Tài liệu Handbook of Mechanical Engineering Calculations P4 pptx

SECTION 4 STEAM GENERATIONEQUIPMENT AND AUXILIARIES Determining Equipment Loading for Generating Steam Efficiently 4.2 Steam Conditions with Two Boilers Supplying the Same Line 4.6 Genera

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

Safety-Valve Steam-Flow Capacity 4.18

Safety-Valve Selection for a Watertube

Fan Analysis at Varying RPM, Pressure,

and Air or Gas Capacity 4.35

Boiler Forced-Draft Fan Horsepower

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 Driven Boiler Fans 4.56

Turbine-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

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 / in2 (gage) (2756 kPa) The plant also has anHRSG capable of generating up to 1000,000 lb / h (45,400 kg / h) of steam in thefired mode at the same pressure How should each steam generator be loaded togenerate 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 primemovers), 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 theplant, 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 / in2 (gage) (2756kPa) from the exhaust of a gas turbine It can be supplementary-fired to generateadditional steam Using the HRSG simulation approach given in another calculationprocedure in this handbook, the HRSG performance at different steam flow ratesshould be developed This may be done manually or by using the HRSG softwaredeveloped by the author

2. Select the gas / steam temperature profile in the design mode

Using a pinch point of 15⬚F (8.33⬚C) and approach point of 17⬚F (9.44⬚C), a perature profile is developed as discussed in the procedure for HRSG simulation.The HRSG exit gas temperature is 319⬚F (159.4⬚C) while generating 25,000 lb / h(11,350 kg / h) of steam at 400 lb / in2(gage) (2756 kPa) using 230⬚F (110⬚C) feed-water

tem-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 cient, as discussed in the procedure on HRSG simulation All the fuel energy goesinto generating steam in single-pressure HRSGs

effi-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 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 ⫽ 50.8 ⫺ 25.4 ⫽ 25.4 MM Btu / h (7.44 MW) If amanual or computer simulation is done on the HRSG, fuel consumption will beseen 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

gen-in Table 1 Note that the exit gas temperature decreases as the steam flow gen-increases.This aspect of an HRSG is discussed in the simulation procedure elsewhere in thishandbook

4. Develop the steam-generator characteristics

Develop the performance of the steam generator at various loads Steam-generatorsuppliers will gladly provide this information in great detail, including plots andtabulations of the boiler’s performance As shown in Table 2, the exit-gas temper-

TABLE 3 Fuel Consumption at Various Steam Loads

Total

steam

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 steamgenerator’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 erator, a steam flow vs fuel table such as that in Table 3 For example, 150,000

gen-lb / h (68,100 kg / h) of steam could be generated in several ways—50,000 gen-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 imizing the HRSG output first is the most efficient way of generating steam because

max-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 significantlywith load as the gas mass flow remains nearly constant The gas pressure dropincreases slightly as the firing temperature increases On the other hand, the steamgenerator fan power consumption vs load increases more in proportion to load

It is also seen that at higher steam capacities the difference in fuel consumptionbetween 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 bekept in mind while developing an operational strategy

If a superheater is used, the performance of the superheater would have to beanalyzed 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 perature increases with firing temperature and can be controlled

tem-Related Calculations. Developing the performance characteristics of eachpiece of equipment as a function of load is the key to determining the mode ofoperation and loading of each type of steam producer For best results, develop aperformance curve for the steam generator, including all operating costs such asfan 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 fuelcost, which is the major cost

This procedure is the work of V Ganapathy, Heat Transfer Specialist, ABCOIndustries, 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 sameshort steam main Steam from boiler No 1 is at 200 lb / in2(1378 kPa) and 420⬚F(215.6⬚C) while steam from boiler No 2 is at 200 lb / in2(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 pressuredrop 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, H3⫽(H1⫹ H2) / 2 Substi-

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, H3⫽(1225⫹ 1164) /

2⫽1194.5 Btu / lb (2783.2 kJ / kg)

2. Find the quality of the mixed steam

Entering the steam tables at 200 lb / in2(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 ⫽Hƒ ⫹x3(H ƒg)⫽ 355.4⫹ x3(843.3); x3⫽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 S1 ⫺ S2 ⫽ 1.575 ⫺ 1.541 ⫽ 0.034 entropy units The lower-temperature

steam gains S3⫺S2⫽1.541 ⫺1.506⫽0.035 units of entropy

Related Calculations. Use this general approach for any mixing of steamflows Where different quantities of steam are being mixed, use the proportion ofeach quantity to the total in computing the enthalpy, quality, and entropy of themixture

GENERATING SATURATED STEAM BY

DESUPERHEATING SUPERHEATED STEAM

Superheated steam generated at 1350 lb / in2(abs) (9301.5 kPa) and 950⬚F (510⬚C)

is to be used in a process as saturated steam at 1000 lb / in2(abs) (6890 kPa) If the

superheated steam is desuperheated continuously by injecting water at 500⬚F(260⬚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 thesuperheated steam at 1350 lb / in2(abs) (9301.5 kPa) and 950⬚F (510⬚C) ⫽ H1 ⫽

1465 Btu / lb (3413.5 kJ / kg); Enthalpy of saturated steam at 1000 lb / in2(abs) (6890kPa) ⫽ H2 ⫽ 1191 Btu / lb (2775 kJ / kg); Enthalpy of water at 500⬚F (260⬚C) ⫽(500⫺ 32)⫽H3⫽488 Btu / lb (1137 kJ / kg)

2. Set up a heat-balance equation and solve it

L X ⫽lb (kg) of water at 500⬚F (260⬚C) required to desuperheat the superheated

steam Then, using the symbols given above, H1⫹ X(H3)⫽ (1 ⫹ X)H2 Solving

for X ⫽ (H1 ⫺ H2) / (H2 ⫺ H3) ⫽ (1465 ⫺ 1191) / (1191 ⫺ 488) ⫽ 0.39 Then,1.0⫹ 0.39⫽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 / in2 (abs) (689 kPa), the amount of superheated steam needed to produce thissaturated steam would be 1000 / 1.39⫽719.4 lb (326.6 kg)

Related Calculations. Desuperheating superheated steam for process and otheruse is popular because it can save purchase and installation of a separate steamgenerator for the lower pressure steam While there is a small loss of energy indesuperheating (from heat losses in the piping and desuperheater), this loss is smallcompared 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) Cel brick, 4-in (10.2-cm) red brick, and 0.25-in (0.64-cm) transite board The ther-

Sil-O-mal conductivity, k, values, Btu / (ft2)(⬚F)(ft) [kJ / (m2)(⬚C)(m)] are as follows: 0.82

at 1800⬚F (982⬚C) for fire brick; 0.125 at 1800⬚F (982⬚C) for Sil-O-Cel; 0.52 at

500⬚F (260⬚C) for transite A temperature of 1800⬚F (982⬚C) exists on the insidewall of the furnace and 200⬚F (93.3⬚C) on the outside wall Determine the heat lossper hour through each 10 ft2(0.929 m2) 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⫽ ⌬t / R, where Q⫽heat transferred, Btu / h (W);⌬t⫽ature difference between the inside of the furnace wall and the outside, ⬚F (⬚C);

temper-R ⫽ resistance of the wall to heat flow ⫽ L / (kXA), where L ⫽ length of path

through which the heat flow, ft (m); k ⫽thermal conductivity, as defined above;

A⫽area of path of heat flow, ft2(m2) Where there is more than one resistance toheat flow, add them to get the total resistance

Substituting, the above values for this furnace wall, remembering that there arethree resistances in series and solving for the heat flow through one square ft(0.0.0929 m2), Q⫽(1800⫺200) / {[(1 / 0.82)(9 / 12)]⫹[(1 / 0.125)(4.5 / 12)]⫹[(1/ 052)(4 / 12)]⫹[(1 / 0.23)(0.25 / 12)]}⫽344 Btu / h ft2(1083.6 W / m2), or 10 (344)

⫽34400 Btu / h for 10 ft2(10,836 W / 10 m2)

2. Compute the temperature within the wall at the stated joint

Use the relation, (⌬t) / (⌬t1) ⫽ (R / R1), where ⌬t ⫽ temperature difference acrossthe wall, ⬚F (⬚C); ⌬t1 ⫽ temperature at the joint being considered, ⬚F (⬚C); R ⫽

total resistance of the wall; R1⫽resistance of the first portion of the wall betweenthe inside and the joint in question

Substituting, (1800⫺200) / (⌬t1)⫽4.646 / 0.915);⌬t1⫽315⬚F (157.2⬚C) Thenthe interface temperature at the between the fire brick and the Sil-O-Cel is 1800⫺

of a furnace wall, the larger the heat loss from the fired vessel Therefore, bothsafety and energy conservation considerations are important in furnace design.Typical interior furnace temperatures encountered in modern steam boilers rangefrom 2400⬚F (1316⬚C) near the fuel burners to 1600⬚F (871⬚C) in the superheaterinterior With today’s emphasis on congeneration and energy conservation, manydifferent fuels are being burned in boilers Thus, a plant in Louisiana burns rice togenerate 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 / ft3 (176 kg / m3) to 20

lb / ft3(320 kg / m3) 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 andproblem

Burning rice hulls in a boiler furnace may create unexpected temperatures bothinside and outside the furnace Hence, it is important that the designer be able toanalyze both the interior and exterior furnace temperatures using the proceduregiven 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 wastewaterplant is burned in a combustor to generate steam for a turbogenerator Not only arefuel requirements for the boiler reduced, there is also significant savings of fuelused to incinerate the sludge in earlier plants Again, the furnace temperature is animportant element in designing such plants

The data present in these comments on new fuels for boilers is from Power

magazine

CONVERTING POWER-GENERATION

POLLUTANTS FROM MASS TO

VOLUMETRIC UNITS

In the power-generation industry, emission levels of pollutants such as CO and NOx

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 relatethese two measures for a gaseous fuel having this analysis: Methane⫽97 percent;Ethane⫽2 percent; Propane⫽1 percent by volume, and excess air⫽10 percent.Ambient air temperature during combustion ⫽80⬚F (26.7⬚C) and relative humid-ity ⫽ 60 percent; fuel higher heating value, HHV ⫽ 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 (ft3of air per ft3of combustible gas)(percent of combustible in the fuel) using

data from Ganapathy, Steam Plant Calculations Manual, Marcel Dekker, Inc thus:

M⫽(9.528⫻97)⫹(16.675⫻2)⫹(23.821⫻1)⫽981.4 moles Then, with 10percent excess air, excess air, EA⫽ 0.1(981.4)⫽ 98.1 moles

The excess oxygen, O2⫽(98.1 moles)(0.21)⫽20.6 moles, where 0.21⫽moles

of oxygen in 1 mole of air The nitrogen, N2, produced by combustion⫽ (1.1 forexcess air)(981.4 moles)(0.79 moles of nitrogen in 1 mole of air)⫽852.8 moles;round to 853 moles for additional calculations

The moisture in the air⫽ (981.4⫹ 98.1)(29⫻ 0.0142 / 18)⫽24.69, say 24.7moles In this computation the values 29 and 18 are the molecular weights of dryair 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, CO2⫽ (1 ⫻ 97) ⫹(2 ⫻ 2) ⫹(3 ⫻ 1) ⫽104 moles For

H2O ⫽ (2 ⫻ 97) ⫹ (3 ⫻ 2) ⫹ (4 ⫻ 1) ⫹ 24.7 ⫽ 228.7 moles From step 1,

N2⫽853 moles; O2⫽ 20.6 moles

Now, the total moles⫽104⫹228.7⫹853⫹20.6⫽1206.3 moles The percentvolume of CO2 ⫽ (104 / 1206.3)(100) ⫽ 8.6; the percent H2O ⫽ (228.7 /1206.1)(100) ⫽ 18.96; the percent N2 ⫽ (853 / 1206.3)(100) ⫽ 70.7; the percent

O2⫽(20.6 / 1206.3)(100)⫽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 NOxor CO produced to ppmv,

we must know the amount of flue gas produced per million Btu (1.055 MJ) Fromstep 2, the molecular weight of the flue gases⫽[(8.68⫻ 44)⫹(18.96 ⫻18) ⫹(70.7⫻28)⫹ (1.71⫻ 32)] / 100⫽27.57

The molecular weight of the fuel⫽[(97⫻16)⫹(2⫻30)⫹(1⫻44)] / 100⫽16.56 Now the ratio of flue gases / fuel⫽(1206.3⫻27.57) / (100⫻16.56)⫽20.08

lb flue gas / lb fuel (9.12 kg / kg) Hence, 1 million Btu fired produces (1,000,000) /23,789⫽42 lb (19.1 kg) fuel⫽ (42)(20.08)⫽844 lb (383 kg) wet flue gases

4. Calculate ppm values for the gases

Let 1 million Btu fired generate N lb (kg) of NO x For emission calculations, NOx

is considered to have a molecular weight of 46 Also, the reference for NOxor COregulations is 3 percent dry oxygen by volume for steam generators Hence, we

have the relation, N N⫽106(Y x )(N / 46)(MW ƒg)[(21⫺3) / (21⫺O2XY)], where V N⫽ppm dry NOx ; Y⫽ 100 / (100 ⫺percent H2O), where percent H2O is the percent

volume water vapor in the flue gases; N⫽lb (kg) of NOx per million Btu (1.055

MJ) fired on an HHV basis, MW ƒg ⫽molecular weight of wet flue gases; W gm⫽amount of wet flue gas produced per million Btu (1.055 MJ) fired

Substituting in the above relation, V N ⫽ 106(N x)[100 / (100 ⫺ 18.96)] (27.57 /844)(21⫺ 3) / [(21⫺ 1.71)(100 / (100⫺18.96)]⫽ 832N

Similarly, V c⫽ppmv CO2generated per million Btu (1.055 MJ) fired⫽1367⬚C,

where C⫽lb (kg) of CO generated per million Btu (1.055 MJ) and V c⫽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

⫽832N and V c⫽1367 for natural gas

Related Calculations. These calculations for oil fuels also to show that V N⫽

S⫽0.0270; ash⫽0.0827; total⫽1.0000 The coal contains 1.61 percent moisture.The boiler-room intake air and the fuel temperature⫽79⬚F (26.1⬚C) dry bulb, 71⬚F(21.7⬚C) wet bulb The flue-gas temperature is 500⬚F (260⬚C), and the analysis ofthe flue gas shows these percentages: CO2 ⫽ 12.8; CO ⫽ 0.4; O2 ⫽ 6.1; N2 ⫽80.7; total⫽100.0 Measured ash and refuse⫽9.42 percent of dry coal; combus-tible in ash and refuse⫽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⫽heating value of fuel⫽ 13,850 Btu / lb (32,215 kJ / kg)

2. Compute the output of the boiler

The output of any boiler ⫽ Btu / lb (kJ / kg) of fuel ⫹ the losses In this step thefirst portion of the output, Btu / lb (kJ / kg) of fuel will be computed The losses will

be computed in step 3

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 ⫽45,340 / 4370⫽ 10.34 lb of steam per lb (4.65 kg / kg) of fuel

Once W s is known, the output h1Btu / lb of fuel can be found from h1⫽W s (h s

⫺h w ), where h s⫽enthalpy of steam leaving the superheater, or boiler if a

super-heater is not used; h w ⫽enthalpy of feedwater, Btu / lb For this boiler with steam

at 125 lb / in2(gage) [⫽139.7 lb / in2(abs)] and 400⬚F (930 kPa, 204⬚C), h s⫽1221.2

Btu / lb (2841 kJ / kg), and h w⫽180.92 Btu / lb (420.8 kJ / kg), from the steam tables

Then h1⫽10.34(1221.2⫺ 180.92)⫽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 h2 Btu / lb (kJ / kg) of fuel is given by h2 ⫽

0.24W g⫻ (T g⫺ T a ), where W g⫽lb of dry flue gas per lb of fuel; T g⫽ flue-gasexit temperature,⬚F; T a⫽intake-air temperature,⬚F

Before W g can be found, however, it must be determined whether any excessair is passing through the boiler Compute the excess air, if any, from excess air,percent ⫽100 (O2⫺1⁄2CO) / [0.264N2⫺ (O2⫺1⁄2CO)], where the symbols refer

to the elements in the flue-gas analysis Substituting values from the flue-gas ysis gives excess air ⫽ 100(6.1 ⫺ 0.2) / [0.264 ⫻ 80.7 ⫺ (6.1 ⫺ 0.2)] ⫽ 38.4percent

anal-Using the method given in earlier calculation procedures, find the air requiredfor complete combustion as 10.557 lb / lb (4.571 kg / kg) of coal With 38.4 percentexcess air, the additional air required ⫽(10.557)(0.384) ⫽ 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⫽11.018 (4.958 kg / kg) Then,the total flue gas at 38.4 percent excess air ⫽ 11.018 ⫹ 4.053 ⫽ 15.071 lb / lb(6.782 kg / kg) of fuel

With a flue-gas temperature of 500⬚F (260⬚C), and an intake-air temperature of

79⬚F (26.1⬚C), h2⫽0.24(15.071)(500⫺70)⫽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 H2O This water is evaporated by heat

in the fuel, and less heat is available for producing steam This loss is h3Btu / lb

of fuel ⫽ 9H(1089 ⫺ Tƒ ⫹ 0.46T g ), where H ⫽ percent H2 in the fuel ⫼ 100;

Tƒ⫽ temperature of fuel before combustion,⬚F; other symbols as before For thisfuel with 5.07 percent H2, h3 ⫽9(5.07 / 100)(1089 ⫺79 ⫹ 0.46 ⫻500) ⫽ 565.8Btu / lb (1316 kJ / kg) of fuel

5. Compute the loss from evaporation of fuel moisture

This loss is h4 Btu / lb of fuel ⫽ W mƒ(1089⫺ Tƒ ⫹ 0.46T g ), where W mƒ ⫽ lb ofmoisture 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⫺1.61)⫽0.0164, or 1.64 percent

Then h4⫽ (1.64 / 100)(1089⫺ 79⫹0.46⫻500)⫽20.34 Btu / lb (47.3 kJ / kg) offuel

6. Compute the loss from moisture in the air

This loss is h5Btu / lb of fuel ⫽ 0.46W ma (T g⫺ T a ), W ma⫽ (lb of water per lb ofdry air)(lb air supplied per lb fuel) From a psychrometric chart, the weight ofmoisture per lb of air at a 79⬚F (26.1⬚C) dry-bulb and 71⬚F (21.7⬚C) wet-bulbtemperature is 0.014 (0.006 kg) The combustion calculation, step 3, shows that the

total air required with 38.4 percent excess air⫽10.557⫹4.053 ⫽14.61 lb of air

per lb (6.575 kg / kg) of fuel Then, W ma⫽(0.014)(14.61)⫽0.2045 lb of moisture

per lb (0.092 kg / kg) of air And h5⫽ (0.46)(0.2045)(500 ⫺ 79) ⫽ 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 h6Btu / lb of fuel⫽[CO / CO⫹CO2)](C)(10.190), where CO and CO2are the percent by volume of these compounds in the flue gas by Orsat analysis;

C⫽lb carbon per lb of coal With the given flue-gas analysis and the coal ultimate

analysis, h6 ⫽ 0.4 / (0.4 ⫹ 12.8)[(77.57) / (100)](10.190) ⫽ 239.5 Btu / lb (557

kJ / kg) of fuel

8. Compute the loss due to unconsumed carbon in the refuse

This loss is h7Btu / lb of fuel⫽W c (14,150), where W c⫽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, h7 ⫽ (9.42 / 100)(32.3 / 100)(14,150)⫽ 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 turer’s engineering data to approximate the radiation loss in the boiler Either sourcewill show that the radiation loss is 1.09 percent of the gross heat input Since thegross heat input is 13,850 Btu / lb (32,215 kJ / kg) of fuel, the radiation loss ⫽(13,850)(1.09 / 100)⫽ 151.0 Btu / lb (351.2 kJ / kg) of fuel

manufac-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 anytype of boiler—watertube or firetube—in any kind of service—power, process, orheating—using any kind of fuel—coal, oil, gas, wood, or refuse Note that step 3shows how to compute excess air from an Orsat flue-gas analysis

