Pressure Drop inside Tubes

Pressure drop inside the tubes is also an important piece of information while design- ing superheaters, air heaters, economizers. Often, consideration of pressure drop and heat transfer coefficient determines the initial cost and cost of operation. Hence, pressure drop data along with the heat transfer coefficient help to optimize a design as will be shown in Chapter 2. The basic equation for pressure drop is

Δ P = 810 × 10 −6 fL e vw 2/ d i 5 (B.6) where ΔP is the pressure drop in KPa (gas-steam-liquids).

390 Appendix B: Tube-Side Heat Transfer Coefficients and Pressure Drop

Friction factor f varies with Reynolds number inside the tube and the tube roughness factor. Typically, flow in boilers, superheaters, economizers, and air heaters is turbulent, and hence, the friction factor is taken as a function of tube inner diameter as shown in Table B.5. Table B.6 shows the pressure drop equations in all systems of units.

TABLE B.5

Friction Factor as a Function of Tube Inner Diameter Tube inner diameter (mm)

126 200 250 Friction factor f

0.0245 0.023 0.021 0.0195 0.018 0.0175 0.0165 0.016 0.014 0.013 Note: Standard pipe sizes are shown in Table B8.

TABLE B.6

Pressure Drop Equations

SI Units

British Units

Metric Units

Δ P = 810 × 10 −6 fL e vw 2 /d i 5 Δ P = 3.36 × 10 −6 fL e vw 2 /d i 5 Δ P = 0.6375 × 10 −12 fL e vw 2 /d i 5 w, kg/s

w, kg/h Δ P, kPa

w, lb/h

Δ P, kg/cm 2 L e ,m

d i ,m Δ P = 0.08262fL e vw 2 /d i 5 Δ P = 93 × 10 −6 fL e vw 2/ d i 5 Δ P = 6.382 × 10 −9 fL e vw 2/ d i 5 Δ P, mm wc in above equation

d i , in.

Δ P, mm wc Notes:

Δ P, in wc

f = friction factor inside the tubes. L e = effective length of tube, m. One may use Table B.7 to determine the effective lengths in a coil or piping with bends and fittings. w = flow per tube = total flow of fluid/ streams, kg/s (if there are multiple streams, then based on effective length of each stream and tube diameter, flow and pressure drop equations must be solved to arrive at the flow in each stream. d i = tube inner

diameter, m. v = specific volume of fluid, m 3 /kg.

TABLE B.7

Effective Lengths for Bends, Valves, Fitting

Fitting

L e /d i 90° square bend

57 45° elbows

15 90° elbows, standard radius

32 90° elbows, medium radius

26 90° elbows, long sweep

20 180° elbows, close return bends

75 180° medium radius return bends

50 Gate valves open

7 Gate valves, one-fourth closed

40 Gate valves, half closed

200 Gate valves, three-fourths closed

800 Globe valves open

Appendix B: Tube-Side Heat Transfer Coefficients and Pressure Drop 391

Example B.6

Determine the pressure drop in a 76 mm schedule 80 line carrying water at 38°C and 9000 kPa if the total equivalent length is 305 m (1000 ft). Flow is 4.8 kg/s (38,085 lb/h)

and specific volume of water is 0.001 m 3 /kg (0.016 ft 3 /lb) from steam tables. Inner diameter of pipe = 76 mm (3 in.).

Solution

Δ P = 810fL e vw 2/ d i 5

= 810 × 10 −6 × 0.0175 × 305 × 0.001 × 4.8 2 /0.076 5 = 39.3 kPa In British units, ΔP = 3.36 × 10 −6 fL e vw 2/ d i 5 = 3.36 × 10 −6 × 0.0175 × 1,000 × 0.016 ×

38,085 2 /3 5 = 5.6 psi. In metric units, ΔP = 0.6375 × 10 −12 fL e vw 2/ d i 5 = 0.6375 × 10 −12 × 0.0175 × 305 × 0.001 ×

17,280 2 /0.076 5 = 0.4 kg/cm 2 .

