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

8.48

Q:

Discuss the simplified approach to designing fire tube boilers.

A:

Engineers often must estimate the size of heat transfer equipment such as heat exchangers, gas coolers, boilers, and economizers for preliminary costing and to check space requirements. With the approach presented here, one can quickly determine one or more configurations to accomplish a certain amount of heat transfer. One can also size equipment so as to limit the pressure drop without performing lengthy calculations. Life-cycle costing can then be applied to select the optimum design.

Two situations will be discussed [8].

1.The tube-side heat transfer coefficient governs the overall heat transfer. Examples: Fire tube boilers; gas coolers; heat exchangers in which a medium such as air or flue gas flows on the tube side and a fluid with a high heat transfer coefficient flows on the outside. Phase changes can also occur on the outside of the tubes.

2.The shell side governs. Examples: Water tube boilers, steam–air exchangers, and gas–liquid heat transfer equipment. See Q8.49.

Tube-Side Transfer Governs

In a fire tube boiler, gas flows inside the tubes and a steam–water mixture flows on the outside. The gas heat transfer coefficient is small, about 10–20 Btu=ft2 h F, compared to the outside coefficient of 2000–3000 Btu=ft2 h F. The metal resistance is also small; hence the gas-side coefficient governs the overall coefficient and the size of the equipment.

The energy transferred is given by

Because the inside coefficient governs U, we can rewrite Eq. (4) as follows (neglecting lower order resistances, such as ho, metal resistance, and fouling factors, which contribute to about 5% of U):

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TABLE 8.35 Factors F1=Cp; F2=Cp; F2, and F3 for Air and

Flue Gasa

aFlue gas is assumed to have 12% water vapor by volume.

Gas Pressure Drop

Now consider gas pressure drop. The equation that relates the geometry to tubeside pressure drop in in.H2O is

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TABLE 8.36 Values of K1 as a Function of Tube Diameter and Length

TABLE 8.37 Values of K2 as a Function of Tube Diameter and Length

K1 and K2 appear in Tables 8.36 and 8.37 respectively, as a function of tube ID and length. In the turbulent range, the friction factor for cold-drawn tubes is a function of inner diameter.

Using Eq. (97), one can quickly figure the tube diameter and length that limit tube pressure drop to a desired value. Any two of the three variables N; L, and di determine thermal performance as well as gas pressure drop. Let us discuss the conventional design procedure:

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1.Assume w, calculate N.

2.Calculate U, using Eqs. (4) and (90).

3.Calculate L after obtaining A from Eq. (89).

4.Calculate DPi from Eq. (95).

If the geometry or pressure drop obtained is unsuitable, repeat steps 1–4. This procedure is lengthy.

Some examples will illustrate the simplified approach. The preceding equations are valid for single-pass design. However, with minor changes one can derive the relationships for multipass units (e.g., use length ¼ L=2 for twopass units).

Example 1

A fire tube waste heat boiler will cool 66,000 lb=h of flue gas from 1160 F to 440 F. Saturation temperature is 350 F. Molecular weight is 28.5, and gas pressure is atmospheric. If L is to be limited to 20 ft due to layout, determine N and DPi for two tube sizes: (1) 2 1.77 in. (2 in. OD, 1.77 in. ID) and (2) 1.75 1.521 in.

Solution. Use Eq. (92) to find N. Use 2 in. tubes. F1=Cp from Table 8.35

Compute DPi using Eq. (95). From Table 8.37, K2 is 0.035. Compute the gas specific volume.

¼ 1:23 in:H2O

Repeat the exercise with 1.75 in. tubes; length remains at 20 ft. From Eq. (92) we note that for the same thermal performance and gas flow, N0:2L=di0:8 ¼

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a constant. The above concept comes in handy when one wants to quickly figure the effect of geometry on performance. Hence,

With smaller tubes, one needs fewer tubes for the same duty. This is due to a higher heat transfer coefficient; however, the gas pressure drop would be higher. From Table 8.37, K2 ¼ 0.076 for 1.521 in. tubes. From Eq. (95),

DPi ¼ 9:3 10 5 66;000 2 333

¼ 8:95 in:H2O

Example 2

Size the heat exchanger for 2.0 in. tubes with a pressure drop of 3.0 in.H2O. For the same thermal performance, determine the geometry.

Solution. The conventional approach would take several trials to arrive at the right combination. However, with Eq. (97), one can determine the geometry

From Table 8.36, we can obtain several combinations of tube diameter and length that have the same K1 value and would yield the same thermal performance and pressure drop. For the 1.77 in. ID tube, L is 21.75 ft. Use Eq. (92) to calculate the number of tubes.

