Tài liệu Handbook of Mechanical Engineering Calculations P4 pptx

4.1
SECTION 4
STEAM GENERATION
EQUIPMENT AND AUXILIARIES
Determining Equipment Loading for
Generating Steam Efficiently
4.2
Steam Conditions with Two Boilers
Supplying the Same Line
4.6
Generating Saturated Steam by
Desuperheating Superheated Steam
4.7
Determining Furnace-Wall Heat Loss
4.8
Converting Power-Generation Pollutants
from Mass to Volumetric Units
4.10
Steam Boiler Heat Balance
Determination
4.11
Steam Boiler, Economizer, and Air-
Heater Efficiency
4.14
Fire-Tube Boiler Analysis and Selection
4.16
Safety-Valve Steam-Flow Capacity
4.18
Safety-Valve Selection for a Watertube
Steam Boiler
4.19

Steam-Quality Determination with a
Throttling Calorimeter
4.24
Steam Pressure Drop in a Boiler
Superheater
4.25
Selection of a Steam Boiler for a Given
Load
4.26
Selecting Boiler Forced- and Induced-
Draft Fans
4.30
Power-Plant Fan Selection from
Capacity Tables
4.33
Fan Analysis at Varying RPM, Pressure,
and Air or Gas Capacity
4.35
Boiler Forced-Draft Fan Horsepower
Determination
4.37
Effect of Boiler Relocation on Draft Fan
Performance
4.38
Analysis of Boiler Air Ducts and Gas
Uptakes
4.38
Determination of the Most Economical
Fan Control
4.44

Smokestack Height and Diameter
Determination
4.46
Power-Plant Coal-Dryer Analysis
4.48
Coal Storage Capacity of Piles and
Bunkers
4.50
Properties of a Mixture of Gases
4.51
Steam Injection in Air Supply
4.52
Boiler Air-Heater Analysis and Selection
4.53
Evaluation of Boiler Blowdown,
Deaeration, Steam and Water Quality
4.55
Heat-Rate Improvement Using Turbine-
Driven Boiler Fans
4.56
Boiler Fuel Conversion from Oil or Gas
to Coal
4.60
Energy Savings from Reduced Boiler
Scale
4.64
Ground Area and Unloading Capacity
Required for Coal Burning
4.66
Heat Recovery from Boiler Blowdown

Systems
4.67
Boiler Blowdown Percentage
4.69
Sizing Flash Tanks to Conserve Energy
4.70
Flash Tank Output
4.71
Determining Waste-Heat Boiler Fuel
Savings
4.74
Figuring Flue-Gas Reynolds Number by
Shortcuts
4.75
Determining the Feasibility of Flue-Gas
Recirculation for No Control in
Packaged Boilers
4.77
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS
4.2
POWER GENERATION
DETERMINING EQUIPMENT LOADING FOR
GENERATING STEAM EFFICIENTLY
A plant has a steam generator capable of delivering up to 1000,000 lb/h (45,400
kg/h) of saturated steam at 400 lb / in
2
(gage) (2756 kPa). The plant also has an

HRSG capable of generating up to 1000,000 lb/h (45,400 kg/h) of steam in the
fired mode at the same pressure. How should each steam generator be loaded to
generate a given quantity of steam most efficiently?
Calculation Procedure:
1. Develop the HRSG characteristics
In cogeneration and combined-cycle steam plants (gas turbine plus other prime
movers), the main objective of supervising engineers is to generate a needed quan-
tity of steam efficiently. Since there may be both HRSGs and steam boilers in the
plant, the key to efficient operation is an understanding of the performance char-
acteristics of each piece of equipment as a function of load.
In this plant, the HRSG generates saturated steam at 400 lb/in
2
(gage) (2756
kPa) from the exhaust of a gas turbine. It can be supplementary-fired to generate
additional steam. Using the HRSG simulation approach given in another calculation
procedure in this handbook, the HRSG performance at different steam flow rates
should be developed. This may be done manually or by using the HRSG software
developed by the author.
2. Select the gas/steam temperature profile in the design mode
Using a pinch point of 15
Њ
F (8.33
Њ
C) and approach point of 17
Њ
F (9.44
Њ
C), a tem-
perature profile is developed as discussed in the procedure for HRSG simulation.
The HRSG exit gas temperature is 319

Њ
F (159.4
Њ
C) while generating 25,000 lb/ h
(11,350 kg/h) of steam at 400 lb /in
2
(gage) (2756 kPa) using 230
Њ
F (110
Њ
C) feed-
water.
3. Prepare the gas/steam temperature profile in the fired mode
A simple approach is to use the fact that supplementary firing is 100 percent effi-
cient, as discussed in the procedure on HRSG simulation. All the fuel energy goes
into generating steam in single-pressure HRSGs.
Compute the duty of the HRSG—i.e., the energy absorbed by the steam—in the
unfired mode, which is 25.4 MM Btu /h (7.44 MW). The energy required to gen-
erate 50,000 lb/ h (22,700 kg /h) of steam is 50.8 MM Btu/ h (14.88 MW). Hence,
the additional fuel required
ϭ
50.8
Ϫ
25.4
ϭ
25.4 MM Btu/h (7.44 MW). If a
manual or computer simulation is done on the HRSG, fuel consumption will be
seen to be 24.5 MM Btu/ h (7.18 MW) on a Lower Heating Value (LHV) basis.
Similarly, the performance at other steam flows is also computed and summarized
in Table 1. Note that the exit gas temperature decreases as the steam flow increases.

This aspect of an HRSG is discussed in the simulation procedure elsewhere in this
handbook.
4. Develop the steam-generator characteristics
Develop the performance of the steam generator at various loads. Steam-generator
suppliers will gladly provide this information in great detail, including plots and
tabulations of the boiler’s performance. As shown in Table 2, the exit-gas temper-
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.3
TABLE 1
Performance of HRSG
Load, % 25 50 75 100
Steam generation, lb/ h (kg / h) 25,000 (11,350) 50,000 (22,700) 75,000 (34,050)
100,000 (45,400)
Duty, MM Btu/ h (MW) 25.4 (7.4) 50.8 (14.9) 76.3 (22.4) 101.6 (29.8)
Exhaust gas flow, lb/ h (kg / h) 152,000 (69,008) 153,140 (69,526) 154,330 (70,066)
155,570 (70,629)
Exit gas temperature,
Њ
F(
Њ
C) 319 (159) 285 (141) 273 (134) 269 (132)
Fuel fired, MM Btu / h LHV basis (MW) 0 (0) 24.50 (7.2) 50.00 (14.7) 76.50 (22.4)
ASME PTC 4.4 efficiency, % 70.80 83.79 88.0 89.53
Boiler pressure
ϭ
400 lb / in
2

(gage) (2756 kPa); feedwater temperature
ϭ
230
Њ
F (110
Њ
C); blowdown
ϭ
5 percent. Fuel
used: natural gas; percent volume C
1
ϭ
97; C
2
ϭ
2; C
3
ϭ
1; HHV
ϭ
1044 Btu / ft
3
(38.9 MJ / m
3
); LHV
ϭ
942 Btu /
ft
3
(35.1 MJ / m

3
).
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.4
TABLE 2
Performance of Steam Generator
Load, % 25 50 75 100
Steam generation, lb/ h (kg / h) 25,000 (11,350) 50,000 (22,700) 75,000 (34,050)
100,000 (45,400)
Duty, MM Btu/ h (MW) 25.4 (7.4) 50.8 (14.9) 76.3 (22.4) 101.6 (29.8)
Excess air, % 30 10 10 10
Flue gas, lb / h (kg/ h) 30,140 (13,684) 50,600 (22,972) 76,150 (34,572)
101,750 (46,195)
Exit gas temperature,
Њ
F(
Њ
C) 265 (129) 280 (138) 300 (149) 320 (160)
Heat losses, %
—Dry gas loss 3.93 3.56 3.91 4.27
—Air moisture loss 0.10 0.09 0.10 0.11
—Fuel moisture loss 10.43 10.49 10.58 10.66
—Radiation loss 2.00 1.00 0.70 0.50
Efficiency, %
—Higher Heating Value basis 83.54 84.86 84.70 84.46
—Lower Heating Value basis 92.58 94.05 93.87 93.60
Fuel fired, MM Btu / h LHV basis (MW) 27.50 (8.06) 54.00 (15.8) 81.30 (23.8) 108.60

(31.8)
Boiler pressure
ϭ
400 lb / in
2
(gage) (2756 kPa); feedwater temperature
ϭ
230
Њ
F (110
Њ
C); blowdown
ϭ
5 percent. Fuel
used: natural gas; percent volume C
1
ϭ
97; C
2
ϭ
2; C
3
ϭ
1; HHV
ϭ
1044 Btu / ft
3
(38.9 MJ / m
3
); LHV

