01 POWER ISLAND / 01 CCPP / V. Ganapathy-Industrial Boilers and HRSG-Design (2003)

FIGURE 8.9 Tubular air heater.

Copyright © 2003 Marcel Dekker, Inc.

where tg and ta refer to the average of gas and air temperatures. Calculate U using

Metal resistance is neglected. Air-and gas-side pressure drops can be computed by Eqs. (26) and (28) of Chapter 7, after surfacing is done:

DPg ¼ 93 10 6 fw2 L þd55di

rg i

DPair ¼ 9:3 10 10 fG2 Nd

rair

It is also good to check for partial load performance to see if dew point corrosion problems are likely. Methods like air bypass or steam-air heating must be considered. Vibration of tube bundles must also be checked.

C and F are given in Table 8.24 for easy reference.

Example

A quantity of 500,000 lb=h of flue gas from a boiler is cooled from 700 F; 400,000 lb=h of air at 80 F is heated to 400 F. Design a suitable tubular air heater. Carbon steel tubes of 2 in. OD and 0.087 in. thickness are available.

Solution. Assume that duct dimensions are not a limitation. Hence, the bundle arrangement is quite flexible. Choose ST =d ¼ 1.5 and SL=d ¼ 1.25 in. inline; use a maximum flue gas velocity of 50 ft=s.

From an energy balance, assuming negligible losses and for a specific heat of 0.25 for gas and 0.24 for the air side,

Q¼ 500;000 0:25 ð700 tÞ

¼400;000 0:24 ð400 80Þ

¼30:7 106 Btu=h

TABLE 8.24 C and F Factors for Calculating hi and ho of Tubular Air

Copyright © 2003 Marcel Dekker, Inc.

Hence, the gas temperature leaving the air heater is 454 F. The average flue gas temperature is (700 þ 454)=2 ¼ 577 F. Let the molecular weight of the flue gas be 30. Then

From Eq. (69),

0:05 500;000

Nt ¼ 1:8262 0:0396 50 ¼ 3800

ST ¼ 3:0 in:; SL ¼ 2:5 in:

Let Nw ¼ 60. Hence, the width of the air heater is

60 312:0 ¼ 15 ft

Nd ¼ 63 because Nt ¼ Nw Nd ; so

Depth ¼ 63 2:5=12 ¼ 13:2 ft

¼ 7:2 Btu=ft2 h F

To estimate ho; G is required. This requires an idea of L. We must assume a value for the length and check later to see if it is sufficient. Hence, it is a trial-and- error approach. Try L ¼ 15 ft:

FGA ¼ ST12 d NwL ¼ 121 60 15 ¼ 75 ft2

G ¼ 400;000=75 ¼ 5333 lb=ft2 h

Average gas and air temperatures are

tg ¼ 577 F; ta ¼ 240 F

tf ¼ 3 577 þ 240 ¼ 492 F 4

Copyright © 2003 Marcel Dekker, Inc.

From Table 8.24, F is 0.105. Then

ho ¼ 0:9 53330:6 0:105=20:4 ¼ 12:3 Btu=ft2 h F

We must calculate FT , the correction factor for DT, for the case of one fluid mixed and the other unmixed. From Fig. 8.10 (single-pass crossflow),

R ¼ 700 454 ¼ 0:77 400 80

P¼ 400 80 ¼ 0:516 700 80

L ¼ 11:95 ft

Hence, the assumed L is not correct. Try L ¼ 11.0 ft.

FGA ¼ 1115 75 ¼ 55 ft2

G ¼ 7272 lb=ft2 h

Copyright © 2003 Marcel Dekker, Inc.

FIGURE 8.10 Crossflow correction factors (From Refs. 1 and 2).

Copyright © 2003 Marcel Dekker, Inc.

The calculated and assumed lengths are close to each other, and the design may be frozen. Check the metal temperature at the exit portion. Because the gasside resistance and air film resistances are 0.152 and 0.067, the metal temperature at the exit of the air heater can be calculated as follows. The drop across the gas film will be

0:152ð454 80Þ ¼ 260 F 0:152 þ 0:067

Metal temperature will be 454 7 260 ¼ 194 F.

If the flue gas contains sulfur, dew point corrosion may occur at the exit. The air-side heat transfer coefficient is high, so the drop across its film is low compared to the gas-side film drop. If we increase the flue gas heat transfer coefficient, the drop across its film will be low and the metal temperature will be higher.

and the air temperature is 80 F. Check the exit gas temperatures of gas and air.

