Free gas area   ( S

 2 Free gas area   ( S

(E.21)

 Total flow area

Factors C 1 –C 6 are given in Table E.2 for solid and serrated fins. These are the revised ESCOA correlations.

Tube Wall and Fin Tip Temperatures

For solid fins, the relation between the tube wall temperature t b and fin tip temperature t f is given by

( t g − t f )  K mr 1 ( e ) × o ( I mr e ) + 1 ( I mr e ) × K mr o ( e )

( (E.22) t g − t b )  K mr 1 ( e ) × o ( o II mr ) + K mr o ( o ) × 1 ( I mr e )

 For serrated fins,

( (E.23)

) cosh mb ()

A good estimate of t f can be obtained for either type of fin using

t f = t b + ( t g − t b ) × ( . 1 42 1 4 − . E )

(E.24)

The fin base temperature t b is estimated as follows:

t b =+ t i qR o ( 3 + R 4 + R 5 (E.25)

where R 3 ,R 4 , and R 5 are the resistances of the inside fluid, fouling layer, and tube wall, respectively.

Heat 2 flux q

o = U o ( t g − t i ) , W/m

(E.26)

(Note that depending on the location in the gas path, one may obtain average heat flux or

maximum heat flux by using appropriate t g and t i values.)

The significance of various resistances and U o was shown in Appendix A.

A /A )

where A i = π. d i

Appendix E: Calculations with Finned Tubes 421

TABLE E.3

F Factors

Carbon steel

Hastelloy B

Note that these correlations have a margin of error, and hence, based on experience and field measurements, suitable correction factors are applied. The author’s experience is that these

correlations are conservative and underpredict actual h c (obtained in the field) by about 10%.

Weight of Finned Tubes

Solid fins w : = . 10 68 × Fbn d ( + h ) × ( h + . 0 03 )

(E.27)

Serrated fins w :† = . 10 68 × Fbnd × ( h + . 0 12

) (E.28)

w is the weight in lb/ft. Factor F is given in Table E.3. Note that d, h, b are in in. in (E.27) and (E.28) and n is fins/in.

Example E.1

A 50.8 × 44 mm solid finned superheater having a fin density of 78.7 fins/m of height 12.7 mm and thickness 1.52 mm is arranged in inline fashion at 101.6 mm square pitch. The supplier has used 18 tubes/row and 6 rows deep, 3.1 m long tubes with 9 streams.

Gas flow is 100,000 kg/h with the following analysis by volume% CO 2 = 3, H 2 O = 7, N 2 = 75, O 2 = 15. The bundle is designed with inlet and exit gas temperatures of 550°C and 479°C; steam flow is 25,000 kg/h (6.94 kg/s) at 369°C at 49 kg/cm 2 g (4,804 kPa) at

superheater exit, and saturated steam enters the superheater in counter-flow fashion at 265°C. Surface area is 188 m 2 . Check the suitability of the design.

Solution

Flue gas data from Appendix F at the average gas temperature of 514°C is C pg = 0.2755 kcal/kg °C, µ g = 0.127 kg/m h, k g = 0.0468 kcal/m h °C

A o = 0.0508 + 2 × 78.7 × 0.00152 × 0.0127 = 0.0538 m 2 /m

A f = π × 78.7 × (2 × .0508 × .0127 + 2 × .0127 × 0.0127 + .00152 × 0.0508 + 2 × .00152 × .0127) = 0.427 m 2 /m

A t = 0.427 + π × .0508 × (1 − 78.7 × .00152) = 0.567 m 2 /m

S = clearance between fins = (1/78.7) − .00152 = 0.0112 m

422 Appendix E: Calculations with Finned Tubes

Convective Heat Transfer Coefficient G= W g /[N w L(S T −A o )] = 100,000/[18 × 3.1 × (0.1016 − 0.0538)] = 37491 kg/m 2 h

= 10.414 kg/m 2 s Re = Gd/µ = 37,491 × .0508/0.127 = 14,996

C 1 = 0.053 × [1.45 − 2.9 × (0.1016/0.0508) −2.3 ] × 14,996 −0.21 = 0.006059 C 3 = 0.2 + 0.65e −.25 × 12.7/11.2 = 0.6900 C 5 = 1.1 − [(0.75 − 1.5e −0.7 × 6 )]e (−2SL/ST) = 1.0016 Assume that average fin temperature is 400°C. After performing a round of calcula-

tions, we will be able to fine-tune this. h c = 0.006059 × 0.690 × 1.0016 × 37,491 × 0.2755 × (0.0468/0.127/0.2755) 0.67 × [(514 + 273)/

