Heat Transfer Inside and Outside Tubes

First the number of tubes is determined based on the following equation. Assumption for

a reasonable tube-side velocity is made (see Equation B.5).

where W g is the flue gas flow, kg/s ρ is the average density of flue gas, kg/m 3

d is the tube inner diameter, m

V is the gas velocity, m/s Then based on flow/tube, the tube-side heat transfer coefficient is determined (see

Appendix B), neglecting nonluminous heat transfer coefficient,

08 . ( C/ p ) k = . 0 0278 C(W /N g )

C = (C p /µ) 0.4 k 0.6 and N is the total number of tubes through which the flue gas (or air) is flowing. N = number of tubes wide × number of tubes deep and W g the gas flow in kg/h. properties C p —kcal/kg °C, µ—kg/m h, k—kcal/m h °C, h c —kcal/m 2 h °C. The heat transfer coefficient outside the tubes may be obtained using correlations discussed in Appendix C. Here too, since the temperatures are not high, nonluminous heat transfer coefficient may be neglected. (A computer program may be written to evaluate the nonluminous coefficient also, but here for illustration, we are limiting to convective heat transfer coefficients.)

03 . Nu = 0 35 . Re Pr (4.26)

where Re = Gd/µ, Pr = µCp/k, Nu = h c d/k. Simplifying, h o = 0.35 FG 0.6 /d 0.4 where F = (C p /µ) 0.3 k 0.7

The gas properties are evaluated at the film temperature, which is approximated as follows: t f = (3t a +t g )/4 when air flows outside the tubes or as (3t g +t a )/4 when flue gas flows outside tubes. Some use the average of gas and air temperatures for preliminary estimates.

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

Example 4.18

At 375°C, 225,000 kg/h of flue gas flows inside the tubes of an air heater, while 200,000 kg/h of air at 25°C flows outside the tubes. Using tubes of size 50.8 × 46.4 mm size a suitable tubular air heater if air is to be heated to 180°C. The flue gas analysis is typical natural gas products of combustion with % volume

CO 2 = . 8 29 , HO 2 = . 18 17 , N 2 = . 71 08 , O 2 = . 2 46 .

MW = . 8 29 × 44 + . 18 17 18 × + 71 .. 08 × 28 + . 2 46 × 32 = .. 27 6

Let tubes be arranged in inline fashion with S T = 82 mm and S L = 70 mm. Fouling factor on air and gas sides is 0.0002 m 2 h °C/kcal.

Solution

The duty of the air heater = 200,000 × 0.2455 × (180 − 25) = 7.61 MM kcal/h = 225,000 ×

0.99 × (375 − t) × 0.2831 or exit flue gas temperature = 254°C. (The specific heats of air and flue gas were obtained from Appendix F at the average air and flue gas temperatures. One percent casing heat loss was assumed and hence the 0.99 factor.)

Computing h i The number of tubes is estimated assuming some flue gas velocity. Let us use 22 m/s.

Density ρ of flue gas at average gas temperature of (375 + 254)/2 or 314°C = 27.6 × 101,325/

(273 + 314)/8314 = 0.573 kg/m 3 . Then, N = 1.246W g /(ρd i 2 V) = 1.246 × 62.5/(0.573 × 0.0464 ×

0.0464 × 22) = 2860. Gas properties at 314°C: C p = 0.2831 kcal/kg °C, µ = 0.0988 kg/m h, k = 0.0371 kcal/m h °C by interpolation from Tables F.6 through F.8.

08 h .

c = h i = . 0 0278 w C p / µ k 06 . / d 18 i . = . 0 0278 CWN g / / d 1 i .. ( 8 ) ( )

2 . / /. 0 0464 . 48 5 kcal/m h° C

Computing h o Approximate film temperature = (3 × 102 + 314)/4 = 155°C using average air temperature

of 102°C and flue gas temperature of 314°C. (One may compute the actual film temperature after estimating h i and h o and revise the calculations if necessary.)

Air properties at 155°C: C p = 0.2477 kcal/kg°C, µ = 0.087 kg/m h, k = 0.0305 kcal/m h °C by interpolation from Tables F.1 through F.3. Let number of tubes wide N w = 50, number deep N d = 2860/50 = 57. This is just the

first trial. We have to check the tube length and gas-, air-side pressure drops and then review this design. These calculations are for illustrative purposes only. We have to assume a height of tubes H; compute h o and then U o and then see if the air heater will transfer the required duty. Let tube length L = 3 m.