ϭ
942 Btu /
ft
3
(35.1 MJ / m
3
).
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.5
TABLE 3
Fuel Consumption at Various Steam Loads
Total
steam
HRSG steam SG steam HRSG fuel Sg fuel Total fuel
lb/h lb/h lb/h MM Btu/h MM Btu/h MM Btu/h
150,000 50,000 100,000 24.50 108.60 133.10
150,000 75,000 75,000 50.00 81.30 131.30
150,000 100,000 50,000 76.50 54.00 130.50
100,000 0 100,000 0 108.60 108.60
100,000 25,000 75,000 0 81.30 81.30
100,000 50,000 50,000 24.50 54.00 78.50
100,000 75,000 25,000 50.00 27.50 77.50
100,000 100,000 0 75.60 0 76.50
50,000 0 50,000 0 54.00 54.00
50,000 25,000 25,000 0 27.50 27.50
50,000 50,000 0 24.50 0 24.50

kg/h kg/h
SI Units
kg/ h MW MW MW
68,100 22,700 45,400 7.2 31.8 38.9
68,100 34,050 34,050 14.7 23.8 38.5
68,100 45,400 45,400 22.4 15.8 38.2
45,400 0 45,400 0 31.8 31.8
45,400 11,350 34,050 0 23.8 23.8
45,400 22,700 22,700 7.2 15.8 23.0
45,400 34,050 11,350 14.7 8.1 22.7
45,400 45,400 0 22.4 0 22.4
22,700 0 22,700 0 15.8 15.8
22,700 11,350 11,350 0 8.1 8.1
22,700 11,350 0 7.2 0 7.2
ature decreases as the load on the steam generator declines. This is because the
ratio of gas/steam is maintained at nearly unity, unlike in an HRSG where the gas
flow remains constant and steam flow alone is varied. Further, the radiation losses
in a steam boiler increase at lower duty, while the exit-gas losses decrease. However,
the boiler’s efficiency falls within a narrow range. Table 3 also shows the steam
generator’s fuel consumption at various loads.
5. Calculate steam vs. fuel data for combined operation of the equipment
The next step is to develop, for combined operation of the HRSG and steam gen-
erator, a steam flow vs. fuel table such as that in Table 3. For example, 150,000
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.6
POWER GENERATION
lb/ h (68,100 kg /h) of steam could be generated in several ways—50,000 lb/h

(22,700 kg /h) in the HRSG and 100,000 lb/h (45,400 kg / h) in the steam generator.
Or each could generate 75,000 lb / h (34,050 kg /h); or 100,000 lb /h (45,400 kg / h)
in the HRSG and the remainder in the steam generator. The table shows that max-
imizing the HRSG output first is the most efficient way of generating steam because
no fuel is required to generate up to 25,000 lb/h (11,350 kg/h) of steam. However,
this may not always be possible because of the plant operating mode, equipment
availability, steam temperature requirements, etc.
Note also that the gas pressure drop in an HRSG does not vary significantly
with load as the gas mass flow remains nearly constant. The gas pressure drop
increases slightly as the firing temperature increases. On the other hand, the steam
generator fan power consumption vs. load increases more in proportion to load.
It is also seen that at higher steam capacities the difference in fuel consumption
between the various modes of operation is small. At 150,000 lb/ h (68,100 kg/ h),
the difference is about 3 MM Btu / h (0.88 MW), while at 100,000 lb/ h (45,400
kg/ h), the difference is 30 MM Btu/ h (8.79 MW). This difference should also be
kept in mind while developing an operational strategy.
If a superheater is used, the performance of the superheater would have to be
analyzed. Steam generators can generally maintain the steam temperature from 40
to 100 percent load, while in HRSGs the range is much larger as the steam tem-
perature increases with firing temperature and can be controlled.
Related Calculations. Developing the performance characteristics of each
piece of equipment as a function of load is the key to determining the mode of
operation and loading of each type of steam producer. For best results, develop a
performance curve for the steam generator, including all operating costs such as
fan power consumption, pump power consumption, and gas-turbine power output
as a function of load. This gives more insight into the total costs in addition to fuel
cost, which is the major cost.
This procedure is the work of V. Ganapathy, Heat Transfer Specialist, ABCO
Industries, Inc. The HRSG software mentioned in this procedure is available from
Mr. Ganapathy.

STEAM CONDITIONS WITH TWO BOILERS
SUPPLYING THE SAME STEAM LINE
Two closely adjacent steam boilers discharge equal amounts of steam into the same
short steam main. Steam from boiler No. 1 is at 200 lb/ in
2
(1378 kPa) and 420
Њ
F
(215.6
Њ
C) while steam from boiler No. 2 is at 200 lb/in
2
(1378 kPa) and 95 percent
quality. (a) What is the equilibrium condition after missing of the steam? (b) What
is the loss of entropy by the higher temperature steam? Assume negligible pressure
drop in the short steam main connecting the boilers.
Calculation Procedure:
1. Determine the enthalpy of the mixed steam
Use the T-S diagram, Fig. 1, to plot the condition of the mixed steam. Then, since
equal amounts of steam are mixed, the final enthalpy, H
3
ϭ
(H
1
ϩ
H
2
)/ 2. Substi-
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.

Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.7
800
˚
F (471
˚
C) 841.8
˚
F (449.8
˚
C) 200 psia (1378 kPa)
FIGURE 1 T-S plot of conditions with two boilers on line.
tuting, using date from the steam tables and Mollier chart, H
3
ϭ
(1225
ϩ
1164)/
2
ϭ
1194.5 Btu/lb (2783.2 kJ/kg).
2. Find the quality of the mixed steam
Entering the steam tables at 200 lb / in
2
(1378 kPa), find the enthalpy of the liquid
as 355.4 Btu/lb (828.1 kJ/ kg) and the enthalpy of vaporization as 843.3 Btu/lb
(1964.9 kJ/kg). Then, using the equation for wet steam with the known enthalpy
of the mixture from Step 1, 1194.5

ϭ
H
ƒ
ϩ
x
3
(H
ƒg
)
ϭ
355.4
ϩ
x
3
(843.3); x
3
ϭ
0.995, or 99.5 percent quality.
3. Find the entropy loss by the higher pressure steam
The entropy loss by the higher-temperature steam, referring to the Mollier chart
plot, is S
1
Ϫ
S
2
ϭ
1.575
Ϫ
1.541
ϭ

0.034 entropy units. The lower-temperature
steam gains S
3
Ϫ
S
2
ϭ
1.541
Ϫ
1.506
ϭ
0.035 units of entropy.
Related Calculations. Use this general approach for any mixing of steam
flows. Where different quantities of steam are being mixed, use the proportion of
each quantity to the total in computing the enthalpy, quality, and entropy of the
mixture.
GENERATING SATURATED STEAM BY
DESUPERHEATING SUPERHEATED STEAM
Superheated steam generated at 1350 lb/ in
2
(abs) (9301.5 kPa) and 950
Њ
F (510
Њ
C)
is to be used in a process as saturated steam at 1000 lb/ in
2
(abs) (6890 kPa). If the
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.

Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.8
POWER GENERATION
superheated steam is desuperheated continuously by injecting water at 500
Њ
F
(260
Њ
C), how many pounds (kg) of saturated steam will be produced per pound
(kg) of superheated steam?
Calculation Procedure:
1. Using the steam tables, determine the steam and water properties
Rounding off the enthalpy and temperature values we find that: Enthalpy of the
superheated steam at 1350 lb/in
2
(abs) (9301.5 kPa) and 950
Њ
F (510
Њ
C)
ϭ
H
1
ϭ
1465 Btu /lb (3413.5 kJ/kg); Enthalpy of saturated steam at 1000 lb/ in
2
(abs) (6890
kPa)
ϭ

H
2
ϭ
1191 Btu/lb (2775 kJ /kg); Enthalpy of water at 500
Њ
F (260
Њ
C)
ϭ
(500
Ϫ
32)
ϭ
H
3
ϭ
488 Btu/lb (1137 kJ/kg).
2. Set up a heat-balance equation and solve it
LX
ϭ
lb (kg) of water at 500
Њ
F (260
Њ
C) required to desuperheat the superheated
steam. Then, using the symbols given above, H
1
ϩ
X(H
3

)
ϭ
(1
ϩ
X)H
2
. Solving
for X
ϭ
(H
1
Ϫ
H
2
)/(H
2
Ϫ
H
3
)
ϭ
(1465
Ϫ
1191)/ (1191
Ϫ
488)
ϭ
0.39. Then,
1.0
ϩ

0.39
ϭ
1.39 lb (0.63 kg) of saturated steam produced per lb (kg) of super-
heated steam. Thus, if the process used 1000 lb (454 kg) of saturated steam at 1000
lb/in
2
(abs) (689 kPa), the amount of superheated steam needed to produce this
saturated steam would be 1000/1.39
ϭ
719.4 lb (326.6 kg).
Related Calculations. Desuperheating superheated steam for process and other
use is popular because it can save purchase and installation of a separate steam
generator for the lower pressure steam. While there is a small loss of energy in
desuperheating (from heat losses in the piping and desuperheater), this loss is small
compared to the savings made. That’s why you’ll find desuperheating being used
in central stations, industrial, commercial and marine plants throughout the world.
DETERMINING FURNACE-WALL HEAT LOSS
A furnace wall consists of 9-in (22.9-cm) thick fire brick, 4.5-in (11.4-cm) Sil-O-
Cel brick, 4-in (10.2-cm) red brick, and 0.25-in (0.64-cm) transite board. The ther-
mal conductivity, k, values, Btu / (ft
2
)(
Њ
F)(ft) [kJ/(m
2
)(
Њ
C)(m)] are as follows: 0.82
at 1800
Њ

F (982
Њ
C) for fire brick; 0.125 at 1800
Њ
F (982
Њ
C) for Sil-O-Cel; 0.52 at
500
Њ
F (260
Њ
C) for transite. A temperature of 1800
Њ
F (982
Њ
C) exists on the inside
wall of the furnace and 200
Њ
F (93.3
Њ
C) on the outside wall. Determine the heat loss
per hour through each 10 ft
2
(0.929 m
2
) of furnace wall. What is the temperature
of the wall at the joint between the fire brick and Sil-O-Cel?
Calculation Procedure:
1. Find the heat loss through a unit area of the furnace wall
Use the relation Q

ϭ ⌬
t/R, where Q
ϭ
heat transferred, Btu /h (W);
⌬
t
ϭ
temper-
ature difference between the inside of the furnace wall and the outside,
Њ
F(
Њ
C);
R
ϭ
resistance of the wall to heat flow
ϭ
L/(kXA), where L
ϭ
length of path
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.9
through which the heat flow, ft (m); k
ϭ
thermal conductivity, as defined above;
A

