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T

2nd Edition By the Editors of POWER magazine

platts 2005 printing

Copyright ©2005 Platts, a Division of The McGraw-Hill Companies. Inc. All rights reserved.

Table of Contents PART L FUNDAMENTALS KEY MEASUREMENTS



1

Symbols, exponents Units of power, work, energy heat Units of volume, area, length Other engineering units

PRACTICAL MATHEMATICS

3

8

Work, force, torque, power Vectors explained Pulleys, gears, levers. linkages Efficiency

STRENGTH OF MATERIALS

10

12

Commercial pipe dimensions Pipe stresses and expansion Water-, air-, and steam-flow charts Friction loss in pipe fittings Figuring tank capacity

13

BASIC WATER CHEMISTRY

How burning starts Combustion, mixing of gases Burning hydrocarbons

FUNDAMENTALS OF HEAT Temperature, heat transfer Heat and specific heat Expansion of solids, liquids Change of state: solid, liquid, gaseous Evaporation, cooling-tower basics

15

22

29

Facts about fluids Water in motion Water power, flow through orifices Pumps and pumping

PIPING, FLOW & TANKS

Water vapor, relative humidity, dewpoint Latent vs sensible heat, using the chart

COMBUSTION THEORY

BASIC ELECTRICITY What electricity is Electrical circuit components Direct-, alternating-current circuits Three-phase circuits, figuring power Finding wire, fuse, conduit sizes Transformers, rectifiers, capacitors

HYDRAULICS & PUMPING

Tension, compression, shear Stress and stretching Beam sizes and formulas

PSYCHROMETRICS

19

Measuring steam Enthalpy defined Saturation pressures, temperatures Superheated steam

Handling formulas Useful geometric facts Trigonometry Areas and volumes Using pocket calculators

MECHANICS & MECHANISMS

STEAM TABLES SIMPLIFIED

33

37

Defining key terms Building-block approach Important elements, acids, acid radicals Understanding cations and anions

NUCLEAR ENERGY Splitting the atom Harnessing nuclear fission Fuels and reactors Critical mass, reactivity control

40

PART II. APPLICATIONS FEEDWATER TREATMENT

1

43

50 HEAT-TRANSFER EQUIPMENT

INSULATION 56

Gaging condenser performance Figuring boiler efficiency Problems and solutions

STEAM TURBINES

60

63

GAS TURBINES

64

OIL AND GAS ENGINES Key elements Basic engine types Fuel-rate curves

66

80

Tips on selecting motors Three motor types Motor characteristics, load requirements Synchronous vs induction motors DC motors, generators Improving power factor Plant system availability

ELECTRIC DISTRIBUTION

Key turbine types Basic cycles and equations Problems and solutions

78

How lubricants work Commercial lube-oil properties Lubricant additives

ELECTRICAL EQUIPMENT

Types of units Selecting hydro units

76

Thermal conductivity, emissivity High-, low-temperature insulation Refractory insulation

LUBRICATION

Basic turbine cycles Calculating turbine performance Heat rate, engine vs thermal efficiency Mollier chart, turbine curves

HYDRAULIC TURBINES

75

Figuring "LMTD'• Tube metals selection

Chemistry of combustion Combustion reactions Evaluating, classifying fuels Fuel analyses, ultimate analysis Short-cut calculations

STEAM GENERATION

68

Air-compressor basics Understanding compressor performance How fans work Making density corrections Nonboiler fan applications Fan laws

The four basic jobs Sources and choices External treatments Internal treatment, chemicals used Boiler blowdown Two key examples

COMBUSTION & FUELS

COMPRESSORS & FANS

87

Low-voItage networks Primary, secondary substations In-plant circuitry

GLOSSARY

89

INDEX

90



PART 1. FUNDAMENTALS

KEY MEASUREMENTS TIIESE tabla arc drawn up in thc units

SYMBOLS USED IN TABLES Sr mido Unit Symbol

EngIIsh Symbol Una

Symbol Unit Á bbl Btu cal

angstrom barre' British thermal unit cabrio

in. kW kWh

cu

cubic

gal

foot gallon gran

mi mM

gr Mg

H,0 hp hr

oz

inch kilowatt k ilow a tthour pound mile minute micron ounce

qt sec sq yd

quarl second square yard

lb

mercury water horsepower hour

h J

hour Joule

kg k Pa

kilOgrarll

L

litre metre second watt

kitopascai

m s W

k

Pretina thousand

M

milhon

p m

MilliOfith

thOuSandth

• An ebbevlation tor "Le Sysleme Inlemehooel Unnes: a workAmcle ~no meammmeM system imwerds whoen mosi countnes are greAtating. NOTE SI convession lactar. nol in tablee Tenmeralios 'C - l e 321,1 8 p ounos o! steam per hput 2,7r 1100'el m rt - 0 284 s'A'

UNITS OF WORK, ENERGY, HEAT moño 1 0.324

252 8.60x 10'

81u/sq B 7;,- SI >233 J. kg -

Slu

ft-b 3.09

1.10x 10' 3.76x 10-1 2.93x 10 ' 1

3.97x 10" 1.28x10-'

