Boiling Heat Transfer Coefficient h o

Numerous correlations are available in the literature for the estimation of h o . It must be understood that the variation among correlations for h o can be wide, but this will not affect U as discussed in Appendix A, as h i is much smaller than h o and hence it is the limiting resistance to energy transfer.

The temperature difference between the tube wall and saturation temperatures is given by the well-known Roshenow correlation:

fg f H  fg   k f  Heat flux q o = ht o ( w − t s ) (4.7)

where

C pf is the specific heat of saturated liquid, kcal/kg °C ρ f ,ρ g is the density of saturated liquid, vapor kg/m 3 σ is the surface tension, kg/m µ f is the viscosity of saturated liquid, kg/m h k f is the thermal conductivity of saturated liquid, kcal/m h °C

ΔH fg is the latent heat, kcal/kg t w ,t s are temperatures of tube wall and steam,°C

q o is the heat flux, kcal/m 2 h

Surface tension, thermal and transport properties, and properties of steam are given in Tables F.22 and F.13.

An iterative procedure is required to determine h o accurately as h o requires q o for its estimation, and to estimate U o (from which q o is calculated), we need to know h o ! Also the equation may appear tedious as steam properties are required. However, an accurate estimate of h o is not that important as U is governed by h i and not by h o .

Let us look at some other simplified correlations for h o . For steam pressure P s in the range of 1–30 atm (a) and heat flux q o from 50,000 kcal/m 2 h (0.058 MW/m 2 ) (18,370 Btu/ft 2 h)

to 1000,000 kcal/m 2 h (1.16 MW/m 2 ) (36,740 Btu/ft 2 h),

logPs h

o =3 07 . . q Ps 0 17 o

(4.8) For P s above 30 atma (atmospheres absolute) and q o from 100,000 to 1 × 10 6 kcal/m 2 h,

. 0 01 h Ps

o =45 . qe o (4.9) For pressures in the range of 1–50 atma and q o from 16,000 to 730,000 kcal/m 2 h,

h o =3 qP o 07 . s 02 . † (4.10)

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

Example 4.7

Determine the size of fire tube boiler required to cool 45,370 kg/h of flue gases from

815°C to 260°C. Flue gas analysis is % volume CO 2 = 12, H 2 O = 12, N 2 = 70, O 2 = 6. Flue gas pressure is atmospheric. Steam pressure is 10.5 kg/cm 2 g. Tubes used: 600 number

of 50.8 × 45 mm carbon steel SA 178a. (Flow per tube is based on experience. For this tube ID, about 75 kg/h is reasonable. Based on actual boiler size and gas pressure drop, one may revise this value. A computer program will be helpful to study the options.)

Fouling factor on gas side is 0.0004 m 2 h °C/kcal (0.002 ft 2 h °F/Btu) and that on steam side is 0.0002 m 2 h °C/kcal (0.001ft 2 h °F/Btu). Tube thermal conductivity = 37 kcal/m h

°C (24.8 Btu/ft h °F). Use a heat loss of 1%, a blowdown of 3%, and a feed water temperature of 104°C. Saturation temperature is 185°C at the operating pressure of steam.

Solution

First determine the tube-side convective heat transfer coefficient h c . Gas properties at

average gas temperature of 538°C may be estimated using the procedure discussed in Appendix F based on the flue gas analysis. The results are as follows:

C p = 0.2876 kcal/kg °C = 1204 J/kg K, µ = 0.1258 kg/m h = 3.484 × 10 −5 kg/m s, k = 0.0476 kcal/m h °C = 0.0552 W/m K. From Appendix B,

. 0 0278 CW 08 .

d 18 . i

w = 45,370/3,600/600 = 0.021 kg/s, d i = 0.045 m. C = (1,204 × 10 5 /3.484) 0.4 × 0.0552 0.6 = 182. h c = 0.0278 × 182 × 0.021 0.8 /0.045 1.8 = 61.1 W/m 2 K (52.55 kcal/m 2 h °C) (10.75 Btu/ft 2 h °F).

