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

8.54b

Q:

Estimate the drop in water temperature of 1 in. thick insulation used in Q8.54a. The water flow is 7500 lb=h.

A:

The total heat loss has been shown to be 33,675 Btu=h. This is lost by the water and can be written as 7500 DT, where DT is the drop in temperature, assuming that the specific heat is 1. Hence

DT ¼ 33;675 ¼ 4:5 F 7500

By equating the heat loss from insulation to the heat lost by the fluid, be it air, oil, steam, or water, one can compute the drop in temperature in the pipe or duct. This calculation is particularly important when oil lines are involved, because viscosity is affected, leading to pumping and atomization problems.

8.55

Q:

In Q8.54 determine the optimum thickness of insulation with the following data.

Cost of energy ¼ $3=MM Btu Cost of operation ¼ $8000=year

Interest and escalation rates ¼ 12% and 7% Life of the plant ¼ 15 years

Total cost of 1 in. thick insulation, including labor and material, ¼ $5200; for 2 in. insulation, $7100; and for 3 in. insulation, $10,500

A:

Let us calculate the capitalization factor F from Q5.22.

F¼ 1:07 1 ð1:07=1:12Þ15 ¼ 10:5 1:12 1 1:07=1:12

Let us calculate the annual heat loss.

For L ¼ 1 in.,

8000

Ca ¼ 33;675 3 106 ¼ $808

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For L ¼ 2 in.,

8000

Ca ¼ 23;157 3 106 ¼ $555 For L ¼ 3 in.,

8000

Ca ¼ 18;604 3 106 ¼ $446

Calculate capitalized cost CaF.

For L ¼ 1 in.,

CaF ¼ 808 10:5 ¼ $8484

For L ¼ 2 in.,

CaF ¼ 555 10:5 ¼ $5827

For L ¼ 3 in.,

CaF ¼ 446 10:5 ¼ $4683

Calculate total capitalized cost or life-cycle cost (LCC):

For L ¼ 1 in.,

LCC ¼ 8484 þ 5200 ¼ $13;684

For L ¼ 2 in., LCC ¼ $12,927; and for L ¼ 3 in., LCC ¼ $15,183.

Hence we see that the optimum thickness is about 2 in. With higher thicknesses, the capital cost becomes more than the benefits from savings in heat loss. A trade-off would be to go for 2 in. thick insulation.

Several factors enter into calculations of this type. If the period of operation were less, probably a lesser thickness would be adequate. If the cost of energy were more, we might have to go for a greater thickness. Thus each case must be evaluated before we decide on the optimum thickness. This example gives only a methodology, and the evaluation can be as detailed as desired by the plant engineering personnel.

If there were no insulation, the annual heat loss would be

8.56

Q:

What is a hot casing? What are its uses?

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A:

Whenever hot gases are contained in an internally refractory-lined (or insulated) duct, the casing temperature can fall below the dew point of acid gases, which can seep through the refractory cracks and cause acid condensation, which is a potential problem. To avoid this, some engineers prefer a ‘‘hot casing’’ design, which ensures that the casing or the vessel or duct containing the gases is maintained at a high enough temperature to minimize or prevent acid condensation. At the same time, the casing is also externally insulated to minimize the heat losses to the ambient (see Fig. 8.21). A ‘‘hot casing’’ is a combination of internal plus external insulation used to maintain the casing at a high enough temperature to avoid acid condensation while ensuring that the heat losses to the atmosphere are low.

Consider the use of a combination of two refractories inside the boiler casing: 4 in. of KS4 and 2 in. of CBM. The hot gases are at 1000 F. Ambient temperature ¼ 60 F, and wind velocity is 100 ft=min. Casing emissivity is 0.9. To keep the boiler casing hot, an external 0.5 in. of mineral fiber is added. Determine the boiler casing temperature, the outer casing temperature, and the heat loss.

One can perform the calculations discussed earlier to arrive at the temperatures and heat loss. For the sake of illustrating the point, a computer printout of the result is shown in Fig. 8.22. It can be seen that the boiler casing is at 392 F, and the outermost casing is at 142 F. The heat loss is 180 Btu=ft2 h. The boiler casing is hot enough to avoid acid condensation, while the heat losses are kept low.

FIGURE 8.21 Arrangement of hot casing.

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FIGURE 8.22 Printout on casing temperatures.

8.57

Q:

What happens if ducts or stacks handling flue gases are not insulated? What would the gas or stack wall temperature be?

A:

This question faces engineers involved in engineering of boiler plants. If ducts and stacks are not insulated, the heat loss from the casing can be substantial. Also, the stack wall temperature can drop low enough to cause acid dew point corrosion.