N0:2

2:197 ¼ 0:606 0:73 21:75 ð66;000Þ0:2 ð1:77Þ0:8

N ¼ 402

Thus, several alternative tube geometries can be arrived at for the same performance, using the preceding approach. One saves a lot of time by not calculating heat transfer coefficients and gas properties.

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Life-Cycle Costing

Such techniques determine the optimum design, given several alternatives. Here, the major operating cost is from moving the gas through the system, and the installed cost is that of the equipment and auxiliaries such as the fan. The lifecycle cost is the sum of the capitalized cost of operation and the installed cost:

LCC ¼ Cco þ Ic

The capitalized cost of operation is

Ca ¼ 0:001 PHCe

where the fan power consumption in kW is

P ¼ 1:9 10 6 Wi DPi rZ

The above procedure is used to evaluate LCC. The alternative with the lowest LCC is usually chosen if the geometry is acceptable. (Ce is cost of electricity.) and H is the number of hours of operation per year.

8.49

Q:

Discuss the simplified approach to designing water tube boilers.

A:

Whenever gas flows outside a tube bundle—as in water tube boilers, economizers, and heat exchangers with high heat transfer coefficients on the tube side— the overall coefficient is governed by the gas-side resistance. Assuming that the other resistances contribute about 5% to the total, and neglecting the effect of nonluminous transfer coefficients, one may write the expression for U as

where the outside coefficient, ho, is obtained from

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Equation (99) is valid for both in-line (square or rectangular pitch) and staggered (triangular pitch) arrangements. For bare tubes, the difference in ho between inline and staggered arrangements at Reynolds numbers and pitches found in practice is 3–5%. For finned tubes, the variation is significant.

Substituting Eqs. (17)–(21) into Eq. (99a) and (99b) and simplifying,

F2 is given in Table 8.35. Gas transport properties are computed at the film temperature.

A ¼ pdoNwNd L=12

Combining the above with Eq. (89) and simplifying gives

Q=DT ¼ UA ¼ p0:9G0:6F2doNwNd L=12do0:4 ¼ 0:235 F2G0:6NwNd Ldo0:6

Substituting for G from Eq. (21),

The above equation relates thermal performance to geometry. When there is a phase change, as in a boiler, further simplification leads to

If the tube diameter and pitch are known, one can estimate Nd or G for a desired thermal performance.

Let us now account for gas pressure drop. The equation that relates the gas pressure drop to G is Eq. (28) of Chapter 7:

DPo ¼ 9:3 10 10 G2f Nd r

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For in-line arrangements, the friction factor is obtained from Eq. (29) of Chapter 7:

f ¼ Re 0:15 X

where

Another form of Eq. (28) of Chapter 7 is

F3 is given in Table 8.35. With Eq. (107), one can easily calculate the geometry for a given tube bank so as to limit the pressure drop to a desired value. An example will illustrate the versatility of the technique.

Example

In a water tube boiler, 66,000 lb=h of flue gas is cooled from 1160 F to 440 F. Saturation temperature is 350 F. Tube outside diameter is 2 in., and an in-line arrangement is used with ST ¼ SL ¼ 4 in. Determine a suitable configuration to limit the gas pressure to 3 in.H2O.

Let us use Eq. (106). Film temperature is 0.5 (800 þ 350) ¼ 575 F. Interpolating from Table 8.35 at 475 F, F3 ¼ 0:643. Gas density at 800 F is 0.031 lb=ft3 from Example 1.

DPo ¼ 4:78 10 10 G2:25 ð4 2Þ

ln 1160 350 440 350

ð0:044 þ 0:08 2Þ

20:75 0:643 0:031

¼128 10 10 G2:25 ¼ 3

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Hence, G ¼ 5200 lb=ft2 h. From Eq. (21) one can choose different combinations of Nw and L:

NwL ¼ 66;000 12=ð2 5200Þ ¼ 76

If Nw ¼ 8, then L ¼ 9.5 ft.

Calculate Nd from Eq. (104):

2:197 ¼ 2:82 0:426Nd =ð52000:4 1 20:4Þ or Nd ¼ 74

Thus, the entire geometry has been arrived at.

8.50

Q:

How is the bundle diameter of heat exchangers or fire tube boilers determined?

A:

Tubes of heat exchangers and fire tube boilers are typically arranged in square or triangular pitch (Fig. 8.20). The ratio of tube pitch to diameter could range from

1.25to 2 depending on the tube size and the manufacturer’s past practice. Looking at the triangular pitch arrangement, we see that half of a tube area

is located within the triangle, whose area is given by

Area of triangle ¼ 0:5 0:866p2 ¼ 0:433p2

If there are N tubes in the bundle, then

Total area occupied ¼ 0:866Np2

If the bundle diameter is D, then 3:14 D2=4 ¼ area of bundle ¼ 0.866Np2, or

In practice, a small clearance is added to the above number for manufacturing purposes.

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