Appendix D: Nonluminous Heat Transfer Calculations

Nonluminous heat transfer plays a significant role in high gas temperature heat transfer equipment such as superheaters or boiler banks or unfired furnaces of waste heat boilers. During combustion of fossil, fuels like natural gas, fuel oil, or coal triatomic gases such

as CO 2 ,H 2 O, and SO 2 are generated, which contribute to radiation. The emissivity pat- tern of these gases has been studied by Hottel, who has developed charts to estimate the emissivity of these gases if the partial pressure and beam length of the tube bank are known.

Net interchange of radiation between gases and surroundings like a water-cooled wall or a tube bundle or a cavity may be written as

g T g – aT 4 go (D.1)

where ∈ g is the emissivity of gases at average gas temperature T g T o is the tube wall temperature

a g is the absorptivity at T o

η w −∆ ∈ (D.2)

4 Computing absorptivity is tedious; also, as T 4 o is typically much smaller than T g , without much loss of accuracy, one can write

g T 4 g T – 4 o = hT n ( g − T o ) (D.3)

σ ∈ 4 g 4  T g – T o

 (D.4a)

where

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

T g ,T o are gas and wall temperatures, K

h n is the nonluminous coefficient, W/m 2 K

408 Appendix D: Nonluminous Heat Transfer Calculations

In British units,

g  T g – 4 T o

 (D.4b)

where σ is the Steffan–Boltzman constant = 0.1714 × 10 −8 T g ,T o in °R

h n in Btu/ft 2 h °F To estimate h n , partial pressures of triatomic gases and beam length L are required. L is a

characteristic dimension that depends on the bundle arrangement and shape of enclosure. For a plain tube bundle,

1 08 2 × TL − d

( SS . 0 785 )

† (D.5)

For a cavity, L is approximately 3.4–3.6 times the volume of the space divided by the surface area of the heat-receiving surface. If a, b, c are the dimensions of the cavity,

34 . abc

2 ( ab + bc + ca )( 1 /a + 1 /b + 1 /c )

(D.6)

For flow inside tubes, the tube inner diameter may be taken as L. As d i is generally small, the beam length and hence the emissivity is small, and h n inside tubes may be neglected if the convective coefficient is large.

For a finned tube bundle,

Beam length L b = ( d + 2 h ) 

2 − . 0 85  (D.7)

Example D.1

Estimate the beam length when 50.8 mm tubes in a boiler bank are arranged with a transverse pitch of 101 mm and longitudinal pitch of 127 mm.

Solution

L = 1.08 × (0.1018 × 0.127 − 0.785 × 0.0508 × 0.0508)/0.0508 = 0.23 m

Example D.2

Flue gases with CO 2 = 12% and H 2 O = 16% by volume are flowing across a superheater

tube bundle. If 50.8 mm tubes are arranged with S T = 101 mm and S L = 127 mm, deter-

mine h n . Flue gas pressure is atmospheric. Average gas temperature T g = 900°C and average wall temperature T o = 300°C.

Appendix D: Nonluminous Heat Transfer Calculations 409

Solution

L = 0.23 m from earlier example. P c , the partial pressure of CO 2 = 0.12 and P w , the partial pressure of water vapor = 0.16

PL c = . 0 23 × . 0 12 = . 0 0276 atm m = . 0 09 atm ft PL w = . 0 23 × . 0 16 = . 0 0368 atm m = 0 12 . atm ft

From Figure D.1a and b at T g = 900 × 1.8 + 32 + 460 = 2112°R and T o = 300 × 1.8 + 32 + 460 = 1032°R, ∈ c = 0.07 and ∈ w = 0.057. In Figure D.1c, corresponding to (P + P w )/2 = 1.16/2 = 0.58 and P w L = 0.12, η = 1.05 and corresponding to P w /(P w +P c ) = 0.16/0.28 = 0.57 and (P c +P w ) L = 0.21, Δv = 0.002 Hence, ∈ g = 0.07 + 1.05 × 0.057 − 0.002 = 0.128 Using (D.2),

h n = 5.67 × 10 −8 × 0.128 × (1173 4 − 573 4 )/(1173 − 573) = 21.60 W/m 2 K Using (D.4b), h n = 0.1714 × 10 −8 × 0.128 × (2112 4 − 1032 4 )/(2112 − 1032) = 3.81 Btu/ft 2 h °F While charts are often used, equations are also available for estimating h n .

g 09 . × ( 1 − e KL ) (D.8)

where attenuation factor

K ( 0816 . + . P w ) × ( − 1 0 00038 . T g ) × ( P c + P = w ) 05 . (D.9)

 ( P c + PL w ) 

T g is in K, L is in m.