(400 + 273)] 0.5 × [(0.0508 + 0.0254)/0.0508] 0.5 = 69.6 kcal/m 2 h °C (80.93 W/m 2 K) Nonluminous Heat Transfer Coefficient

Beam length L b = (d + 2h)[(S T ×S L) /(d + 2h) 2 − 0.85] = (0.0508 + 0.0254) (0.1016 × 0.1016/0.0762 2 − 0.85) = 0.0707 m. Note that beam length is smaller with finned tubes, and hence, the nonluminous heat transfer coefficient will be small for similar gas conditions.

From Appendix D, for p c = 0.03, p w = 0.07, L b = 0.0707,

K = (0.8 + 1.6 × .07) (1 − .38 × 0.787) × 0.1/(0.1 × 0.0707) 0.5 = 0.76 ∈ g = 0.9 × (1 − e −0.76 × 0.0707 ) = 0.047. From Appendix D, h n = 5.67 × 10 −8 × 0.047 × (787 4 − 673 4 )/(787 − 673) = 4.17 W/m 2 K = 3.58 kcal/m 2 h °C

(Average gas temperature is 514°C, and average fin temperature is assumed as 400°C.)

h o = 69.55 + 3.58 = 73.13 kcal/m 2 h °C (85 W/m 2 K) (14.96 Btu/ft 2 h °F) m = [2 × (73.13)/30/0.00152] 0.5 = 56.6 1/m (fin thermal conductivity of 30 kcal/m h °C

(25.8 W/m K) (20.15 Btu/ft h °F) is used.

mh = 56.6 × 0.0127 = 0.719

Using the simplified approach, E = 1/[1 + 0.33 × 0.719 2 × (0.0762/0.0508) 0.5 ] = 0.827 Effectiveness η = 1 − [(1 − 0.827) × 0.427/0.567] = 0.87

Using the Bessel functions (Table E.4), r e = 0.0381, r o = 0.0254, mr e = 56.6 × 0.0381 = 2.156, mr o = 56.6 × 0.0254 = 1.437 E = [2r o /m/(r e 2 −r o 2 )] × [I 1 (mr e )K 1 (mr o )–K 1 (mr e )I1(mr o )]/[I o (mr o )K 1 (mr e )+I 1 (mr e )K o (mr o )] I 1 R4 (2.156) = 1.843, K 1 (2.156) = 0.115, K 1 (1.437) = 0.306, I 1 (1.437) = 0.849, I o (1.437)

= 1.516, K o (1.437) = 0.234

Appendix E: Calculations with Finned Tubes 423

TABLE E.4

I o ,I 1 ,K o ,K 1 Values for Various Arguments

Source: Ganapathy, V., Applied Heat Transfer, Pennwell Books, Tulsa, OK, 1982, p501.

E = [2 × .0254/56.6/(0.0381 2 − 0.0254 2 )] × [(1.843 × 0.306 − 0.115 × 0.849)/(1.516 × 0.115 + 1.843 × 0.234)] = 0.856

η = 1 − [1 − (1.856) × 0.427/0.567] = 0.89 Ratio of external to internal area A t /A i = 0.567/π/0.044 = 4.1

Tube-side coefficient h i from Appendix B: C = 320 from Table B.4 at 5000 kPa (51 kg/cm 2 ) and 318°C. Mass flow/tube = 25,000/9 = 2778 kg/h = 0.772 kg/s (9 streams used)

18 = 2000 WmK / . 2 ( 1720 kcal/m h C 2 °

) ( 3 5 . Btu/ft h F 51 7 ° )

424 Appendix E: Calculations with Finned Tubes

1/U o = 1/(0.89 × 73.13) + 0.0002 + (0.0508/2 × 30) × ln(0.0508/0.044) × (0.567/π/0.044)

= 0.01926 or U o = 51.9 kcal/m 2 h °C (60.35 W/m 2 K) (10.61 Btu/ft 2 h °F) Superheater Duty

Q = 100 000 0 99 0 2755 , × . × . × ( 550 − 479 ) = . 1 936 MM kcal/h ( . 2 25 MW ) ( . 7 682 M MM Btu/h )

Enthalpy absorbed by steam = 1.936 × 10 6 /25,000 = 77.45 kcal/kg (139.4 Btu/lb) (324.3 kJ/kg). Exit steam enthalpy = 744.65 kcal/kg or temperature = 369°C from steam tables.