More stringent environmental laws are requiring larger investments in boiler pollution-control equipment throughout the world To control sulfur emis-

steam-sions, expensive scrubbers are required on large boilers Without such scrubbers thesulfur emissions can lead to acid rain, smog, and reduced visibility in the area ofthe 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 increasedattention is because not all countries have the same environmental control re-quirements When a country with less stringent requirements pollutes an adjacentcountry having more stringent pollution regulations, both political and regulatoryproblems 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 timesweaker 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 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 duringthe next 12 months To stay in compliance with such a large number of regulationsrequires constant attention to those regulations applicable to boiler plants

regula-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 / in2(abs) (4137.0 kPa) 800⬚F (426.7⬚C) steam The boiler is tinuously blown down at the rate of 2500 lb / h (0.31 kg / s) Feedwater enters theeconomizer at 300⬚F (148.9⬚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 H2, 8.9 percent O2, 1.2 percent N2, 3.2 percent S,8.7 percent ash, and 4.5 percent moisture Air enters the boiler at 63⬚F (17.2⬚C)dry-bulb and 56⬚F (13.3⬚C) wet-bulb temperature, with 56 gr of vapor per lb (123.5

con-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⬚F (187.8⬚C) Flue gas entersthe economizer at 850⬚F (454.4⬚C) and has an analysis of 15.8 percent CO2, 2.8percent O2, and 81.4 percent N2 Air enters the air heater at 63⬚F (17.2⬚C) with 56

gr / lb (123.5 gr / kg) of dry air; air leaves the heater at 480⬚F (248.9⬚C) Gas entersthe air heater at 570⬚F (298.9⬚C), and 14 percent of the air to the furnace comesfrom the mill fan Determine the steam generator overall efficiency, economizerefficiency, and air-heater efficiency Figure 2 shows the steam generator and theflow factors that must be considered

Calculation Procedure:

1. Determine the boiler output

The boiler output⫽S(h g⫺hƒ1)⫹S r (h g3⫺h g2)⫹B(hƒ3 ⫺hƒ1), where S⫽steam

generated, lb / h; h g⫽ enthalpy of the generated steam, Btu / lb; hƒ1 ⫽enthalpy of

inlet feedwater; S r⫽ reheated steam flow, lb / h (if any); h g3 ⫽ outlet enthalpy of

reheated steam; h g2 ⫽inlet enthalpy of reheated steam; B⫽ blowoff, lb / h; hƒ3 ⫽blowoff enthalpy, where all enthalpies are in Btu / lb Using the appropriate steam

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 boileroutput ⫽ 56,000(1407.7 ⫺ 269.6) ⫹ 2500(471.6 ⫺ 269.6) ⫽ 64,238,600 Btu / h(18,826.5 kW)

2. Compute the heat input to the boiler

The boiler input⫽FH, where F⫽fuel input, lb / h (as fired); H⫽higher heatingvalue, Btu / lb (as fired) Or, boiler input ⫽ 5958(13,100) ⫽ 78,049,800 Btu / h(22,874.1 kW)

3. Compute the boiler efficiency

The boiler efficiency⫽(output, Btu / h) / (input, Btu / h)⫽64,238,600 / 78,049,800⫽0.822, or 82.2 percent

4. Determine the heat absorbed by the economizer

The heat absorbed by the economizer, Btu / h ⫽w w (hƒ2 ⫺hƒ1), where w w⫽

feed-water flow, lb / h; hƒ1 and hƒ2 ⫽ enthalpies of feedwater leaving and entering theeconomizer, respectively, Btu / lb For this economizer, with the feedwater leavingthe economizer at 370⬚F (187.8⬚C) and entering at 300⬚F (148.9⬚C), heat absorbed

⫽(56,000⫹ 2500)(342.79⫺269.59)⫽4.283,000 Btu / h (1255.2 kW) Note that

the total feedwater flow w wis the sum of the steam generated and the continuousblowdown rate

5. Compute the heat available to the economizer

The heat available to the economizer, Btu / h⫽H g F, where H g⫽heat available influe gas, Btu / lb of fuel ⫽ heat available in dry gas ⫹ heat available in flue-gasvapor, Btu / lb of fuel ⫽ (t;3 ⫺ tƒ1)(0.24G) ⫹ (t3 ⫺ tƒ1)(0.46){Mƒ ⫹ 8.94H2 ⫹

M a [G⫺ Cb⫺ N2 ⫺ 7.94(H2⫺ O2/ 8)]}, where G ⫽{[11CO2 ⫹ 8O2⫹ 7(N2⫹CO)] / [3(CO2 ⫹CO)]}(Cb ⫹S / 2.67) ⫹S / 1.60; Mƒ ⫽lb of moisture per lb fuel

burned; M a⫽lb of moisture per lb of dry air to furnace; Cb⫽lb of carbon burnedper lb of fuel burned⫽C⫽RC r ; C r⫽lb of combustible per lb of refuse; R⫽lb

of refuse per lb of fuel; H2, N2, C, O2, S ⫽lb of each element per lb of fuel, asfired; CO2, CO, O2, N2⫽percentage parts of volumetric analysis of dry combustiongas entering the economizer Substituting gives Cb⫽0.685⫺(0.093)(0.07)⫽0.678

lb / lb (0.678 kg / kg) fuel; G ⫽ [11(0.158) ⫹ 8(0.028) ⫹ 7(0.814)] / [3(0.158)] ⫻(0.678 ⫹ 0.032 / 2.67) ⫹ 0.032 / 1.60; G⫽ 11.18 lb / lb (11.18 kg / kg) fuel H g ⫽(800 ⫺ 300)(0.24) ⫻ (11.18) ⫹ (800 ⫺ 300)(0.46){0.045 ⫹ (8.9)(0.05) ⫹ 56 /7000[11.18 ⫺ 0.678 ⫺ 0.012 ⫺ 7.94 ⫻ (0.05 ⫺ 0.089 / 8)]}; H g ⫽ 1473 Btu / lb(3426.2 kJ / kg) fuel Heat available ⫽ H g F ⫽ (1473)(5958) ⫽ 8,770,000 Btu / h(2570.2 kW)

6. Compute the economizer efficiency

The economizer efficiency ⫽ (heat absorbed, Btu / h) / (heat available, Btu / h) ⫽4,283,000 / 8,770,000⫽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,⫽A h (t2⫺t1)(0.24⫹0.46M a),

where A h⫽air flow through heater, lb / lb fuel⫽A⫺A m ; A⫽total air to furnace,

lb / lb fuel⫽ G ⫺ C b⫺ N2⫺ 7.94(H2⫺ O2/ 8); G ⫽ similar to economizer but

based on gas at the furnace exit; A m⫽external air supplied by the mill fan or other

source, lb / lb of fuel Substituting shows G ⫽ [11(0.16) ⫹ 8(0.26) ⫹ 7(0.184)] /[3(0.16)](0.678⫹0.032 / 2.67)⫹0.032 / 1.60; G⫽11.03 lb / lb (11.03 kg / kg) fuel;

A⫽11.03⫺ 0.69⫺0.012⫺ 7.94(0.05⫺0.089 / 8); A⫽ 10.02 lb / lb (10.02 kg /kg) fuel Heat absorbed⫽(1⫺0.15)(10.02)(480⫺63)(0.24⫹56 / 7000)⫽865.5Btu / lb (2013.2 kJ / kg fuel

8. Compute the heat available to the air heater

The heat available to the air heater, Btu / h⫽(t5⫺t1)0.24G ⫹(t5⫺t1)0.46(Mƒ⫹8.94H2⫹ 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

⫽[11(0.15)⫹8(0.036)⫹7(0.814)] / [3(0.15)](0.678⫹0.032 / 2.67)⫹0.032 / 1.60;

G⫽11.72 lb / lb (11.72 kg / kg) fuel And A⫽11.72 ⫺0.69⫺0.012⫺7.94(0.05

⫺ 0.089 / 8) ⫽ 10.71 lb / lb (10.71 kg / kg) fuel Heat available ⫽ (570 ⫺3)(0.24)(11.72)⫹(570⫺63)(0.46)[0.045⫹8.94(0.05)⫹56 / 7000(10.71)]⫽1561Btu / lb (3630.9 kJ / kg)

9. Compute the air-heater efficiency

The air-heater efficiency ⫽ (heat absorbed, Btu / lb fuel) / (heat available, Btu / lbfuel)⫽865.5 / 1561⫽ 0.554, or 55.4 percent

Related Calculations. The above procedure is valid for all types of steam erators, regardless of the kind of fuel used Where oil or gas is the fuel, alter thecombustion calculations to reflect the differences between the fuels Further, thisprocedure is also valid for marine and portable boilers

gen-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

shell ends are heat-insulated How much steam is generated if the boiler evaporates34.5 lb / h of water per 12 ft2[3.9 g / (m2䡠s)] of heating surface? How much heat isadded by the boiler if it operates at 200 lb / in2 (abs) (1379.0 kPa) with 200⬚F(93.3⬚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 thewater?

Calculation Procedure:

1. Compute the shell area exposed to furnace gas

Shell area⫽␲DL(1 ⫺0.25), where D ⫽boiler diameter, ft; L⫽shell length, ft;

1 ⫺ 0.25 is the portion of the shell in contact with the furnace gas Then shellarea⫽␲(84 / 12)(18)(0.75)⫽297 ft2(27.0 m2)

2. Compute the tube area exposed to furnace gas

Tube area⫽␲dLN, where⫽tube ID, ft; L⫽tube length, ft; N⫽number of tubes

in boiler Substituting gives tube area⫽ ␲(4 / 12)(18)(84)⫽1583 ft2(147.1 m2)

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⫺portion covered by insulation, expressed as a decimal) Substituting,

we get 2␲D2/ 4 ⫺ (2)(84)␲d2/ 4 ⫽ 2␲/ 4(84 / 12)2(0.75) ⫺ (2)(84)␲(4 / 12)2/ 4 ⫽head area⫽43.1 ft2(4.0 m2)

4. Find the total heating surface

The total heating surface of any fire-tube boiler is the sum of the shell, tube, andhead areas, or 297.0⫹1583⫹43.1⫽1923 ft2(178.7 m2), total heating surface

5. Compute the quantity of steam generated

Since the boiler evaporates 34.5 lb / h of water per 12 ft2[3.9 g / (m2䡠s)] of heatingsurface, the quantity of steam generated⫽ 34.5 (total heating surface, ft2) / 12 ⫽34.5(1923.1) / 12⫽5200 lb / h (0.66 kg / s)

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 paring the performance of boilers A term used in lieu of boiler horsepower, with

com-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 ⫽ h g ⫺ hƒ1; from steam table values 1198.4 ⫺167.99⫽ 1030.41 Btu / lb (2396.7 kJ / kg) An alternative way of computing heat

added is h g⫺(feedwater temperature,⬚F, ⫺32), where 32 is the freezing ature of water on the Fahrenheit scale By this method, heat added ⫽ 1198.4 ⫺(200 ⫺ 32) ⫽ 1030.4 Btu / lb (2396.7 kJ / kg) Thus, both methods give the sameresults in this case In general, however, use of steam table values is preferred

temper-7. Compute the factor of evaporation

The factor of evaporation is used to convert from the actual to the equivalent oration, defined earlier Or, factor of evaporation ⫽ (heat added by boiler,Btu / lb) / 970.3, where 970.3 Btu / lb (2256.9 kJ / kg) is the heat added to develop 1

evap-boiler hp (bhp) (0.75 kW) Thus, the factor of evaporation for this evap-boiler ⫽1030.4 / 970.3⫽1.066.

