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

5.08

Q:

What is meant by boiler horsepower? How is it related to steam generation at different steam parameters?

A:

Packaged fire tube boilers are traditionally rated and purchased in terms of boiler horsepower (BHP). BHP refers to a steam capacity of 34.5 lb=h of steam at atmospheric pressure with feedwater at 212 F. However, a boiler plant operates at different pressures and with different feedwater temperatures. Hence conversion between BHP and steam generation becomes necessary.

where

hg ¼ enthalpy of saturated steam at operating steam pressure, Btu=lb hf ¼ enthalpy of saturated liquid, Btu=lb

hfw ¼ enthalpy of feedwater, Btu=lb BD ¼ blowdown fraction

For example, if a 500 BHP boiler generates saturated steam at 125 psig with 5% blowdown and with feedwater at 230 F, the steam generation at 125 psig will be

500 33;475

W¼ ð1193 198Þ þ 0:05 ð325 198Þ

¼ 16;714 lb=h

where 1193, 198, and 325 are the enthalpies of saturated steam, feedwater, and saturated liquid, respectively, obtained from steam tables. (See Appendix.)

5.09A

Q:

Why do we need to know the amount of moisture in air?

A:

In combustion calculations (Chap. 6) we estimate the quantity of dry air required to burn a given amount of fuel. In reality, atmospheric air is never dry; it consists of some moisture, depending on the relative humidity and dry bulb temperature. To compute the partial pressure of water vapor in the flue gas, which is required

Copyright © 2003 Marcel Dekker, Inc.

TABLE 5.3 Dimensions of Iron Steel Pipe (IPS)

aCommonly known as standard. bCommonly known as extra heavy. cApproximately.

Copyright © 2003 Marcel Dekker, Inc.

for calculating nonluminous heat transfer, we need to know the total quantity of water vapor in flue gases, a part of which comes from combustion air.

Also, when atmospheric air is compressed, the saturated vapor pressure (SVP) of water increases, and if the air is cooled below the corresponding water dew point temperature, water can condense. The amount of moisture in air or gas fixes the water dew point, so it is important to know the amount of water vapor in air or flue gas.

5.09B

Q:

Estimate the pounds of water vapor to pounds of dry air when the dry bulb temperature is 80 F and the relative humidity is 65%.

A:

Use the equation

where

M ¼ lb water vapor=lb dry air

pw ¼ partial pressure of water vapor in air, psia

This may be estimated as the vol% of water vapor total air pressure or as the product of relative humidity and the saturated vapor pressure (SVP). From the steam tables we note that at 80 F, SVP ¼ 0.5069 psia (at 212 F, SVP ¼ 14.7 psia). Hence pw ¼ 0.65 0.5069.

0:5069

M ¼ 0:622 0:65 14:7 0:65 0:5069 ¼ 0:0142

Hence, if we needed 1000 lb of dry air for combustion, we would size the fan to deliver 1000 1.0142 ¼ 1014.2 lb of atmospheric air.

5.10A

Q:

What is the water dew point of the flue gases discussed in Q5.05?

A:

The partial pressure of water vapor when the vol% is 8 and total pressure is 14.7 psia will be

pw ¼ 0:08 14:7 ¼ 1:19 psia

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From the steam tables, we note that the saturation temperature corresponding to 1.19 psia is 107 F. This is also the water dew point. If the gases are cooled below this temperature, water can condense, causing problems.

5.10B

Q:

What is the water dew point of compressed air when ambient air at 80 F, 14.7 psia, and a relative humidity of 65% is compressed to 35 psia?

A:

Use the following expression to get the partial pressure of water vapor after compression:

pw2 ¼ pw1 P2 ð14Þ

P1

where

pw ¼ partial pressure, psia P ¼ total pressure, psia

The subscripts 1 and 2 stand for initial and final conditions. From Q5.09b, pw1 ¼ 0.65 0.5069.

35

pw2 ¼ 0:65 0:5069 14:7 ¼ 0:784 psia

From the steam tables, we note that corresponding to 0.784 psia, the saturation temperature is 93 F. This is also the dew point after compression. Cooling the air to below 93 F would result in its condensation.

5.11A

Q:

Calculate the energy absorbed by steam in a boiler if 400,000 lb=h of superheated steam at 1600 psia and 900 F is generated with feedwater at 250 F. What is the energy absorbed, in megawatts?

A:

The energy absorbed is given by

Q ¼ W ðh2 h1Þ (neglecting blowdown) ð15Þ where

W ¼ steam flow, lb=h

h2; h1 ¼ steam enthalpy and water enthalpy, Btu=lb Q ¼ duty, Btu=h

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From the steam tables, h2 ¼ 1425.3 Btu=lb and h1 ¼ 224 Btu=lb.

Q¼ 400;000 ð1425:3 224Þ

¼480:5 106 Btu=h

¼480:5 million Btu/h (MM Btu/h)

Using the fact that 3413 Btu=h ¼ 1 kW, we have

106

Q ¼ 480:5 3413 103 ¼ 141 MW

5.11B

Q:

Estimate the energy absorbed by wet steam at 80% quality in a boiler at 1600 psia when the feedwater temperature is 250 F.