G , = 200 000

= , 42 735 kg/m h 2

. 0 0305 Nu = . 0 35 24 95 × , 33 06 . × . 0 7065 03 . = 137 1 . = o h d/k .

h o = 137 1 0 0305 0 0508 82 3 . × = . / . . kca ll/m h C 2 °

Waste Heat Boilers 255

Tube wall resistance is computed as shown in Appendix A. 1/U o = 1/82.3 +0.0002 + 0.000066 + 0.0002 × 50.8/46.4 + (1/48.5) × 0.0508/0.0464

= 0.01215 + 0.0002 + 0.000066 + 0.000219 + 0.02257 = 0.0352 or U o = 28.4 kcal/m 2 h °C (0.000066 is the resistance of tube wall)

To calculate the corrected LMTD, use the correction factor from Figure A.1. R = (25 − 180)/(254 − 375) = 1.28. P = (254 − 375)/(25 − 375) = 0.345. F = 0.93 for single-pass cross- flow exchanger.

LMTD =

  211 5 C . Corrected LMTD = 0 .9 × 93 211 5 . = 197 ° C .

  ln ( 229 195 / ) 

Required surface area A = 7.61 × 10 6 /197/28.4 = 1360 m 2 = 3.14 × 0.0508 × 2860 × L or L = 2.98 m. Provided height is 3 m. Hence, the exchanger can transfer the duty of 7.61 MM kcal/h.

Pressure Drop outside Tubes

For cross-flow outside tubes, ΔP = 0.204fG 2 N d /ρ a from Appendix C. G = 42,375/3,600 = 11.77 kg/m 2 s. Density ρ a at 103°C is 28.74 × 101,325/8,314 /(273 +

102) = 0.934 kg/m 3 .N d = 57 rows deep. S T /d = 82/50.8 = 1.614 S L /d = 70/50.8 = 1.378

Reynolds number computed at average air temperature of 103°C: Re = 42,735 × 0.0508 /0.0806 = 26,934.

Friction factor f (see Appendix C) for inline arrangement = Re −0.15 [0.044 + 0.08 S L /d/ {(S T /d − 1) (0.43+1.13d/SL) } = 26,934 −0.15 [0.044 + 0.08 × 1.378/{(1.614 − 1) (0.43+1.13//1.378) } =0.053.

∆P = . 0 204 11 77 × . 2 × × 57 0 053 0 934 . / . = 91 mm wc Pressure Drop inside Tubes

Pressure drop inside tubes involves estimation of inlet and exit losses due to gas velocity and the friction loss inside the tubes (Appendix B).

ΔP = 0.08262fL e vw 2 /d i 5 , where w is the flow per tube in kg/s, L the tube length in m, di the tube inner diameter in m, and ΔP is in mm wc. The average density of flue gas = 0.573 kg/m 3 from above at 314°C. Specific volume v = 1.745 m 3 /kg. f may be taken as 0.02 from Appendix B. w = 225,000/2,860/3,600 = 0.0218 kg/s.

∆P

. 0 02862 0 02 3 1 745 = 2 × . ×× . × ( . 0 0218 ) 5 /. 0 0464 = 7 mm wc

Density at 375°C = 27.6 × 101,325/(273 + 375)/8,314 = 0.519 kg/m 3 , and at an exit temperature of 254°C, density = 0.638 kg/m 3 by taking ratio of temperatures. Gas inlet velocity = V 1 = 1.246 W g /ρ/N/d i 2 = 1.246 × (225,000/3,600)/0.519/2,860/.0464 2 = 24.4 m/s. Velocity head VH 1 = 24.4 × 24.4 × 0.519/2/9.8 = 15.8 mm wc. Inlet loss = 0.5 VH 1 = 8 mm wc. Exit gas velocity at 254°C = 24.4 × 0.519/0.638 = 19.8 m/s. VH 2 = 19.8 × 19.8 × 0.638/2/9.8 = 12.8 mm wc. Exit loss = 1 velocity head = 12.8 mm wc. Total pressure drop = 8 + 7 + 13 = 28 mm wc.

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

Checking the Design Using the NTU Method Appendix A describes the elegant NTU method, which has been used in several perfor-

mance evaluation calculations throughout the book.