8. Compute the boiler hp output

Boiler hp⫽(actual evaporation, lb / h) (factor of evaporation) / 34.5 In this relation,the actual evaporation must be computed first Since the furnace delivers 7,000,000Btu / h (2051.5 kW) to the boiler water and the water absorbs 1030.4 Btu / lb (2396.7

kJ / kg) to produce 200-lb / in2 (abs) (1379.0-kPa) steam with 200⬚F (93.3⬚C) water, the steam generated, lb / h ⫽ (total heat delivered, Btu / h) / (heat absorbed,Btu / lb) ⫽ 7,000,000 / 1030.4 ⫽ 6670 lb / h (0.85 kg / s) Then boiler hp ⫽(6760)(1.066) / 34.5⫽ 209 hp (155.9 kW)

feed-The rated hp output of horizontal fire-tube boilers with separate supporting walls

is based on 12 ft2(1.1 m2) of heating surface per boiler hp Thus, the rated hp ofthe boiler ⫽ 1923.1 / 12 ⫽ 160 hp (119.3 kW) When producing 209 hp (155.9kW), the boiler is operating at 209 / 160, or 1,305 times its normal rating, or(100)(1.305)⫽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 / in2(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 issuperheated 100⬚F (55.6⬚C) above its saturation temperature?

Calculation Procedure:

1. Determine the area of the valve annulus

Annulus area, in2⫽ A⫽␲DL, where D⫽ valve diameter, in; L⫽ valve lift, in.Annulus area⫽␲(2.5)(0.25)⫽1.966 in2(12.7 cm2)

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⫽p0.97s A / 60, where p s ⫽ saturated-steam pressure, lb / in2 (abs) For this valve, F i ⫽ (150)0.97(1.966) / 60⫽4.24 lb / s (1.9 kg / s)

3. Compute the actual flow through the valve

Actual flow F a⫽F i c d⫽(4.24)(0.75)⫽3.18 lb / s (1.4 kg / s)⫽(3.18)(3600 s / h)⫽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⫽p0.97s A / [60(1⫹0.0065t s)], where

t s ⫽ superheated temperature, above saturation temperature,⬚F The F is ⫽(150)0.97(1.966) / [60(1⫹ 0.0065⫻100)] ⫽ 3.96 lb / s (1.8 kg / s) The actual flow

m2) 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 tions of the Code require that the safety valve have a minimum relieving capacity

edi-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 / (ft2䡠h) ofsteam [13.6 g / (m2䡠s)] of boiler heating surface, and 16 lb / (ft2䡠h) of steam [21.9

g / (m2䡠s)] of waterwall surface Thus, the minimum safety-valve relieving capacityfor this boiler, based on total heating surface, would be (8200)(10)⫹(1000)(16)⫽92,000 lb / h (11.6 kg / s) In this equation, 1000 ft2(92.9 m2) of waterwall surface

is deducted from the total heating surface of 9200 ft2(854.7 m2) to obtain the boilerheating surface of 8200 ft2(761.8 m2)

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 capacitymust be 100,000 lb / h (12.6 kg / s), because this is the maximum capacity of theboiler 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 ft2(46.5 m2)

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

valves made by placing individual valves on Y bases, or duplex valves having twovalves in the same body, shall be of equal sizes.’’ Also, ‘‘when not more thantwo valves of different sizes are mounted singly, the relieving capacity of thesmaller 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 thesteam drum of this boiler Two or more equal-size valves are usually chosen forthe steam drum of a watertube boiler

Since this boiler handles superheated steam, check the Code requirements garding superheaters The Code states that ‘‘every attached superheater shall have

re-one or more safety valves near the outlet.’’ Also, ‘‘the discharge capacity of thesafety valve, or valves, on an attached superheater may be included in determiningthe number and size of the safety valves for the boiler, provided there are nointervening valves between the superheater safety valve and the boiler, and providedthe discharge capacity of the safety valve, or valves, on the boiler, as distinct fromthe 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 capacitymust then be at least 100,000 ⫺ 80,000 ⫽ 20,000 lb / h (2.5 kg / s) (Use a fewsuperheater safety valves as possible, because this simplifies the installation andreduces cost.) With this arrangement, each steam-drum valve must handle 80,000 /

2⫽40,000 lb / h (5.0 kg / s) of steam, since there are two safety valves on the steamdrum

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 modernboilers, the maximum allowable working pressure is usually 1.5, or more, timesthe rated operating pressure in the lower [under 1000 lb / in2(abs) or 6895.0 kPa]pressure ranges To prevent unnecessary operation of the safety valve and to reducesteam losses, the lowest safety-valve setting is usually about 5 percent higher thanthe boiler operating pressure For this boiler, the lowest pressure setting would be

800⫹800(0.05) ⫽840 lb / in2(abs) (5791.8 kPa) Round this to 850 lb / in2(abs)(5860.8 kPa, or 6.25 percent) for ease of selection from the usual safety-valve ratingtables The usual safety-valve pressure setting is between 5 and 10 percent higherthan 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 thatthe superheater safety valve opens first when overpressure occurs This providessteam 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/ in2(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 decidedearlier, 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 pressureloss that varies from one boiler to another The boiler manufacturer supplies aperformance chart showing the drum outlet pressure for various percentages of themaximum continuous steaming capacity of the boiler This chart also shows thesuperheater outlet pressure for the same capacities The difference between the drumand superheater outlet pressure for any given load is the superheater pressure loss.Obtain this pressure loss from the performance chart

Assume, for this boiler, that the superheater pressure loss, plus any pressurelosses in the nonreturn valve and dry pipe, at maximum rating, is 60 lb / in2(abs)(413.7 kPa) The steam-drum operating pressure will then be superheater outletpressure⫹superheater pressure loss⫽800⫹60⫽860 lb / in2(abs) (5929.7 kPa).

As with the superheater safety valve, the steam-drum safety valve is usually set toopen at about 5 percent above the drum operating pressure at maximum steamoutput For this boiler then, the drum safety-valve set pressure ⫽ 860 ⫹860(0.05)⫽903 lb / in2(abs) (6226.2 kPa) Round this to 900 lb / in2(abs) (6205.5kPa) to simplify valve selection

Some designers add the drum safety-valve blowdown or blowback pressure ference between the valve opening and closing pressures, lb / in2) to the total ob-tained above to find the drum operating pressure However, the 5 percent allowanceused above is sufficient to allow for the blowdown in boilers operating at less than

(dif-1000 lb / in2(abs) (6895.0 kPa) At pressures of 1000 lb / in2(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 atvarious working pressures For the two steam-drum valves, enter the table at 900

lb / in2(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 / in2(abs) (6205.5 kPa) a valve with an orifice area of 0.944 in2(6.4

cm2) will have a capacity of 42,200 lb / h (5.3 kg / s) of saturated steam This is 5.5percent 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 desiredcapacity Provided that the valve chosen has a greater steam relieving capacity thanrequired, there is no danger of overpressure in the steam drum Be careful to notethat 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 / in2(abs) (5860.8-kPa) steam at 900⬚F (482.2⬚C) Safety valves handling superheatedsteam have a smaller capacity than when handling saturated steam To obtain thecapacity 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 beused here because it is more direct

Obtain the correction factor from the safety-valve manufacturer’s engineeringdata by entering at the steam pressure and projecting to the steam temperature, asshow below

Thus, at 850 lb / in2(abs) (5860.8 kPa) and 900⬚F (482.2⬚C), the correction factor

is 0.80 The required saturated steam capacity then is 20,000 / 0.80⫽25,000 lb / h(3.1 kg / s)

Refer to the manufacturer’s saturated-steam capacity table as before, and at 850

lb / in2(abs) (5860.8 kPa) find the closest capacity as 31,500 lb / h (4.0 kg / s) for a0.785-in2 (5.1-cm2) orifice As with the steam-drum valves, the actual capacity ofthe safety valve is somewhat greater than the required capacity In general, it isdifficult 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 valveconstruction features For the steam-drum valves having 0.994-in2(6.4-cm2) orificeareas, the engineering data show, for 900-lb / in2(abs) (6205.5-kPa) service, eachvalve is 11⁄2-in (3.8-cm) unit rated for temperatures up to 1050⬚F (565.6⬚C) Theinlet is 900-lb / in2(6205.5-kPa) 11⁄2-in (3.8-cm) flanged connection, and the outlet

is a 150-lb / in2 (1034.3-kPa) 3-in (7.6-cm) flanged connection Materials used inthe valve include: body, cast carbon steel; disk seat, stainless steel AISI 321 Theoverall height is 277⁄8in (70.8 cm); dismantled height is 323⁄4in (83.2 cm)

Similar data for the superheated steam valve show, for a maximum pressure of

900 lb / in2(abs) (6205.5 kPa), that it is a 11⁄2-in (3.8-cm) unit rated for temperatures

up to 1000⬚F (537.8⬚C) The inlet is a 900-lb / in2 (6205.5-kPa) 11⁄2-in (3.8-cm)flanged connection, and the outlet is a 150-lb / in2 (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, stainlesssteel Overall height is 213⁄8in (54.3 cm); dismantled height is 251⁄4in (64.1 cm)

Checking the Code shows that ‘‘every safety valve used on a superheater

discharg-ing superheated steam at a temperature over 450⬚F (232.2⬚C) shall have a casing,including the base, body, bonnet and spindle, of steel, steel alloy, or equivalentheat-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 notless than the combined areas of inlet connections of all safety valves with which

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⫺(900)(0.04)⫽865 lb / in2(abs) (5964.2 kPa) The superheater safety valvewill close at 850⫺(850)(0.04)⫽816 lb / in2(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 boilersafety valves with escape pipes to carry the steam out of the building and awayfrom personnel Extend the escape pipe to at least 6 ft (1.8 m) above the roof ofthe 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 recommendincreasing the escape-pipe diameter by1⁄2in (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 maythen chatter excessively

Support the escape pipe independently of the safety valve Fit a drain to thevalve body and rip pan as shown in Fig 3 This prevents freezing of the condensate

and also eliminates the possibility of condensate in the escape pipe raising the valveopening pressure When a muffler is fitted to the escape pipe, the inlet diameter ofthe muffler should be the same as, or larger than, the escape-pipe diameter Theoutlet 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 andheating surface Thus, with an operating pressure of 200 lb / in2(gage) (1379.0 kPa)and 1800 ft2 (167.2 m2) of heating surface, the Code table shows that the safety-

valve connection should have an area of at least 9.148 in2 (59.0 cm2) A 31⁄2-in(8.9-cm) connection would provide this area; or two smaller connections could beused provided that the sum of their areas exceeded 9.148 in2(59.0 cm2)

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 anunapproved valve can lead to its rejection by the bureau or other agency controllingboiler installation and operation

STEAM-QUALITY DETERMINATION WITH A

THROTTLING CALORIMETER

Steam leaves an industrial boiler at 120 lb / in2 (abs) (827.4 kPa) and 341.25⬚F(171.8⬚C) A portion of the steam is passed through a throttling calorimeter and isexhausted to the atmosphere when the barometric pressure is 14.7 lb / in2 (abs)(101.4 kPa) How much moisture does the steam leaving the boiler contain if thetemperature of the steam at the calorimeter is 240⬚F (115.6⬚C)?