Example B.7

Determine the gas pressure drop of 0.021 kg/s (166.6 lb/h) of flue gas with a molecular weight of 29 at atmospheric pressure and at an average gas temperature of 500°C, which flows inside a tube of inner diameter 45 mm (1.77 in.). Tube total effective length = 10 m. (It is common practice in boiler industry to express pressure drop of air or flue gas in mm wc as it can be easily measured in this unit.)

The density of gas at atmospheric pressure = MW × 273/(273 + 500)/22.4 = 0.457 kg/m 3 . Hence, specific volume = 1/density = 2.19 m 3 /kg (35 ft 3 /lb). Friction factor from Table B.5 may be taken as 0.02.

∆P = . 0 08262 fLvw e 2 / d i 5 = . 0 08262 0 02 10 2 19 0 021 × . × × . × .

= . mm wc In British units, ΔP = 93 × 10 −6 fL e vw 2 /d i 5 = 93 × 10 −6 × 0.02 × 35 × 32.8 × 166.6 2 /1.77 5 =

3.41 in. wc (Table B.8).

TABLE B.8

Thickness of Standard Steel Pipe

Pipe Size (in.)

OD (in.)

Sch 40

Sch 80

392 Appendix B: Tube-Side Heat Transfer Coefficients and Pressure Drop

References

1. P. Abbrecht and S. Churchill, The thermal entrance region in fully developed turbulent flow, AIChE Journal , 6, 268, June 1960.

2. V. Ganapathy, Applied Heat Transfer, Pennwell Books, Tulsa, OK, 1982, p600. 3. V. Ganapathy, Industrial Boilers and HRSGs, CRC Press, 2003, Boca Raton, FL, p531.

Appendix C: Heat Transfer Coefficients Outside Plain Tubes

In order to determine the performance of a plain tube bundle such as boiler bank tubes or superheater tube bundle (Figure C.1), one must determine the outside heat transfer coefficient

h o , which consists of h c , the convective heat transfer coefficient, and h n , the nonluminous heat transfer coefficient, which is significant above 700°C. This appendix explains how h c and the gas side pressure drop ΔP g may be estimated. There are numerous correlations, but the few commonly used are cited in the following text. (For finned tubes, see Appendix E.) Fishenden and Saunders correlation for h c for cross-flow of gases over plain tube banks takes the following form:

03 Nu . = 0 35

h F Re Pr (C.1)

F h depends on the tube geometry, whether staggered or inline, and is shown in Table C.1 (Figure C.2). If one reviews this table, the correction factor for inline and staggered arrangement for industrial boilers will be nearly the same for tube spacing typically seen in boiler practice, namely, S T /d of 2–3 and S L /d of 2–3. In the case of shell and tube heat exchangers where tubes sizes are very small (on the order of 0.375–1 in. versus 1.5–2.5 in. in boilers), smaller longi-

tudinal spacing is used in which case staggered arrangement may give a slightly higher h c value over inline arrangement. Hence, from heat transfer point of view, not much is gained by using staggered arrangement in boilers or heaters. On the other hand, as shown later, the gas pressure drop across the staggered tube bundle is much larger than an inline bundle, and hence, staggered plain tubes are generally avoided in boiler practice. One may also note from Table C.1 that as we reduce S L in inline arrangement, the correction factor or the heat transfer coefficient decreases, while in staggered arrangement, it increases! One cannot decrease S L as it will affect the ligament efficiency and increase the thickness of the drum or headers.

A conservative correlation for both inline and staggered arrangements is as follows [1]:

= 0 33 . Re 06 Pr 0 33 (C.2) where

Nu .

Reynolds number Re = Gd/µ

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

G=W g /[N w L(S T – d)] W g is the flow over tubes, kg/s

d is the tube outer diameter, m N w is the number of tubes/row or tubes wide L is the effective length of tube, m S T ,S L are transverse and longitudinal pitch, m µ is the viscosity of gas, kg/m s or Pa s

Nusselt number Nu = h c d/k, where h c , the convective heat transfer coefficient, is in W/m 2 K;

d is in m; and k, the thermal conductivity of the gas, is in W/m K Prandtl number Pr = µC p /k, where C p , the gas specific heat, is in J/kg K

394 Appendix C: Heat Transfer Coefficients Outside Plain Tubes

To process

From economizer or deaerator

Steam out

Air out

Air in

Steam in

FIGURE C.1

Typical tube banks in boilers. (a) Evaporator; (b) tubular air heater; (c) superheater.

