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

where Cc is the cost of equipment and F is a factor that capitalizes the operating cost over the life of the equipment. It can be shown [4,5] that

1 1 þ e T

1 þ i

1 1 þ e

1 þ i

The annual cost of operation is given by

Ca ¼ PCe N

where P is the electric power consumed, kW.

P ¼ 1:17 10 4 qHw

Zf

where

Hw ¼ head, in. WC

Zf ¼ efficiency, fraction q ¼ flow, acfm

Let us use the subscripts 1 and 2 for bids 1 and 2.

P1 ¼ 1:17 10 4 10;000 0:860 ¼ 15:6 kW

P2 ¼ 1:17 10 4 10;000 0:875 ¼ 12:48 kW

From Eq. (33), substituting e ¼ 0.08, i ¼ 0.13, and T ¼ 15, we get F ¼ 10.64. Calculate Ca from Eq. (34):

Ca1 ¼ 15:6 8000 0:025 ¼ $3120

Ca2 ¼ 12:48 8000 0:025 ¼ $2500

Using Eq. (32), calculate the life-cycle cost.

LCC1 ¼ 17;000 þ 3120 10:64 ¼ $50;196

LCC2 ¼ 21;000 þ 2500 10:64 ¼ $47;600

We note that bid 2 has a lower LCC and thus may be chosen. However, we have to analyze other factors such as period of operation, future cost of energy, and so on, before deciding. If N were lower, it is likely that bid 1 would be better.

Hence, the choice of equipment should not be based only on the initial investment but on an evaluation of the life-cycle cost, especially as the cost of energy is continually increasing.

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5.23

Q:

A process kiln omits 50,000 lb=h of flue gas at 800 F. Two bids were received for heat recovery systems, as follows:

If the plant operates for 6000 h=year and interest, escalation rates, and life of plant are as in Q5.22, evaluate the two bids if energy costs $4=MM Btu.

A:

Let us calculate the capitalized savings and compare them with the investments. For bid 1:

Energy recovered ¼ 50;000 0:25 ð800 450Þ

¼ 4:375 MM Btu=h

This energy is worth

4:375 4 ¼ $17:5=h

Annual savings ¼ 6000 17:5 ¼ $105;000

The capitalization factor from Q5.22 is 10.64. Hence capitalized savings (savings throughout the life of the plant) ¼ 105,000 10.64 ¼ $1.12 106. A similar calculation for bid 2 shows that the capitalized savings will be $1.6 106. The difference in capitalized savings of $0.48 106, or $480,000, exceeds the difference in the investment of $235,000. Hence bid 2 is more attractive.

If, however, energy costs $3=MM Btu and the plant works for 2500 h=year, capitalized savings on bid 1 will be $465,000 and that of bid 2 $665,000. The difference of $200,000 is less than the difference in investment of $235,000. Hence under these conditions, bid 1 is better.

The cost of energy and period of operation are important factors in arriving at the best choice.

5.24

Q:

Determine the thickness of the tubes required for a boiler super-heater. The material is SA 213 T11; the metal temperature is 900 F (see Q8.16a for a

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From Table 5.7, Sa is 13,100. Substituting into Eq. (36) yields

1000 1:75

tw ¼ 2 13;100 þ 1000 þ 0:005 1:75 ¼ 0:073 in:

The tube with the next higher thickness would be chosen. A corrosion allowance, if required, may be added to tw.

5.25

Q:

Determine the maximum pressure that an SA 53 B carbon steel pipe of size 3 in. schedule 80 can be subjected to at a metal temperature of 550 F. Use a corrosion allowance of 0.02 in.

A:

By the ASME Code, Sec. 1, 1980, p. 27, the formula for determining allowable pressures or thickness of pipes, drums, and headers is

where

E ¼ ligament efficiency, 1 for seamless pipes c ¼ corrosion allowance

From Table 5.3, a 3 in. schedule 80 pipe has an outer diameter of 3.5 in. and a nominal wall thickness of 0.3 in. Considering the manufacturing tolerance of 12.5%, the minimum thickness available is 0.875 0.3 ¼ 0.2625 in.

Substituting Sa ¼ 15,000 psi (Table 5.7) and c ¼ 0:02 into Eq. (38), we

have

3:5P

0:2625 ¼ 2 15;000 þ 0:8P þ 0:02

Solving for P, we have P ¼ 2200 psig.