Let the flue gas flow be W lb=h at a temperature of tg1 at the inlet to the duct or stack (Fig. 8.23). The heat loss from the casing wall is given by Eq. (110),

The temperature drop across the gas film is given by

tg tw1 ¼ q do=di hc

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where

hc ¼ convective heat transfer coefficient Btu=ft2 h F do; di ¼ outer and inner diameter of the stack, in.

where, from Eq. (12),

C ¼ Cp 0:4 k0:6

m

The duct wall temperature drop is given by Eq. (111), which can be rearranged to give

tw1 two ¼ qdo lnðdo=diÞ

24Km

where tw1; two are the inner and outer wall temperatures, F.

The total heat loss from the duct or stack is Q ¼ 3:14do H=12 where H is

FIGURE 8.23 Stack wall temperature.

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The above equations have to be solved iteratively. A trial value for tg2 is assumed, and the gas properties are computed at the average gas temperature. The casing temperature is also obtained through an iterative process. The total heat loss is computed and tg2 is again evaluated. If the assumed and calculated tg2 values agree, then iteration stops. A computer program can be developed to obtain accurate results, particularly if the stack is tall and calculations are better done in several segments.

Example

110,000 lb=h of flue gases at 410 F enter a 48 in. ID stack that is 50 ft long and 1 in. thick. If the ambient temperature is 70 F and wind velocity is 125 ft=min, determine the casing temperature, total heat loss, and exit gas temperature.

Flue gas properties can be assumed to be as follows at 400 F (or com-

¼ 601 Btu=ft2 h

Gas temperature drop across gas film ¼ 601=4.5 ¼ 134 F.

Temperature drop across the stack wall ¼

601 50 lnð50=48Þ 24 25

Hence stack wall outer temperature ¼ 400 7 134 7 2 ¼ 264 F.

It can be shown that at a casing or wall temperature of 256 F, the heat loss through gas film matches the loss through the stack wall. The heat loss ¼ 629 Btu=ft2 h, and total heat loss ¼ 411,400 Btu=h.

411;400 Gas temperature drop ¼ 110;000 0:265

The average gas temperature ¼ 410 7 14 ¼ 396 F, which is close to the 400 F assumed. With a computer program, one can fine-tune the calculations to include fouling factors.

Copyright © 2003 Marcel Dekker, Inc.

8.58

Q:

What are the effects of wind velocity and casing emissivity on heat loss and casing temperature?

A:

Using the method described earlier, the casing temperature and heat loss were determined for the case of an insulated surface at 600 F using 3 in. of mineral fiber insulation. (Aluminum casing has an emissivity of about 0.15, and oxidized steel, 0.9.) The results are shown in Table 8.40.

It can be seen that the wind velocity does not result in reduction of heat losses though the casing temperature is significantly reduced. Also, the use of lower emissivity casing does not affect the heat loss, though the casing temperature is increased, particularly at low wind velocity.

8.59a

Q:

How does one check heat transfer equipment for possible noise and vibration problems?

A:

A detailed procedure is outlined in Refs. 1 and 8. Here only a brief reference to the methodology will be made.

Whenever a fluid flows across a tube bundle such as boiler tubes in an economizer, air heater, or superheater (see Fig. 8.24), vortices are formed and shed in the wake beyond the tubes. This shedding on alternate sides of the tubes causes a harmonically varying force on the tube perpendicular to the normal flow of the fluid. It is a self-excited vibration. If the frequency of the von Karman vortices, as they are called, coincides with the natural frequency of vibration of the tubes, resonance occurs and the tubes vibrate, leading to leakage and damage

TABLE 8.40 Results of Insulation Performance

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FIGURE 8.24 Crossflow of gas over tube bundles. (a) Water tube boiler design;

(b) air heater; (c) superheater.

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at supports. Vortex shedding is more prevalent in the range of Reynolds numbers from 300 to 2 105. This is the range in which many boilers, economizers, and superheaters operate. Another mechanism associated with vortex shedding is acoustic oscillation, which is normal to both fluid flow and tube length. This is observed only with gases and vapors. These oscillations coupled with vortex shedding lead to resonance and excessive noise. Standing waves are formed inside the duct.

Hence in order to analyze tube bundle vibration and noise, three frequencies must be computed: natural frequency of vibration of tubes, vortex shedding frequency, and acoustic frequency. When these are apart by at least 20%, vibration and noise may be absent. Q8.59b–Q8.59e show how these values are computed and evaluated.

8.59b

Q:

How is the natural frequency of vibration of a tube bundle determined?

A:

The natural frequency of transverse vibrations of a uniform beam supported at each end is given by

where do and di are in inches.

Table 8.41 gives C for various end conditions.

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and air, sonic velocity is obtained by substituting 32 for g0, 1.4 for n, and 1546=MW for R, where the molecular weight for flue gases is nearly 29. Hence,

Wavelength l ¼ 2W =n, where W is the duct width, ft, and n is the mode of vibration.

do ¼ tube outer diameter, in. V ¼ gas velocity, ft=s

S is available in the form of charts for various tube pitches; it typically ranges from 0.2 to 0.3 (see Fig. 8.25) [1].

Q8.59e shows how a tube bundle is analyzed for noise and vibration.

Copyright © 2003 Marcel Dekker, Inc.