Calculation Procedure:

1. Plot the throttling process on the Mollier diagram

Begin with the endpoint, 14.7 lb / in2 (abs) (101.4 kPa) and 240⬚F (115.6⬚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 / in2(abs) (101.4kPa) has a temperature of 212⬚F (100.0⬚C), whereas the steam in this calorimeterhas a temperature of 240⬚F (115.6⬚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 / in2

(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 / in2(abs) (827.4-kPa) pressure line, and the 3 percent moistureline 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 atatmospheric pressure, connect the calorimeter outlet to an area at a pressure lessthan atmospheric Expand the steam from the source, and read the temperature atthe calorimeter If the steam temperature is greater than that corresponding to the

FIGURE 4 Mollier-diagram plot of a throttling-calorimeter process.

absolute pressure of the vacuum area—for example, a temperature greater than133.76⬚F (56.5⬚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 themoisture content as before

STEAM PRESSURE DROP IN A

BOILER SUPERHEATER

What is the pressure loss in a boiler superheater handling w s⫽200,000 lb / h (25.2

kg / s) of saturated steam at 500 lb / in2 (abs) (3447.5 kPa) if the desired outlettemperature is 750⬚F (398.9⬚C)? The steam free-flow area through the superheater

tubes A s ft2 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 Bƒis 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 thesteamv g and the mass-flow ratio w s / A smust be known From the steam table,v g⫽

0.9278 ft3/ lb (0.058 m3/ kg) at 500 lb / in2(abs) (3447.5 kPa) saturated The

mass-flow ratio w s / A s⫽200,000 / 0.500 ⫽400,000

2. Compute the superheater entrance and exit pressure loss

Entrance and exit pressure loss p E lb / in2 ⫽ vƒ/ 8(0.00001w s / A s) ⫽ 0.9278 /8[(0.00001)⫻(400,000)]2⫽1.856 lb / in2(12.8 kPa)

3. Compute the pressure loss in the straight tubes

Straight-tube pressure loss p s lb / in2 ⫽ vƒƒ / ID(0.00001w s / A s)2 ⫽ 0.9278(150) ⫻(0.025) / 2.125[(0.00001)(400,000)]2⫽26.2 lb / in2(abs) (180.6 kPa)

4. Compute the pressure loss in the superheater bends

Bend pressure loss p b ⫽ 0.0833Bƒ(0.00001w s / A s)2 ⫽ 0.0833(12.0)[(0.00001) ⫻(400,000)]2⫽16.0 lb / in2(110.3 kPa)

5. Compute the total pressure loss

The total pressure loss in any superheater is the sum of the entrance, straight-tube,bend, and exit-pressure losses These losses were computed in steps 2, 3, and 4

above Therefore, total pressure loss p t⫽ 1.856 ⫹ 26.2⫹ 16.0 ⫽ 44.056 lb / in2(303.8 kPa)

Note: Data for superheater pressure-loss calculations are best obtained from the

boiler manufacturer Several manufacturers have useful publications discussing perheater pressure losses These are listed in the references at the beginning of thissection

su-SELECTION OF A STEAM BOILER FOR A

GIVEN LOAD

Choose a steam boiler, or boilers, to deliver up to 250,000 lb / h (31.5 kg / s) ofsuperheated steam at 800 lb / in2(abs) (5516 kPa) and 900⬚F (482.2⬚C) Determinethe type or types of boilers to use, the capacity, type of firing, feedwater-qualityrequirements, and best fuel if coal, oil, and gas are all available The normal con-tinuous steam requirement is 200,000 lb / h (25.2 kg / s)

Calculation Procedure:

1. Select type of steam generator

Use Fig 5 as a guide to the usual types of steam generators chosen for variouscapacities and different pressure and temperature conditions Enter Fig 5 at the left

at 800 lb / in2(abs) (5516 kPa), and project horizontally to the right, along AB, until the 250,000-lb / h (31.5-kg / s) capacity ordinate BC is intersected At B, the oper-

ating point of this boiler, Fig 5 shows that a watertube boiler should be used.Boiler units presently available can deliver steam at the desired temperature of

900⬚F (482.2⬚C) The required capacity of 250,000 lb / h (31.5 kg / s) is beyond the

range of packaged watertube boilers—defined by the American Boiler

Manufac-turer Association as ‘‘a boiler equipped and shipped complete with fuel-burningequipment, mechanical-draft equipment, automatic controls, and accessories.’’

Shop-assembled boilers are larger units, where all assembly is handled in the

builder’s plant but with some leeway in the selection of controls and auxiliaries

FIGURE 5 Typical pressure and capacity relationships from steam generators (Power.)

The current maximum capacity of shop-assembled boilers is about 100,000 lb / h(12.6 kg / s) Thus, a standard-design, larger-capacity boiler is required

Study manufacturers’ engineering data to determine which types of watertubeboilers are available for the required capacity, pressure, and temperature This studyreveals that, for this installation, a standard, field-assembled, welded-steel-cased,bent-tube, single-steam-drum boiler with a completely water-cooled furnace would

be suitable This type of boiler is usually fitted with an air heater, and an economizermight also be used The induced- and forced-draft fans are not integral with theboiler Capacities of this type of boiler usually available range from 50,000 to350,000 lb / h (6.3 to 44.1 kg / s); pressure from 160 to 1050 lb / in2(1103.2 to 7239.8kPa); steam temperature from saturation to 950⬚F (510.0⬚C); fuels—pulverized coal,oil, gas, or a combination; controls—manual to completely automatic;efficiency—to 90 percent

2. Determine the number of boilers required

The normal continuous steam requirement is 200,000 lb / h (25.2 kg / s) If a250,000-lb / h (31.5-kg / s) boiler were chosen to meet the maximum required output,the boiler would normally operate at 2000,000 / 250,000, or 80 percent capacity.Obtain the performance chart, Fig 6, from the manufacturer and study it This chartshows that at 80 percent load, the boiler efficiency is about equal to that at 100percent load Thus, there will not be any significant efficiency loss when the unit

is operated at its normal continuous output The total losses in the boiler are lower

at 80 percent load than at full (100 percent) load

FIGURE 6 Typical watertube steam-generator losses and efficiency.

Since there is not a large efficiency decrease at the normal continuous load, andsince there are not other factors that require or make more than one boiler desirable,

a single boiler unit would be most suitable for this installation One boiler is moredesirable than two or more because installation of a single unit is simpler andmaintenance costs are lower However, where the load fluctuates widely and two

or more boilers could best serve the steam demand, the savings in installation andmaintenance costs would be insignificant compared with the extra cost of operating

a relatively large boiler installed in place of two or more smaller boilers Therefore,each installation must be carefully analyzed and a decision made on the basis ofthe existing conditions

3. Determine the required boiler capacity

The stated steam load is 250,000 lb / h (31.5 kg / s) at maximum demand Study theinstallation to determine whether the steam demand will increase in the future Try

to determine the rate of increase in the steam demand; for example, installation ofseveral steam-using process units each year during the next few years will increasethe steam demand by a predictable amount every year By using these data, the rate

of growth and total steam demand can be estimated for each year Where the growthwill exceed the allowable overload capacity of the boiler—which can vary from 0

to 50 percent of the full-load rating, depending on the type of unit chosen—considerinstalling a larger-capacity boiler now to meet future load growth Where the futureload is unpredictable or where no load growth is anticipated, a unit sized to meettoday’s load would be satisfactory If this situation existed in this plant, a 250,000-

lb / h unit (31.5-kg / s) would be chosen for the load Any small temporary overloadscould be handled by operating the boiler at a higher output for short periods.Alternatively, assume that a load of 25,000 lb / h (3.1 kg / s) will be added to themaximum demand on this boiler each year for the next 5 years This means that

in 5 years the maximum demand will be 250,000 ⫹ 25,000(5) ⫽ 375,000 lb / h(47.2 kg / s) This is an overload of (375,000 ⫺ 250,000) / 250,000 ⫽ 0.50, or 50percent It is unlikely that the boiler could carry a continuous overload of 50 per-cent Therefore it might be wise to install a 375,000-lb / h (47.2-kg / s) boiler to meetpresent and future demands Base this decision on the accuracy of the future-demand prediction and the economic advantages or disadvantages of investing moremoney now for a demand that will not occur until some future date Refer to thesection on engineering economics for procedures to follow in economics calcula-tions of this type

Thus, with no increase in the future load, a 250,000-lb / h (31.5-kg / s) unit would

be chosen With the load increase specified, a 375,000-lb / h (47.2-kg / s) unit would

be the choice, if there were no major economic disadvantages

4. Choose the type of fuel to use

Watertube boilers of the type being considered will economically burn the threefuels available—coal, oil, or gas—either singly or in combination In the designconsidered here, the furnace watercooled surfaces and boiler surfaces are integralparts of each other For this reason the boiler is well suited for pulverized-coalfiring in the 50,000- to 300,000-lb / h (6.3- to 37.8-kg / s) capacity range Thus, if a250,000-lb / h (31.5-kg / s) unit were chosen, it could be fired by pulverized coal.With a larger unit of 375,000 lb / h (47.2 kg / s), pulverized coal, oil, or gas firingmight be used Use an economic comparison to determine which fuel would givethe lowest overall operating coast for the life of the boiler