In addition to convective heat transfer coefficient, the nonluminous heat transfer coefficient has to be estimated. However, it will be small as the beam length is the tube inner diameter. The nonluminous heat transfer coefficient is worked out in Example

4.5, Appendix D. h n = 3.93 W/m 2 K. Hence, h i =h c +h n = 65 W/m 2 K = 55.9 kcal/m 2 h (11.4 Btu/ft 2 h °F). Let us use U i = 0.95 × 55.9 = 53.1 kcal/m 2 h °C or U o = 53.1 × 45/50.8 = 47 kcal/m 2 h °C in order to estimate the average heat flux and then h o .

q = 47 × ( 538 185 − ) = , 16 591 kcal/m h 2 o . Using (4.10), h o = 3 × 16,121 0.7 × 11.1 0.2 = 4279 kcal/m 2 h °C.

Let us estimate q o using (4.6). Obtain steam properties from Appendix F. σ = 0.002813 lb/ft = 0.004186 kg/m from Table F.22. From steam tables and Table F.13,

µ f = 0.365 lb/ft h = 0.543 kg/m h. ΔH fg = 858 Btu/lb = 476.6 kcal/kg. ρ g ,ρ f = 5.78 and 882 kg/m 3 . C pf = 1.05 kcal/kg °C, k f = 0.389 Btu/ft h °F = 0.5788 kcal/kg °C. Using (4.6),

1.05(t w –t s )/476.6 = 0.013[(16,591/0.543 × 476.6) × (0.004186/876.2) 0.5 ] 0.33 × (0.543 × 1.05 /0.5788) = 0.00663 or (t w –t s ) = 3.0°C

or h o = 16,591/3.0 = 5530 kcal/m 2 h °C

1/U o = (50.8/45)/55.9 + 0.0004 × 50.8/45 + (0.0508/2/37) × ln(50.8/45) + 0.0002 + 1/5530 = 0.02 + 0.000452 + 0.000083 + 0.0002 + 0.0001808 = 0.0209 or U o = 47.8 kcal/m 2 h °C

q o = 47.8 × (538 − 185) = 16,882 kcal/m 2 h, close to that obtained earlier. The various

resistances are in m 2 h °C/kcal:

Gas-side film: 0.02 Gas-side fouling: 0.000452 Tube wall resistance: 0.000083

Waste Heat Boilers 233

Steam-side fouling: 0.0002

Steam-side heat transfer = 0.0001808 (resistances in m 2 h °C/kcal) Duty = 45,370 × 0.99 × 0.2876 × (815 − 260) = 7.17 MM kcal/h = 8337.2 kW. From steam tables,

saturated steam enthalpy = 664.4 kcal/kg, saturated liquid enthalpy = 187.8 kcal/kg, and

feed water enthalpy = 104.3 kcal/kg. Steam generation = 7.17 × 10 6 /[(664.4 − 104.3) + 0.03 × (187.8 − 104.3)] = 12,744 kg/h

LMTD =  (

)  = 261 °C C.

ln  ( 815 185 − )( / 260 − 185 )

Surface area required (based on tube OD) = A o = 7,170,000/261/47.8 = 575 m 2 (A i = 575 × 45/50.8 = 509 m 2 ). With 600 tubes, the length required = 575/(3.14 × 0.0508 × 600) = 6.0 m. One may add some margin if required.

Let us now compute the gas-side pressure drop inside the boiler tubes. This consists of three parts: the inlet loss at the tube entrance, the exit loss at tube exit, and the friction loss inside the tubes.

Gas Pressure Drop inside Tubes To estimate the pressure drop, density of flue gas is required, which in turn requires the

molecular weight (MW).

MW = 0.12 × 44 +0.12 × 18 +0.70 × 28 +0.06 × 32 = 28.96. Density at gas inlet at atmo-

spheric pressure = ρ g1 = (28.96/22.4) × 273 /(273 + 815) = 0.324 kg/m 3 . At exit, density

ρ g2 = 0.324 × (273 + 815)/(273 + 260) = 0.662 kg/m 3 . Gas velocity at inlet = 4 × (45,270/600/3,600)/0.324/(π × 0.045 × 0.045) = 40.7 m/s; at exit,

gas velocity = 19.87, say, 20 m/s (taking ratio of gas densities). Inlet loss = 0.5 × velocity head = 0.5 × ρ g1 ×V 2 /2/9.8 = 0.5 × 0.324 × 40.6 2 /2/9.8 = 13.6 kg/m 2 = 13.6 mm wc (9.8 m/s 2 is acceleration due to gravity). Exit loss = 1 × VH = 0.661 × 20 2 /2/9.8 = 13.5 mm wc. Friction loss in tubes: ΔP g = 810 × 10 −6 fLvw 2 /d i 5 (from Table B.6). Flow w is in kg/s, L and d i in m, v the specific volume in m 3 /kg, and ΔP g in

kPa. Gas density at average gas temperature of 537.5°C = (28.96/22.4) × 273/(273 + 537.5) = 0.435 kg/m 3 or v = 2.296 m 3 /kg.