LMTD = [(550 − 369) – (479 − 265)]/ln[(550 − 369)/(479 − 265)] = 197°C.

Surface area required = 1.936 × 10 6 /(197 × 51.9) = 189 m 2 and matches the value shown by the boiler supplier. Hence, the design seems reasonable. Estimation of heat flux q o = 51.9 × (550 − 369) = 9394 kcal/m 2 h (10.92 kW/m 2 ) (3464 Btu/ft 2 h) (the U o value should have computed at the gas inlet temperature and steam exit temperature conditions and will show a higher value. However, here we are inter- ested in showing the methodology for computing heat flux and the tube and fin tip temperatures.)

Fin base or maximum tube wall temperature t b = 369 + 9394 × (0.00238 + 0.00082 +

0.0005) = 404°C. Suitable correction factors or margins may be used for tube and fin material selection.

One may use Equation E.22 to estimate the fin tip temperature. (t g –t f )/(t g –t b ) = [0.115 × 2.28 + 1.591 × 0.114]/[0.14 × 1.513 + 0.2625 × 1.591] = 0.795 (550 − t f )/(550 − 404) = 0.795 or t f = 433°C. Average fin temperature at the hot end = (404 +

433)/2 = 418C. Similarly, at the cold end, one has to check the average fin temperature

and use that value for t a in the equation for h c . However, the difference will be small and hence neglected in this exercise. Using (E.24), t f = 405 + (550 − 405) × (1.42 − 1.4 × .856) = 432°C If local heat transfer coefficients were used, t f and t b values would be higher, and a

computer program would help perform these calculations more accurately. Note that the tube-side coefficient at higher steam temperature will also be lower increasing the tube wall and fin tip temperatures. A computer program will be helpful in performing these iterations.

Gas Pressure Drop

2 = 2  ( T − A o ) / S T  =  ( . 0 1016 − . 0 0538 ) /. 0 1016  = . 0 221

a = (1 + B 2 )(t g1 −t g2 )/(4N d T g ) = (1 + 0.221 2 ) (479 − 550)/(4 × 6 × 787) = −0.00394

Appendix E: Calculations with Finned Tubes 425

Density at an average gas temperature of 514°C is ρ g = 12.17 MW/T g = 12.17 × 28.38/787 = 0.4388 kg/m 3

= . 0 205 × ( . 0 0676 0 00394 − . ) × . 10 414 × ∆P 6 g = . 19 3 mm wc .

Steam-Side Pressure Drop Let us estimate the total developed length. From Appendix B, L e = 3.1 × 6 × 18/9 + 32 ×

.044 × 18/9 (9 streams) = 54 m. (Standard bend data from Table B.7 were used). Friction

factor is 0.02. From steam tables, the specific volume of steam at 50 kg/cm 2 g and 315°C

is obtained as 0.04868 m 3 /kg.

Hence, using the equation from Table B.1, ΔP = 0.6375 × 10 −12 × 0.02 × 54 × (25,000/9) 2 ×

0.04868/0.044 5 = 1.57 kg/cm 2 (22.3 psi) (154 kPa).

( 25000 9 / ) × . let velocity of steam = 0 0394 = .m 19 6 m/s

Exit velocity = 19.6 × 0.0554/0.0394 = 27.5 m/s (specific volumes of steam at inlet and exit are 0.0394 and 0.0554 m 3 /kg, respectively)

Inlet velocity head = 19.6 × 19.6/(2 × 9.8 × 0.0394) = 497 kg/cm 2 . Loss = 0.5 × 497 (half

the velocity head) = 248 kg/m 2 = 0.025 kg/cm 2 .

Exit velocity head = 27.5 × 27.5/(2 × 9.8 × 0.0554) = 696 kg/m 2 . Exit loss = 0.07 kg/cm 2 (one velocity head). Total pressure drop = 1.57 + 0.025 + 0.07 = 1.67 kg/cm 2 . It may be seen that these

calculations are tedious. Hence, a computer program developed for the purpose is used.