In our case, the flue gas is unmixed while air can be considered mixed. From Appendix A, the effectiveness factor from Table A.4 is

ε = 1 − exp{ − 1 /[ C 1 − exp( − NTC C × )]}

WC min = 200 000 0 2455 49 100 , × . = , WC max = 225 000 0 99 0 2831 63 06 , × . × . = , 00

C = , 49 100 63 060 0 7786 /, = . NTU = UA/WC min = . 28 4 1360 49 100 0 78 × /, = . 666

∈= − 1 exp { −  1 − exp ( − . 0 7866 0 7786 × . )  /0 . 7786 } = . 0 445

Q = 0.445 × 49,100 × (375 − 25) = 7.647 MM kcal/h. Agrees closely with the design duty. Tube Wall Temperature

One has to check if the lowest tube wall temperature is below or close to the acid or water dew point temperature to ensure that corrosion does not occur due to condensa- tion of water or acid vapor in which case suitable materials as discussed in Chapter 3 should be used. At the cold end, Figure 4.43 shows the temperature distribution.

One should estimate the actual h i and h o values corresponding to 254°C flue gas and 25°C air temperatures and use that U o to compute the heat flux and lowest tube wall temperatures. However, for illustrating the procedure, let us use the same h i and h o as earlier.

Heat flux q o = 28.4 × (254 − 25) = 6503 kcal/m 2 h.

Temperature drop across the gas film = 6503 × 0.0227 = 147.6°C. Temperature drop across the inside fouling layer = 6503 × 0.000219 = 1.4°C Temperature drop across tube wall = 6503 × 0.000066 = 0.4°C. Temperature drop across air fouling layer = 6503 × 0.0002 = 1.3°C Temperature drop across air film = 6503 × 0.01215 = 79°C Hence, tube outer wall temperature = 25 + 79 + 1.3 = 105.3°C.

Flue gas 0.0227

Temperature distribution at the cold end of air heater. Note: (A) Gas film, (B) gas-fouled layer, (C) tube wall, (D) air-fouled layer, and (E) air film.

Waste Heat Boilers 257

Air film temperature is 25 + 79 = 104°C. One may revise the runs and compute h i and h o based on corrected air film and average gas temperatures. However, tube wall will

be at about 105°C, which is above the water dew point which for 18.17% volume water vapor corresponds to 58°C or 136°F from steam tables. If sulfuric acid vapor had been present, then we would have to increase the air inlet temperature; steam–air exchangers are commonly used for increasing the cold-end temperature as at part load conditions, the tube wall temperature will be even lower.

Part Load Performance Example 4.19

Determine the performance of the aforementioned air heater when 130,000 kg/h of flue gas with same analysis flows inside the tubes at 330°C while air flow is 100,000 kg/h at 25°C.

Solution

The NTU method may be applied as it is direct and gives the duty from which the exit gas and air temperatures may be computed.

U o value may be obtained as a first approximation as follows: With a computer program, one may obtain accurate U values accounting for variations in gas properties with temperature. Let us use the values of h i and h o calculated earlier and apply multi- plication factors for an initial estimate.

h i = 48.5 × (130,000/225,000) 0.8 = 31.3 kcal/m 2 h °C. Flue gas–side resistance corrected for tube OD = 50.8/46.4/31.3 = 0.035 m 2 h °C/kcal.

06 h .

o = . 82 3 × ( 100 000 200 000 , / , ) = . 54 3 kcal/m h C /h 2 . 1 o = . 0 0184 mh ° 2 °°C/kcal .

1 /U o = . 0 0184 + . 0 0002 + . 0 000066 + . 0 000219 + . 0 035 = . 0 0538 8 8 or U

18 56 kcal/m h C o 2 = . °

WC min = 100,000 × .24 = 24,000. WC max = 130,000 × 0.275 = 35,750. C = 24,000/35,750 = 0.671 (Specific heat values were assumed but have to be corrected based on actual gas and film temperatures.)

NTU = . 18 56 1360 240 000 1 0517 × / , = .

∈=− 1 exp  −  { 1 − exp ( − . 1 0517 0 671 × . ) }} /. 0 671   = . 0 53

Q = . 0 53 24 000 × , × 330 25 − = . 3 88 MM kcal/h . ( )

Exit air temperature = 25 + 3.88 × 10 6 /24,000 = 187°C and exit gas temperature = 330 − 3.88 × 10 6 /35,750 = 221°C. With a computer program, one can compute the air and gas properties at the corrected average film and gas temperatures. This exercise is left as an exercise to the readers. The tube wall temperature will also be lower now.

Heat flux at cold end (using the same U o though the actual U o will be lower at the cold end) = 18.56 × (221 − 25) = 3628 kcal/m 2 h. Low-end tube wall temperature = 25 + 3628 × (0.0002 + 0.0184) = 92.5°C. Hence, we see that the tube wall temperature is much lower at lower loads.

Appendix E illustrates with an example why fins are not used in air heaters or gas to gas exchangers where the tube-side and gas-side heat transfer coefficients are in the same range of values.

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

Heat Transfer Inside and Outside Tubes