ϭ
area of path of heat flow, ft
2
(m
2
). Where there is more than one resistance to
heat flow, add them to get the total resistance.
Substituting, the above values for this furnace wall, remembering that there are
three resistances in series and solving for the heat flow through one square ft
(0.0.0929 m
2
), Q
ϭ
(1800
Ϫ
200)/ {[(1 /0.82)(9/12)]
ϩ
[(1/ 0.125)(4.5 /12)]
ϩ
[(1
/052)(4 / 12)]
ϩ
[(1/ 0.23)(0.25 /12)]}
ϭ
344 Btu / h ft
2
(1083.6 W /m
2
), or 10 (344)
ϭ

34400 Btu/h for 10 ft
2
(10,836 W/10 m
2
).
2. Compute the temperature within the wall at the stated joint
Use the relation, (
⌬
t)/(
⌬
t
1
)
ϭ
(R/R
1
), where
⌬
t
ϭ
temperature difference across
the wall,
Њ
F(
Њ
C);
⌬
t
1
ϭ

temperature at the joint being considered,
Њ
F(
Њ
C); R
ϭ
total resistance of the wall; R
1
ϭ
resistance of the first portion of the wall between
the inside and the joint in question.
Substituting, (1800
Ϫ
200)/ (
⌬
t
1
)
ϭ
4.646/ 0.915);
⌬
t
1
ϭ
315
Њ
F (157.2
Њ
C). Then
the interface temperature at the between the fire brick and the Sil-O-Cel is 1800

Ϫ
315
ϭ
1485
Њ
F (807.2
Њ
C)
Related Calculations. The coefficient of thermal conductivity given here,
Btu/ (ft
2
)(
Њ
F)(ft) is sometimes expressed in terms of per inch of thickness, instead
of per foot. Either way, the conversion is simple. In SI units, this coefficient is
expressed in kJ/ (m
2
)(
Њ
C)(m), or cm
2
and cm.
The exterior temperature of a furnace wall is an important considered in boiler
and process unit design from a human safety standpoint. Excess exterior tempera-
tures can cause injury to plant workers. Further, the higher the exterior temperature
of a furnace wall, the larger the heat loss from the fired vessel. Therefore, both
safety and energy conservation considerations are important in furnace design.
Typical interior furnace temperatures encountered in modern steam boilers range
from 2400
Њ

F (1316
Њ
C) near the fuel burners to 1600
Њ
F (871
Њ
C) in the superheater
interior. With today’s emphasis on congeneration and energy conservation, many
different fuels are being burned in boilers. Thus, a plant in Louisiana burns rice to
generate electricity while disposing of a process waste material.
Rice hulls, which comprise 20 percent of harvested rice, are normally processed
in a hammermill to increase their bulk from about 11 lb/ft
3
(176 kg/m
3
)to20
lb/ft
3
(320 kg /m
3
). Then they are spread or piled on land adjacent to the rice mill.
The hulls often smolder in the fields, like mine tailings from coal production. Con-
tinuous, uncontrolled burning may result, creating an environmental hazard and
problem.
Burning rice hulls in a boiler furnace may create unexpected temperatures both
inside and outside the furnace. Hence, it is important that the designer be able to
analyze both the interior and exterior furnace temperatures using the procedure
given here.
Another modern application of waste usage for power generation is the burning
of sludge in a heat-recovery boiler to generate electricity. Sludge from a wastewater

plant is burned in a combustor to generate steam for a turbogenerator. Not only are
fuel requirements for the boiler reduced, there is also significant savings of fuel
used to incinerate the sludge in earlier plants. Again, the furnace temperature is an
important element in designing such plants.
The data present in these comments on new fuels for boilers is from Power
magazine.
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.10
POWER GENERATION
CONVERTING POWER-GENERATION
POLLUTANTS FROM MASS TO
VOLUMETRIC UNITS
In the power-generation industry, emission levels of pollutants such as CO and NO
x
are often specified in mass units such as pounds per million Btu (kg per 1.055 MJ)
and volumetric units such as ppm (parts per million) volume. Show how to relate
these two measures for a gaseous fuel having this analysis: Methane
ϭ
97 percent;
Ethane
ϭ
2 percent; Propane
ϭ
1 percent by volume, and excess air
ϭ
10 percent.
Ambient air temperature during combustion

ϭ
80
Њ
F (26.7
Њ
C) and relative humid-
ity
ϭ
60 percent; fuel higher heating value, HHV
ϭ
23,759 Btu/ lb (55,358 kJ/
kg); 100 moles of fuel gas is the basis of the flue gas analysis.
Calculation Procedure:
1. Find the theoretical dry air required, and the moisture in the actual air
The theoretical dry air requirements, in M moles, can be computed from the sum
of (ft
3
of air per ft
3
of combustible gas)(percent of combustible in the fuel) using
data from Ganapathy, Steam Plant Calculations Manual, Marcel Dekker, Inc. thus:
M
ϭ
(9.528
ϫ
97)
ϩ
(16.675
ϫ
2)

ϩ
(23.821
ϫ
1)
ϭ
981.4 moles. Then, with 10
percent excess air, excess air, EA
ϭ
0.1(981.4)
ϭ
98.1 moles.
The excess oxygen, O
2
ϭ
(98.1 moles)(0.21)
ϭ
20.6 moles, where 0.21
ϭ
moles
of oxygen in 1 mole of air. The nitrogen, N
2
, produced by combustion
ϭ
(1.1 for
excess air)(981.4 moles)(0.79 moles of nitrogen in 1 mole of air)
ϭ
852.8 moles;
round to 853 moles for additional calculations.
The moisture in the air
ϭ

(981.4
ϩ
98.1)(29
ϫ
0.0142/ 18)
ϭ
24.69, say 24.7
moles. In this computation the values 29 and 18 are the molecular weights of dry
air and water vapor, respectively, while 0.0142 is the lb (0.0064 kg) moisture per
lb of dry air as shown in the previous procedure.
2. Compute the flue gas analysis for the combustion
Using the given data, CO
2
ϭ
(1
ϫ
97)
ϩ
(2
ϫ
2)
ϩ
(3
ϫ
1)
ϭ
104 moles. For
H
2
O

ϭ
(2
ϫ
97)
ϩ
(3
ϫ
2)
ϩ
(4
ϫ
1)
ϩ
24.7
ϭ
228.7 moles. From step 1,
N
2
ϭ
853 moles; O
2
ϭ
20.6 moles.
Now, the total moles
ϭ
104
ϩ
228.7
ϩ
853

ϩ
20.6
ϭ
1206.3 moles. The percent
volume of CO
2
ϭ
(104/ 1206.3)(100)
ϭ
8.6; the percent H
2
O
ϭ
(228.7/
1206.1)(100)
ϭ
18.96; the percent N
2
ϭ
(853/ 1206.3)(100)
ϭ
70.7; the percent
O
2
ϭ
(20.6/ 1206.3)(100)
ϭ
1.71.
3. Find the amount of flue gas produced per million Btu (1.055 MJ)
To relate the pounds per million Btu (1.055 MJ) of NO

x
or CO produced to ppmv,
we must know the amount of flue gas produced per million Btu (1.055 MJ). From
step 2, the molecular weight of the flue gases
ϭ
[(8.68
ϫ
44)
ϩ
(18.96
ϫ
18)
ϩ
(70.7
ϫ
28)
ϩ
(1.71
ϫ
32)]/ 100
ϭ
27.57.
The molecular weight of the fuel
ϭ
[(97
ϫ
16)
ϩ
(2
ϫ

30)
ϩ
(1
ϫ
44)]/ 100
ϭ
16.56. Now the ratio of flue gases/ fuel
ϭ
(1206.3
ϫ
27.57)/ (100
ϫ
16.56)
ϭ
20.08
lb flue gas/lb fuel (9.12 kg /kg). Hence, 1 million Btu fired produces (1,000,000)/
23,789
ϭ
42 lb (19.1 kg) fuel
ϭ
(42)(20.08)
ϭ
844 lb (383 kg) wet flue gases.
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.11
4. Calculate ppm values for the gases

Let 1 million Btu fired generate N lb (kg) of NO
x
. For emission calculations, NO
x
is considered to have a molecular weight of 46. Also, the reference for NO
x
or CO
regulations is 3 percent dry oxygen by volume for steam generators. Hence, we
have the relation, N
N
ϭ
10
6
(Y
x
)(N/ 46)(MW
ƒg
)[(21
Ϫ
3)/ (21
Ϫ
O
2
XY)], where V
N
ϭ
ppm dry NO
x
; Y
ϭ

100/ (100
Ϫ
percent H
2
O), where percent H
2
O is the percent
volume water vapor in the flue gases; N
ϭ
lb (kg) of NO
x
per million Btu (1.055
MJ) fired on an HHV basis, MW
ƒg
ϭ
molecular weight of wet flue gases; W
gm
ϭ
amount of wet flue gas produced per million Btu (1.055 MJ) fired.
Substituting in the above relation, V
N
ϭ
10
6
(N
x
)[100/ (100
Ϫ
18.96)] (27.57/
844)(21

Ϫ
3)/ [(21
Ϫ
1.71)(100/ (100
Ϫ
18.96)]
ϭ
832N.
Similarly, V
c
ϭ
ppmv CO
2
generated per million Btu (1.055 MJ) fired
ϭ
1367
Њ
C,
where C
ϭ
lb (kg) of CO generated per million Btu (1.055 MJ) and V
c
ϭ
amount
in ppmvd (dry). The effect of excess air on these calculations is not at all significant.
One may perform these calculations at 30 percent excess air and still show that V
N
ϭ
832N and V
c

