01 POWER ISLAND / 01 CCPP / V. Ganapathy-Industrial Boilers and HRSG-Design (2003)
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Simplifying and solving for flows, we have
W1 ¼ 5353 lb=h; W2 ¼ 4054 lb=h; W3 ¼ 5591 lb=h
This type of calculation is done to check if each pass receives adequate steam flow to cool it. Note that pass 2 had the least flow, and a metal temperature check must be performed. If the metal temperature is high, the tube length or tube sizes must be modified to ensure that the tubes are protected from overheating.
7.26
Q:
How is the equivalent length of a piping system determined? 100 ft of a piping system has three globe valves, a check valve, and three 90 bends. If the line size is 2 in., determine the total equivalent length.
A:
The total equivalent length is the sum of the developed length of the piping plus the equivalent lengths of valves, fittings, and bends. Table 7.10 gives the equivalent length of valves and fittings. A globe valve has 58.6 ft, a check valve has 17.2 ft, and a 90 bend has 5.17 ft of equivalent length. The equivalent length of all valves and fittings is
3 58:6 17:2 þ 3 5:17 ¼ 208:5 ft
Hence the total equivalent length is ð100 þ 208:5Þ ¼ 308:5 ft.
TABLE 7.10 Equivalent Length Le for Valves and
Fittingsa
Pipe size (in.) |
1b |
2b |
3b |
4b |
|
|
|
|
|
1 |
0.70 |
8.70 |
30.00 |
2.60 |
2 |
1.40 |
17.20 |
60.00 |
5.20 |
3 |
2.00 |
25.50 |
87.00 |
7.70 |
4 |
2.70 |
33.50 |
114.00 |
10.00 |
6 |
4.00 |
50.50 |
172.00 |
15.20 |
8 |
5.30 |
33.00 |
225.00 |
20.00 |
10 |
6.70 |
41.80 |
284.00 |
25.00 |
12 |
8.00 |
50.00 |
338.00 |
30.00 |
16 |
10.00 |
62.50 |
425.00 |
37.50 |
20 |
12.50 |
78.40 |
533.00 |
47.00 |
aLe ¼ Kdi =12f , where di is the pipe inner diameter (in.) and K is the number of velocity heads (adapted from Crane Technical Paper 410). f is the Darcy friction factor.
b1, Gate valve, fully open; 2, swing check valve, fully open; 3, globe valve, fully open; 4, 90 elbow.
Copyright © 2003 Marcel Dekker, Inc.
7.27
Q:
Determine the pressure drop of flue gases and air flowing over a tube bundle under the following conditions:
Gas mass velocity ¼ 7000 lb=ft2 h
Tube size ¼ 2 in. OD
Transverse pitch ¼ 4.0 in.
Longitudinal pitch ¼ 3.6 in.
Arrangement: in-line
Average gas temperature ¼ 800 F
Number of rows deep ¼ 30
A:
The following procedure may be used to determine gas pressure drop over tube bundles in in-line and staggered arrangements [11].
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NH |
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DPg ¼ 9:3 10 10 fG2 |
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ð28Þ |
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rg |
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where |
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G ¼ gas mass velocity, lb=ft2 h |
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DPg ¼ gas pressure drop, in. WC |
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f ¼ friction factor |
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rg ¼ gas density, lb=cu ft |
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NH ¼ number of rows deep |
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For an in-line arrangement for ST =d ¼ 1.5–4.0 and for |
2000 < |
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Re < 40;000 [12], |
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f ¼ Re 0:15 0:044 þ ðST =d 1Þ0:43þ1:13d=SL ! |
ð29Þ |
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0:08SL=d |
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where ST is the transverse pitch and SL is the longitudinal pitch, in. For a staggered arrangement for ST =d ¼ 1.5–4.0,
f ¼ Re 0:16 |
0:25 þ ðST =d 1Þ1:08 |
ð30Þ |
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0:1175 |
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In the absence of information on gas properties, use a molecular weight of 30 for flue gas. Then, from Chapter 5,
492
rg ¼ 30 359 ð460 þ 800Þ ¼ 0:0326 lb=cu ft
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The viscosity is to be estimated at the gas film temperature. However, it can be computed at the average gas temperature, and the difference is not significant for Reynolds number computations.
