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

TABLE 5.4 Suggested Water Quality Limitsa

Boiler type: Industrial water tube, high duty, primary fuel fired, drum type

Makeup water percentage: Up to 100% of feedwater

Conditions: Includes superheater, turbine drives, or process restriction on steam purity

(mmho=cm) @ 25 C without neutralization

Copyright © 2003 Marcel Dekker, Inc.

n.d. ¼ not detectable; n.s. ¼ not specified.

aNo values are given for saturated steam purity target because steam purity achievable depends upon many variables, including boiler water total alkalinity and specific conductance as well as design of boiler, steam drum internals, and operating conditions (see footnote i). Because boilers in this category require a relatively high degree of steam purity, other operating parameters must be set as low as necessary to achieve this high purity for protection of the superheaters and turbines and=or to avoid process contamination.

bWith local heat fluxes > 473.2 kW=m2 ( > 150,000 Btu=h ft2), use values for the next higher pressure range.

cBoilers below 6.21 MPa (900 psig) with large furnaces, large steam release space, and internal chelant, polymer, and=or antifoam treatment can sometimes tolerate higher levels of feedwater impurities than those in the table and still achieve adequate deposition control and steam purity. Removal of these impurities by external pretreatment is always a more positive solution. Alternatives must be evaluated as to practicality and economics in each case.

dValues in table assume the existence of a deaerator.

eNonvolatile TOCs are the organic carbon not intentionally added as part of the water treatment regime.

fMaximum total alkalinity consistent with acceptable steam purity. If necessary, should override conductance as blowdown control parameter. If makeup is demineralized water at 4.14–6.89 MPa (600–1000 psig), boiler water alkalinity and conductance should be that in table for 6.90– 10.34 MPa (1001–1500 psig) range.

g‘‘Not detectable’’ in these cases refers to free sodium or potassium hydroxide alkalinity. Some small variable amount of total alkalinity will be present and measurable with the assumed congruent or coordinated phosphate pH control or volatile treatment employed at these high pressure ranges.

hMinimum level of OH alkalinity in boilers below 6.21 MPa (900 psig) must be individually specified with regard to silica solubility and other components of internal treatment.

iMaximum values are often not achievable without exceeding suggested maximum total alkalinity values, especially in boilers below 6.21 MPa (900 psig) with > 20% makeup of water whose total alkalinity is > 20% of TDS naturally or after pretreatment with soda lime or sodium cycle ionexchange softening. Actual permissible conductance values to achieve any desired steam purity must be established for each case by careful steam purity measurements. Relationship between conductance and steam purity is affected by too many variables to allow its reduction to a simple list of tabulated values.

Source: Adapted from ASME 1979 Consensus.

Copyright © 2003 Marcel Dekker, Inc.

Example

A boiler generates 50,000 lb=h of saturated steam at 300 psia, out of which 10,000 lb=h is taken for process and returns to the deaerator as condensate at 180 F. The rest is consumed. Makeup water enters the deaerator at 70 F, and steam is available at 300 psia for deaeration. The deaerator operates at a pressure of 25 psia. The blowdown has a total dissolved solids (TDS) of 1500 ppm, and the makeup has 100 ppm TDS.

TABLE 5.5 Recommended Boiler Water Limits and Associated Steam Purity at Steady-State Full Load Operation—Water Tube Drum-Type Boilers

Actual values within the range reflect the TDS in the feedwater. Higher values are for high solids in the feedwater, lower values for low solids.

bActual values within the range are directly proportional to the actual value of TDS of boiler water. Higher values are for the high solids in the boiler water, lower values for low solids. cThese values are exclusive of silica.

dDictated by boiler water treatment.

Source: American Boiler Manufacturers Association, 1982.

Copyright © 2003 Marcel Dekker, Inc.

From an energy balance around the deaerator,

10;000 148 þ 1202:8 D þ M 38 ¼ 209 F ¼ 209 ð50;000 þ BÞ

ð22Þ

From a balance of solids concentration,

In Eq. (22), 1202.8 is the enthalpy of the steam used for deaeration, 209 the enthalpy of boiler feedwater, 148 the enthalpy of the condensate return, and 38 that of the makeup, all in Btu=lb. The equation assumes that the amount of solids in returning condensate and steam is negligible, which is true. Steam usually has a TDS of 1 ppm or less, and so does the condensate. Hence, for practical purposes we can neglect it. The net solids enter the system in the form of makeup water and leave as blowdown. There are three unknowns—D, M, and B—and three equations. From Eq. (21),

Substituting (23) into (24),

D þ 15B ¼ 40;000 þ B2

or

From (22),

1;480;000 þ 1202:8D þ 38 15B

¼ 209 50;000 þ 209B

Solving this equation, we have B ¼ 2375 lb=h, D ¼ 6750 lb=h, M ¼ 35,625 lb=h, and F ¼ 52,375 lb=h. Considering venting of steam from the deaerator to expel dissolved gases and the heat losses, 1–3% more steam may be consumed.

5.18

Q:

How can the boiler blowdown be utilized? A 600 psia boiler operates for 6000 h annually and discharges 4000 lb=h of blowdown. If this is flashed to steam at 100 psia, how much steam is generated? If the cost of the blowdown system is $8000, how long does payback take? Assume that the cost of steam is $2=1000 lb.

Copyright © 2003 Marcel Dekker, Inc.