778 2.66x 10,

3413

4 19J 1 36J 1.06 1/J 3.60 MJ

1

1

T.,/ 4 kJ..m'

that engincers in power and process use most frequently. For each unit, usually no more Iban threc significant figures are included, which is accurate enough for powerplant calculations. (Significan) figures are the total digits in a numbcr, exccpt zcroes before and alter, regardIcss uf decimal point. Both 102,000 and 0.00102 have threc %ignilieant figures.) The febles are easy lo use: Merely select unit ol interest at top ol column, then move down to horizontal row in which "1" apocare; all other numbera in row are conversion factors to unit shown at top of coturno. Shaded columne show metric conversion factor* for units at "1" positions. As an example. in Units of Power

table, to lind Btu/hr in I horscpower, select -hp" column. movc to "1" pasi(ion, scan right tu "Blu/hr" column, read 2540. To convert 5 hp (for example) lo Btu/hr, multiply 5 x 2540 = 12,700. An explanation of symbols in the tabla is givcn at kft. Rcmcmber. the SI metric system distinguishes between mass and force; kg, kilogram, is a mass

UNITS OF POWER SI kW

SO 1. 34

1 1 0.746 t.82x 10' ' 1.36x 10 , 2 26x 10- • 3.03x 10 , 5.05x IV 3.77x 10 , 142 1.055 1.76x 10 ' 0.236 2.93x10- • 3.93x10-'

thlb/mc

11.4 /mIn

738 550

44,200 33.000 1 60 1 1.67x10-, 2.78 x 10 4 1.67 x 10' 46,700 778 778 12.98 0.216 13

I1-10/hr

2.65x10' 1.98 x 10' 3600 60 1 2.8 x 10' 46,700 778

Btu/sec Btu/min 56.9 0948 42.4 0 707 7.71x10' 1.28x10-, 1.29x10' 2.14x104 2.14x10'` 3.57 x 10- , 1 60 1.67 x 10 4 1 0.167 2.78 x ir

Stuas

3410 2540 4.61 7.71x10_. 1.29x 10 ' 3600

(wo114) 1000 746 /.36

60 1

2.26x 10 3.77 x101055 17.6 0.293

ce-11 1.33 x 10 ' 2.30x 10 6.20x 10 : 2.40x 10 , 7.67x 10 3.07 x 10 ' 1.29x 10 ' 1

SI 1.64 ml 28.3 L 0.765 m' 29.6 ml 0.9461. 3.79 1. 159 1. 1230 in,

UNITS OF VOLUME (INCLUDES CAPACITY) briol

cu In 1 1728 4.67x10' 1.8 57.7 231 9700 7.52x10'

POMA IlanObC0k

cm ft 5.79x10 ' 1 17 1.04 x 10 , 334/00 ' 0./34 5.61 4.36x10'

ou yd 2.14 x 10-1 3.70x 10' 1 3.86x 10 • 1.24 x 10 , 4.95x 10 , 0.208 1610

oz

0.554 957 2.59 x 10• 1 32 128 5370

4.17x10'

61 1.73x10-' 29.93 808 3.10 x 10 1 1 4 168 1 30x 10'

gol 4.33x 10 , 7.48 202 7.81x104 0.25 1 42 3.26x10'

02 960 1.03 x 10- • 0 177 4.81 186x10' 5.94x 10- , 2.38x10' 1 7760



UNITS OF PRESSURE mm Mg I 25.3 187 22.4 3.23 51.7 0.359

in. Hg 0 019 1 0 0735

1

ft 11,0 0.045 / 13 0.0833

12 .73 7.7 .192

0.144 2.31 0.016

14,0 '.536 3.6

0.882 0 128 2.04 0.0141

1

16

1

lb/sq ft 2.75 70.7 5.20 61 4 9 144

0.111

0.00694

1

lb/sq in. (914 0 0193 1491 0.0361 0 433 0.0625

orín in.

0.309 786 0.577 693 1



SI (k9a) 0 133

3.38 O 249

299 043/ 6.89 0.0479

UNITS OF TIME sec MM nr 1 1.67x10' 2 78x 10 • 60 1 1.67x10' 60 3600 1 1440 8.64 x 10' 24 1.01x 10' 6.05 X 10' 168 2.63 x 10, 4 313x 10' 730 31.5x 10' 5.26x 10' 8760 _.NOTE: SI mees ceo usape ol Ihe minio descat•agno e% sane

dm wssIt 1.16x 10- , 1.65x10'• 694x10• 9.92x 10.> 0.417 5.90x10" 1 0 143 7 1 30.4 4.34 315 57 1

menlh 381x10 2.28x 10 > 1 37 x 10 ' 0.329 0.230 1 12

mar

3 17x 10 I 1.90x10• 1.14x 10 ' 2.74 x 10> 0.0192 0.0833 1

une is

UNITS OF VELOCITY in./ me 1 12 0.20 3,30 x 10 ' 1056 17.61

I

11/m1n 9/1w ml/ mm ft/mc muhr (mph) 8.33 X 10-' 5 568X10 , 300 9.47 x 10 ' 1 60 3600 1.14 x 10- , 0.682 1.67 x 10- , 1 60 1.89x10• 1, /4% 10-/ 2,78%10' 1.67% 10 1 1 3.16 x 10 • 1.89x10-' 88 5280 3.17x 10, 1 60 1.47 1 5280 1.67x 10- , 1 88

SI (m/') 2.54 x 10 , 0.305 5.08 X /0// 8.47x 10-' 26 82 0.441:'...;f4,

UNITS OF LENGTH micros Inch ysrd nide Ingstrom tont l - 1 1 00 x 10 ' - 10.000 1 3.94 x 10-1 3.28 x 10 .4 1.09 x 10-4 2.54 x 10' 1 1.58 x 10 2.54 x 10' 8.33 x 10-, 2.78 x 10' 1.89 x 10 ' 3.05 x 10, 3.05 X 10" 12 1 0.333 9.14 x 10* 9.14 x 106 36 3 1 5.68 x 10 • 1 1.61x10" 1.81x10 5280 1760 6.34 X 10

_ sq in.

• SI (m) 1.00x 10

7.54x10' 0.305 0.914 1609

• _•.