From Table 7.7, m ¼ 0:08 lb=ft h. From Eq. (27),
Re ¼ Gd ¼ 7000 2 ¼ 14;580 12m 12 0:08
From Eq. (29),
f ¼ ð14;580Þ 0:15 0:044 þ 0:08 2 ¼ 0:0484 1
DPg ¼ 9:3 10 0 0:0484 70002 0:032630 ¼ 2:03 in: WC
Similarly, using Eq. (30) we can estimate DPg for a staggered arrangement. Note: The foregoing procedure may be used in the absence of field-tested
data or correlation.
7.28
Q:
Determine the gas pressure drop over a bundle of circumferentially finned tubes in an economizer when
Gas mass velocity of flue gas ¼ 6000 lb=ft2 h
(The method of computing G for plain and finned tubes is discussed in Chapter 8.)
Average gas temperature ¼ 800 F
Tube size ¼ 2.0 in.
Transverse pitch ST ¼ 4.0 in.
Longitudinal pitch SL ¼ 3.6 in.
Number of rows deep NH ¼ 10
A:
The equation of Robinson and Briggs [11] may be used in the absence of site-proven data or correlation provided by the manufacturer for staggered arrangement:
D |
P |
g ¼ |
1:58 10 8 G1:684 d0:611 m0:316 ð460 þ tÞ NH |
ð |
31 |
Þ |
|
ST0:412 SL0:515 MW |
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|
Copyright © 2003 Marcel Dekker, Inc.
where
G ¼ gas mass velocity, lb=ft h MW ¼ gas molecular weight
d ¼ tube outer diameter, in.
FðtÞ ¼ m0:316 ð460 þ tÞ
ST ; SL ¼ transverse and longitudinal pitch, in.
FðtÞ is given as a function of gas temperature in Table 7.11. Substituting into Eq. (31) gives us
DPg ¼ 1:58 10 8 60001:684 20:611 556 10
40:412 3:60:515 30
¼3:0 in: WC
7.29
Q:
What is boiler circulation, and how is it determined?
A:
The motive force driving the steam–water mixture through boiler tubes (water tube boilers) or over tubes (in fire tube boilers) is often the difference in density between the cooler water in the downcomer circuits and the steam–water mixture in the riser tubes (Fig. 7.4). A thermal head is developed because of this difference, which forces a certain amount of steam–water mixture through the system. This head overcomes several losses in the system such as
Friction loss in the downcomers
Friction loss and flow acceleration loss in the risers and connecting pipes to the drum
TABLE 7.11 FðtÞ
Versus t for Air or Flue
Gases
tðF Þ |
FðtÞ |
200 |
251 |
400 |
348 |
600 |
450 |
800 |
556 |
1000 |
664 |
1200 |
776 |
1600 |
1003 |
|
|
Copyright © 2003 Marcel Dekker, Inc.
FIGURE 7.4 Scheme of natural circulation boiler showing furnace, drum, riser, and downcomer circuits.
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Gravity loss in the evaporator tubes and the riser system
Losses in the drum internals
Generally, the higher the drum operating pressure, the less the difference between the densities of water and the steam–water mixture, and hence the lower the circulation rate.
Circulation ratio (CR) is defined as the ratio between the mass of the steam–water mixture flowing through the system and the mass of the steam generated. If CR ¼ 15, then a boiler generating 10,000 lb=h of steam would have 150,000 lb=h of steam–water mixture flowing through the downcomers, risers, internals, etc. The quality of steam at the exit of the riser ¼ 1=CR, or 0.067 if CR ¼ 15. In other words, 6.7% would be the average wetness of steam in the mixture. Low pressure systems have an average CR ranging from 10 to 40. If there are several parallel circuits for the steam–water mixture, each would have a different resistance to flow, and hence CR would vary from circuit to circuit. For natural circulation systems, CR is usually arrived at by trial and error or by iterative calculation, which first assumes a CR and computes all the losses and then balances the losses with the available thermal head. This computation is iterated until the available head and the losses balance.
Sometimes the difference in density between the water and the steam–water mixture is inadequate to circulate the mixture through the system. In such cases, a circulation pump is installed at the bottom of the steam drum, which circulates a desired quantity of mixture through the system (Fig. 7.5). This system is called a forced circulation system. One has to ensure that there are an adequate number of
FIGURE 7.5 Scheme of forced circulation boiler.
