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

Let us compute U. Because it is based on tube outside surface, let us call it Uo.

¼0:10 þ 0:001 þ 0:00226 þ 0:00041 þ 0:0005

¼0:10417

If U is computed on the basis of tube inner surface area, then Ui is given by the expression

Ai Ui ¼ Ao Uo

Hence,

Ui ¼ 9:6 1:277 ¼ 10:85 Btu=ft2 h F

Log-mean temperature difference is

DT ¼ ð1500 366Þ ð500 366Þ ¼ 468 F ln½ð1500 366Þ=ð500 366Þ&

Hence

Ao ¼ 28:13 106 ¼ 6261 ft2 468 9:6

¼ 3:14 2 600 12L

so required length L of the tubes ¼ 19.93 ft. Use 20 ft. Then Ao ¼ 3:14 2 600 2012 ¼ 6280 ft2

Ai ¼ 5558 ft2

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Let us compute the gas pressure drop using Eq. (12) of Chapter 7.

DPg ¼ 93 10 6 w2f Le dv5

i

Friction factor f depends on tube inner diameter and can be taken as 0.02. The equivalent length Le can be approximated by L þ 5di to include the tube inlet and exit losses.

Specific volume v obtained as 1=density, or v ¼ 1=r. Gas density at the average gas temperature of 1000 F is rg ¼ 39=1460 ¼ 0.0267 lb=cu ft. Therefore,

DPg ¼ 93 10 6 1672 0:02

20þ 5 1:77

0:0267 ð1:77Þ5 ¼ 3:23 in: WC

This is only one design. Several variables such as tube size and mass flow could be changed to arrive at several options that could be reviewed for optimum operating and installed costs.

8.11

Q:

What is the effect of tube size and gas velocity on boiler size? Is surface area the sole criterion for boiler selection?

A:

Surface area should not be used as the sole criterion for selecting or purchasing boilers, because tube size and gas velocity affect this variable.

Shown in Table 8.7 are the design options for the same boiler duty using different gas velocities and tube sizes; the procedure described in Q8.10 was used to arrive at these options. The purpose behind this example is to bring out the fact that surface area can vary by as much as 50% for the same duty.

1.As the gas velocity increases, the surface area required decreases, which is obvious.

2.The smaller the tubes, the higher the heat transfer coefficient for the same gas velocity, which also decreases the surface area.

3.For the same gas pressure drop, the tube length is smaller if the tube size is smaller. This fact helps when we try to fit a boiler into a small space.

4.For the same tube size, increasing the gas velocity results in a longer boiler, a greater gas pressure drop, but smaller surface area.

In the case of water tube boilers, more variables such as tube spacing and in-line or staggered arrangement in addition to gas velocity and tube size can affect surface area. This is discussed elsewhere.

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TABLE 8.7 Effect of Tube Size and Gas Velocity on Fire Tube Boiler Design

blowdown ¼ 5%; steam ¼ 28,950 lb=h; gas analysis (vol%): CO2 ¼ 7; H2O ¼ 12; N2 ¼ 75; O2 ¼ 6; boiler duty ¼ 29.4 MM Btu=h.

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8.12

Q:

How is the tube wall temperature in fire tube boilers evaluated? Discuss the importance of heat flux.

A:

To compute the tube wall temperatures, heat flux must be known.

qo ¼ heat flux outside tubes ¼ Uo ðtg tiÞ Btu=ft2 h

Similarly, qi (heat flux inside the tube) would be Ui ðtg tiÞ. However, heat flux outside the tubes is relevant in fire tube boilers because boiling occurs outside the tubes, whereas in water tube boilers the heat flux inside the tubes would be relevant. A high heat flux can result in a condition called departure from nucleate boiling (DNB), which will result in overheating of the tubes. It is preferable to keep the actual maximum heat flux below the critical heat flux, which varies from 150,000 to 250,000 Btu=ft2 h depending on steam quality, pressure, and tube condition [1].