ϭ
1367 for natural gas.
Related Calculations. These calculations for oil fuels also to show that V
N
ϭ
783N and V
c
ϭ
1286C.
This procedure is the work of V. Ganapathy, Heat Transfer Specialist, ABCO
Industries, Inc.
STEAM BOILER HEAT BALANCE
DETERMINATION
A steam generator having a maximum rated capacity of 60,000 lb/ h (27,000 kg/
h) is operating at 45,340 lb/h (20,403 kg /h), delivering 125-lb/in
2
(gage) 400
Њ
F
(862-kPa, 204
Њ
C) steam with a feedwater temperature of 181
Њ
F (82.8
Њ
C). At this
generating rate, the boiler requires 4370 lb/h (1967 kg/ h) of West Virginia bitu-
minous coal having a heating value of 13,850 Btu/lb (32,215 kJ/ kg) on a dry basis.
The ultimate fuel analysis is: C
ϭ

0.7757; H
2
ϭ
0.0507; O
2
ϭ
0.0519; N
2
ϭ
0.0120;
S
ϭ
0.0270; ash
ϭ
0.0827; total
ϭ
1.0000. The coal contains 1.61 percent moisture.
The boiler-room intake air and the fuel temperature
ϭ
79
Њ
F (26.1
Њ
C) dry bulb, 71
Њ
F
(21.7
Њ
C) wet bulb. The flue-gas temperature is 500
Њ

F (260
Њ
C), and the analysis of
the flue gas shows these percentages: CO
2
ϭ
12.8; CO
ϭ
0.4; O
2
ϭ
6.1; N
2
ϭ
80.7; total
ϭ
100.0. Measured ash and refuse
ϭ
9.42 percent of dry coal; combus-
tible in ash and refuse
ϭ
32.3 percent. Compute a heat balance for this boiler based
on these test data. The boiler has four water-cooled furnace walls.
Calculation Procedure:
1. Determine the heat input to the boiler
In a boiler heat balance the input is usually stated in Btu per pound of fuel as fired.
Therefore, input
ϭ
heating value of fuel
ϭ

13,850 Btu/lb (32,215 kJ/kg).
2. Compute the output of the boiler
The output of any boiler
ϭ
Btu/ lb (kJ/ kg) of fuel
ϩ
the losses. In this step the
first portion of the output, Btu/ lb (kJ/kg) of fuel will be computed. The losses will
be computed in step 3.
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.12
POWER GENERATION
First find W
s
, lb of steam produced per lb of fuel fired. Since 45,340 lb/h (20,403
kg/ h) of steam is produced when 4370 lb/ h (1967 kg /h) of fuel is fired, W
s
ϭ
45,340/ 4370
ϭ
10.34 lb of steam per lb (4.65 kg/kg) of fuel.
Once W
s
is known, the output h
1
Btu/ lb of fuel can be found from h
1

ϭ
W
s
(h
s
Ϫ
h
w
), where h
s
ϭ
enthalpy of steam leaving the superheater, or boiler if a super-
heater is not used; h
w
ϭ
enthalpy of feedwater, Btu/ lb. For this boiler with steam
at 125 lb/ in
2
(gage) [
ϭ
139.7 lb/ in
2
(abs)] and 400
Њ
F (930 kPa, 204
Њ
C), h
s
ϭ
1221.2

Btu/ lb (2841 kJ/kg), and h
w
ϭ
180.92 Btu /lb (420.8 kJ/ kg), from the steam tables.
Then h
1
ϭ
10.34(1221.2
Ϫ
180.92)
ϭ
10,766.5 Btu/lb (25,043 kJ/kg) of coal.
3. Compute the dry flue-gas loss
For any boiler, the dry flue-gas loss h
2
Btu/ lb (kJ/kg) of fuel is given by h
2
ϭ
0.24W
g
ϫ
(T
g
Ϫ
T
a
), where W
g
ϭ
lb of dry flue gas per lb of fuel; T

g
ϭ
flue-gas
exit temperature,
Њ
F; T
a
ϭ
intake-air temperature,
Њ
F.
Before W
g
can be found, however, it must be determined whether any excess
air is passing through the boiler. Compute the excess air, if any, from excess air,
percent
ϭ
100 (O
2
Ϫ
1
⁄
2
CO)/ [0.264N
2
Ϫ
(O
2
Ϫ
1

⁄
2
CO)], where the symbols refer
to the elements in the flue-gas analysis. Substituting values from the flue-gas anal-
ysis gives excess air
ϭ
100(6.1
Ϫ
0.2)/ [0.264
ϫ
80.7
Ϫ
(6.1
Ϫ
0.2)]
ϭ
38.4
percent.
Using the method given in earlier calculation procedures, find the air required
for complete combustion as 10.557 lb/lb (4.571 kg /kg) of coal. With 38.4 percent
excess air, the additional air required
ϭ
(10.557)(0.384)
ϭ
4.053 lb/lb (1.82 kg /
kg) of fuel.
From the same computation in which the air required for complete combustion
was determined, the lb of dry flue gas per lb of fuel
ϭ
11.018 (4.958 kg/kg). Then,

the total flue gas at 38.4 percent excess air
ϭ
11.018
ϩ
4.053
ϭ
15.071 lb/lb
(6.782 kg/kg) of fuel.
With a flue-gas temperature of 500
Њ
F (260
Њ
C), and an intake-air temperature of
79
Њ
F (26.1
Њ
C), h
2
ϭ
0.24(15.071)(500
Ϫ
70)
ϭ
1524 Btu/ lb (3545 kJ/kg) of fuel.
4. Compute the loss due to evaporation of hydrogen-formed water
Hydrogen in the fuel is burned in forming H
2
O. This water is evaporated by heat
in the fuel, and less heat is available for producing steam. This loss is h

3
Btu/lb
of fuel
ϭ
9H(1089
Ϫ
T
ƒ
ϩ
0.46T
g
), where H
ϭ
percent H
2
in the fuel
Ϭ
100;
T
ƒ
ϭ
temperature of fuel before combustion,
Њ
F; other symbols as before. For this
fuel with 5.07 percent H
2
, h
3
ϭ
9(5.07/ 100)(1089

Ϫ
79
ϩ
0.46
ϫ
500)
ϭ
565.8
Btu/ lb (1316 kJ/ kg) of fuel.
5. Compute the loss from evaporation of fuel moisture
This loss is h
4
Btu/ lb of fuel
ϭ
W
mƒ
(1089
Ϫ
T
ƒ
ϩ
0.46T
g
), where W
mƒ
ϭ
lb of
moisture per lb of fuel; other symbols as before. Since the fuel contains 1.61 percent
moisture, in terms of dry coal this is (1.61)/(100
Ϫ

1.61)
ϭ
0.0164, or 1.64 percent.
Then h
4
ϭ
(1.64/ 100)(1089
Ϫ
79
ϩ
0.46
ϫ
500)
ϭ
20.34 Btu/ lb (47.3 kJ /kg) of
fuel.
6. Compute the loss from moisture in the air
This loss is h
5
Btu/ lb of fuel
ϭ
0.46W
ma
(T
g
Ϫ
T
a
), W
ma

ϭ
(lb of water per lb of
dry air)(lb air supplied per lb fuel). From a psychrometric chart, the weight of
moisture per lb of air at a 79
Њ
F (26.1
Њ
C) dry-bulb and 71
Њ
F (21.7
Њ
C) wet-bulb
temperature is 0.014 (0.006 kg). The combustion calculation, step 3, shows that the
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.13
total air required with 38.4 percent excess air
ϭ
10.557
ϩ
4.053
ϭ
14.61 lb of air
per lb (6.575 kg/ kg) of fuel. Then, W
ma
ϭ
(0.014)(14.61)

ϭ
0.2045 lb of moisture
per lb (0.092 kg/ kg) of air. And h
5
ϭ
(0.46)(0.2045)(500
Ϫ
79)
ϭ
39.6 Btu/ lb
(92.1 kJ/kg) of fuel.
7. Compute the loss from incomplete combustion of C to CO
2
in the stack
This loss is h
6
Btu/ lb of fuel
ϭ
[CO/CO
ϩ
CO
2
)](C)(10.190), where CO and CO
2
are the percent by volume of these compounds in the flue gas by Orsat analysis;
C
ϭ
lb carbon per lb of coal. With the given flue-gas analysis and the coal ultimate
analysis, h
6

ϭ
0.4/ (0.4
ϩ
12.8)[(77.57)/ (100)](10.190)
ϭ
239.5 Btu / lb (557
kJ/ kg) of fuel.
8. Compute the loss due to unconsumed carbon in the refuse
This loss is h
7
Btu/ lb of fuel
ϭ
W
c
(14,150), where W
c
ϭ
lb of unconsumed carbon
in refuse per lb of fuel fired. With an ash and refuse of 9.42 percent of the dry
coal and combustible in the ash and refuse of 32.3 percent, h
7
ϭ
(9.42/ 100)
(32.3/ 100)(14,150)
ϭ
430.2 Btu/lb (1006 kJ/kg) of fuel.
9. Find the radiation loss in the boiler furnace
Use the American Boiler and Affiliated Industries (ABAI) chart, or the manufac-
turer’s engineering data to approximate the radiation loss in the boiler. Either source
will show that the radiation loss is 1.09 percent of the gross heat input. Since the

gross heat input is 13,850 Btu/lb (32,215 kJ/kg) of fuel, the radiation loss
ϭ
(13,850)(1.09/ 100)
ϭ
151.0 Btu/lb (351.2 kJ/kg) of fuel.
10. Summarize the losses; find the unaccounted-for loss
Set up a tabulation thus, entering the various losses computed earlier.
The unaccounted-for loss is found by summing all the other losses, 3 through
9, and subtracting from 100.00.
Related Calculations. Use this method to compute the heat balance for any
type of boiler—watertube or firetube—in any kind of service—power, process, or
heating—using any kind of fuel—coal, oil, gas, wood, or refuse. Note that step 3
shows how to compute excess air from an Orsat flue-gas analysis.
More stringent environmental laws are requiring larger investments in steam-
boiler pollution-control equipment throughout the world. To control sulfur emis-
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.14
POWER GENERATION
sions, expensive scrubbers are required on large boilers. Without such scrubbers the
sulfur emissions can lead to acid rain, smog, and reduced visibility in the area of
the plant and downwind from it.
With the increased number of free-trade agreements between adjacent countries,
cross-border pollution is receiving greater attention. The reason for this increased
attention is because not all countries have the same environmental control re-
quirements. When a country with less stringent requirements pollutes an adjacent
country having more stringent pollution regulations, both political and regulatory
problems can arise.

