Determination of Heat Transfer Coefficient h c Convective Heat Transfer Coefficient

Determination of Heat Transfer Coefficient h c Convective Heat Transfer Coefficient

The ESCOA correlations that were revised in 1993 [2,3] for solid and serrated fins are widely used in the industry (Table E.2). For an inline bundle,

For a staggered bundle,

.  0 25 ( d + 2 h )   T

 NLS w ( T − A o ) 

where

A o =+2 d nbh (E.4)

A o is the obstruction area, m 2 /m

C p is the gas specific heat at average temperature, J/kg K

h c is the convective heat transfer coefficient, W/m 2 K

k is the thermal conductivity of gas, W/m K µ is the viscosity, kg/m s (Gas properties are computed at the average gas temperature.)

T g ,T f are absolute temperature of gas and fin, K

G is the gas mass velocity, kg/m 2 s

d is the outer diameter of tube, m N w is the tubes/row or number of tubes wide L is the effective length, m

h is the fin height, m

b is the fin thickness, m n is the fin density, fins/m W g is the gas flow, kg/s S T ,S L are transverse and longitudinal pitch, m S is the spacing between fins = 1/n-b, m

Appendix E: Calculations with Finned Tubes 417

TABLE E.2

ESCOA Revised Correlations: Factors C 1 –C 6 for Solid and Serrated Fins in Inline and Staggered Arrangements

Solid fins Inline

5 = 11 . − (. 0 75 1 5 − . e − 07 . N d ) e − 20 . SL/S T C 6 = 16 . − (. 0 75 1 5 − . e − 07 . N d ) e e −2 0 .( SL/ST )

J = CCC 1 3 5 [(d 2h)/d] [(t + 0.5 g + 460)/(t a + 460 )] 05 .

2 C 2 5 = 07 . + (. 0708 − . e − . 0 15 N d )[ e − 10 . SL/S T ] C 6 = 11 . + (. 1821 − . e − . N N 0 15 d ) e − 20 .( SL/S T ) − [. 0708 − . e − . N 0 15 2 d ] e − 06 .( SL/S T )

JCCCd = 1 3 5 [( + 2 h)/d ] [( 05 . t g + 460 )/( t a + 460 )] . 0 25

f = CCC 2 4 6 [( d + 2 h)/d ] [( 05 . t g + 460 )/( t a + 460 )] − . 0 25

Serrated fins Inline

C 5 = 11 . − (. 0 75 1 5 − . e − 07 . N d ) e − 20 . SL/S T C 6 = 16 . − (. 0 75 1 5 − . e − 07 . N d ) e e −2 0 .( SL/ST ) 2

2 C 2 5 = 07 . + (. 0708 − . e − . 0 15 N d )[ e − 10 . SL/S T ] C 6 = 11 . + (. 1821 − . e − . N N 0 15 d ) e − 20 .( SL/S T ) − 2 [. 0708 − . e − . N 0 15 d ] e − 06 .( SL/S T )

J = CCC 1 3 5 [( d + 2 h)/d ] [( 05 . t g + 460 )/( t a + 460 )] . 0 25

fCCC = 2 4 6 [( d + 2 h)/d ] [( 05 . t g + 460 )/( t a + 460 )] − . 0 25

Source: Fintube Technologies, Tulsa, OK.

418 Appendix E: Calculations with Finned Tubes

Fin Efficiency and Effectiveness For Both Solid and Serrated Fins

1 − ( 1EA − ) f

A (E.5)

where A f ,A t are area of fins and total area per unit length, m 2 /m.

For Solid Fins

A f = π n × ( 2 dh + 2 h 2 + bd + 2 bh )

(E.6)

A t = A f + π d ( 1 − nb

(E.7)

Fin efficiency E is given by the following formula using Bessel functions:

( e o ) ×

E =  2 r /m/ r 2 − r 2   ( )( ) ( )( ) 

1 I mr K mr e 1 o − K 1 mr I mr e 1 o

(E.8)

 I o ( mr K mr m o )( 1 e ) + 1 ( I mr K e )( o mr o ) 

where

Bessel functions I 1 ,I o ,K 1 ,K o are shown in Table E.4

r e = (d + 2h)/2 and r o = d/2, m

A simpler formula for E is

2  (E.9) 1 + . 33 mh

{ ( d + 2 h /d } 

K m is the thermal conductivity of fin, W/m K.

For Serrated Fins

π dn  2 h × ( ws b + ) +× b ws

d ( 1 − nb )

(E.12)

Appendix E: Calculations with Finned Tubes 419

E = tanh mh ( )

m =√  h o ( b + ws /K /b/ws ) m

(E.14)

ws is the width of serration, m.

Nonluminous Heat Transfer Coefficient

With finned tubes, due to the large surface area for a given volume, the beam length will

be much smaller than a comparable plain tube bundle at the same tube spacing, and hence, the nonluminous heat transfer coefficient is not high. The beam length may be approxi- mated as

(E.15)

Using the procedure discussed in Appendix D, the gas emissivity and the nonluminous heat transfer coefficient h n may be estimated. Generally, this is quite small compared to plain tubes due to the larger surface area in a given volume.

Gas Pressure Drop across Finned Tube Bundles

. 2 0 205 faGN +

(E.16)

where ΔP g is the gas pressure drop, mm wc N d is the number of rows deep ρ g is the density of gas at average gas temperature, kg/m 3

For staggered arrangement,

.  0 25 − ( d + 2 h )   T

f = friction factor = CCC 2 4 6 

(E.18)

For inline arrangement,

.  0 25 ( d + 2 h )   T

f = CCC 2 4 6 

(E.19)

420 Appendix E: Calculations with Finned Tubes

(E.20)

 4 N

dx ( t g + 273 ) 