An electrical analogy can be used in determining the tube wall temperatures. Heat flux is analogous to current, electrical resistance to thermal resistance, and voltage drop to temperature drop. Using the example worked in Q8.10, we have that at average gas conditions the product of current (heat flux) and resistance (thermal resistance) gives the voltage drop (temperature drop):

qo ¼ heat flux ¼ 9:6 ð1000 366Þ ¼ 6086 Btu=ft2 h

Temperature drop across gas film ¼ 6086 0:1 ¼ 609 F

Temperature drop across gas-side fouling ¼ 6086 0:00226 ¼ 14 F

Temperature drop across tube wall ¼ 6086 0:00041 ¼ 3 F

Temperature drop across steam-side fouling ¼ 6086 0:001 ¼ 6 F

Temperature drop across steam film ¼ 6085 0:0005 ¼ 3 F

Hence,

Average inside tube wall temperature ¼ 1000 609 14 ¼ 377 F

Outside tube wall temperature ¼ 377 3 ¼ 347 F:

The same results are obtained working from the steam side.

Outside tube wall temperature ¼ 366 þ 6 þ 3 ¼ 375 F

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One can also compute the maximum tube wall temperature by obtaining the heat flux at the hot gas inlet end.

8.13

Q:

What is the effect of scale formation on tube wall temperatures?

A:

If nonsoluble salts such as calcium or magnesium salts or silica are present in the feedwater, they can deposit in a thin layer on tube surfaces during evaporation, thereby resulting in higher tube wall temperatures.

Table 8.8 lists the thermal conductivity k of a few scales. Outside fouling factor ff o can be obtained if the scale information is available.

ffo ¼ thickness of scale conductivity

Let us use the same example as in Q8.10 and check the effect of ff o on boiler duty and tube wall temperatures. Let a silicate scale of thickness 0.03 in. be formed. Then,

ffo ¼ 00::036 ¼ 0:05 ft2 h F=Btu

TABLE 8.8 Thermal Conductivities of Scale

Materials

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Assume that other resistances have not changed. (Because of different duty and gas temperature profile, the gas-side heat transfer coefficient will be slightly different. However, for the sake of illustration, we neglect this.) We have

1 ¼ 0:10 þ 0:00226 þ 0:00041 þ 0:05 þ 0:0005

Uo

¼ 0:15317

Hence, Uo ¼ 6.52 Btu=ft2 h F

Heat flux qo ¼ 6:52 ð1000 366Þ ¼ 4133 Btu=ft2 h

Temperature drop across outside steam film ¼ 0:0005 4133 ¼ 2 F

Temperature drop across steam-side fouling layer or scale ¼ 4133 0:05 ¼ 207 F

Temperature drop across tube wall ¼ 4133 0:00041 ¼ 2 F

We see that average tube wall temperature has risen to 366 þ 2 þ 207 þ 2 ¼ 577 F from an earlier value of about 375 F. Scale formation is a serious problem. Note that the heat flux is now lower, but that does not help. At the front end, where the heat flux is higher, the tubes would be much hotter.

Hence tg2 ¼ 630 F compared to 500 F earlier. The reason for tg2 going up is the lower Uo caused by scale formation.

Hence new duty ¼ 100,000 0.98 0.287 (1500 7 630) ¼ 24.47 106 Btu=h. The decrease in duty is 28.13 7 24.47 ¼ 3.66 MM Btu=h. Even assuming

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a modest energy cost of $3=MM Btu, the annual loss due to increased fouling is 3.66 3 8000 ¼ $87,800. The steam production in turn gets reduced.

Plant engineers should check the performance of their heat transfer equipment periodically to see if the exit gas temperature rises for the same inlet gas flow and temperature. If it does, then it is likely due to fouling on either the gas or steam side, which can be checked. Fouling on the gas side affects only the duty and steam production, but fouling on the steam side increases the tube wall temperature in addition to reducing the duty and steam production.

To ensure that variations in exit gas temperature are not due to fouling but are due to changes in gas flow or temperature, one can use simulation methods. For example, if, for the same gas flow, the inlet gas temperature is 1800 F, we can expect the exit gas temperature to rise. Under clean conditions, this can be estimated using the equation (32)

Now if, in operation, the exit gas temperature were 570–600 F, then fouling could be suspected; but if the gas temperature were only about 535 F, this would only be due to the increased gas inlet temperature. Similarly, one can consider the effect of gas flow and saturation temperature.