For alloy steels, the factor 0.8 in the denominator would be different. The ASME Code may be referred to for details [6]. Table 5.8 gives the maximum allowable pressures for carbon steel pipes up to a temperature of 650 F [7].

5.26

Q:

How is the maximum allowable external pressure for boiler tubes determined?

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where Sb is the lesser of 2 times the maximum allowable stress values at design metal temperature from the code stress tables or 1.8 times the yield strength of the material at design metal temperature. Then the smaller of the Pa1 or Pa2 is used for Pa.

Example

Determine the maximum allowable external pressure at 600 F for 120 in. SA 192 tubes of outer diameter 2 in. and length 15 ft used in fire tube boilers.

From Fig. 5.3 factor A ¼ 0.004. From Fig. 5.4, B ¼ 9500. Since do=t > 10,

5.27

Q:

What is a decibel? How is it expressed?

A:

The decibel (dB) is the unit of measure used in noise evaluation. It is a ratio (not an absolute value) of a sound level to a reference level and is stated as a sound pressure level (SPL) or a sound power level (PWL). The reference level for SPL is 0.0002 mbar. A human ear can detect from about 20 dB to sound pressures 100,000 times higher, 120 dB.

Audible frequencies are divided into octave bands for analysis. The center frequencies in hertz (Hz) of the octave bands are 31.5, 63, 125, 250, 500, 1000, 2000, 4000, and 8000 Hz. The human ear is sensitive to frequencies between 500 and 3000 Hz and less sensitive to very high and low frequencies. At 1000 Hz, for example, 90 dB is louder than it is at 500 Hz.

The sound meter used in noise evaluation has three scales, A, B, and C, which selectively discriminate against low and high frequencies. The A scale (dBA) is the most heavily weighted scale and approximates the human ear’s response to noise (500–6000 Hz). It is used in industry and in regulations regarding the evaluation of noise. Table 5.9 gives typical dBA levels of various

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FIGURE 5.3 Factor A for use in external pressure calculation [9].

noise sources, and Table 5.10 gives the permissible Occupational Safety and Health Act (OSHA) noise exposure values.

5.28

Q:

How are decibels added? A noise source has the following dB values at center frequencies:

What is the overall noise level?

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FIGURE 5.4 Factor B for use in external pressure calculation (SA 178A, SA 192 tubes) [9].

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TABLE 5.9 Typical A-Weighted Sound Levels

A:

Decibels are added logarithmically and not algebraically. 97 dB plus 97 dB is not 194 dB but 100 dB.

P¼ 10 logð10P1=10 þ 10P2=10 þ 10P3=10 þ Þ

¼10 logð109:7 þ 109:7 þ 109:5 þ 109:1 þ 108:4 þ 108:2 þ 108

þ108:5 þ 108:5Þ

¼102 dB

TABLE 5.10 Permissible Noise Exposures (OSHA)

Copyright © 2003 Marcel Dekker, Inc.

5.29

Q:

What are SPL and PWL?

A:

SPL is sound pressure level, which is dependent on the distance and environment and is easily measured with a sound level meter. SPL values should be referred to distance. PWL is sound power level and is a measure of the total acoustic power radiated by a given source. It is defined as

PWL is a constant for a given source and is independent of the environment. It cannot be measured directly but must be calculated. PWL can be roughly described as being equal to the wattage rating of a bulb. Manufacturers of fans and gas turbines publish the values of PWL of their machines. When selecting silencers for these equipment, PWL may be converted to SPL depending on distance, and the attenuation desired at various frequencies may be obtained. A silencer that gives the desired attenuation can then be chosen.

5.30

Q:

A sound level of 120 dB is measured at a distance of 3 ft from a source. Find the value at 100 ft.

A:

The following formula relates the PWL and SPL with distance:

where L ¼ distance, ft.

PWL is a constant for a given source. Hence

SPL þ 20 log L ¼ a constant

120 þ 20 log 3 ¼ SPL2 þ 20 log 100

Hence

SPL2 ¼ 89:5 dB

Thus we see that SPL has decreased by 30 dB with a change from 3 ft to 100 ft. When selecting silencers, one should be aware of the desired SPL at the desired distance. Neglecting the effect of distance can lead to specifying a larger and more costly silencer than necessary.

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