5. Determine the feedwater-quality requirements

Watertube boilers of all types require careful control of feedwater quality to preventscale and sludge deposits in tubes and drums Corrosion of the interior boiler sur-faces must be controlled Where all condensate is returned to the boiler, the makeupwater must be treated to prevent the conditions just cited Therefore, a comprehen-sive water-treating system must be planned for, particularly if the raw-water supply

is poor

6. Estimate the boiler space requirements

The space occupied by steam-generating units is an important consideration inplants in municipal areas and where power-plant buildings are presently crowded

by existing equipment The manufacturer’s engineering data for this boiler showthat for pulverized-coal firing, the hopper-type furnace bottom is best The dataalso show that the smallest boiler with a hopper bottom occupies a space 21 ft (6.4m) wide, 31 ft (9.4 m) high, and 14 ft (4.3 m) front to rear The largest boileroccupies a space 21 ft (6.4 m) wide, 55 ft (16.8 m) high, and 36 ft (11.0 m) front

to rear Check these dimensions against the available space to determine whetherthe chosen boiler can be installed without major structural changes The steel wallspermit outdoor or indoor installation with top or bottom support of the boiler op-tional in either method of installation

Related Calculations. Use this general procedure to select boilers for trial, central-station, process, and marine applications

indus-Where a boiler is to burn hazardous industrial waste as a fuel, the designer mustcarefully observe two waste laws: the 1980 Superfund law and the 1976 Resource

Conservation and Recovery Act These laws regulate the firing of hazardous wastes

in boilers to control air pollution and explosion dangers

Since hazardous wastes from industrial operations can vary in composition, it isimportant that the designer know what variables might be met during actual firing.Without correct analysis of the wastes, air pollution can become a severe problem

in the plant locale

The Environmental Protection Agency (EPA) and state regulatory agenciesshould be carefully consulted before any final design decisions are made for a newand expanded boiler plants While the firing of hazardous wastes can be a conven-ient way to dispose of them, the potential impact on the environment must beconsidered before any design is finalized

SELECTING BOILER FORCED- AND

INDUCED-DRAFT FANS

Combustion calculations show that an oil-fired watertube boiler requires 200,000

lb / h (25.2 kg / s) of air for combustion at maximum load Select forced- andinduced-draft fans for this boiler if the average temperature of the inlet is 75⬚F(23.9⬚C) and the average temperature of the combustion gas leaving the air heater

is 350⬚F (176.7⬚C) with an ambient barometric pressure of 29.9 inHg (101.0 kPa).Pressure losses on the air-inlet side are as follows, in inH2O: air heater, 1.5 (0.37kPa); air-supply ducts, 0.75 (0.19 kPa); boiler windbox, 1.75 (0.44 kPa); burners,1.25 (0.31 kPa) Draft losses in the boiler and related equipment are as follows, ininH2O: furnace pressure, 0.20 (0.05 kPa); boiler, 3.0 (0.75 kPa); superheater 1.0(0.25 kPa); economizer, 1.50 (0.37 kPa); air heater, 2.00 (0.50 kPa); uptake ductsand dampers, 1.25 (0.31 kPa) Determine the fan discharge pressure and hp input.The boiler burns 18,000 lb / h (2.3 kg / s) of oil at full load

Calculation Procedure:

1. Compute the quantity of air required for combustion

The combustion calculations show that 200,000 lb / h (25.2 kg / s) of air is ically required for combustion in this boiler To this theoretical requirement must

theoret-be added allowances for excess air at the burner and leakage out of the air heaterand furnace Allow 25 percent excess air for this boiler The exact allowance for agiven installation depends on the type of fuel burned However, a 25 percent excess-air allowance is an average used by power-plant designers for coal, oil, and gasfiring With this allowance, the required excess air ⫽ 200,000(0.25) ⫽ 50,000

lb / h (6.3 kg / s)

Air-heater air leakage varies from about 1 to 2 percent of the theoretically quired airflow Using 2 percent, we see the air-heater leakage allowance ⫽200,000(0.02)⫽ 4,000 lb / h (0.5 kg / s)

re-Furnace air leakage ranges from 5 to 10 percent of the theoretically requiredairflow With 7.5 percent, the furnace leakage allowance ⫽ 200,000(0.075) ⫽15,000 lb / h (1.9 kg / s)

The total airflow required is the sum of the theoretical requirement, excess air,and leakage Or, 200,000 ⫹ 50,000 ⫹ 4000 ⫹ 15,000 ⫽ 269,000 lb / h (33.9

kg / s) The forced-draft fan must supply at least this quantity of air to the boiler.Usual practice is to allow a 10 to 20 percent safety factor for fan capacity to ensure

an adequate air supply at all operating conditions This factor of safety is applied

to the total airflow required Using a 10 percent factor of safety, we see that fancapacity ⫽ 269,000 ⫹ 269,000(0.1) ⫽ 295,900 lb / h (37.3 kg / s) Round this to269,000-lb / h (37.3 kg / s) fan capacity.

2. Express the required airflow in cubic feet per minute

Convert the required flow in pounds per hour to cubic feet per minute To do this,apply a factor of safety to the ambient air temperature to ensure an adequate airsupply during times of high ambient temperature At such times, the density of theair is lower, and the fan discharges less air to the boiler The usual practice is toapply a factor of safety of 20 to 25 percent to the known ambient air temperature.Using 20 percent, we see the ambient temperature for fan selection ⫽ 75 ⫹75(0.20)⫽90⬚F (32.2⬚C) The density of air at 90⬚F (32.2⬚C) is 0.0717 lb / ft3(1.15

kg / m3), found in Baumeister and Marks—Standard Handbook for Mechanical

69,400 ft3/ min (32.8 m3/ s) This is the minimum capacity the forced-draft fan mayhave

3. Determine the forced-draft discharge pressure

The total resistance between the forced-draft fan outlet and furnace is the sum ofthe losses in the air heater, air-supply ducts, boiler windbox, and burners For thisboiler, the total resistance, inH2O ⫽1.5⫹0.75⫹1.75⫹1.25⫽5.25 inH2O (1.3kPa) Apply a 15 to 30 percent factor of safety to the required discharge pressure

to ensure adequate airflow at all times Or, fan discharge pressure, with a 20 percentfactor of safety⫽5.25⫹5.25(0.20)⫽6.30 inH2O (1.6 kPa) The fan must there-fore deliver at least 69,400 ft3/ min (32.8 m3/ s) at 6.30 inH2O (1.6 kPa)

4. Compute the power required to drive the forced-draft fan

The air hp for any fan⫽0.0001753Hƒ⫽total head developed by fan, inH2O; C⫽airflow, ft3/ min For this fan, air hp ⫽ 0.0001753(6.3)(69,400) ⫽ 76.5 hp (57.0kW) Assume or obtain the fan and fan-driver efficiencies at the rated capacity(69,400 ft3/ min, or 32.8 m3/ s) and pressure (6.30 inH2O, or 1.6 kPa) With a fanefficiency of 75 percent and assuming the fan is driven by an electric motor having

an efficiency of 90 percent, we find the overall efficiency of the fan-motor bination is (0.75)(0.90) ⫽0.675, or 67.5 percent Then the motor horsepower re-quired⫽air hp / overall efficiency⫽76.5 / 0.675⫽113.2 hp (84.4 kW) A 125-hp(93.2-kW) motor would be chosen because it is the nearest, next larger unit readily

com-available Usual practice is to choose a larger driver capacity when the computed

capacity is lower than a standard capacity The next larger standard capacity isgenerally chosen, except for extremely large fans where a special motor may beordered

5. Compute the quantity of flue gas handled

The quantity of gas reaching the induced-draft fan is the sum of the actual airrequired for combustion from step 1, air leakage in the boiler and furnace, and theweight of fuel burned With an air leakage of 10 percent in the boiler and furnace(this is a typical leakage factor applied in practice), the gas flow is as follows:

Determine from combustion calculations for the boiler the density of the fluegas Assume that the combustion calculations for this boiler show that the flue-gasdensity is 0.045 lb / ft3(0.72 kg / m3) at the exit-gas temperature To determine theexit-gas temperature, apply a 10 percent factor of safety to the given exit temper-ature, 350⬚F (176.6⬚C) Hence, exit-gas temperature⫽ 350⫹ 350(0.10)⫽ 385⬚F(196.1⬚C) Then flue-gas flow, ft3/ min⫽(flue-gas flow, lb / h) / (60)(flue-gas density,

lb / ft3)⫽ 343,600 / [(60)(0.045)]⫽127,000 ft3/ min (59.9 m3/ s) Apply a 10 to 25percent factor of safety to the flue-gas quantity to allow for increased gas flow.With a 20 percent factor of safety, the actual flue-gas flow the fan must handle⫽127,000⫹127,000(0.20)⫽152,400 ft3/ min (71.8 m3/ s), say 152,500 ft3/ min (71.9

m3/ s) for fan-selection purposes

6. Compute the induced-draft fan discharge pressure

Find the sum of the draft losses from the burner outlet to the induced-draft faninlet These losses are as follows for this boiler:

Allow a 10 to 25 percent factor of safety to ensure adequate pressure during allboiler loads and furnace conditions With a 20 percent factor of safety for this fan,the total actual pressure loss⫽8.95⫹8.95(0.20)⫽10.74 inH2O (2.7 kPa) Roundthis to 11.0 inH2O (2.7 kPa) for fan-selection purposes

7. Compute the power required to drive the induced-draft fan

As with the forced-draft fan, air hp ⫽ 0.0001753HƒC ⫽ 0.0001753(11.0) ⫻(127,000) ⫽ 245 hp (182.7 kW) If the combined efficiency of the fan and itsdriver, assumed to be an electric motor, is 68 percent, the motor hp required ⫽

245 / 0.68⫽ 360.5 hp (268.8 kW) A 375-hp (279.6-kW) motor would be chosenfor the fan driver