8.14

Q:

How is the size of a water tube boiler determined?

A:

The starting point in the design of an evaporator (Fig. 8.3) is the estimation of the overall heat transfer coefficient U. The cross-sectional data such as the number of tubes wide, spacing, and length of tubes are assumed. From the duty and logmean temperature difference, the surface area is obtained. Then the number of rows deep is estimated. Tube wall temperature calculations and gas pressure drop evaluation then follow. A computer program is recommended to perform these tedious calculations, particularly if several alternatives have to be evaluated.

Example

200,000 lb=h of clean flue gas from an incinerator must be cooled from 1100 F to 600 F in a bare tube evaporator. Steam pressure ¼ 250 psig saturated. Feedwater temperature ¼ 230 F. Blowdown ¼ 5%. Fouling factors on steamand gas-side ¼ 0.001 ft2 h F=Btu. Gas analysis (vol%): CO2 ¼ 7; H2O ¼ 12; N2 ¼ 75; O2 ¼ 6. Let heat loss from casing ¼ 1%.

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FIGURE 8.3 Boiler evaporator bundle.

Solution: Use 2 1.773 in carbon steel tubes; number wide ¼ 24; length ¼ 10 ft; tube spacing ¼ 4 in. square.

Average gas temperature ¼ 0:5 ð1100 þ 600Þ ¼ 850 F

Steam temperature inside tubes ¼ 406 F. Assume tube wall temperature ¼ 410 F (this should be checked again later).

Film temperature ¼ 0:5 ð850 þ 410Þ ¼ 630 F

Gas properties at film temperature are (from Appendix) Cp ¼ 0.2741, m ¼ 0.0693, k ¼ 0.0255

Cp at average gas temperature ¼ 0.282

Duty Q ¼ 200,000 0.99 0.282 (1100 7 600) ¼ 27.92 MM Btu=h Steam enthalpy change ¼ (1201.7 7 199) þ 0.05 (381.4 7 199) ¼

1011.82 Btu=lb

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Gas emissivity eg ¼ 0:9 ð1 e 0:746 0:176Þ ¼ 0:1107

Assuming that the tube wall is at 420 F (to be checked later)

13:14 8:84 hN ¼ 0:173 0:9 0:1107 1310 880

Using a conservative boiling heat transfer coefficient of 2000 Btu=ft2 h and a tube thermal conductivity of 25 Btu=ft h F; we have

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or

U ¼ 12:3 Btu=ft2 h F

Log-mean temperature difference ¼ ð1100 406Þ ð600 406Þ ln½ð1100 406Þ=ð600 406Þ&

¼ 393 F

Surface area required A ¼ 27:92 106 ¼ 5776 ft2 12:3 393

A ¼ 3:14 2 Nd 24 12=12 ¼ 5776; or Nd ¼ 38:4

Use 40 rows deep. Surface provided ¼ 6016 ft2. Let us estimate the gas pressure drop.

The average heat flux on tube ID basis is

q ¼ 12:3 ð850 406Þ 2=1:773 ¼ 6160 Btu=ft2 h

Temperature drop across inside fouling layer ¼ 6160 0.001 ¼ 62 F Temperature drop across inside film coefficient ¼ 6160=2000 ¼ 3.1 F Drop across tube wall ¼ 0.0004 1.773 6160=2 ¼ 2.2 F

Hence tube outer wall temperature ¼ 406 þ 6.2 þ 3.1 þ 2.2 ¼ 418 F. Since this is close to the assumed value another iteration is not necessary.

Note that this is only the average tube wall temperature. The maximum heat flux is at the gas inlet, and one has to redo these calculations to obtain the maximum tube wall temperature. A computer program would help speed up these calculations.

8.15a

Q:

How is the off-design performance of a boiler evaluated? Predict the performance of the boiler designed earlier under the following conditions: Gas flow ¼ 230,000 lb=h; gas inlet temperature ¼ 1050 F; steam pressure ¼ 200 psig. Gas analysis remains the same.

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