For example, two adjacent countries are currently discussing pollution problems
of a cross-border type. One country’s standard for particulate emissions is 10 times
weaker than the adjacent country’s, while its sulfur dioxide limit is 8 times weaker.
With such a wide divergence in pollution requirements, cross-border flows of pol-
lutants can be especially vexing.
All boiler-plant designers must keep up to date on the latest pollution regula-
tions. Today there are some 90,000 environmental regulations at the federal, state,
and local levels, and more than 40 percent of these regulations will change during
the next 12 months. To stay in compliance with such a large number of regulations
requires constant attention to those regulations applicable to boiler plants.
STEAM BOILER, ECONOMIZER, AND
AIR-HEATER EFFICIENCY
Determine the overall efficiency of a steam boiler generating 56,00 lb / h (7.1
kg/s) of 600 lb/in
2
(abs) (4137.0 kPa) 800
Њ
F (426.7
Њ
C) steam. The boiler is con-
tinuously blown down at the rate of 2500 lb/h (0.31 kg / s). Feedwater enters the
economizer at 300
Њ
F (148.9
Њ
C). The furnace burns 5958 lb/ h (0.75 kg/s) of 13,100-
Btu/ lb (30,470.6-kJ /kg), HHV (higher heating value) coal having an ultimate anal-
ysis of 68.5 percent C, 5 percent H
2
, 8.9 percent O

2
, 1.2 percent N
2
, 3.2 percent S,
8.7 percent ash, and 4.5 percent moisture. Air enters the boiler at 63
Њ
F (17.2
Њ
C)
dry-bulb and 56
Њ
F (13.3
Њ
C) wet-bulb temperature, with 56 gr of vapor per lb (123.5
gr/ kg) of dry air. Carbon in the fuel refuse is 7 percent, refuse is 0.093 lb/lb (0.2
kg/ kg) of fuel. Feedwater leaves the economizer at 370
Њ
F (187.8
Њ
C). Flue gas enters
the economizer at 850
Њ
F (454.4
Њ
C) and has an analysis of 15.8 percent CO
2
, 2.8
percent O
2
, and 81.4 percent N

2
. Air enters the air heater at 63
Њ
F (17.2
Њ
C) with 56
gr/ lb (123.5 gr/kg) of dry air; air leaves the heater at 480
Њ
F (248.9
Њ
C). Gas enters
the air heater at 570
Њ
F (298.9
Њ
C), and 14 percent of the air to the furnace comes
from the mill fan. Determine the steam generator overall efficiency, economizer
efficiency, and air-heater efficiency. Figure 2 shows the steam generator and the
flow factors that must be considered.
Calculation Procedure:
1. Determine the boiler output
The boiler output
ϭ
S(h
g
Ϫ
h
ƒ1
)
ϩ

S
r
(h
g3
Ϫ
h
g2
)
ϩ
B(h
ƒ3
Ϫ
h
ƒ1
), where S
ϭ
steam
generated, lb/ h; h
g
ϭ
enthalpy of the generated steam, Btu/lb; h
ƒ1
ϭ
enthalpy of
inlet feedwater; S
r
ϭ
reheated steam flow, lb/h (if any); h
g3
ϭ

outlet enthalpy of
reheated steam; h
g2
ϭ
inlet enthalpy of reheated steam; B
ϭ
blowoff, lb/h; h
ƒ3
ϭ
blowoff enthalpy, where all enthalpies are in Btu / lb. Using the appropriate steam
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.15
FIGURE 2 Points in a steam generator where temperatures and enthalpies
are measured in determining the boiler efficiency.
table and deleting the reheat factor because there is no reheat, we get boiler
output
ϭ
56,000(1407.7
Ϫ
269.6)
ϩ
2500(471.6
Ϫ
269.6)
ϭ
64,238,600 Btu / h

(18,826.5 kW).
2. Compute the heat input to the boiler
The boiler input
ϭ
FH, where F
ϭ
fuel input, lb/ h (as fired); H
ϭ
higher heating
value, Btu/lb (as fired). Or, boiler input
ϭ
5958(13,100)
ϭ
78,049,800 Btu/ h
(22,874.1 kW).
3. Compute the boiler efficiency
The boiler efficiency
ϭ
(output, Btu/h) / (input, Btu /h)
ϭ
64,238,600/ 78,049,800
ϭ
0.822, or 82.2 percent.
4. Determine the heat absorbed by the economizer
The heat absorbed by the economizer, Btu/ h
ϭ
w
w
(h
ƒ2

Ϫ
h
ƒ1
), where w
w
ϭ
feed-
water flow, lb/h; h
ƒ1
and h
ƒ2
ϭ
enthalpies of feedwater leaving and entering the
economizer, respectively, Btu/lb. For this economizer, with the feedwater leaving
the economizer at 370
Њ
F (187.8
Њ
C) and entering at 300
Њ
F (148.9
Њ
C), heat absorbed
ϭ
(56,000
ϩ
2500)(342.79
Ϫ
269.59)
ϭ

4.283,000 Btu/h (1255.2 kW). Note that
the total feedwater flow w
w
is the sum of the steam generated and the continuous
blowdown rate.
5. Compute the heat available to the economizer
The heat available to the economizer, Btu/h
ϭ
H
g
F, where H
g
ϭ
heat available in
flue gas, Btu / lb of fuel
ϭ
heat available in dry gas
ϩ
heat available in flue-gas
vapor, Btu/ lb of fuel
ϭ
(t;
3
Ϫ
t
ƒ1
)(0.24G)
ϩ
(t
3

Ϫ
t
ƒ1
)(0.46){M
ƒ
ϩ
8.94H
2
ϩ
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.16
POWER GENERATION
M
a
[G
Ϫ
C
b
Ϫ
N
2
Ϫ
7.94(H
2
Ϫ
O
2

/8)]}, where G
ϭ
{[11CO
2
ϩ
8O
2
ϩ
7(N
2
ϩ
CO)]/ [3(CO
2
ϩ
CO)]}(C
b
ϩ
S/ 2.67)
ϩ
S/ 1.60; M
ƒ
ϭ
lb of moisture per lb fuel
burned; M
a
ϭ
lb of moisture per lb of dry air to furnace; C
b
ϭ
lb of carbon burned

per lb of fuel burned
ϭ
C
ϭ
RC
r
; C
r
ϭ
lb of combustible per lb of refuse; R
ϭ
lb
of refuse per lb of fuel; H
2
,N
2
,C,O
2
,S
ϭ
lb of each element per lb of fuel, as
fired; CO
2
, CO, O
2
,N
2
ϭ
percentage parts of volumetric analysis of dry combustion
gas entering the economizer. Substituting gives C

b
ϭ
0.685
Ϫ
(0.093)(0.07)
ϭ
0.678
lb/ lb (0.678 kg/kg) fuel; G
ϭ
[11(0.158)
ϩ
8(0.028)
ϩ
7(0.814)]/ [3(0.158)]
ϫ
(0.678
ϩ
0.032/ 2.67)
ϩ
0.032/ 1.60; G
ϭ
11.18 lb/lb (11.18 kg/kg) fuel. H
g
ϭ
(800
Ϫ
300)(0.24)
ϫ
(11.18)
ϩ

(800
Ϫ
300)(0.46){0.045
ϩ
(8.9)(0.05)
ϩ
56/
7000[11.18
Ϫ
0.678
Ϫ
0.012
Ϫ
7.94
ϫ
(0.05
Ϫ
0.089/ 8)]}; H
g
ϭ
1473 Btu/ lb
(3426.2 kJ/kg) fuel. Heat available
ϭ
H
g
F
ϭ
(1473)(5958)
ϭ
8,770,000 Btu / h

(2570.2 kW).
6. Compute the economizer efficiency
The economizer efficiency
ϭ
(heat absorbed, Btu / h) /(heat available, Btu /h)
ϭ
4,283,000/ 8,770,000
ϭ
0.488, or 48.8 percent.
7. Compute the heat absorbed by air heater
The heat absorbed by the air heater, Btu/ lb of fuel,
ϭ
A
h
(t
2
Ϫ
t
1
)(0.24
ϩ
0.46M
a
),
where A
h
ϭ
air flow through heater, lb/ lb fuel
ϭ
A

Ϫ
A
m
; A
ϭ
total air to furnace,
lb/ lb fuel
ϭ
G
Ϫ
C
b
Ϫ
N
2
Ϫ
7.94(H
2
Ϫ
O
2
/8); G
ϭ
similar to economizer but
based on gas at the furnace exit; A
m
ϭ
external air supplied by the mill fan or other
source, lb/ lb of fuel. Substituting shows G
ϭ

[11(0.16)
ϩ
8(0.26)
ϩ
7(0.184)]/
[3(0.16)](0.678
ϩ
0.032/ 2.67)
ϩ
0.032/ 1.60; G
ϭ
11.03 lb/lb (11.03 kg /kg) fuel;
A
ϭ
11.03
Ϫ
0.69
Ϫ
0.012
Ϫ
7.94(0.05
Ϫ
0.089/ 8); A
ϭ
10.02 lb/lb (10.02 kg /
kg) fuel. Heat absorbed
ϭ
(1
Ϫ
0.15)(10.02)(480

Ϫ
63)(0.24
ϩ
56/ 7000)
ϭ
865.5
Btu/ lb (2013.2 kJ/ kg fuel.
8. Compute the heat available to the air heater
The heat available to the air heater, Btu / h
ϭ
(t
5
Ϫ
t
1
)0.24G
ϩ
(t
5
Ϫ
t
1
)0.46(M
ƒ
ϩ
8.94H
2
ϩ
M
a