A:

Assume the gas leaves the air heater at 400 F. Then

Q ¼ 300;000 0:25 ð620 400Þ

¼ 250;000 0:24ðt 80Þ ¼ 16:5 106

Air temperature leaving ¼ 355 F

Copyright © 2003 Marcel Dekker, Inc.

From Fig. 8.10,

P¼ 355 80 ¼ 0:51 620 80

R ¼ 620 400 ¼ 0:8 355 80

FT ¼ 0:9

Since the air heater can transfer more energy, assume a higher air temperature, 390 F, at the exit:

Q¼ 250;000 0:24ð390 80Þ ¼ 18:6 106

¼ 300;000 0:25 ð620 tÞ

Transferred Q ¼ 4:2 23;640 213

¼ 21:1 106 Btu=h

Copyright © 2003 Marcel Dekker, Inc.

Again, they don’t tally. Next, try Q ¼ 20 106 Btu=h.

Air temperature leaving ¼ 410 F

Gas temperature leaving ¼ 353 F

FT ¼ 0.75, DT ¼ 0.75 242 ¼ 182 F

Transferred Q ¼ 4:2 23;640 182 ¼ 18 106 Btu=h

Again, try an exit air temperature at 400 F. Then

Q¼ 250;000 0:24 ð400 80Þ ¼ 19:2 106 Btu=h

Transferred Q ¼ 4:2 193 23;640 ¼ 19:16 106 Btu=h

Q ¼ 19:2 106 Btu=h

The gas leaves at 364 F against 454 F at full load.

Metal temperature can be computed as before. At lower loads, metal temperature is lower, and the air heater should be given some protection. This protection may take two forms: Bypass part of the air or use steam to heat the air entering the heater to 100–120 F. Either of these will increase the average metal temperature of the air heater. In the first case, the air-side heat transfer coefficient will fall. Because U decreases, the gas temperature leaving the air heater will increase and less Q will be transferred. Hence, metal temperature will increase. In the second case, air temperature entering increases, so protection of the metal is ensured. Again, the gas temperature differential at the exit will be higher, causing a higher exit gas temperature.

Example

Solve the problem using the NTU method.

Copyright © 2003 Marcel Dekker, Inc.

Solution. Often the NTU method is convenient when trial-and-error

¼1 expf 1:25½1 expð 1:65 0:8Þ&g

¼0:59

Effectiveness ¼ 0:59 ¼ air temperature rise 620 80

Air temperature rise ¼ 319 F

Air temperature leaving ¼ 319 þ 80 ¼ 399 F

This compares well with the answer of 400 F. When U does not change much, this method is very handy.

8.30

Q:

Predict the exit gas and water temperatures and the energy transferred in an economizer under the following conditions:

tg1 ¼ gas temperature in ¼ 1000 F tw1 ¼ water temperature in ¼ 250 F

A ¼ surface area ¼ 6000 ft2 Wg ¼ gas flow ¼ 75,000 lb=h Ww ¼ water flow ¼ 67,000 lb=h

U ¼ overall heat transfer coefficient ¼ 8 Btu=ft2 h F Cpg ¼ gas specific heat ¼ 0.265 Btu=lb F

Cpw ¼ water specific heat ¼ 1 Btu=lb F

A:

Figure 8.11 shows the arrangement of an economizer. A trial-and-error method is usually adopted to solve for the duty of any heat transfer equipment if the surface area is known. This procedure is detailed in Q8.29. Alternatively, the number of transfer units (NTU) method predicts the exit temperatures and duty. For more on

Copyright © 2003 Marcel Dekker, Inc.

FIGURE 8.11 Economizer.

this theory, the reader is referred to any textbook on heat transfer [2]. Basically, the duty Q is given by

where e depends on the type of flow, whether counterflow, parallel flow, or crossflow. In economizers, usually a counterflow arrangement is adopted. e for this is given by

ðWCpÞmin ¼ 75;000 0:265 ¼ 19;875 ðWCpÞmax ¼ 67;000 1 ¼ 67;000

C¼ 19;875 ¼ 0:3 67;000

NTU ¼ 8 6000 ¼ 2:42 19;875

Substituting into Eq. (74) yields

1 expð 2:42 0:7Þ

e ¼ 1 0:3 expð 2:42 0:7Þ ¼ 0:86

Copyright © 2003 Marcel Dekker, Inc.