with different fuels, and Figures 2.7 and 2.8 show how the furnace exit gas temperature varies with net heat input for different fuels. Corrections have to be made for various factors such as excess air, FGR, and burner location based on field data and experience.

Example 2.1

Determine the furnace exit gas temperature when natural gas (methane = 97%, ethane 3% by volume) is fired in a boiler furnace at 15% excess air. The net heat release rate per

effective area is 250 kW/m 2 . Furnace dimensions are approximately 2.44 m wide, 3.35 m high, and 9.75 m long (EPRS = 127 m 2 ). Flue gas analysis % volume CO 2 = 8, H 2 O = 18, N 2 = 71.5, O 2 = 2.5. The tube walls may be assumed to be at 660 K. Its method of estimation is discussed later.

Steam Generator Furnace Design

4 erature, °C 1200 5

rnace exit temp 900 Fu 800

Net heat input, kcal/m 2 h

FIGURE 2.7

Typical furnace exit gas temperatures for some fuels: (1) pulverized coal, (2) natural gas, (3) fuel oils, (4) grate fired bagasse, (5) grate fired coal. (1 is the topmost curve and 5 is the bottommost curve.)

°C T, 1200 Gas FO

Net heat release rate, kW/m 2

FIGURE 2.8

Furnace exit gas temperature for oil and natural gas. (From Ganapathy, V., Industrial Boilers and HRSGs, CRC Press, Boca Raton, FL, 2001, p112.)

The calculation of net heat release rate itself involves the following steps: • Calculation of boiler duty (energy absorbed by steam) in kW.

• Calculation of boiler efficiency (see Chapter 1), from which the fuel flow is

obtained. • Net heat input (fuel flow × LHV). • Net heat input divided by furnace effective projected radiant surface area

(water-cooled surfaces). (For surfaces covered with refractory, a multiplication factor of 0.1–0.2 is used.)

Assume one has done all the calculations given earlier. From combustion calculations (Chapter 1), it can be shown that the HHV of the fuel gas = 55,345 kJ/kg and its LHV = 49,915 kJ/kg.

It may be shown that each GJ of natural gas fired on HHV basis requires a theoretical amount of 314 kg of dry air (Table 1.2). This amount of heat input requires

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

1 × 10 6 /55,345 = 18.07 kg of fuel. So ratio of dry air to fuel is 314/18.07 = 17.37, or at 15% excess air, dry air per kg fuel is 1.15 × 17.37 = 19.97 or wet air about 20 kg/kg fuel as air always contains some moisture. Hence, the ratio of flue gas to fuel is 19.97 + 1 = 20.97.

If wet air is considered, the ratio of flue gas to fuel W g /W f is about 21. From energy balance in the furnace, we can write the following: From (2.1), Q f =A p ∈ w ∈ f σ (T g 4 −T o 4 )=W f LHV – W g h e (heat losses are on the order of 0.5% in oil- and gas-fired boilers and hence neglected). Dividing by W f , we have

A p ∈ w ∈ f σ (T g 4 −T o 4 )/W f = LHV − W g h e /W f or Q f /A p = 250,000 W/m 2 =W f × LHV/A p. Rearranging the terms, A p /W f = LHV/250,000 = 49,915,029/250,000 = 199.7 m 2 s/kg

P c = 0.08, P w = 0.18. Assume furnace exit gas temperature T e = 1150°C. h g = 1430.49 kJ/kg (see Appendix F). Use T g = 1150 + 170 = 1320°C = 1593 K (170°C more than the FEGT [furnace exit gas temperature])

K = ( 0 . 8 + . 1 6P w ) ×− ( 1 P c + P . w 0 000 38Te ) × ( ) 0 . 5  ( P c + P w ) L 