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pumps to ensure circulation, because the failure of the pump would mean starvation of flow in the evaporator tubes. Because we are forcing the mixture through the tubes, the CR is preselected, and the circulating pump is chosen accordingly. A CR of 3–10 is typical. This system is usually used when the pressure drop through the evaporator is likely to be high such as when horizontal tubes are used. When horizontal tubes are used, the critical heat flux to avoid DNB (departure from nucleate boiling) conditions is lower, so forced circulation helps to ensure adequate flow inside the tubes. Circulating pumps are also used when the boiler pressure is high owing to the lower difference in density between the water and the steam–water mixture.
7.30
Q:
What is the main purpose of determining CR?
A:
Determination of CR is not the end in itself. The CR value is used to determine whether a given circuit in the boiler has all the conditions necessary to avoid departure from nucleate boiling (DNB) problems. For each pressure and quality (or CR) there is a particular heat flux beyond which the type of boiling may change from nucleate boiling, which is preferred, to film boiling, which is to be avoided because it can cause the tube wall temperatures to rise significantly, resulting in tube failure. DNB occurs at heat fluxes of 100,000–400,000 Btu=ft2 h depending on size and orientation of tubes, pressure, mass velocity, quality, and tube roughness. DNB occurs at a much lower heat flux in a horizontal tube than in an equivalent vertical tube because the steam bubble formation and release occurs more freely and rapidly in vertical tubes than in horizontal tubes, where there is a possibility of bubbles adhering to the top of the tube and causing overheating. More information on DNB and circulation can be found in references cited in Refs. 11 and 14.
Note that the heat flux in finned tubes is much higher than in bare tubes owing to the large ratio of external to internal surface area; this aspect is also discussed elsewhere. Hence one has to be careful in designing boilers with extended surfaces to ensure that the heat flux in the finned tubes does not reach critical levels or cause DNB. That is why boilers with very high gas inlet temperatures are designed with a few rows of bare tubes followed by a few rows of low-fin-density tubes and then high-fin-density tubes. As the gas cools, the heat flux decreases.
7.31a
Q:
Describe the procedure for analyzing the circulation system for the water tube boiler furnace shown in Fig. 7.4.
Copyright © 2003 Marcel Dekker, Inc.
A:
First, the thermal data such as energy absorbed, steam generated, pressure, and geometry of downcomers, evaporator tubes, and risers should be known. These are obtained from an analysis of furnace performance (see example in Chap. 8). The circulation ratio (CR) is assumed; then the flow through the system is computed, followed by estimation of various pressure losses. Thom’s method is used for evaluating two-phase flow losses [15, 16].
The losses can be estimated as follows. DPf , the friction loss in two-phase flow (evaporators=risers), is given by
D |
P |
f ¼ |
4 |
|
10 10 |
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v |
|
f L |
G2 r |
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ð |
32 |
Þ |
f |
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3 |
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di i |
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The factor r3 is shown in Fig. 7.6. Gi is the tube-side mass velocity in lb=ft2 h. The friction factor used is that of Fanning, which is 0.25 times the Moody friction factor.
FIGURE 7.6 Thom’s two-phase multiplication factor for friction loss. (See Refs. 11, 15, and 16.)
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DPg, the gravity loss in the heated riser=evaporator, is given by
DPg ¼ 6:95 10 3 |
r4 |
ð33Þ |
L vf |
where r4 is obtained from Fig. 7.7.
DPa, the acceleration loss, which is significant at lower pressures and at
high mass velocities, is given by |
|
DPa ¼ 1:664 10 11 vf Gi2 r2 |
ð34Þ |
Figure 7.8 gives r2.
Single-phase pressure losses such as losses in downcomers are obtained
from
V 2
DP ¼ 12 f Le r 2g di
FIGURE 7.7 Thom’s two-phase multiplication factor for gravity loss. (See Refs. 11, 15, and 16.)
Copyright © 2003 Marcel Dekker, Inc.
FIGURE 7.8 Thom’s two-phase multiplication factor for acceleration loss. (See Refs. 11, 15, and 16.)
Copyright © 2003 Marcel Dekker, Inc.