UNITS OF AREA

acre SI (4/1') sq ft 1141 yd 1 6 94 X 10/ 1 7.72 X 10 ' 1.6 X 10- ' 6 45 x 10 ' 144 1 0.111 2.3x 10- 4 00979 1296 9 1 2.07 X 10-1 0 836 6.27X 10• 4.38X 10' 4047 4840 1 dr Iceculrol md le Te toa el a arch, oí 1/10001 in dlamtle . t - 14. .nini u-41 n! 1

cir mil • sq in. 0./85 x 10-• 1 1.27x 10, 1

UNITS OF DENSITY yr /sysl lb/gsl 1 1.43x 10' 7000 1 62 x 104 231 935 0.134 1 86 x 10' 267 9.90 68,900

lb/cu in. lb/cu ft 0.619x 10 • 1.07 x 10 ' 4.33 x 10- > 7.48 1 1728 1 5.79 x 10 ' 1.16 2000 0.0429 74.1

ton/ cu ft 5.37 x 10 ' 3 74 x 10 ' 0.864 5.0x10-' 1 0.037

Ion/cu yd 1.45 x /O • 0.101 23.3 0.0135 27 1

SI (kg/Ma) ¡

0.0171 120.0 2 77x 10' 160 3.20 x 10' 1187 POWCI

. Handbook

UNITS OF VOLUME-FLOW RATES eu ft /mc

cu fi irnin(cfm)

cu ft /hr

gal/aec

gal , rnm (gpm)

1 0.0167 2.78 x 10 1 0.134 2.23x 10- 3 3.71x10- 1

60

3600

7.48

449

1 0 0167 8.02 0.134 7 73 x 10 '

60

0 125 2.08x 10 ' 1 0.0167 2 78x 10 '

7.48 0 125 60 1

gal/hr 2.69x 10' 449 7 48 3600 60

e 01e7

/

1

481 8.02 0 114

EXPONENTS USED

UNITS OF WEIGHT 9 r

oz•

Ib'

1 2.29 x 10 1 1.43x 10 • 437.5 1 0.0625 1 7000 16 14 x 10• 3.2x10' 2000

toa' 7.18x 10- 1 3.22X10-'

6.0x10- • 1_

IN TABLES

( 8 9) ' 6.48 X 10-1

0 0283 0.454 907

• Avdirdupole oz and lo. bbor Ion 01 2000 lo

unit whilc N. newton, is the force unit. Also. in SI, usagc of the minare unit is discouraged for technical calculations. Explanation of exponents is shown abo y e. Examples are: 2,660,000 = 2.66 X 0.000293 - 2.93 x 10'. Following are severa] ways to apply the Cables: EXAMPLE 1. How much heat per hour is generated by a brakc applicd to drag 40 Ib on che rim of a pulley moving at 2000 rumio? SOLUTION. Work done in (ool-pounds per minute is 40 X 2000 = 80,000. Units (f Power shows that I ft-lb/min = 0.0771 Btu/hr. llourlv !wat generated, thercfore. is 80.000 X 0.0771 = 6168, which can be rounded off to 6170 Rtu. EXAMPLE 2. A 19-oz wcight on a 50-sq in. soft diaphragm balances air pressure on the other side of che diaphragm. What is the air pressure in inches of

/31(L /e) 28.3 0.472 7.87x 10-1 3.79 0.0631 1.05x 10-,

10' 10 ,

= 1.000.000 = 100.000 101 = 10 000 10' - 1000 10( = 100

water? In SI pressure units? SOLUTION. Pressure is 19/50 = 0.38 oz/sq in. Units of Pressure shows that 1 oz/sq in. = 1.73 in. H,0. So pressure = 0.38 X 1.73 = 0.66 in. 1-1 20. From the same cable. multiply 0.66 x 0.249 = 0.164 k Pa. EXAMPLE 3. What kilowatt power is required (neglecting losscs) to hoist 1500 lb at the vate of 20 ft/scc? SOLUTION. Foot-pounds of work per seeond are 20 x 1500 = 30,000. From Units of Power, 1 ft-Ib/seo = 0.00136 kW. Therefore, theoretical power = 30.000 x 0.00136 = 40.8 kW. EXAMPLE 4. A lcaky valve tests une gallon of wastc in fivc minutos? What is the vearly waste? SOLUTION. Gallons per minuto is 0.2. Unis«sf Tinte shows 5.26 X 10' min/yr. Thus, the yearly water loss is 526,000 X

10'

0.000001

10 - * =

0.00001 10 -• - 0.0001 10- 1 = 0.001 10 - ' = 0.01

0.20 = 105.200, or 105,000 gal. EXAMPLE 5. A rectangular tank nteasuring 15 X 12 X 8 ft is brin) full of water. How long will it cake to drain at the rato of 40 gpm? SOLUTION. Volumc of water = 15 X 12 x 8 = 1440 cu ft. L'nirsof Volante shows 7.48 gal/cu ft. Volumc in gallons = 1440 x 7.48 = 10,770 gal. At 40 gpm, time needed to drain the tank = 10,770/40 = 269 min. or 41/2 hr. EXAMPLE 6. A piston is 10 in. in diameter. llow much volume in cubic leer does it displace if it has a I 4-in. stroke? In cubie mores? SOLUTION. Arca of the piston is 1(10/2)' = 78.5 sq in., or 0.545 sq ft. Piston stroke is 14/12 = 1.17 ft. Volumc displaced is 0.545 x 1.17 = 0.64 cu ft. From Units of Volusne. I cu ft = 0.0283 m'. so 0.64 X 0.0283 = 0.018 nt'.