8. Choose the fans from a manufacturer’s engineering data

Use the next calculation procedure to select the fans from the engineering data of

an acceptable manufacturer For larger boiler units, the forced-draft fan is usually

a backward-curved blade centrifugal-type unit Where two fans are chosen to erate in parallel, the pressure curve of each fan should decrease at the same ratenear shutoff so that the fans divide the load equally Be certain that forced-draftfans are heavy-duty units designed for continuous operations with well-balancedrotors Choose high-efficiency units with self-limiting power characteristics to pre-vent overloading the driving motor Airflow is usually controlled by dampers onthe fan discharge

op-Induced-draft fans handle hot, dusty combustion products For this reason, treme care must be taken to choose units specifically designed for induced-draftservice The usual choice for large boilers is a centrifugal-type unit with forward-

ex-TABLE 4 Fan Correction Factors

or backward-curved, or flat blades, depending on the type of gas handled Flatblades are popular when the flue gas contains large quantities of dust Fan bearingsare generally water-cooled

Related Calculations. Use the procedure given above for the selection of draftfans for all types of boilers—fire-tube, packaged, portable, marine, and stationary.Obtain draft losses from the boiler manufacturer Compute duct pressure losses byusing the methods given in later procedures in this handbook

POWER-PLANT FAN SELECTION FROM

CAPACITY TABLES

Choose a forced-draft fan to handle 69,400 ft3/ min (32.8 m3/ s) of 90⬚F (32.2⬚C)air at 6.30-inH2O (1.6-kPa) static pressure and an induced-draft fan to handle152,500 ft3/ min (72.0 m3/ s) of 385⬚F (196.1⬚C) gas at 11.0-inH2O (2.7-kPa) staticpressure The boiler that these fans serve is installed at an elevation of 5000 ft.(1524 m) above sea level Use commercially available capacity tables for makingthe fan choice The flue-gas density is 0.045 lb / ft3(0.72 kg / m3) at 385⬚F (196.1⬚C)

Calculation Procedure:

1. Compute the correction factors for the forced-draft fan

Commercial fan-capacity tables are based on fans handling standard air at 70⬚F(21.1⬚C) at a barometric pressure of 29.92 inHg (101.0 kPa) and having a density

of 0.075 lb / ft3(1.2 kg / m3) Where different conditions exist, the fan flow rate must

be corrected for temperature and altitude

Obtain the engineering data for commercially available forced-draft fans, andturn to the temperature and altitude correction-factor tables Pick the appropriatecorrection factors from these tables for the prevailing temperature and altitude ofthe installation Thus, in Table 4, select the correction factors for 90⬚F (32.2⬚C) air

and 5000-ft (1524.0-m) altitude These correction factors are C T⫽1.018 for 90⬚F(32.2⬚C) air and C A⫽1.095 for 5000-ft (1524.0-m) altitude

Find the composite correction factor (CCF) by taking the product of the perature and altitude correction factors Or, CCF⫽(1.018)(1.095)⫽1.1147 Nowdivide the given cubic feet per minute (cfm) by the correction factor to find the

tem-TABLE 5 Typical Fan Capacities

capacity-table ft3/ min Or, capacity-table ft3/ min ⫽ 69,400 / 1.147 ⫽ 62,250 ft3/min (29.4 m3/ s)

2. Choose the fan size from the capacity table

Turn to the fan-capacity table in the engineering data, and look for a fan delivering62,250 ft3/ min (29.4 m3/ s) at 6.3-inH2O (1.6-kPa) static pressure Inspection of thetable shows that the capacities are tabulated for 6.0- and 6.5-inH2O (1.5- and 1.6-kPa) static pressure There is no tabulation for 6.3-inH2O (1.57-kPa) static pressure.Enter the table at the nearest capacity to that required, 62,250 ft3/ min (29.4

m3/ s), as shown in Table 5 This table, excerpted with permission from the ican Standard Inc engineering data, shows that the nearest capacity of this particulartype of fan is 62,595 ft3/ min (29.5 m3/ s) The difference, or 62,595⫺ 62,250⫽

Amer-345 ft3/ min (0.16 m3/ s), is only 345 / 62,250 ⫽0.0055, or 0.55 percent This is anegligible difference, and the 62,595-ft3/ min (29.5-m3/ s) fan is well suited for itsintended use The extra static pressure of 6.5 ⫺ 6.3 ⫽ 0.2 inH2O (0.05 kPa) isdesirable in a forced-draft fan because furnace or duct resistance may increaseduring the life of the boiler Also, the extra static pressure is so small that it willnot markedly increase the fan power consumption

3. Compute the fan speed and power input

Multiply the capacity-table rpm and brake hp (bhp) by the composite factor todetermine the actual rpm and bhp Thus, with data from Table 5, the actual rpm⫽(1096)(1.1147)⫽1221.7 r / min Actual bhp⫽(99.08)(1.1147)⫽110.5 bhp (82.4kW) This is the hp input required to drive the fan and is close to the 113.2 hp(84.4 kW) computed in the previous calculation procedure The actual motor hpwould be the same in each case because a standard-size motor would be chosen.The difference of 113.2 ⫺ 110.5 ⫽ 2.7 hp (2.0 kW) results from the assumedefficiencies that depart from the actual values Also, a sea-level attitude was as-sumed in the previous calculation procedure However, the two methods used showhow accurately fan capacity and hp input can be estimated by judicious evaluation

of variables

4. Compute the correction factors for the induced-draft fan

The flue-gas density is 0.045 lb / ft3(0.72 kg / m3) at 385⬚F (196.1⬚C) Interpolate inthe temperature correction-factor table because a value of 385⬚F (196.1⬚C) is nottabulated Find the correction factor for 285⬚F (196.1⬚C) thus: [(Actual temperature

⫺ lower temperature) / (higher temperature⫺ lower temperature)] ⫻ (higher perature correction factor⫺ lower temperature correction factor)⫹lower temper-ature correction factor Or, [(385⫺375) / (400⫺375)](1.273⫺1.255)⫹1.255⫽1.262

tem-The altitude correction factor is 1.095 for an elevation of 5000 ft (1524.0 m),

as shown in Table 4

As for the forced-draft fan, CCF⫽ C T C A⫽(1.262)(1.095)⫽ 1.3819 Use theCCF to find the capacity-table ft3/ min in the same manner as for the forced-draftfan Or, capacity-table ft3/ min ⫽ (given ft3/ min) / CCF ⫽ 152,500 / 1.3819 ⫽110,355 ft3/ min (52.1 m3/ s)

5. Choose the fan size from the capacity table

Check the capacity table to be sure that it lists fans suitable for induced-draft(elevated-temperature) service Turn to the 11-inH2O (2.7-kPa) static-pressure ca-pacity table, and find a capacity equal to 110,355 ft3/ min (52.1 m3/ s) In the en-gineering data used for this fan, the nearest capacity at 11-inH2O (2.7-kPa) staticpressure is 110,467 ft3/ min (52.1 m3/ s), with an outlet velocity of 4400 ft / min(22.4 m / s), an outlet velocity pressure of 1.210 inH2O (0.30 kPa), a speed of 1222

r / min, and an input hp of 255.5 bhp (190.5 kW) The tabulation of these quantities

is of the same form as that given for the forced-draft fan, step 2 The selectedcapacity of 110,467 ft3/ min (52.1 m3/ s) is entirely satisfactory because it is only110,467 ⫺ 110,355 / 110,355 ⫽ 0.00101, to 0.1 percent, higher than the desiredcapacity

6. Compute the fan speed and power input

Multiply the capacity-table rpm and brake hp by the CCF to determine the actualrpm and brake hp Thus, the actual rpm ⫽(1222)(1.3819)⫽ 1690 r / min Actualbrake hp⫽(255.5)(1.3819)⫽353.5 bhp (263.6 kW) This is the hp input required

to drive the fan and is close to the 360.5 hp (268.8 kW) computed in the previouscalculation procedure The actual motor horsepower would be the same in eachcase because a standard-size motor would be chosen The difference in hp of 360.5

⫺353.5 ⫽7.0 hp (5.2 kW) results from the same factors discussed in step 3

Note: The static pressure is normally used in most fan-selection procedures

be-cause this pressure value is used in computing pressure and draft losses in boilers,economizers, air heaters, and ducts In any fan system, the total air pressure⫽staticpressure ⫹velocity pressure However, the velocity pressure at the fan discharge

is not considered in draft calculations unless there are factors requiring its tion These requirements are generally related to pressure losses in the fan-controldevices

evalua-Related Calculations. Use the fan-capacity table to obtain these additional tails of the fan: outlet inside dimensions (length and width), fan-wheel diameterand circumference, fan maximum bhp, inlet area, fan-wheel peripheral velocity,NAFM fan class, and fan arrangement Use the engineering data containing the fan-capacity table to find the fan dimensions, rotation and discharge designations, ship-ping weight, and, for some manufacturers, prices

de-FAN ANALYSIS AT VARYING RPM, PRESSURE,

AND AIR OR GAS CAPACITY

A fan delivers 12,000 ft3/ min (339.6 m3/ min) at a static pressure of 1 in (0.39 cm)

WG at 70⬚F (21.1⬚C) when operating at 400 r / min; required power input is 4 hp

(2.98 kW) (a) If in the same installation, 15,000 ft3/ min (424.5 m3/ min) are

re-quired, what will be the new fan speed, static pressure, and power input? (b) If the

air temperature is increased to 200⬚F (93.3⬚C) and the fan speed remains at 400 r/ min, what will be the new static pressure and power input with a flow rate of12,000 ft3/ min (339.5 m3/ min)? (c) If the speed of the fan is increased to deliver

1 in (0.39 cm) WG at 200⬚F (93.3⬚C), what will be the new speed, capacity, and

power input? (d ) If the speed of the fan is increased so as to deliver the same

weight of air at 200⬚F (93.3⬚C) as at 70⬚F (21.1⬚C), what will be the new speed,capacity, static pressure, and power input?

higher capacity⫽400(15,000 / 12,000)⫽500 r / min Hence, the fan speed must be

increased by 25 percent, i.e., 100 r / min—to have the fan handle 25 percent more

air This verifies the first fan law that capacity varies directly as fan speed

Use the second fan law to determine the new static pressure This law states:

Fan pressure (static, velocity, and total) varies as the square of the fan speed Thus,

the new static pressure with the larger flow rate⫽1(500 / 400)2⫽1.5625 in (3.97cm)

Find the new required power input at the higher flow rate and higher discharge

pressure by using the third fan law, which states: Power demand of a fan varies as

kW)

2. Determine the new static pressure and power

(b) When the density of air or gas handled by a fan changes, three other fan laws apply The first of these laws is: At constant fan speed; i.e., rpm, and capacity; i.e.,

is 0.075 lb / ft3(1.2 kg / m3); at 200⬚F (93.3⬚C) the air density is 0.06018 lb / ft3(0.963

kg / m3) Then, new static pressure⫽1.0(0.06018 / 0.075)⫽0.80 in (2.04 cm) Thenew power is found from 4(0.06018 / 0.075)⫽3.21 hp (2.39 kW)

3. Find speed, capacity, and power input at the new pressure

(c) We now have a constant-pressure output Under these conditions, with a varying air or gas density, the fan law states that: At constant pressure the speed, capacity,

400 (0.075 / 0.06018)0.5⫽446.5 r / min

The new capacity at the 1-in (0.39-cm) static pressure ⫽ 12,000(0.075 /0.06018)0.5 ⫽ 13,396 r / min at 200⬚F (93.3⬚C) The new power ⫽ 4 (0.075 /0.06018)0.5⫽ 4.46 hp (3.33 kW)

4. Compute the new speed, capacity, static pressure, and power at the

The new capacity⫽12,000(0.075 / 0.06018)⫽14,955 ft3/ min (423.5 m3/ min).Likewise, the new static pressure ⫽ 1.0(0.075 / 0.06018) ⫽ 1.246 in (3.17 cm).Finally, the new power⫽4 (0.075 / 0.06018)2⫽6.21 hp (4.63 kW).