A). In this relation, all symbols are the same as for the economizer
except that G and A are based on the gas entering the heater. Substituting gives G
ϭ
[11(0.15)
ϩ
8(0.036)
ϩ
7(0.814)]/ [3(0.15)](0.678
ϩ
0.032/ 2.67)
ϩ
0.032/ 1.60;
G
ϭ
11.72 lb/ lb (11.72 kg/ kg) fuel. And A
ϭ
11.72
Ϫ
0.69
Ϫ
0.012
Ϫ
7.94(0.05
Ϫ
0.089/ 8)
ϭ
10.71 lb/lb (10.71 kg/kg) fuel. Heat available
ϭ
(570
Ϫ

3)(0.24)(11.72)
ϩ
(570
Ϫ
63)(0.46)[0.045
ϩ
8.94(0.05)
ϩ
56/ 7000(10.71)]
ϭ
1561
Btu/ lb (3630.9 kJ/ kg).
9. Compute the air-heater efficiency
The air-heater efficiency
ϭ
(heat absorbed, Btu/ lb fuel)/(heat available, Btu /lb
fuel)
ϭ
865.5/ 1561
ϭ
0.554, or 55.4 percent.
Related Calculations. The above procedure is valid for all types of steam gen-
erators, regardless of the kind of fuel used. Where oil or gas is the fuel, alter the
combustion calculations to reflect the differences between the fuels. Further, this
procedure is also valid for marine and portable boilers.
FIRE-TUBE BOILER ANALYSIS AND SELECTION
Determine the heating surface in an 84-in (213.4-cm) diameter fire-tube boiler 18
ft (5.5 m) long having 84 tubes of 4-in (10.2-cm) ID if 25 percent of the upper
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.

Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.17
shell ends are heat-insulated. How much steam is generated if the boiler evaporates
34.5 lb / h of water per 12 ft
2
[3.9 g /(m
2
⅐
s)] of heating surface? How much heat is
added by the boiler if it operates at 200 lb/ in
2
(abs) (1379.0 kPa) with 200
Њ
F
(93.3
Њ
C) feedwater? What is the factor of evaporation for this boiler? How much
hp is developed by the boiler if 7,000,000 Btu / h (2051.4 kW) is delivered to the
water?
Calculation Procedure:
1. Compute the shell area exposed to furnace gas
Shell area
ϭ
␲
DL(1
Ϫ
0.25), where D
ϭ

boiler diameter, ft; L
ϭ
shell length, ft;
1
Ϫ
0.25 is the portion of the shell in contact with the furnace gas. Then shell
area
ϭ
␲
(84/ 12)(18)(0.75)
ϭ
297 ft
2
(27.0 m
2
).
2. Compute the tube area exposed to furnace gas
Tube area
ϭ
␲
dLN, where
ϭ
tube ID, ft; L
ϭ
tube length, ft; N
ϭ
number of tubes
in boiler. Substituting gives tube area
ϭ
␲

(4/ 12)(18)(84)
ϭ
1583 ft
2
(147.1 m
2
).
3. Compute the head area exposed to furnace gas
The area exposed to furnace gas is twice (since there are two heads) the exposed
head area minus twice the area occupied by the tubes. The exposed head area is
(total area)(1
Ϫ
portion covered by insulation, expressed as a decimal). Substituting,
we get 2
␲
D
2
/4
Ϫ
(2)(84)
␲
d
2
/4
ϭ
2
␲
/4(84 / 12)
2
(0.75)

Ϫ
(2)(84)
␲
(4/ 12)
2
/4
ϭ
head area
ϭ
43.1 ft
2
(4.0 m
2
).
4. Find the total heating surface
The total heating surface of any fire-tube boiler is the sum of the shell, tube, and
head areas, or 297.0
ϩ
1583
ϩ
43.1
ϭ
1923 ft
2
(178.7 m
2
), total heating surface.
5. Compute the quantity of steam generated
Since the boiler evaporates 34.5 lb/h of water per 12 ft
2

[3.9 g / (m
2
⅐
s)] of heating
surface, the quantity of steam generated
ϭ
34.5 (total heating surface, ft
2
)/12
ϭ
34.5(1923.1)/ 12
ϭ
5200 lb/h (0.66 kg /s).
Note: Evaporation of 34.5 lb /h (0.0043 kg/s) from and at 212
Њ
F (100.0
Њ
C) is
the definition of the now-discarded term boiler hp. However, this term is still met
in some engineering examinations and is used by some manufacturers when com-
paring the performance of boilers. A term used in lieu of boiler horsepower, with
the same definition, is equivalent evaporation. Both terms are falling into disuse,
but they are included here because they still find some use today.
6. Determine the heat added by the boiler
Heat added, Btu/lb of steam
ϭ
h
g
Ϫ
h

ƒ1
; from steam table values 1198.4
Ϫ
167.99
ϭ
1030.41 Btu / lb (2396.7 kJ / kg). An alternative way of computing heat
added is h
g
Ϫ
(feedwater temperature,
Њ
F,
Ϫ
32), where 32 is the freezing temper-
ature of water on the Fahrenheit scale. By this method, heat added
ϭ
1198.4
Ϫ
(200
Ϫ
32)
ϭ
1030.4 Btu / lb (2396.7 kJ/kg). Thus, both methods give the same
results in this case. In general, however, use of steam table values is preferred.
7. Compute the factor of evaporation
The factor of evaporation is used to convert from the actual to the equivalent evap-
oration, defined earlier. Or, factor of evaporation
ϭ
(heat added by boiler,
Btu/ lb) /970.3, where 970.3 Btu/lb (2256.9 kJ/kg) is the heat added to develop 1

Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.18
POWER GENERATION
boiler hp (bhp) (0.75 kW). Thus, the factor of evaporation for this boiler
ϭ
1030.4/ 970.3
ϭ
1.066.
8. Compute the boiler hp output
Boiler hp
ϭ
(actual evaporation, lb /h) (factor of evaporation)/ 34.5. In this relation,
the actual evaporation must be computed first. Since the furnace delivers 7,000,000
Btu/ h (2051.5 kW) to the boiler water and the water absorbs 1030.4 Btu/lb (2396.7
kJ/ kg) to produce 200-lb/in
2
(abs) (1379.0-kPa) steam with 200
Њ
F (93.3
Њ
C) feed-
water, the steam generated, lb/h
ϭ
(total heat delivered, Btu /h)/(heat absorbed,
Btu/ lb)
ϭ
7,000,000/ 1030.4

ϭ
6670 lb / h (0.85 kg / s). Then boiler hp
ϭ
(6760)(1.066)/ 34.5
ϭ
209 hp (155.9 kW).
The rated hp output of horizontal fire-tube boilers with separate supporting walls
is based on 12 ft
2
(1.1 m
2
) of heating surface per boiler hp. Thus, the rated hp of
the boiler
ϭ
1923.1/ 12
ϭ
160 hp (119.3 kW). When producing 209 hp (155.9
kW), the boiler is operating at 209/ 160, or 1,305 times its normal rating, or
(100)(1.305)
ϭ
130.5 percent of normal rating.
Note: Today most boiler manufacturers rate their boilers in terms of pounds per
hour of steam generated at a stated pressure. Use this measure of boiler output
whenever possible. Inclusion of the term boiler hp in this handbook does not in-
dicate that the editor favors or recommends its use. Instead, the term was included
to make the handbook as helpful as possible to users who might encounter the term
in their work.
SAFETY-VALVE STEAM-FLOW CAPACITY
How much saturated steam at 150 lb/in
2

(abs) (1034.3 kPa) can a 2.5-in (6.4-cm)
diameter safety valve having a 0.25-in (0.6-cm) lift pass if the discharge coefficient
of the valve c
d
is 0.75? What is the capacity of the same valve if the steam is
superheated 100
Њ
F (55.6
Њ
C) above its saturation temperature?
Calculation Procedure:
1. Determine the area of the valve annulus
Annulus area, in
2
ϭ
A
ϭ
␲
DL, where D
ϭ
valve diameter, in; L
ϭ
valve lift, in.
Annulus area
ϭ
␲
(2.5)(0.25)
ϭ
1.966 in
2

(12.7 cm
2
).
2. Compute the ideal flow for this safety valve
Ideal flow F
i
lb/ s for any safety valve handling saturated steam is F
i
ϭ
A/60,
0.97
p
s
where p
s
ϭ
saturated-steam pressure, lb / in
2
(abs). For this valve, F
i
ϭ
(150)
0.97
(1.966)/ 60
ϭ
4.24 lb/s (1.9 kg/s).
3. Compute the actual flow through the valve
Actual flow F
a
ϭ

F
i
c
d
ϭ
(4.24)(0.75)
ϭ
3.18 lb/ s (1.4 kg / s)
ϭ
(3.18)(3600 s / h)
ϭ
11,448 lb/h (1.44 kg /s).
4. Determine the superheated-steam flow rate
The ideal superheated-steam flow F
is
lb/s is F
is
ϭ
A/ [60(1
ϩ
0.0065t
s
)], where
0.97
p
s
t
s
ϭ
superheated temperature, above saturation temperature,

Њ
F. The F
is
ϭ
(150)
0.97
(1.966)/ [60(1
ϩ
0.0065
ϫ
100)]
ϭ
3.96 lb/ s (1.8 kg /s). The actual flow
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.19
is F
as
ϭ
F
is
c
d
ϭ
(3.96)(0.75)
ϭ
2.97 lb/s (1.4 kg/s)

ϭ
(2.97)(3600)
ϭ
10,700 lb/
h (1.4 kg / s).
Related Calculations. Use this procedure for safety valves serving any type of
stationary or marine boiler.
SAFETY-VALVE SELECTION FOR A WATERTUBE
STEAM BOILER
Select a safety valve for a watertube steam boiler having a maximum rating of
100,000 lb/ h (12.6 kg /s) at 800 lb/ in
2
(abs) (5516.0 kPa) and 900
Њ
F (482.2
Њ
C).
Determine the valve diameter, size of boiler connection for the valve, opening
pressure, closing pressure, type of connection, and valve material. The boiler is oil-
fired and has a total heating surface of 9200 ft
2
(854.7 m
2
) of which 1000 ft
2
(92.9
m
2
) is in waterwall surface. Use the ASME Boiler and Pressure Vessel Code rules
when selecting the valve. Sketch the escape-pipe arrangement for the safety valve.