PRACTICAL MATHEMATICS HANDLING FORMULAS Most rulos for engincering ealculations arc set down in formulas, which are nothi ng more Man mathematical shorthand. usually taking the forro of cquations. lIcre arc somc functions you can perform with equation. Remember that cach leiter or symbol in an equation stands (or a quantity which. likc weights on a sea le. must balance out. ADDITION, SUBTRACTION. If cquals are addcd lo equals, die results are equal. II. a = b and e = d. then a + e = b + d. or a-e=b-d. Quantities may be changed fmm one side of an equation to the other by changing "sign. - Thus, if a + b = e. thcn a = e - b. Power HandboOk



equals are multiplica( by equals. the results are equal. If a = h ande = d, then a x e = b X d. or ale = bid. Quantities may be changcd from onc side of the equation to the other by multiplying or dividing bolh sides by the quantity. I f a/h = e. then h x (a/h) = h x (e), ora-bxe. PARENTHESES. A quantity outside a parcnthcsis multiplics all quantitics inside. Thus. !Mb + e - cf) - ah + oc ad. (Note: a x b = ah.) POWERS, ROOTS. Both sides of an equa'ion may be raised to any powcr. If a = Is, thcn a' = = b. Also, if a' = b. then a = p/b. MULTIPLICATION. DIVISION.

for proper units or dimensions in equation. If IV = f x d Bul and f is lb and d isfr, thcn W constants can have dimensions, toa. If W = 12 X f X d, then IV is in.-lb because 12 most likely stands for 12 in./ft. Most formulas are writlen for specific units. LOGARITHMS. A logarithm is an exponent. The equation V = 8 expresscs a relationship among the !lamber 8, the batel and the exponen! 3. Stated another way: 3 is thc logaralun of the number = p. then by 8 tu the base 2. Thus, if definition x is che logarithm of p lo the p. lIcre's how to use base b. or x logarithms: UNITS. Watch out

3



Multiplica:ion. To multiply two num bers, look up thcir "logs" in a table cf common logarithms. Add thc logs, thc n find thc number corresponding to ths logarithm in thc tablc. It is the produ it of thc two numbcrs. Division is simila .

oxeen' the log of the dcnominator is subtrarted from the log of Ihe numerator. Thc number corresponding lo this lagarhin is thc desired quoticnt. Raising a number ro a poner. To do this, look up thc lag of the number in a tabla.

Multiply this log by thc cxponcnt of thc powcr. Thcn. using the tablc, find thc desired power of the given number. Finding a root is similar, except thc log of the number is divided by the index of the resol

USEFUL GEOMETRIC FACTS A, + 8, - 180". A 2 + B2 ,4 180'

VERTICAL ANGLES. When straight fines cross. opposite Or "vertical" anglos are equal. Adjacent anglas add up to 180 deg

.43r,

A

A2

B, = 132

A 191

82 A l Al

/12 =- A,

A2

13, -

- B3 n

82 B4

A3 133 84 A.

PARALLEL LINES. When straight line crosses parallel fines, indicated anglos are equal

BISECTING ANCLE. Swing arc from point O to give points A. Swing equal ares from points A. Draw line from O through intersection at 8

O

ERECTING PERPENDICULAR. From poto, O swing ares to give points A. From points A swing two equal ares. Draw line through intersections

A

Equal &vana,

A + 8 + C - 180' INTERIOR ANCLES. The sum of the interior anglos of any triangle always equals 180 deg

49 5 SO in ENLARGING AREAS. As size increases. the arca increases as the square of any dimension

5 in

7.5/ 5 / 5 x 1.5 2 25 x 22

C- 3.1416x()

8,

CIRCUMFERENCE OF CIRCLE. Circumference, or line bounding circle, equals 3.1416 x diameter. 6.2832 X radius

n

ENLARGING VOLUMES. Both volume and weight (if same material) vary as dimension cubed

82

B3 SUBTENDED ANCLES. Angle subtended by arc al center of circle is twice anglo subtended by arc lo points anywhere on circle's circumference

15 2 25 49 5

b

23.4 lb

5/4 a 1.25 1.25x 1.25 x 1.25 x. 1.95 23.4 lb 195 x 12

RIGHT TRIANGLES. The square of the hypotenuse C equals the sum of Ihe squares of the two legs A and 8

ANCLES IN SEMICIRCLES. Any anglo drawn in a semicircle from points A and 8 will always be 90 deg

DIVIDING LINE TO SCALE. To divide line AB into equal segments, draw line from A at convenient anglo and mark off equal segments lo convenient measure, creating line AC. Connect 8 to C. Draw fines parallel to that is. 5-5, 4-4, 3-3, etc .1

POwer HandDOOk

TRIGONOMETRY The sidcs and anglos of a right triangle bcar definite rclationships to cach other. Thcsc are callcd sinos, cosines. and tantenis of the angla (seo sketch). By using these functions, it's possiblc to figure the Iengths of all sides and the degrees of both anglos in a right triangle whcn only one anglo and one sido are known. The values of sinos, cosines, and tangcnts for anglos from 0 to 90 deg are listed bclow. Following are examples of ways w uso these values. The bcst approach to solving "trig" problems is te draw a simple sketch and label all known dimensions. EXAMPLE 1. A pipo is pitchcd downward 5 deg from horizontal. How far does it fall per 10 ft of length? SOLUTION. Use formula a = e X sin A. From the Cable, sin 5 deg is 0.0872. The pipo falls 10 x 0.0872 = 0.872 ft. EXAMPLE 2. A 6-ft ladder leaning against a wall is 3 ft out at the bottom. What anglo does the ladder make with the floor? SOLUTION. Use formula cos A = Cos A = 3/6 = 0.5. In the Cable, the anglo whosc cosinc is 0.5 is 60 deg. .EXAMPLE 3. The front wall of a room 20-ft long is 2 ft higher (han the back

