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

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

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

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-
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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

3
).
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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

ϭ
942 Btu /
ft
3
(35.1 MJ / m
3
).
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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

(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.

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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

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
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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

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

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
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STEAM GENERATION EQUIPMENT AND AUXILIARIES
4.9
through which the heat flow, ft (m); k
ϭ
thermal conductivity, as defined above;
A

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

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

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.
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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

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

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

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

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

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

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

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.
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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

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

⁄
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,

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

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
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

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-
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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.

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

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

(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

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

/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

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

(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

[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

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

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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

(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)

[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

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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

(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

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,

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.

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

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.

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

(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.
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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

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

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

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.
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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

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

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

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
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