Note that all thermal and transport properties for heat transfer coefficient for plain tubes are estimated at the gas film temperature, which is approximately the average of the tube wall and gas temperature.

Substituting for Nu, Re, and Pr in (C.2) and simplifying, we have

h = . 0 33

F G /d 06 . 04 . where F k = . 0 67 C . 0 33 / µ . c 0 27 ( ) p

(C.3)

Grimson’s correlation is widely used in boiler design practice: Nu N = B Re (C.4)

Appendix C: Heat Transfer Coefficients Outside Plain Tubes 395

TABLE C.1

F h for Inline and Staggered Arrangements

S L /d 1.25 1.5 2 3 1.25 1.5 2 3 S T /d

Inline Bank

Staggered Bank

FIGURE C.2

(a) Inline and (b) staggered tube banks.

B and N have been arrived at for Reynolds number varying from 2,000 to 40,000. Table C.2 shows the B and N factors, while Table C.3 shows the correction factor for the number of rows deep while Table C.4 shows the correction factor for angle of attack.

NO T E : Based on his experience and analysis of field data of numerous boilers, the author has found that these correlations underpredict the heat transfer coefficient by about 10%. Hence, a correction factor of 1.1 may be used if one wants to be less conservative.

396 Appendix C: Heat Transfer Coefficients Outside Plain Tubes

TABLE C.2

Grimson’s Values of B and N

S T /d 1.25 1.5 2.0 3.0 S L /d

TABLE C.3

Correction F n Factor for Rows Deep

No. of Rows Deep

TABLE C.4

Correction for Angle of Attack Degree

90 80 70 60 50 40 30 20 10 F n

Example C.1

Flue gases of 20 kg/s (158,688 lb/h) from the combustion of natural gas (analysis:

% volume CO 2 = 8, H 2 O = 18, N 2 = 71, O 2 = 3) at 400°C (752°F) average film tem-

perature flow over a tube bundle with more than 10 rows deep. Tube OD = 50.8 mm (2.0 in.), and transverse and longitudinal pitch 101.8 and 101.8 mm (4.0 in.), respec-

tively. Effective tube length is 3.5 m (11.5 ft), and there are 12 tubes/row. What is h c

for inline and staggered arrangements? From the table of gas properties in Appendix F,

Appendix C: Heat Transfer Coefficients Outside Plain Tubes 397

We can see that at 400°C, specific heat C p = 0.289 kcal/kg °C (1210 J/kg K), viscosity µ = 0.1088 kg/m h (0.0000302 kg/m s), and thermal conductivity k = 0.0414 kcal/m h °C (0.0481 W/m K).

20 = . 9 39 kg/m s 2 ( , 6 898 lb/ft h 2

R Re Gd/ = =

Using Grimson’s correlation, B = 0.482, N = 0.556 for staggered, and B = 0.229, N = 0.632 for inline arrangement for S T /d = S L /d = 2.

For inline arrangement, Nu = 0.229 × 15,795 0.632 = 103.1 or h c = 103.1 × 0.0481/.0508 = 97.6

W/m 2 K (83.9 kcal/m 2 h °C) (17.2 Btu/ft 2 h °F).

For staggered arrangement, Nu = 0.482 × 15,795 0.556 = 104 or h c = 98.56 W/m 2 K

(84.9 kcal/m 2 h °C) (17.36 Btu/ft 2 h °F).

Using Fishenden and Saunders equation,

F h for inline arrangement is 1.0, while for staggered, it is about 1.02.

0 . × 35 15,795 0 . 6 × . 759 0 . 3 = 164 0 . = h c ×

00 . 481 or

h c = 1 00. 8 W/m K 86 7 kcal/m h C 2 2 °

) ( . 17 75 Btu/ft h F ° )

h c is about 2% more for staggered arrangement. Hence, it may be seen that the difference in h c between staggered and inline arrange-

ments is not significant. Hence, gas pressure drop over plain tubes should be estimated to see which arrangement is better. Gas pressure drop gives an idea of the incremental addition to the fan power consumption. This is discussed later.

Pressure Drop inside Tubes