Sin A = ate a = c x Sin A c a/Sin A Cos A= bíc b= c x Cos A = b/Cos A Tan A= a/b a = b x Tan A b= a/Tan A wall. What anglo does the ceiling make

with the from wall'? SOLUTION. Use formula tan A = a/b. Tan A = 20/2 = 10. Nearcst tangcnt to 10 in the tablc is 9.514, or 84 deg. EXAMPLE 4. Two cylindrical tanks are 10-ft apart on ccnters. How long must a drain connection be from the centerline of either tank to a discharge point bclow, assuming the anglo to either centerlinc is 20 deg? SOLUTION. The anglo of interest is 90 20 = 70 deg. Use formula r = bicos A. From the Labio, e = 5/cos 70 = 5/0.3420 = 14.6 ft. HELPFUL HINTS. The tablc below lists values only for whole degrees. For tables

SINES, COSINES, TANGENTS Den

Sin

Tan

Coa

Don

O 1

0.0000

0.0000

1.00130

3 4

0.0175 0 . 0349 0.0523 00699

0.0175 0 . 0349 0.0524 00699

0 9998 0.9994 09986 0.9976

30 31 32 33 34

Sin 0 5000 0 5150 0.5299 0.5446 0.5592

Tan 0.5774 0.6009 0.6249 0 . 6494 0 . 6745

5 6 7 8 9

0.0872 0.1045 01219 0 1392 O 1564

00875 0.1061 O /228 0 1405 0 1584

0.9962 0.9945 0.9925 0.9903 09877

35 36 37 38 39

0.5736 0.5878 0.6018 0.6157 0.6293

0 7002 0 8192 07265 0.8090 0.7536 0.7986 07813 0.7880 08098 0.7771

10 11 12 13 14

O 1736 0 1906 0.2079 0.2250 0.2419

0 1763 0 1944 02126 0.2309 02493

0.9848 09816 09781 0.9744 09703

40 41 43 44

0.6428 0.8391 0.6561 0 8693 06691 0.9004 0.6820 0.9325 06947 0.9657

15 16 17 18 19

12588 0.2756 02924 03090 03256

0.2679 0.2867 0.3057 0.3249 0.3443

0.9659 0.9613 0.9563 0.9511 0.9455

45 46 47 48 49

0.7071 0.7193 07314 0.7431 0.7547

20 21 22 23 24

0 3420 03584 0 3746 0 3907 0 4067

0.3640

25 26 27 28 29

0.4226 0.4384 0 4540 0 . 4695 0.4848

42

0.9397

50

0 3839 0.9336

SI

04040 09272 04245 09205 0 4452 09135

52 53 54

0.7660 O 7771 0.7880 0.7986 0 . 8090

0.4663 0.4877 0.5096 0 5317 0.5543

55 56 57 Se 59

08192 08290 08387 08480 08572

Power Handbook

0 9063 0.8988 0.8910 0.8829 0 8746

Coa Din Sin 0.8660 60 0.8660 0 8672 61 0.8746 0 nao 62 0.8829 08387 63 08910 0.8290 64 08968

Tan

Cu

1.732 05000

1 . 804 1.880 1.962 2.050

0.4848 0.4695 0.4540 0.4384

65 66 67 68 69

0 91363 0.9135 0.9205 09272 0.9336

2 145 2 246 2356 2475 2606

0 4226 041367 03907 03746 0 3584

0.7660 0 7547 07431 07314 0.7193

70 71 72 73 74

0 9937 0.9455 0.9611 0.9563 0.9613

2 748 2.934 3.078 3.271 3.487

O 3420 0.3256 0.3090 0.2924 0.2756

1.0300 1.0355 1.0724 1.1106 1.1504

07071 0.6947 0.6820 0.6691 0.6561

75 76 77 78 79

0.9659 0.9703 0.9744 0.9781 09816

3.732 4011 4.331 4.705 5.145

0 2588 0 2419 0.2250 02079 0.1908

1.1918 1 2349 1 2799 1 3270 1 3764

0.6428 0.6293 0 6157 0 6018 0.5878

80 81 82 83 84 85

091348 09877 0.9903 09925 09945 0.9962

5671 6314 7 115 8 144 9 514 11.430

01736 0 1564 01292 0.1219 01015 00672

1 4281 1.4826 1.5399 1.0003 1.6643

0.5736 05592 05446 05299 0 5150

86 87 88 89 90

0 9976 14.301 00698 09986 19.081 00523 0.9994 28.636 00349 0.9998 57.290 00175 10000 0.0000

also listing anglos in tcnths or in minutes, seo standard handbooks. Avoid tables marked lagarithnis of functions; use only thosc markcd Mg or natural functions. In some formulas the functions cotangcnt (col), secant (seo), and cosccant (cscc) are used. These are merely reciprocals of tangent, cosinc, and sine: tan = 1/col, cos = 1/sec, and sin = 1/csec. Anglos over 90 deg can be convencd to anglos between O and 90 for solving problems. Subtract anglos between 90 and 180 from 180. Subtract 180 from anglos between 180 and 270. Subtract anglos over 270 from 360. Thus 60. 120. 240, and 300 deg have thc same sine, cosinc. and tangent.

