Condensation over Finned Tubes

Finned tubes are used in condensing economizer service as a more compact tube bundle with lower gas pressure drop is obtained. The following equation is recommended by Beatty and Katz for condensation heat transfer [2,3]:

A f is the fin area, m 2 /m Δ

H is the latent heat, kcal/kg η f is the fin efficiency, fraction

h is the fin height, m

A ef 2 is the effective area, η f A f +A t, m /m

A f ,A t are fin and tube areas, m 2 /m

d, d h are tube diameter, effective diameter, m

Example 6.3

Solid finned tubes are used in the condensing economizer application provided earlier. Tube size = 50.8 × 44 mm, fins/m = 117, fin height = 15 mm, fin thickness = 1.5 mm. Transverse spacing S T = 85 mm, S L = 80 mm. Arrangement is staggered. Total exter-

nal area = 0.901 m 2 /m, fin surface area = 0.77 m 2 /m. Plain tube area = 0.131 m 2 /m. Determine the size of the economizer for the same duty as Example 6.2.

Solution

Let us use the properties of the condensate at 40°C as before. Calculations for finned tubes are shown in Appendix E. Using the procedure described, the fin efficiency may

be shown to be 0.8. The convective heat transfer coefficient = 82 kcal/m 2 h °C. .  0 25 1 

 †. = 13 × ( . 0 77 0 80 0 901 0 015 × . / . / . . 0 25 ) + . 0 131 0 901 / . / /0 0508 . . 0 25 = . 1 953 + . 0 305 = . d 2 258 h

  The condensing heat transfer coefficient h c

= . 0 689   × . 2 258 = = 8937 kcal/m h C 2  

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

Assume the number of rows deep is 12; the correction factor = 12 −0.167 = 0.66 or

h c = 8937 × 0.66 = 5898 kcal/m 2 h °C. The tube-side coefficient is the same as before.

The weighted average outside heat transfer coefficient = 3976/[(3542/5898) +

(434/0.77/82)] = 531 kcal/m 2 h °C. [0.77 is the fin effectiveness as discussed in Appendix E.] The overall heat transfer coefficient is

U o = / 1 531 0 0002 + . + ( . 0 901 0 1595 / . ) × . 0 0508 × ln ( . 50 8 44 / ) // 2 37 + 00 .0 002 × . 50 8 44 / + ( . 50 8 44 / )( × / 1 4632 ) = . 0 00188 + . 0 0002 + . 0 000553 + 0 . 0 00023 + . 0 000249 = . 0 00311

or U o = 321 kcal/m h C 2 °

Surface area required = (3976 × 860)/24/321 = 444 m 2 = 30 × 3 × 0.901 × N d or N d = 5.47. Use six rows deep. The corrected outside coefficient = 8937 × 0.74 = 6613 kcal/m 2 h °C. The weighted average outside coefficient = 3976/[(3542/6613) + (434/.77/82)] = 536 kcal/m 2 h °C (not much different from the earlier run). Hence, six rows should be fine.

Figure 6.3 shows the scheme of a condensing economizer. Tube and fin materials are of stainless steel as dilute acid is formed. Chapter 1 discusses the acid dew point correlations. More care should be taken when the flue gas contains sulfuric acid vapor as it will con- dense first and then the water vapor. Provision is made for proper drainage and handling of corrosive condensate while designing condensing economizers.

Miscellaneous Boiler Calculations 317

Wall Temperature of Uninsulated Duct, Stack

If in a steam plant, boiler ducts or stacks are not insulated, the heat loss from the casing can

be substantial. However, a more important issue is the low stack wall temperature, which can cause acid dew point corrosion problems if the flue gas contains sulfur components. Water dew point temperature may also be reached at low loads of boiler, causing casing corrosion in the long run. Hence, plant engineers should have an idea of the casing wall temperatures when stacks are not insulated. These calculations assume significance particularly at low ambient and high wind velocity conditions. The calculations that follow show how one may estimate the stack wall temperature (Figure 6.4).

Let W kg/s of flue gas at a temperature of t g1 °C enter the stack. The heat loss from the

casing to the environment in kW/m 2 is given by

 ( V ++ 21 

q = . 5 67 × 10 ∈ ( 273 + t c ) − ( 273 + a  + . 0 00195 ×

where q is the heat loss, kW/m 2 t a ,t c are ambient and casing temperatures, °C

V is the wind velocity, m/min ∈ is the emissivity of casing

The temperature drop across the gas film is given by t g −t w1 = 1000q (d o /d i )/h i (6.6)

where

h i is the flue gas convective heat transfer coefficient inside the stack, W/m 2 K

d o ,d i are outside and inside stack diameters, m t g ,t w1 are average gas temperature and inner wall temperature, °C

Tg2 Detail A

Stack wall Insulation V,T 0 Stack 1 2

Stack wall temperature. Note: (1) T g , (2) T w1 , (3) T w0 , (4) T c , and (5) T 0 .

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

As shown in Appendix B, the convective heat transfer coefficient h i is given by

. 0 0278 CW 08 .

(6.7a) where W = flue gas flow, kg/s

C p , µ, k are the flue gas specific heat, viscosity, and thermal conductivity at the average film temperature of 0.5(t g +t w1 ). C is given in Table B.2 for a few common gas streams. The temperature drop across the stack wall is given by

qd ln d /d o ( o i )

(This drop is small and can be neglected, or we can assume t w1 =t w2 =t c ) The total heat loss Q = 3 14 . o d Hq , (6.9) where H is the stack height, m. The exit gas temperature from stack t g 2 = t g 1 − Q/ W C g p

) (6.10) Average gas temperature t g = 05 . t g 1 + t g 2 . (6.11)

The determination of stack wall temperature is an iterative process as explained later.