Calculation Procedure:
1. Determine the minimum valve relieving capacity
Refer to the latest edition of the Code for the relieving-capacity rules. Recent edi-
tions of the Code require that the safety valve have a minimum relieving capacity
based on the pounds of steam generated per hour per square foot of boiler heating
surface and waterwall heating surface. In the edition of the Code used in preparing
this handbook, the relieving requirement for oil-fired boilers was 10 lb/ (ft
2
⅐
h) of
steam [13.6 g /(m
2
⅐
s)] of boiler heating surface, and 16 lb/(ft
2
⅐
h) of steam [21.9
g/(m
2
⅐
s)] of waterwall surface. Thus, the minimum safety-valve relieving capacity
for this boiler, based on total heating surface, would be (8200)(10)
ϩ
(1000)(16)
ϭ
92,000 lb / h (11.6 kg /s). In this equation, 1000 ft
2
(92.9 m
2
) of waterwall surface

is deducted from the total heating surface of 9200 ft
2
(854.7 m
2
) to obtain the boiler
heating surface of 8200 ft
2
(761.8 m
2
).
The minimum relieving capacity based on total heating surface is 92,000 lb / h
(11.6 kg/s); the maximum rated capacity of the boiler is 100,000 lb/h (12.6
kg/ s). Since the Code also requires that ‘‘the safety valve or valves will discharge
all the steam that can be generated by the boiler,’’ the minimum relieving capacity
must be 100,000 lb/h (12.6 kg/ s), because this is the maximum capacity of the
boiler and it exceeds the valve capacity based on the heating-surface calculation.
If the valve capacity based on the heating-surface steam generation were larger than
the stated maximum capacity of the boiler, the Code heating-surface valve capacity
would be used in safety-valve selection.
2. Determine the number of safety valves needed
Study the latest edition of the Code to determine the requirements for the number
of safety valves. The edition of the Code used here requires that ‘‘each boiler shall
have at least one safety valve and if it [the boiler] has more than 500 ft
2
(46.5 m
2
)
of water heating surface, it shall have two or more safety valves.’’ Thus, at least
two safety valves are needed for this boiler. The Code further specifies, in the
edition used, that ‘‘when two or more safety valves are used on a boiler, they may

be mounted either separately or as twin valves made by placing individual valves
on Y bases or duplex valves having two valves in the same body casing. Twin
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.20
POWER GENERATION
valves made by placing individual valves on Y bases, or duplex valves having two
valves in the same body, shall be of equal sizes.’’ Also, ‘‘when not more than
two valves of different sizes are mounted singly, the relieving capacity of the
smaller valve shall not be less than 50 percent of that of the larger valve.’’
Assume that two equal-size valves mounted on a Y base will be used on the
steam drum of this boiler. Two or more equal-size valves are usually chosen for
the steam drum of a watertube boiler.
Since this boiler handles superheated steam, check the Code requirements re-
garding superheaters. The Code states that ‘‘every attached superheater shall have
one or more safety valves near the outlet.’’ Also, ‘‘the discharge capacity of the
safety valve, or valves, on an attached superheater may be included in determining
the number and size of the safety valves for the boiler, provided there are no
intervening valves between the superheater safety valve and the boiler, and provided
the discharge capacity of the safety valve, or valves, on the boiler, as distinct from
the superheater, is at least 75 percent of the aggregate valve capacity required.’’
Since the safety valves used must handle 100,000 lb /h (12.6 kg/s), and one or
more superheater safety valves are required by the Code, assume that the two steam-
drum valves will handle, in accordance with the above requirement, 80,000 lb/h
(10.1 kg/ s). Assume that one superheater safety valve will be used. Its capacity
must then be at least 100,000
Ϫ
80,000

ϭ
20,000 lb / h (2.5 kg/s). (Use a few
superheater safety valves as possible, because this simplifies the installation and
reduces cost.) With this arrangement, each steam-drum valve must handle 80,000 /
2
ϭ
40,000 lb / h (5.0 kg/s) of steam, since there are two safety valves on the steam
drum.
3. Determine the valve pressure settings
Consult the Code. It requires that ‘‘one or more safety valves on the boiler proper
shall be set at or below the maximum allowable working pressure.’’ For modern
boilers, the maximum allowable working pressure is usually 1.5, or more, times
the rated operating pressure in the lower [under 1000 lb/in
2
(abs) or 6895.0 kPa]
pressure ranges. To prevent unnecessary operation of the safety valve and to reduce
steam losses, the lowest safety-valve setting is usually about 5 percent higher than
the boiler operating pressure. For this boiler, the lowest pressure setting would be
800
ϩ
800(0.05)
ϭ
840 lb / in
2
(abs) (5791.8 kPa). Round this to 850 lb/in
2
(abs)
(5860.8 kPa, or 6.25 percent) for ease of selection from the usual safety-valve rating
tables. The usual safety-valve pressure setting is between 5 and 10 percent higher
than the rated operating pressure of the boiler.

Boilers fitted with superheaters usually have the superheater safety valve set at
a lower pressure than the steam-drum safety valve. This arrangement ensures that
the superheater safety valve opens first when overpressure occurs. This provides
steam flow through the superheater tubes at all times, preventing tube burnout.
Therefore, the superheater safety valve in this boiler will be set to open at 850 lb
/in
2
(abs) (5860.8 kPa), the lowest opening pressure for the safety valves chosen.
The steam-drum safety valves will be set to open at a higher pressure. As decided
earlier, the superheater safety valve will have a capacity of 20,000 lb/ h (2.5 kg / s).
Between the steam drum and the superheater safety valve, there is a pressure
loss that varies from one boiler to another. The boiler manufacturer supplies a
performance chart showing the drum outlet pressure for various percentages of the
maximum continuous steaming capacity of the boiler. This chart also shows the
superheater outlet pressure for the same capacities. The difference between the drum
and superheater outlet pressure for any given load is the superheater pressure loss.
Obtain this pressure loss from the performance chart.
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.21
Assume, for this boiler, that the superheater pressure loss, plus any pressure
losses in the nonreturn valve and dry pipe, at maximum rating, is 60 lb/in
2
(abs)
(413.7 kPa). The steam-drum operating pressure will then be superheater outlet
pressure
ϩ

superheater pressure loss
ϭ
800
ϩ
60
ϭ
860 lb/in
2
(abs) (5929.7 kPa).
As with the superheater safety valve, the steam-drum safety valve is usually set to
open at about 5 percent above the drum operating pressure at maximum steam
output. For this boiler then, the drum safety-valve set pressure
ϭ
860
ϩ
860(0.05)
ϭ
903 lb/in
2
(abs) (6226.2 kPa). Round this to 900 lb/ in
2
(abs) (6205.5
kPa) to simplify valve selection.
Some designers add the drum safety-valve blowdown or blowback pressure (dif-
ference between the valve opening and closing pressures, lb/ in
2
) to the total ob-
tained above to find the drum operating pressure. However, the 5 percent allowance
used above is sufficient to allow for the blowdown in boilers operating at less than
1000 lb/in

2
(abs) (6895.0 kPa). At pressures of 1000 lb/ in
2
(abs) (6895.0 kPa) and
higher, add the drum safety-valve blowdown and the 5 percent allowance to the
superheater outlet pressure and pressure loss to find the drum pressure.
4. Determine the required valve orifice discharge area
Refer to a safety-valve manufacturer’s engineering data listing valve capacities at
various working pressures. For the two steam-drum valves, enter the table at 900
lb/in
2
(abs) (6205.5 kPa), and project horizontally until a capacity of 40,000 lb / h
(5.0 kg/ s), or more, is intersected. Here is an excerpt from a typical manufacturer’s
capacity table for safety valves handling saturated steam:
Thus, at 900 lb/ in
2
(abs) (6205.5 kPa) a valve with an orifice area of 0.944 in
2
(6.4
cm
2
) will have a capacity of 42,200 lb/ h (5.3 kg / s) of saturated steam. This is 5.5
percent greater than the required capacity of 40,000 lb /h (5.0 kg/s) for each steam-
drum valve. However, the usual selection cannot be made at exactly the desired
capacity. Provided that the valve chosen has a greater steam relieving capacity than
required, there is no danger of overpressure in the steam drum. Be careful to note
that safety valves for saturated steam are chosen for the steam drum because su-
perheating of the steam does not occur in the steam drum.
The superheater safety valve must handle 20,000 lb /h (2.5 kg/ s) of 850 lb/in
2

(abs) (5860.8-kPa) steam at 900
Њ
F (482.2
Њ
C). Safety valves handling superheated
steam have a smaller capacity than when handling saturated steam. To obtain the
capacity of a safety valve handling superheated steam, the saturated steam capacity
is multiplied by a correction factor that is less than 1.00. An alternative procedure
is to divide the required superheated-steam capacity by the same correction factor
to obtain the saturated-steam capacity of the valve. The latter procedure will be
used here because it is more direct.
Obtain the correction factor from the safety-valve manufacturer’s engineering
data by entering at the steam pressure and projecting to the steam temperature, as
show below.
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.22
POWER GENERATION
Thus, at 850 lb/ in
2
(abs) (5860.8 kPa) and 900
Њ
F (482.2
Њ
C), the correction factor
is 0.80. The required saturated steam capacity then is 20,000/0.80
ϭ
25,000 lb /h