USING POCKET CALCULATORS Pockct calculators are available in a

rango of computing capacities and prices. The more complcx devices are recommended if you intend to work with trigomeric or logarithmic functions, or nced extensive memory capacity. Otherwise a calculator costing only a few dollars will handl° most engincering calculations. Even 24-key devices. however, pack considerable computing powcr. Besidcs addition, subtraction, multiplication, and division, they will also figure percentages, squares, and squarc roots at the touch of a button. Also, sum of products, product of sums, product/quotient of sucos, and sum of quoticnts can be found by calculating pan of the problem, putting the result finto memory, then recalling it to complete the calculation. To find square roots, oven of complex cxpressions, is easy. For example, to find the square root of 16' + 33 on a typical small calculator, dcpress thesc keys in sequence: "1" and "6" "Multiplication (X)" and "equal (=)." which squares 16 "Plus (+)" and "3" (twice), which adds 33 to 16' "Equal" and "square root (4/~)," which displays 289 (sum of 16' + 33) and 17 (square root of 289). As another example, tu solve Chis expression, (2 + 1)

(7 + 5) X (6 + 4) dcpress these keys in sequence: "7: "plus," and "5" "Equal" and "put iota memory;' 5

which displays thc sum of 7 and 5 an puts it in the calculator's memory "6." "plus." and "4" kcys "Multiplication," "mcmory recall.' and "equal." which multiplies the sum o 6 + 4 by the sum recalled from mentor. 1 12). and displays the result (120) "Citar memory." "pul into memo ry," which cicars the memore of ol e

storage (12) and enters neu storage (120) "2.." "plus." and "1" "Division," "memory tecali." and "cual," which divides the sum of 2 + I by the sum recalled from memory (120). and displays the resol', or 0.025 l'weit more complex expressions can be handled In ;mune) ditterete tlertion. into

memory. solving another ponlion. com • bining it (as necessary) with the first portion. clearing the memory, putting the combination into memory. solving another portion, combining it with the value recalled from mcmory. clearing the memory, etc. With a link practice, you will lind it easy (and logical) to work with a pocket calculator.

TYPICAL PROBLEMS Thc foliowing exainples show practica applications for material presented s for, using a pocket calculator. EXAMPLE 1. How many cubic feet in block 6-% x 4.% x 22/1. in.? SOLUTION. Convert incites co feet an inultiply, rcferring tu thc formula abov for %tilinte oí a rectangular solid. pocha calculator can be usad lo mak the conversion in a continuous sequenc of stens: 6'/. in. = 27/4 dividixl by 12 0.563 ft. which is storcd in memory 4% in. = 37/8 divided by 12 0.385 ft, which is mulliplied by 0.56 (recalled from memory) and the produ is storcd in memory 2%. in. = 37/16 divided by 12 0.193. This is multiplied by the produ' recalled from memory, co give

roundcd-olf answcr of 0.042 cu ft. EXAMPLE 2. What is the weight per foot ()I' wroughi-iron hex rod ineasuring two incites across its flats? Wrought iron wcighs 0.28 lb/cu in. SOLUTION. From "Arcas of Common Figures." facing page, thc seation arca of a hexagon is 0.8660' or 0.866 x 2' = 3.46 sy in. Volume per foot is 12 X 3.46 = 41.5 cu in. Weight per foot is 0.28 X 41.5 = 11,6 lb. EXAMPLE 3. A circular pit, 10 ro deep and 3 ft across (inside dimensión), is lincd wilh a concrete wat] 6 in. thick. How many cubic feet of concrete are there in 1he wall? SOLUTION. Cross-sectional arca of pit is /r/Y/4 = tr9/4. Cross-scctional arca of pit plus wall = rlY /4 = r16/4. Crosssectional aren of wall r 16/4 - tr9/4

= w7/4. Volume of concrete = 1.751r X 10 = 55 cu ft. EXAMPLE O. What is the volunte of a right circular cone. 10 ft across at thc base. if thc cone slopes up at an angla of 50 dce from the horizontal? SOLUTION. First. lind the height. using a = b X tan A. From the table, tan 50 dee is 1.19: b is hall the diameter. or 5 ft. Thus thc height is 5 x 1.19 zw 5.95 ft. From "Common Volumen" right cone has volume of 0.2620W, or 0.262 x 10' x 5.95 = 156 cu ft. EXAMPLE 5. How long is thc guy wire strctching from the top of a 30-ft pole to an anchor 15 ft from the pole's base? Here. the guy wire forms the hypocenuse of a right triangle. so c = 15' + 30' and e = 1 /1125. or 33.5 ft, which can be round from a pockct calculator.

CIRCUMFERENCES AND AREAS OF CIRCLES ()bimotora 10 to 100

Dlameters 1 to 10 by tenths DA

Gro" Área

lO I1 12 3.142 3.456 3.770 0 785 0.950 1.131 2.1 6.597 3.464

13 4.084 1 327

1.6 5.027 2.011

1.7 1.8 1.9 5.341 5.655 5.969 2.270 2.545 2.835

Ora Caceen Atea

10 11 12 31 42 34.56 37.70 78.54 95.03 113 1

13 40.84 132 7

14 43.98 153 9

15 47.12 176 7

24 7.540 4.524

28 29 26 27 8168 8 482 8 796 9 111 5 309 5 726 6 158 6605

Da Crrcum Area

20 62.83 314 2

23 72.26 415.5

24 75.40 452.4

27 25 26 78.54 81.68 84.82 490.9 530.9 572.6

3.8 37 11.62 11.94 10.75 1134

3.9 12 25 1195

Do Crroon Asea

30 31 94 25 97 39 706.9 754.8

4.7 4.8 49 14.77 15.08 15.39 17.35 18.10 18.86

Ora Cacen Afea

40 125.7 1257.