1. Assume an average gas temperature t g .

2. Assume an average gas casing t c .

3. Calculate q using (6.5).

4. Calculate Q using (6.9).

5. Calculate t g2 using (6.10) and t g from (6.6).

6. Calculate h i using a film temperature of 0.5(t g +t c ).

7. Check t g using (6.6).

8. If t g calculated in steps 5 and 7 agree, the assumed casing temperature is correct; else, another iteration is warranted.

Example 6.4

13.86 kg/s (110,000 lb/h) of flue gases from the combustion of natural gas enter an uninsulated boiler stack with an outer diameter of 1.27 m (50 in.) and an inner diameter

of 1.2195 m (48 in.) at 210°C (410°F). Flue gas analysis is % volume CO 2 = 8, H 2 O = 18, N 2 = 71, O 2 = 3. If ambient temperature is 21°C (70°F) and wind velocity is 38 m/min (125 ft/min), determine the stack wall temperature, gas temperature leaving the stack, and the total heat loss. Casing emissivity is 0.9. Stack wall thermal conductivity is

43 W/m K (25 Btu/ft h °F). Stack height = 15 m (49 ft).

Miscellaneous Boiler Calculations 319

Solution

As the first trial value, let us assume that the casing temperature is the average of ambient and gas temperatures to start with = 0.5 × (210 + 21) = 116°C

Casing loss q = 5.67 × 10 −11 ∈ [(273 + t c ) 4 − (273 + t a ) 4 ] + 0.00195 × (t c −t a ) 1.25 × [(V + 21)/21] 0.5

q= . 5 67 10 × − 11 × 09 . ×  273 116 + 4 − 273 21 + 4  + . 0 00195 × 116 21 − . × 11 25 / 59 21 05 .

= . 0 787 0 969 1 756 + . = . kW/m 2 ( 555 Btu/ft h 2

The drop across the stack wall can be shown to be = 1756 × 1.27 × ln(1.27/1.2195)/2/43 = 1.05°C and can be neglected; hence, t w1 ,t w2 need not be calculated.

Estimate h i at a gas film temperature of 0.5 × (117 + 210) = 163.5°C. From Table B.2, C for natural gas products of combustion in the following equation is 159.4.

h i = 0.0278C W 0.8 /d i 1.8 = 0.0278 × 159.4 × 13.86 0.8 /1.2195 1.8 = 25.4 W/m 2 K (4.47 Btu/ft 2 h °F). Gas temperature drop = 1756/25.4 = 69°C. Hence, t g = 117 + 69 = 186º C .

The heat loss in the stack = 3.14 × 1,756 × 2.7 × 15 = 223,310 W = 223.31 kW (0.762 MM Btu/h). The exit gas temperature = 210 − 223.310/13.86/1.153 = 196°C (using average gas

specific heat of 1.153 kJ/kg°C from Appendix F). Average gas temperature t g is 203°C. The difference in the computed value of t g (203°C versus 186°C) is large, and hence, a few

more iterations are required to arrive at t g and hence t c .

Use 121°C for casing wall temperature in the next iteration. =

q Casing loss − = . 5 67 10 × 11 ∈ [( 273 + t

c ) 4 − ( 273 + t a )] 4 + . 0 00195 × ( t c − tt a ) . 1 25 × [ ( V + ) / 21 21 ] 05 . = 0.848 + 1.033 = 1.881 kW/m 2 = 1881 W/m 2 . Inner wall temperature = 121 + 1881 × 1.27 ln(1.27/1.2195)/2/43 = 122°C. Hence, (t g −t c ) = 1881/25.36 = 74.2°C or t g = 122 + 74.2 =

196.2°C. Correcting for film temperature of (122 + 196.2)/2 = 159°C, h i will be unchanged

at 25.36 W/m 2 K. Heat loss in stack = 3.14 × 1.881 × 2.7 × 15 = 239.2 kW. Gas temperature

drop in stack = 239.2/13.86/1.153 = 15°C. Hence, exit gas temperature = 210 − 15 = 195°C. Average gas temperature = 0.5(195 + 210) = 202.5°C. The difference has come down but is still large. Try 127°C as casing temperature.

q = . 2 2 0367 kW/m Gas temperature drop .† = 2036 7 25 36 80 . / . = ° Ct .

g = 1 227 80 207 + = °C. Correcting for film temperature of 0.5 × (127 + 207) = 167°C, h i = 25.5 W/m 2 K

Heat loss in stack = 3.14 × 2.0367 × 2.7 × 15 = 259 kW. Gas temperature drop = 259/13.86/1.153 = 16°C. Stack exit gas temperature = 210 − 16 = 194°C. Hence, average gas

temperature t g = 202°C, which agrees somewhat closely with the estimated temperature

of 207°C. Hence, casing temperature is about 124°C–126°C, and exit gas temperature from stack about 202°C. With a computer program, one can fine-tune the results con- sidering variations in gas properties. Hence, plant engineers should be concerned about uninsulated stacks or ducts at gas temperatures particularly when the flue gas contains sulfur compounds. At low boiler loads, the casing temperature will be even lower as

h i decreases further; at low ambient and high wind velocity conditions, the condition worsens, and it may be better to insulate the stack.

Flue gas recirculation is commonly used in steam generators for NO x reduction. Depending on the fuel analysis, combustion temperature and emission levels of NO x anywhere from 10% to 25% of flue gases are recirculated from economizer exit to the forced draft fan suction in gas-fired units. If this duct is not insulated, the condensation of water vapor can occur in the recirculation duct causing corrosion and operational problems for the fan.

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

Condensation over Finned Tubes