Related Calculations. The fan laws, as given here, are powerful in the analysis

of the speed, capacity, and pressure of any fan handling air or gases These lawscan be used for fans employed in air conditioning, ventilation, forced and induced

draft, kitchen and hood exhausts, etc The fan can be used in stationary, mobile,

marine, aircraft, and similar applications The fan laws apply equally well

BOILER FORCED-DRAFT FAN HORSEPOWER

DETERMINATION

Find the motor turbine hp needed to provide forced-draft service to a boiler thatburns coal at a rate of 10 tons (9080 kg) / h The boiler requires 59,000 ft3/ min(5481 cum / min) of air under 6 in (15.2 cm) water gage (WG) from the fan whichhas a mechanical efficiency of 60 percent The air is delivered at a total pressure

of 6 in (15.2-cm) WG by the fan What would be the effect on the required power

to this fan if the total pressure were doubled to 12 in (30.5 cm) WG? If the requiredair delivery was increased to 75,000 ft3/ min (2123 m3/ min), at 6 in (15.2 cm) WG,what input hp would be required?

Calculation Procedure:

1. Find the required power input to the fan

Use the relation, fan hp⫽ft3/ min(total pressure developed, lb / ft2) / 33,000(fan ficiency) To apply this equation we must convert the water gage pressure to lb / ft2

ef-by (in WG / 12)(62.4 lb / ft3water density) Or (6 / 12)(62.4) 31.2 lb / ft2(1.49 kPa).Substituting, hp⫽ 59,000(31.2) / 33,000(0.60)⫽ 92.96 hp (69.4 kW) Use a 100-

hp (75 kW) motor or turbine to drive this induced-draft fan

2. Determine the required power input at the higher delivery pressure

Use the same relation as in Step 1 to find, hp⫽59,000(62.4) / 33,000(0.60)⫽185.9

hp (138.7 kW) Thus, the required power input doubles as the developed pressuredoubles

The sharp increase in the power input is a graphic example of why the pressurerequirements for any type of fan must be carefully analyzed before the final choice

is made Since the cost of a fan does not rise in direct proportion to its deliverypressure, the engineer should apply a factor of safety to the computed power input

to take care of possible future overloads

3. Find the required power input at the higher flow rate

Using the same relation, hp ⫽118.2 (88.2 kW) Again, the required power inputrises as the output from the fan is increased This further illustrates the strong need

to explore the maximum output requirements before making a final equipmentchoice

Related Calculations. This approach can be used for any fan used in plant, HVAC, and similar applications The key point to observe is the rise in powerrequirements as the fan pressure of air volume delivered increases

power-EFFECT OF BOILER RELOCATION ON DRAFT

Calculation Procedure:

1. Determine the new inlet condition for the fan

When a fan is required to handle air or gas at conditions other than standard, acorrection must be made in the static pressure and hp (kW) Since a fan is essen-tially a constant-volume machine, the ft3/ min (m3/ min) delivered will not changematerially if the speed and system configuration do not change, regardless of theair or gas density

The static pressure, however, changes directly with density Hence, the staticpressure must be carefully calculated for specified conditions For the situationdescribed here, assume a gas molecular weight of 28, a typical value The densitycorrection factor can be computed from the ratio of the new-location barometricpressure to the first-location barometric pressure, both expressed inHg (cmHg) Or,density correction factor⫽24 in / 30 in⫽ 0.80

There is no temperature correction factor because the air temperature remainsthe same Therefore, at the new elevated location of the boiler the intake conditionfor the fan will be (10 in WG)(0.8 correction factor)⫽ 8 in (20.3 cm) WG

2. Compute the new volume condition

Use the relation Volume flow⫽(lb / h)(molecular weight of gas)(cfm at inlet ditions)[(absolute temperature of flue gas) / (‘‘standard’’ air temperature of 60⬚F inabsolute terms)][(reduced barometric pressure) / (reduced barometric pressure⫺newsuction condition)] Substituting, new volume condition⫽(600,000 / 28)(379)[(460

con-⫹290) / (60⫹460)][(24) / (24⫺8)]⫽ 17.57⫻106ft3/ h (0.497 m3/ h⫻ 106)

Related Calculations. With used power-plant equipment becoming more ular throughout the world (see the classified section of any major engineering mag-azine) it is important that the engineer be able to determine the performance of re-used equipment at different locations

pop-ANALYSIS OF BOILER AIR DUCTS AND

GAS UPTAKES

Three oil-fired boilers are supplied air through the breeching shown in Fig 7a.

Each boiler will burn 13,600 lb / h (1.71 kg / s) of fuel oil at full load The draft loss

FIGURE 7 (a) Boiler intake-air duct; (b) boiler uptake ducts.

through each boiler is 8 inH2O (2.0 kPa) Uptakes from the three boilers are

con-nected as shown in Fig 7b Determine the draft loss through the entire system if a

50-ft (15.2-m) high metal stack is used and the gas temperature at the stack inlet

is 400⬚F (204.4⬚C)

Calculation Procedure:

1. Determine the airflow through the breeching

Compute the airflow required, cubic feet per pound of oil burned, using the methodsgiven in earlier calculation procedures For this installation, assume that the com-bustion calculation shows that 250 ft3/ lb (15.6 m3/ kg) of oil burned is required.Then the total airflow required ⫽ (number of boilers)(lb / h oil burned perboiler)(ft3/ lb oil) / (60 min / h) ⫽ (3)(13,600)(250) / 60 ⫽ 170,000 ft3/ min (80.2

m3/ s)

2. Select the dimensions for each length of breeching duct

With the airflow rate of 170,000 ft3/ min (80.2 m3/ s) known, the duct area can be

determined by assuming an air velocity and computing the duct area A dft2from

A d⫽(airflow rate, ft3/ min) / (air velocity, ft / min) Once the area is known, the ductcan be sized to give this area Thus, if 9 ft2(0.8 m2) is the required duct area, aduct 3⫻3 ft (0.9⫻0.9 m) or 2⫻4.5 ft (0.6⫻1.4 m) would provide the requiredarea

In the usual power plant, the room available for ducts limits the maximumallowable duct size So the designer must try to fit a duct of the required area intothe available space This is done by changing the duct height and width until aduct of suitable area fitting the available space is found If the duct area is reducedbelow that required, compute the actual air velocity to determine whether it exceedsrecommended limits

In this power plant, the space available in the open area between A and C, Fig.

7, is a square 11 ⫻ 11 ft (3.4 ⫻ 3.4 m) By allowing a 3-in (7.6-cm) clearancearound the outside of the duct and using a square duct, its dimensions would be10.5 ⫻10.5 ft (3.2 ⫻ 3.2 m), or a cross-sectional area of (10.5)(10.5)⫽ 110 ft2(10.2 m2), closely With 170,000 ft3/ min (80.2 m3/ s) flowing through the duct, theair velocityvft / min⫽ft3/ min / A d⫽170,000 / 110⫽1545 ft / min (7.8 m / s) This

is a satisfactory air velocity because the usual plant air system velocity is 1200 to

3600 ft / min (6.1 to 18.3 m / s)

Between C and D the open area in this power plant is 10 ft 9 in (3.3 m) by 14

ft (4.3 m) Using the same 3-in (7.6-cm) clearance all around the duct, we find the

dimensions of the vertical duct CD are 10.25⫻ 13 ft (3.1 ⫻ 4.0 m), or a sectional area of 10.25⫻13 ⫽133 ft2(12.5 m2), closely The air velocity in thissection of the duct is v ⫽ 170,000 / 133 ⫽ 1275 ft / min (6.5 m / s) Since it isdesirable to maintain, if possible, a constant velocity in all sections of the ductwhere space permits, the size of this duct might be changed so it equals that of

be high because the limited space available would require alteration of the

power-plant structure Also, the velocity is section CD is above the usual minimum value

of 1200 ft / min (6.1 m / s) For these reasons, the duct will be installed in the 10.25

⫻13 ft (3.1⫻4.0 m) size

Between E and F the vertical distance available for installation of the duct is

3.5 ft (1.1 m), and the horizontal distance is 8.5 ft (2.6 m) Using the same 3-in(7.6-cm) clearance as before gives a 3⫻8 ft (0.9⫻2.4 m) duct size, or a cross-sectional area of (3)(8)⫽24 ft2(2.2 m2) At E the duct divides into three equal-

size branches, one for each boiler, and the same area, 24 ft2(2.2 m2), is availablefor each branch duct The flow in any branch duct is then 170,000 / 3 ⫽ 56,700

ft3/ min (26.8 m3/ s) The velocity in any of the three equal branches is v ⫽56,700 / 24⫽2360 ft / min (12.0 m / s) When a duct system has two or more equal-size branches, compute the pressure loss in one branch only because the losses in

the other branches will be the same The velocity in branch EF is acceptable cause it is within the limits normally used in power-plant practice At F the air

be-enters a large plenum chamber, and its velocity becomes negligible because of thelarge flow area The boiler forced-draft fan intakes are connected to the plenumchamber Each of the three ducts feeds into the plenum chamber

3. Compute the pressure loss in each duct section

Begin the pressure-loss calculations at the system inlet, point A, and work through

each section to the stack outlet This procedure reduces the possibility of error and