Example D.3

Compute h n in Example D.2 using Equations D.8 and D.9.

K = ( . 0 8 1 6 0 16 + . × . )( × 1 − . 00038 1173 × ) × . 0 28 05 . = . 0 645 ( . 0 28 0 23 × . )

g g = 09 . × 1 − e . 0 645 0 23 × ( . ) = . 0 124

h ° n = . 20 9 W/m K 2 ( . 3 68 Btu/ft h F 2

Now, let us see how the performance of an unfired furnace may be evaluated. This fea- ture is seen in several waste heat boilers (Figure D.2).

Example D.4

How is the duty of an unfired waste heat boiler furnace determined? Waste gases of

27.8 kg/s at 1000°C enter a waste heat boiler furnace that is completely water cooled and is generating steam at 16 barg. Saturation temperature is 202°C. Furnace dimensions are

3 m wide × 4 m deep × 8 m long. Flue gas analysis is % vol CO 2 = 12, H 2 O = 15, N 2 = 67,

O 2 = 6. Effective cooling area = 2 × (3 × 4 + 4 × 8 + 3 × 8) = 136 m 2 .

410 Appendix D: Nonluminous Heat Transfer Calculations

0.2 P

0.5 P w L = 20 atm. f

c L = 5.0 atm. f

0.15 c 0.09 0.03 0.06 w 0.08

Absolute temperature (T *R)

(b)

Absolute temperature (*R)

1.8 ft 1.6 L = 0.005 atm. 0.25 0.50

η 1.0 P w L = 10.0 atm. ft

1/2 (P + P w ) (atm)

Correction factor for emissivity of water vapor. 7,2

P c L+P w L = 5 atm. ft P c L+P w L = 5 atm. ft 0.05

Є 0.04 c L+P w L = 5 atm. ft

FIGURE D.1

Emissivity of (a) carbon dioxide and (b) water vapor. Correction factor for (c) the emissivity of water vapor and (d) for the presence of water vapor and carbon dioxide.

Appendix D: Nonluminous Heat Transfer Calculations 411

FIGURE D.2

Waste heat boiler with furnace section.

Solution

Assume the exit gas temperature of 800°C = 1073 K. Use saturation temperature plus 18°C for tube outer wall temperature = 220°C = 493 K. Average specific heat of the gas is 1323 J/kg K between 1000°C and 800°C.

At 900°C = 1173 K average gas temperature, the emissivity of the flue gases may be computed. The beam length of the furnace is 1.7/(1/3 + 1/4 + 1/8) = 2.41 m

( 08 . + 16 . × . 15 1 − . 00038 × K 1173 = )( ) × . 0 27

Gas emissivity = 0.9 × (1 − e −0.193 × 2.41 ) = 0.334 Hence, radiation to furnace walls = 136 × 0.334 × 0.9 × 5.67 × 10 −8 × (1173 4 − 493 4 ) = 4,251 kW

Energy from flue gases = 27.8 × 1323 × (1,000 − 800) × .99 = 7,282,300 W = 7,282 kW So the assumed exit gas temperature is too low. Try exit gas temperature = 900°C or average gas temperature = 950°C = 1,223 K Gas emissivity = (0.8 + 1.6 × 0.15) × (1 − 0.00038 × 1223) × 0.27/(0.27 × 2.41)0.5 = 0.186

Gas emissivity = 0.9 × (1 − e −0.186 × 2.41 ) = 0.325

Hence, radiation to furnace walls = 0.325 × 0.9 × 5.67 × 10 −8 × (1223 4 − 493 4 ) × 136 = 4912 kW Energy from flue gases = 27.8 × 1,323 × (1,000 − 900) = 7,355,880 W = 3,677 kW Try exit gas temperature = 870°C average gas temperature = 935°C = 1,208 K Using the same emissivity,

Radiation to furnace = 0.325 × 0.9 × 5.67 × 10 −8 × (1208 4 − 494 4 ) × 136 = 4,668 kW Energy from flue gases = 27.8 × 1323 × 130 × 0.99 = 4,733 kW. Hence, flue gas exit temperature is approximately 870°C. With a computer program,

one can study the effect of varying the furnace dimensions on the furnace duty and exit gas temperature. For improved accuracy, one may split up the furnace into zones and arrive at the exit gas temperature and duty more accurately.

412 Appendix D: Nonluminous Heat Transfer Calculations

Appendix D: Nonluminous Heat Transfer Calculations