Simplified Procedure for Evaluating Performance of Plain Tube Bundles

There are numerous parameters involved in the evaluation of design and performance of plain tube bundles. A computer program would be the best tool for this exercise as numerous cases can be evaluated and the optimum chosen. However, as an academic exercise, a simplified procedure for convective plain tube bundles is presented later using which one can arrive at the geometry of a tube bundle for a specified duty and gas pressure drop. It is assumed that the nonluminous heat transfer coefficient is negligible, and only convective heat transfer is predominant. Such derivations also help plant engineers quickly visualize the effect of tube geometry or other factors on boiler performance. Keep in mind that in the following derivation, the effect of nonluminous heat transfer coefficient is neglected; this gives reasonable results up to 600°C–700°C.

From Appendix A, it was shown that U = 0.95h o for boilers as other resistances may be neglected.

h o is obtained from Nu = 0.35Re 0.6 Pr 0.3

Nu = h o d/k, Re = Gd/µ, Pr = µCp/k; all the properties in consistent units.

 NLS w ( T − d ) 

(W g , kg/h; h o , kcal/m 2 h °C; d, m; k, kcal/m h °C; G, kg/m 2 h; µ, kg/m h; C p , kcal/kg °C). Substituting for the various terms and simplifying, we have

where F 2 = 

 k (C.8b)

0 . 3325 G 0 . 6 F 2

(C.9)

(gas properties are computed at the average film temperature) Q

 . 0 3325 G 06 . F 2 

 π dN N L w d = . 1 044 FG 2 . NNL w d ∆ (C.10) T d

= 06 UA = 

Appendix C: Heat Transfer Coefficients Outside Plain Tubes 405

Substituting for G from (C.8a), we have

06 Q .

= UA = . 1 044 FWNNL 2 g w d

(C.11)

NL 0606 . .

( S /d 1 T − )

( S /d T − 1 )

This equation may be used to relate the tube geometry with duty. However, when there is

a phase change as in a boiler evaporator, the equation may be simplified further. ln [(T 1 −T s )/(T 2 −T s )] = UA/(W g C p ) (as seen in Appendix A)

0404 . 044 . g F W Nw L N / S /d d ( T − 1 06 .

( WC g p )

040 FNL . 4 044 06 .. .

N S /d ( T − 1

WC g 04 ( . p )

. 1 044 2 F /C N p d

(C.13a)

 If the tube geometry is known, one can estimate N d or G for a specific thermal perfor-

G 04 . ( S /d T − 1 ) d 04 .

mance. Now introducing the equation for gas pressure drop (inline which is widely used),

∆ = . 0 204 fGN 2 − 8 d 2 2 ∆ = . 1 574 10 × fGN d P 2 g ( if G is in kg/m s ) or P g ( G in k g g/m h )

(C.13b)

f = −0 15 . Re X (C.13c) where

X = . 0 044

. 0 08 S /d

(. 0 43 1 13 + . dS / L ( ) S /d

Substituting for Re = Gd/µ in above equation C.13c and for N d from (C.13), we have

 ( T 1 − T s )  ( S T − dX )

∆ P g = . 1 5077 10 × × G × ln 

)  Fd 3 ρ g

This is a very important equation linking gas pressure drop with thermal performance and duty. While a computer program would yield close results, equations such as this help one visualize the impact of variables on boiler design and performance.

406 Appendix C: Heat Transfer Coefficients Outside Plain Tubes

Example C.5

Flue gases of 29,945 kg/h from a steam generator are cooled from 627°C to 227°C in an

evaporator generating saturated steam at 177°C. Flue gas analysis is CO 2 = 8.29, H 2 O = 18.1, N 2 = 71.08, O 2 = 2.46% volume. S T =S L = 101.6 mm and d = 50.8 mm. Gas film temperature is about 300°C, and density at average gas temperature of 427°C = 0.48 kg/m 3 . From

Appendix F, C p = 0.2821 kcal/kg°C, µ = 0.0972 kg/m h, and k = 0.0364 kcal/m h°C at film temperature of 300°C. Determine the boiler geometry for a gas pressure drop of 76 mm wc.

First one has to determine G from (C.14). X = 0.044 + 0.08 × 2 = 0.204. F 3 = 0.0364 0.7 /.2821 0.7 /0.0972 0.7 = 0.681 Density ρ g = 27.58 × 273/22.4/(273 + 427) = 0.48 kg/m 3 . Hence,

G 24,812 kg/m h 2

24,812 = G =

or NL w = . 23 75 [ .

N Lx w 000 . 58 ]

If N w = 8, then L = 2.97 m. Let us get N d from (C.13). F 2 = (Cp/µ) 0.3 k 0.7 = (0.2821/.0972) 0.3 × 0.0364 0.7 = 0.135

Ln [T T/T

1 − s )( 2 − T s )  = . 1 044 ( 2 F /C N / G p ) d [ ( S /d T − 1 ) d ]

Thus, the entire geometry is arrived at. One can also see the effect of changing tube spacing, tube diameter, and so on.

Example C.6

If the number of rows deep is limited to 50, what happens to the gas pressure drop and exit gas temperature? Assume same cross section and tube spacing and gas inlet conditions.

Substituting in (C.16), Ln[[(T 1 −T s )/(T 2 −T s )] = 1.044 × (0.135/.2821) × 50/(24812 0.4 × 0.0508 0.4 ) = 1.441 or

[(T 1 −T s )/(T 2 −T s )] = 4.22 = (627 − 177)/(T2 − 177) or T 2 = 284°C

Using (C.14), we have ΔP g = 1.5077 × 10 −8 ×G 2.25 × ln[[(T 1 −T s )/(T 2 −T s )] × (S T − d)X/[F 3 d 0.75 ρ g ]

00 . 24 = 0 5 mm wc

One can fine-tune for the lower density due to the higher exit gas temperature; however, these equations may be used to get some quick ideas of the variables involved in design or performance of convective tube plain tube bundles.

Reference

1. V. Ganapathy, Applied Heat Transfer, Pennwell Books, Tulsa, OK, 1982, p130.

Simplified Procedure for Evaluating Performance of Plain Tube Bundles