A:

The enthalpy of wet steam can be computed as

where h is the enthalpy in Btu=lb. The subscripts g and f stand for saturated vapor and liquid at the referenced pressure, obtained from saturated steam properties. x is the steam quality fraction.

From the steam tables, hg ¼ 1163 Btu=lb and hf ¼ 624 Btu=lb at 1600 psia. The enthalpy of feedwater at 250 F is 226 Btu=lb.

h2 ¼ 0:8 1163 þ 0:2 624 ¼ 1054 Btu=lb

h1 ¼ 226 Btu=lb

Q ¼ 1054 226 ¼ 828 Btu=lb

If steam flow were 400,000 lb=h, then

Q ¼ 400;000 828 ¼ 331 106 ¼ 331 MM Btu=h

5.12

Q:

How is the wetness in steam specified? How do we convert steam by volume (SBV) to steam by weight?

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A:

A steam–water mixture is described by the term quality, x, or dryness fraction. x ¼ 80% means that in 1 lb of wet steam, 0.8 lb is steam and 0.2 lb is water. To relate these two terms, we use the expression

A CR of 4 means that the steam quality is 0.25 or 25%; in other words, 1 lb of mixture would have 0.25 lb of steam and the remainder would be water.

5.13A

Q:

How is the quality of steam determined using a throttling calorimeter?

A:

Throttling calorimeters (Fig. 5.1) are widely used in low pressure steam boilers for determining the moisture or wetness (quality) of steam. A sampling nozzle is located preferably in the vertical section of the saturated steam line far from bends or fittings. Steam enters the calorimeter through a throttling orifice and passes into a well-insulated expansion chamber. Knowing that throttling is an isoenthalpic process, we can rewrite Eq. (16) for enthalpy balance as

hs ¼ hm ¼ xhg þ ð1 xÞhf

where

hs; hm; hf ; hg ¼ enthalpies of steam, mixture, saturated liquid, and saturated steam, respectively

x ¼ steam quality fraction

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FIGURE 5.1 Throttling calorimeter.

The steam temperature after throttling is measured at atmospheric pressure, and then the enthalpy is obtained with the help of steam tables. The steam is usually in superheated condition after throttling.

Example

A throttling calorimeter measures a steam temperature of 250 F when connected to a boiler operating at 100 psia. Determine the steam quality.

Solution. hs at atmospheric pressure and at 250 F ¼ 1168.8 Btu=lb from steam tables; hg ¼ 1187.2 and hf ¼ 298.5 Btu=lb, also from steam tables. Hence

1168:8 ¼ 1187:2x þ ð1 xÞ298:5

or

x ¼ 0:979 or 97:9% quality

5.13B

Q:

How is steam quality related to steam purity?

Copyright © 2003 Marcel Dekker, Inc.

5.15

Q:

How is the water requirement for cooling a gas stream estimated? Estimate the water quantity required to cool 100,000 lb=h of flue gas from 900 F to 400 F. What is the final volume of the gas?

A:

From an energy balance it can be shown [1] that

q ¼ 5:39 10 4 ðt1 t2Þ

W

1090 þ 0:45 ðt2 150Þ

where

q ¼ water required, gpm

t1; t2 ¼ initial and final gas temperatures, F W ¼ gas flow entering the cooler, lb=h

Substitution yields

q ¼ 5:39 10 4 ð900 400Þ

100;0001090 þ 0:45 ð400 150Þ

¼ 23 gpm

The final gas volume is given by the expression

W

ð460 þ t2Þ 2361 þ 0:341 The final volume is 43,000 acfm.

5.16

Q:

In selecting silencers for vents or safety valves, we need to figure the volume of steam after the throttling process. Estimate the volume of steam when 60,000 lb=h of superheated steam at 650 psia and 800 F is blown to the atmosphere through a safety valve.

A:

We have to find the final temperature of steam after throttling, which may be considered an isoenthalpic process; that is, the steam enthalpy remains the same at 650 and 15 psia.

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From the steam tables, at 650 psia and 800 F, h ¼ 1402 Btu=lb. At 15 psia (atmospheric conditions), the temperature corresponding to an enthalpy of 1402 Btu=lb is 745 F. Again from the steam tables, at a pressure of 15 psia and a temperature of 745 F, the specific volume of steam is 48 cu ft=lb. The total volume of steam is 60,000 48 ¼ 2,880,000 cu ft=h.

5.17

Q:

How do we determine the steam required for deaeration and boiler blowdown water requirements?

A:

Steam plant engineers have to frequently perform energy and mass balance calculations around the deaerator and boiler to obtain the values of makeup water, blowdown, or deaeration steam flows. Boiler blowdown quantity depends on the total dissolved solids (TDS) of boiler water and the incoming makeup water. Figure 5.2 shows the scheme around a simple deaerator. Note that there could be several condensate returns. This analysis does not consider venting of steam from the deaerator or the heating of makeup using the blowdown water. These refinements can be done later to fine-tune the results.

The American Boiler Manufacturers Association (ABMA) and ASME provide guidelines on the TDS of boiler water as a function of pressure (see Tables 5.4 and 5.5. The drum solids concentration can be at or less than the value shown in these tables. Plant water chemists usually set these values after reviewing the complete plant chemistry.

FIGURE 5.2 Scheme of deaeration system.

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