2.5 7.854 4.909

Mea

20 6.2E3 3.142

da Ceceen Ares

3.0 31 3.2 3.3 9.425 9.739 10.05 10 37 7.069 7.548 8.042 8.533

De Cenan Area

41 40 12.57 12 88 12.57 13.20

4.2 4.3 44 4.5 13 19 13.51 13.82 14.14 13 85 1452 15.21 15.90

[ha Decore Atea

5.0 15 71 19 63

5.1 1602 2043

5.2 16.34 21.24

5.3 1665 22.06

5.4 16 96 22 90

5.5 58 56 57 17 28 17.59 17.91 18.22 23.76 24.63 25.52 26.41

59 18.54 27.34

Ora Gro" Atea

50 157 1 1963.

Ora Drarre Area

60 18 85 28.27

61 6.2 19 16 19.18 29.22 30 19

6.3 19 79 31 17

6.8 6.4 67 6.5 6.6 20.11 20.42 20.73 11.05 21.36 32 17 33.18 34 21 35.26 3632

6.9 21.68 37 39

Da

60 188.5 2827

Da

71 70 21.99 22 31 3848 39 59

7.2 22 62 40 72

C icuta Área

8.0 25.13 50 27

8.4 8.2 8.3 25.76 26.08 26.39 52.81 54 11 55 42

Pa Orcum Ares

91 93 9.0 92 28.27 28.59 28.90 29 22 63 62 65.04 66 48 67 93

Osa Cecear

CITUIrl

Área Da

6

8.1 25 45 51.53

22 23 6912 7.226 3 801 4 155

I4 1.5 4 398 4.712 1 539 1.767

34 3.5 36 1068 11.00 11.31 9.079 9.621 10.18 4.6 14 45 1662

Gran Área

21 65 97 3464

22 69 12 380 1

16 50.27 201.1

17 53.41 2270

18 56.55 254.5

19 59.69 283.5

28 29 87.96 91.11 615.8 660.5

34 35 36 32 33 100.5 103.7 1068 110.0 113.1 804.2 855.3 9079 962 1 1018.

37 116.2 1075.

38 119.4 1134.

39 122.5 1195.

41 42 43 128.8 1319 135.1 1320. 1385. 1452.

44 138.2 1521.

45 141 4 1593

46 144.5 1662

47 147 7 1735.

48 150 8 1810

49 153.9 11E16

51 160 2 1043

52 1634 2124

53 166.5 2206

51 169.6 2290

55 56 172.8 175.9 2376 2463

57 179.1 2552

58 182,2 2642

59 185.4 2734.

61 191.6 2922.

62 194.8 3019.

63 197.9 3117

64 201.1 3217.

65 66 204.2 X173 3318 3421

67 210.5 3526

68 69 213.6 216.8 3632 3739.

7.9 24.82 49.02

Dream Área

70 71 219.9 123.1 3848 3959

72 73 74 226.2 229.3 232 5 4072 4185 4301

75 235 6 4418

76 238 8 4536

77 78 2419 2450 4657 4778

79 218.2 4902.

8.5 86 87 88 89 26.70 27.02 27.33 27 65 27% 56.75 5809 59.45 60.82 62.21

da Croen Area

80 251 3 5027

81 254 5 5153

82 83 7576 260 8 5281. 5411

84 263.9 5542.

85 267.0 5675

86 270.2 5809

87 273.3 5945.

88 276 5 6082.

89 279.6 6221.

94 9.5 96 97 9.8 9.9 29.53 29.85 30.16 30.47 30.79 31.10 73.90 75.43 76.98 6940 70.88 72.38

Da Croen/ Área

90 282. 7 6362

91 285 9 6504

92 2890 66.48

93 292 2 6793

94 295.3 6940

97 95 96 298 5 301 6 304.7 7088 7238. 7390

98 307.9 7543

99 311.1 7698

73 744 7.5 22.93 23.25 23.56 41.85 43.01 44.18

7.6 23 88 45 36

7.7 24 19 46. S 7

7.8 24.50 1778

Dla

vote, Handbook

AREAS OF COMMON FIGURES RECTANGLE Area - AB

PARALLELOGRAM

OCTAGON

Yr ROUND

Area = Oil

Área - O 828D

Area O 785R'

TRIANGLE

Afea = AR

TRAPEZOID Arca - Va 8 (8, +

90' FILLET

CIRCLE Ama 0.7850

RIGHT TRIANGLE

AU

Area Sector - Triangle

Area 0.8660

Area = BH

E,

SEGMENT

HEXAGON

B,)

4. \

Area = 0 215R-

RING Area O 785102

ELLIPSE Area = O 785AB

-

SECTOR Area !VA/ 115

PARABOLIC SECTION Area - 2/1 XY

COMMON VOLUMES

COMPLEX AREAS FIGURE METHOD Divide the complex figure rito simple figures and add up the ares of each. Here, add up areas of the linee Mangles and subtracf area of the cacle

CURE Volume = A'

RECTANGULAR SOLID

SPHERE Volume = 0.524D

Volume - ABC

; H

SOUARE METHOD Transfer shape to squared paper or place transparent squared paper over shape. Count the squares within figure, add estamated Iractional squares

Hrt

41^ CYLINDER Volume = 0.785D2H

CONE Volume = 0.262D'H

•

FRUSTRUM Volume = 0.262H (0,,

+0,D, + D?)

STRIP METHOD Divide shape into parallel sinos el equal width Area lapproximate) is sum of the lengths of all strips limes width of one strip Power Handbook

ANY PRISM OF CYLINDER Volui

H x Area

ANY CONE OR PYRAMID Volume - , H x Base aren

EXAMPLE B. Roughly how many squar

a

leca uf he:o-transfer arca are there in shcll-and-coil heat exchanger malle from 30 closely spaced turns of O (outsidc diameter) tubing if the coil ha

SOLUTION. Thc length of each turn will

eenlerline

diameter of 1 ft?