A:

To estimate the flash steam produced we may use the expression

From the steam tables, at 600 psia, h ¼ 471:6, and at 100 psia, hg ¼ 1187 and hf ¼ 298, all in Btu=lb. Using Eq. (26), we have

471:6 ¼ 1187x þ ð1 xÞ 298

or

x ¼ 0:195

About 20% of the initial blowdown is converted to flash steam, the quantity being 0.2 4000 ¼ 800 lb=h. This 800 lb=h of 100 psia steam can be used for process. The resulting savings annually will be

800 2 60001000 ¼ $9600

Simple payback will be 8000=9600 ¼ 0.8 year or about 10 months.

Tables are available that give the flash steam produced if the initial and flash pressures are known. Table 5.6 is one such table.

5.19A

Q:

Estimate the leakage of steam through a hole 1=8 in. in diameter in a pressure vessel at 100 psia, the steam being in a saturated condition.

A:

The hourly loss of steam in lb=h is given by [2]

where

W ¼ steam leakage, lb=h A ¼ hole area, in.2

P ¼ steam pressure, psia

t; tsat ¼ steam temperature and saturated steam temperature, F

Copyright © 2003 Marcel Dekker, Inc.

TABLE 5.6 Steam Flash and Heat Content at Differential Temperatures

W ¼ 50 3:14 18 2 14 100 ¼ 61 lb=h

If the steam were superheated and at 900 F, then

61

W ¼ 1 þ 0:00065 ð900 544Þ ¼ 50 lb=h

544 F is the saturation temperature at 1000 psia. If the steam were wet with a quality of 80%, then

61

W ¼ p ¼ 68 lb=h 0:8

Copyright © 2003 Marcel Dekker, Inc.

5.19B

Q:

How is the discharge flow of air from high pressure to atmospheric pressure determined?

A:

Critical flow conditions for air are found in several industrial applications such as flow through soot blower nozzles, spray guns, and safety valves and leakage through holes in pressure vessels. The expression that relates the variables is [8]

What is the leakage air flow from a pressure vessel at 40 psia if the hole is 0.25 in. in diameter? Air is at 60 F.

A ¼ 3:14 0:25 0:425 ¼ 0:049 in:2

Hence,

W ¼ 356 0:049 40 28:9 0:5¼ 164 lb=h 520

5.20A

Q:

Derive an expression for the leakage of gas across a damper, stating the assumptions made.

A:

Most of the dampers used for isolation of gas or air in ducts are not 100% leakproof. They have a certain percentage of leakage area, which causes a flow of

Copyright © 2003 Marcel Dekker, Inc.

The hourly heat loss can be obtained from

Q¼ WCpðt taÞ ¼ 2040 0:26 ð540 80Þ

¼ 240;000 Btu=h ¼ 0:24 MM Btu=h

where Cp is the gas specific heat, Btu=lb F. Values of 0.25–0.28 can be used for quick estimates, depending on gas temperature. ta is the ambient temperature inF. 80 F was assumed in this case. The cost of this leakage ¼ 0.24 3 ¼ $0.72=h.

5.20C

Q:

How is the sealing efficiency of a damper defined?

A:

The sealing efficiency of a damper is defined on the basis of the area of cross section of the damper and also as a percentage of flow. The latter method of definition is a function of the actual gas flow condition.

In Q5.20b, the damper had an efficiency of 99.5% on an area basis. Assume that the actual gas flow was 230,000 lb=h. Then, on a flow basis, the efficiency would be

100 2040 ¼ 99:12%

230;000

If the flow were 115,000 lb=h and the differential pressure were maintained, the efficiency on an area basis would still be 99.5%, whereas on a flow basis it would be

100 2040 ¼ 98:24%

115;000

Plant engineers should be aware of these two methods of stating the efficiency of dampers.

5.21

Q:

50,000 lb=h of flue gas flows from a boiler at 800 F. If a waste heat recovery system is added to reduce its temperature to 350 F, how much energy is saved? If energy costs $3=MM Btu and the plant operates for 6000 h=year, what is the annual savings? If the cost of the heat recovery system is $115,000, what is the simple payback?

Copyright © 2003 Marcel Dekker, Inc.

A:

The energy savings Q ¼ WCpðt1 t2Þ; where t1 and t2 are gas temperatures before and after installation of the heat recovery system, F. Cp is the gas specific heat, Btu=lb F. Use a value of 0.265 when the gas temperature is in the range of 400–600 F.

Q¼ 50;000 0:265 ð800 350Þ ¼ 5:85 106

¼ 5:85 MM Btu=h

Annual savings ¼ 5:85 6000 3 ¼ $105;000

Hence

Simple payback ¼ 115;000 ¼ 1:1 years, or 13 months 105;000

5.22

Q:

What is life-cycle costing? Two bids are received for a fan as shown below. Which bid is better?

A:

Life-cycle costing is a methodology the computes the total cost of owning and operating the equipment over its life. Several financing methods and tax factors would make this a complicated evaluation. However, let us use a simple approach to illustrate the concept. To begin with, the following data should be obtained.

Cost of electricity, Ce ¼ $0.25=kWh

Annual period of operation, N ¼ 8000 h

Life of equipment, T ¼ 15 years

Interest rate, i ¼ 0.13 (13%)

Escalation rate, e ¼ 0.08 (8%)

Copyright © 2003 Marcel Dekker, Inc.