(3.1 kg/s).
Refer to the manufacturer’s saturated-steam capacity table as before, and at 850
lb/in
2
(abs) (5860.8 kPa) find the closest capacity as 31,500 lb/h (4.0 kg/ s) for a
0.785-in
2
(5.1-cm
2
) orifice. As with the steam-drum valves, the actual capacity of
the safety valve is somewhat greater than the required capacity. In general, it is
difficult to find a valve with exactly the required steam relieving capacity.
5. Determine the valve nominal size and construction details
Turn to the data section of the safety-valve engineering manual to find the valve
construction features. For the steam-drum valves having 0.994-in
2
(6.4-cm
2
) orifice
areas, the engineering data show, for 900-lb/in
2
(abs) (6205.5-kPa) service, each
valveis1
1
⁄
2
-in (3.8-cm) unit rated for temperatures up to 1050
Њ
F (565.6
Њ

C). The
inlet is 900-lb /in
2
(6205.5-kPa) 1
1
⁄
2
-in (3.8-cm) flanged connection, and the outlet
is a 150-lb/ in
2
(1034.3-kPa) 3-in (7.6-cm) flanged connection. Materials used in
the valve include: body, cast carbon steel; disk seat, stainless steel AISI 321. The
overall height is 27
7
⁄
8
in (70.8 cm); dismantled height is 32
3
⁄
4
in (83.2 cm).
Similar data for the superheated steam valve show, for a maximum pressure of
900 lb /in
2
(abs) (6205.5 kPa), that it is a 1
1
⁄
2
-in (3.8-cm) unit rated for temperatures
up to 1000

Њ
F (537.8
Њ
C). The inlet is a 900-lb/in
2
(6205.5-kPa) 1
1
⁄
2
-in (3.8-cm)
flanged connection, and the outlet is a 150-lb/in
2
(1034.3-kPa) 3-in (7.6-cm)
flanged connection. Materials used in the valve include: body, cast alloy steel,
ASTM 217-WC6; spindle, stainless steel; spring, alloy steel; disk seat, stainless
steel. Overall height is 21
3
⁄
8
in (54.3 cm); dismantled height is 25
1
⁄
4
in (64.1 cm).
Checking the Code shows that ‘‘every safety valve used on a superheater discharg-
ing superheated steam at a temperature over 450
Њ
F (232.2
Њ
C) shall have a casing,

including the base, body, bonnet and spindle, of steel, steel alloy, or equivalent
heat-resisting material. The valve shall have a flanged inlet connection.’’
Thus, the superheater valve selected is satisfactory.
6. Compute the steam-drum connection size
The Code requires that ‘‘when a boiler is fitted with two or more safety valves on
one connection, this connection to the boiler shall have a cross-sectional area not
less than the combined areas of inlet connections of all safety valves with which
it connects.’’
The inlet area for each valve
ϭ
␲
D
2
/4
ϭ
␲
(1.5)
2
/4
ϭ
1.77 in
2
(11.4 cm
2
). For
two valves, the total inlet area
ϭ
2(1.77)
ϭ
3.54 in

2
(22.8 cm
2
). The required
minimum diameter of the boiler connection is d
ϭ
2(A/
␲
)
0.5
, where A
ϭ
inlet area.
Or, d
ϭ
2(3.54/
␲
)
0.5
ϭ
2.12 in (5.4 cm). Select a 2
1
⁄
2
ϫ
1
1
⁄
2
ϫ

1
1
⁄
2
in (6.4
ϫ
3.8
ϫ
3.8 cm). Y for the two steam-drum valves and a 2
1
⁄
2
-in (6.4-cm) steam-drum outlet
connection.
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.23
FIGURE 3 Typical boiler safety-valve discharge elbow and
drip-pan connection. (Industrial Valve and Instrument Division of
Dresser Industries Inc.)
7. Compute the safety-valve closing pressure
The Code requires safety valves to ‘‘close after blowing down not more than 4
percent of the set pressure.’’ For the steam-drum valves the closing pressure will
be 900
Ϫ
(900)(0.04)
ϭ

865 lb/in
2
(abs) (5964.2 kPa). The superheater safety valve
will close at 850
Ϫ
(850)(0.04)
ϭ
816 lb/in
2
(abs) (5626.3 kPa).
8. Sketch the discharge elbow and drip pan
Figure 3 shows a typical discharge elbow and drip-pan connection. Fit all boiler
safety valves with escape pipes to carry the steam out of the building and away
from personnel. Extend the escape pipe to at least 6 ft (1.8 m) above the roof of
the building. Use an escape pipe having a diameter equal to the valve outlet size.
When the escape pipe is more than 12 ft (3.7 m) long, some authorities recommend
increasing the escape-pipe diameter by
1
⁄
2
in (1.3 cm) for each additional 12-ft (3.7-
m) length. Excessive escape-pipe length without an increase in diameter can cause
a backpressure on the safety valve because of flow friction. The safety valve may
then chatter excessively.
Support the escape pipe independently of the safety valve. Fit a drain to the
valve body and rip pan as shown in Fig. 3. This prevents freezing of the condensate
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES

4.24
POWER GENERATION
and also eliminates the possibility of condensate in the escape pipe raising the valve
opening pressure. When a muffler is fitted to the escape pipe, the inlet diameter of
the muffler should be the same as, or larger than, the escape-pipe diameter. The
outlet area should be greater than the inlet area of the muffler.
Related Calculations. Compute the safety-valve size for fire-tube boilers in the
same way as described above, except that the Code gives a tabulation of the required
area for safety-valve boiler connections based on boiler operating pressure and
heating surface. Thus, with an operating pressure of 200 lb /in
2
(gage) (1379.0 kPa)
and 1800 ft
2
(167.2 m
2
) of heating surface, the Code table shows that the safety-
valve connection should have an area of at least 9.148 in
2
(59.0 cm
2
).A3
1
⁄
2
-in
(8.9-cm) connection would provide this area; or two smaller connections could be
used provided that the sum of their areas exceeded 9.148 in
2
(59.0 cm

2
)
Note: Be sure to select safety valves approved for use under the Code or local
low governing boilers in the area in which the boiler will be used. Choice of an
unapproved valve can lead to its rejection by the bureau or other agency controlling
boiler installation and operation.
STEAM-QUALITY DETERMINATION WITH A
THROTTLING CALORIMETER
Steam leaves an industrial boiler at 120 lb / in
2
(abs) (827.4 kPa) and 341.25
Њ
F
(171.8
Њ
C). A portion of the steam is passed through a throttling calorimeter and is
exhausted to the atmosphere when the barometric pressure is 14.7 lb/in
2
(abs)
(101.4 kPa). How much moisture does the steam leaving the boiler contain if the
temperature of the steam at the calorimeter is 240
Њ
F (115.6
Њ
C)?
Calculation Procedure:
1. Plot the throttling process on the Mollier diagram
Begin with the endpoint, 14.7 lb /in
2
(abs) (101.4 kPa) and 240

Њ
F (115.6
Њ
C). Plot
this point on the Mollier diagram as point A, Fig 4. Note that this point is in the
superheat region of the Mollier diagram, because steam at 14.7 lb/in
2
(abs) (101.4
kPa) has a temperature of 212
Њ
F (100.0
Њ
C), whereas the steam in this calorimeter
has a temperature of 240
Њ
F (115.6
Њ
C). The enthalpy of the calorimeter steam is,
from the Mollier diagram, 1164 Btu/lb (2707.5 kJ/kg).
2. Trace the throttling process on the Mollier diagram
In a throttling process, the steam expands at constant enthalpy. Draw a straight,
horizontal line from point A to the left on the Mollier diagram until the 120-lb /in
2
(abs) (827.4-kPa) pressure curve is intersected, point B, Fig. 4. Read the moisture
content of the steam as 3 percent where the 1164-Btu/ lb (2707.5-kJ/kg) horizontal
trace AB, the 120-lb/ in
2
(abs) (827.4-kPa) pressure line, and the 3 percent moisture
line intersect.
Related Calculations. A throttling calorimeter must produce superheated steam

at the existing atmospheric pressure if the moisture content of the supply steam is
to be found. Where the throttling calorimeter cannot produce superheated steam at
atmospheric pressure, connect the calorimeter outlet to an area at a pressure less
than atmospheric. Expand the steam from the source, and read the temperature at
the calorimeter. If the steam temperature is greater than that corresponding to the
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.
Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES
STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.25
FIGURE 4 Mollier-diagram plot of a throttling-calorimeter process.
absolute pressure of the vacuum area—for example, a temperature greater than
133.76
Њ
F (56.5
Њ
C) in an area of 5 inHg (16.9 kPa) absolute pressure—follow the
same procedure as given above. Point A would then be in the below-atmospheric
area of the Mollier diagram. Trace to the left to the origin pressure, and read the
moisture content as before.
STEAM PRESSURE DROP IN A
BOILER SUPERHEATER
What is the pressure loss in a boiler superheater handling w
s
ϭ
200,000 lb/h (25.2
kg/ s) of saturated steam at 500 lb / in
2
(abs) (3447.5 kPa) if the desired outlet

temperature is 750
Њ
F (398.9
Њ
C)? The steam free-flow area through the superheater
tubes A
s
ft
2
is 0.500, friction factor ƒ is 0.025, tube ID is 2.125 in (5.4 cm),
developed length l of a tube in one circuit is 150 in (381.0 cm), and the tube bend
factor B
ƒ
is 12.0.
Calculation Procedure:
1. Determine the initial conditions of the steam
To compute the pressure loss in a superheater, the initial specific volume of the
steam
v
g
and the mass-flow ratio w
s
/A
s
must be known. From the steam table,
v
g
ϭ
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
Copyright © 2006 The McGraw-Hill Companies. All rights reserved.

Any use is subject to the Terms of Use as given at the website.
STEAM GENERATION EQUIPMENT AND AUXILIARIES