17 L . =

= 29m ( .0

K = ( . 8 + 16 . × . 18 ) ×− ( 1 . 000 38 × 1423 ) × ( ) = 0 . 176

∈ f = (1 − e −.176 × 2.09 ) = 0.308. Let emissivity of walls ∈ w = 0.9 From (2.3), LHS (left-hand side) = 199.7 × 0.9 × .308 × 5.67 × 10 –8 × (1593 4 – 660 4 ) = 19.61 × 10 6 W

RHS = 49,915,029−21 × 1,430,490 = 19.93 × 10 6 W

Hence, furnace exit gas temperature is close to 1150°C. It may be noted that furnace duty estimation is an iterative process. We have assumed

an efficiency to obtain the fuel input. In reality, one has to assume the furnace size, obtain the furnace duty, and then design the heating surfaces downstream of the furnace to obtain the desired exit gas temperature that matches the efficiency used or use the efficiency obtained in the earlier calculations. A well-written computer program is the solution. The earlier example is to explain the methodology only.

The furnace net heat input = 250 × 127 = 31,750 kW. Fuel input = 31,750/49,915 = 0.636 kg/s. Flue gas generated = 0.0636 × 21 = 13.357 kg/s

The furnace duty = 31,750−13.357 × 1430.49 = 12,642 kW. The average heat flux =

12,642/127 = 99.5 kW/m 2

Example 2.2

In the same furnace if no. 2 oil at 15% excess air is fired, then determine the exit gas tempera-

ture. Flue gas analysis is CO 2 = 12, H 2 O = 12, N 2 = 73.5, O 2 = 2.5. Fuel oil LHV = 43,050 kJ/kg and HHV = 45,770 kJ/kg. Net heat input is the same as before, namely, 250 kW/m 2 .

Solution

A p /W f = LHV/250,000 = 43,050 × 10 3 /250,000 = 172.2 m 2 s/kg

Assume furnace exit gas temperature is 1050°C = 1323 K.

0 . K 5 = ( 0 . 8 + 16 . × 0 . 12 ) ×− ( 1 . 0 000 × 38 1323 ) × 0 . 24/ ( 0 . × 24 2 9 . 0 ) = 0 . 1 67 6

Steam Generator Furnace Design

For theoretical combustion of fuel oil, 1 GJ heat put requires 320 kg of dry air (see

Chapter 1). Or 1 × 10 6 /45,770 = 21.85 kg of fuel requires 320 kg dry air. Or ratio of flue gas to fuel at 15% excess air = 1.15 × 320/21.85 + 1 = 17.84 or say 18 kg/kg wet air.

For a luminous flame, the emissivity consists partly of luminous and a nonluminous

portion. Then ∈ f = 0.55 × ∈ l + .45 × ∈ nl

K = . 00 16 × 1323 − . 5 = . 1 616 . ∈= l 0 . 75 × 1 − e − . 1 616 2 × . 0 ( 9 = 0 . 724

∈= nl (. 0 816 + . × . 12 )( ×− 1 . 000 × 38 1323 ). × 24 / (. × 24 2 9 . 00 ). 5 = 0 . 167 .

Hence, ∈ f = 0.55 × 0.724 + .45 × 0.167 = 0.473 LHS = 172.2 × 0.473 × 0.9 × 5.67 × (14.93 4 − 6.6 4 ) = 19.86 × 10 6 W

(Here again, we added 170°C to the assume FEGT.)

RHS = 43.05 × 10 6 −18 × 1,296,280 = 19.72 × 10 6 W

Hence, furnace exit gas temperature is close to 1050°C.

This is only an estimate. It is preferable to use charts developed based on field data and experience as shown in Figure 2.8. Furnace exit gas temperature is a function of so many variables as mentioned earlier. Hence, experience and field testing can provide more reliable data for a particular furnace configuration. One may also work backward

from economizer exit and use the field data to establish T e as discussed in Appendix A.

Once the furnace exit gas temperature is estimated, the furnace duty has to be com- puted. This is discussed later.

Furnace Exit Gas Temperature Evaluation