( 45370 600 3600 / ∆P / )

(0.02 if the friction factor [see Appendix B]. 6.1 m is the approximate tube length in the earlier equation including tube sheet thickness.)

Total gas pressure drop = 13.6 + 55 + 13.3 = 82 mm wc. One has to add the inlet and exit vestibule losses and any duct losses. The pressure drop value 82 mm wc is only for the losses inside the tubes. If ferrules are used, the gas inlet velocity will be much higher, and the inlet loss will be much higher depending on the ferrule inner diameter.

Evaluating Tube Wall Temperatures The maximum heat flux outside the tubes q o is required to determine the maximum

tube wall temperatures. q o is also important as it gives an idea if DNB is a concern. (In fire tube boilers, heat flux outside the tubes is important as steam is outside the tubes, while for water tube boiler, q i is important as steam is inside the tubes). DNB is more likely in reformed gas boilers in hydrogen plants where the tube-side heat transfer coefficients can be much higher than the heat transfer coefficients with typical flue gases, about six to seven times. This we shall see later.

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

The maximum heat flux is generally used to determine the maximum tube wall temperature. U o has to be computed at the gas inlet temperature. From Appendix F, at 815°C, C p = 0.3050 kcal/kg °C = 1277 J/kg K, µ = 0.1534 kg/m h = 4.261 × 10 −5 kg/m s

k = . 0 0613 kcal/mh C °= . 0 07128 WmK / h c = .0278 × 0.021 0.8 × (1277 × 10 5 /4.261) 0.4 × 0.07128 0.6 /0.045 1.8 = 67.3 W/m 2 K

= 57.8 kcal/m 2 h °C

h n using charts and equations from Appendix D = 7.7 W/m 2 K; hence, h i = 75 W/m 2 K = 64.5 kcal/m 2 h °C. Let us assume U i = 0.95 × 64.5 = 61.3 kcal/m 2 h °C or U o = 54.27 kcal/m 2 h °C. (U i × ratio of tube inner to outer diameter = U o ).

q o = 5427 . × ( 815 − 185 ) = 34190 kcal m h / 2

σ = 0.002813 lb/ft = 0.004186 kg/m from Appendix F. From steam tables in Appendix F, µ f = 0.365 lb/ft h = 0.543 kg/m h. ΔH fg = 858 Btu/lb = 476.6 kcal/kg ρ g ,ρ f = 5.78 and 882 kg/m 3 .C pf = 1.05 kcal/kg °C

k f = 0.389 Btu/ft h °F = 0.5788 kcal/kg °C. Using (4.6), 1.05(t w –t s )/476.6 = 0.013[(34,190/0.543 × 476.6) × (0.004186/876.2) 0.5 ] 0.33

× (0.543 × 1.05 /0.5788) = 0.008416 or (t w –t s ) = 3.8°C or

h o = 34,190/3.8 = 9,000 kcal/m 2 h °C

1/U o = (50.8/45)/64.5 + 0.0004 × 50.8/45 +(0.0508/2/37) × ln(50.8/45) + 0.0002 + 1/8997 = 0.0175 + 0.000452 + 0.000083 + 0.0002 + 0.0001111 = 0.01834 or

U o = 54.5 kcal/m 2 h °C

This is close to U o estimated earlier. Hence, let us proceed with this.

q o = . 54 5 × 815 185 −

( 2 ) = , 34 348 kcal/m h .

Temperature drop across gas film = 34,348 × 0.0175 = 601°C Temperature drop across gas fouling layer = 34,348 × 0.000452 = 15.5°C Temperature drop across tube wall = 34,348 × 0.000083 = 2.8°C Temperature drop across steam-side fouling layer = 34,348 × 0.0002 = 6.9°C Temperature drop across steam film = 34,348 × 0.0001111 = 3.8°C

Hence, tube inner wall temperature = 815 − 601 − 15.5 = 198.5°C and outer wall temperature = 195.7°C. For design purposes, one may use, say, 10–15°C margin.

Boiling Heat Transfer Coefficient h o