Iength is then 30 x 3.2 = 96 ft. Circumference of tubing is r X 0.25/12 = 0.065 ft. Outside arca of tubing is 96 x 0.065 = 6.24 sq ft of heat-transfcr

be a hule more iban the centerline circuroference of a 1-ft-diameter circle (rD, or r X 1), or about 3.2 ft. Coil

arca.

MECHANICS & MECHANISMS WORK AND FORCE When a force acts on a moving body, thcl force (pounds) times distante thc body moves (feet) equals work done (footpounds). Sketch at ncar right illustratcs ibis: Force of 11 lb at cnd of rope lifts wcight 3 ft. Work done is 33 ft-lb. Neglecting friction, work pul in equals work delivcrcd. Some mechanicál devices incrcasc force at the expense of distante moved. Scrcw jack at middle musa turn 'bar times to lift load 1 in. Force applicd at end of handle moves 502 in. in those four turns. Assuming no friction. work in equals work out. Thus (orces applied are inverscly proportional to distantes. and I lb at end of handle will theoretically lift 502 lb. This ratio of 502:1 is called the inerhanical advantage, or MA. Thc familiar block and tal) at far right is another device for increasing force at the expense of distante moved. An ea'v

way to find MA is to count the number of ropcs leading to (he lowcr block. Thus. MA in the sketch is 4. so a 100-lb pul) will lift 400 lb.

204 0 !V

MECHANICAL ADVANTAGE of jack (middle) and block-and-fall (right) is obtained by "trading •• distante moved to mercase force

'00 lo

111

WORK DONE (lett) m lifting 11 lb three feet is 33 ft-lb Puu

111b NOM done a 3 x 11 a 33 ft Ib

31! :_t_

Atch V. pn To lir/ load one triCh, hand moves 20 x 2 x 3.14 x 4

Foco Top, suppo.1 403 rr

502 in

400 Ib

MA 502

TORQUE AND POWER Energy can be delined as thc capacity for doing work. It is measured in thc samc units as work and for all practica) purposes may be considcred interchangeable with work. Energy takes two forms: potential and kinetic. Weight in sketch aboye has 33 ft-lb of potential energy when it is in raised position, because it would do that much work if allowed to fall 3 ft. In the course of such a fall, potential energy diminishes as height aboye cable &creases. Just bcforc it hits, all potential energy is gonc and the wcight thcn has 33 ft-lb of kinctic energy because of its motion. Neglecting losses, it will deliver that much work when it bits. While work and energy can be interchanged without much harm, the tern) power dilters and the difieren& should be elurly grasped. Power is the rate of doing work. Look again at the sketch of the weight. If it takes one minuto lo raisc, the power necdcd is 33 ft-lb/min. If the job takes 10 min, only one-tenth as much power (3.3 ft-lb/min) is needed.

Power may be incasurcd in ft-lb/rnin or ft-lb/sec. Engineers, however, are more familiar with horsepower (hp) and kilowatis (kW). One horsepower mcans doing work at the tate of 550 ft-lb/sec or 33,000 ft-lb/min. In example abo ye, 33 ft-lb/min amounts to 0.001 hp. Also. I kW = 1.34 hp and 1 hp = 0.746 kW. Terciar, or bending 'nomen:, is a twisting effort, usually measured in poundinches. For example, hand pulling 50 lb on the end of a I 2-in. wrench handlc exerts a torquc of 600 Ib-in. For rotating machinery, a time factor is introduced to relate torque and horsepower. As middlc sketch shows, hp = 27rXNXR F/33,000. To obtain torque from a givcn speed and horsepower. multiply hp by 100.000 and divide by 19 times the rpm. In sketch at lowcr right, engine turning at 190 rpm is loadcd with a prony brakc. Brakc load is 50 Ib at 24-in. (2-ft) radios. Torque is 50 X 2, or 100 lb-ft. Ilorsepower is 19 x 190 x 2 X 50/100,000 = 3.61 hp.

50-0 puf?

TOROUE, or Iwisting effort. 15 force times the lever arm over which it aCtS F Forte. lb R a R9déus, h

HORSEPOWER formula may be expressed as hp = 2xXNxRX

P33,000

'Pm

Power bland bOok

VECTORS EXPLAINED A VECTOR is an arrow showing direction and amount el a movement, velocity. or force. Afrow's length, which represents amount. may be lo any convenient state. For example. vector abo ye might mean a five-mile walk slighlly north of casi

Crane problcm below typifies enginecr. i ng use of vcctors. Forces acting at poi nt P must balance, sincc it does not ~ve. Resultant must be zar). so vector diagram is a closed triangle. Knowing direclion of torees and amount of load, draw diagram to scalc. Valucs of tensión and thrust can aten be picked off to the sane %cale.

#

o 595°7 #



ve.

Resultan! TO ADD VECTORS (aboye),

e

11942.

rearrange them end lo end (below) with arrows running same way. Resultanl is line OD (dotted) from heel of first vector lo point of las! 8 A O 10

Thrust = 3400 ib

TO FIND TENSION in be rod. compression in lib, draw 2000-lb vector down to represent load. Draw other vectors In direction of lie and pb, and scale off values C

PULLEYS, GEARS "

400 (pa?

DRNEN vary inversely as number of leeth Tooth ratio abo ye is 60/40. or 3/2, so rpm rabo is 2 lo 3. For pulleys. rpms vary inversely as diameiers GEAR RPMs

16 If1

DRNFR BELT SPEED is orcumference of either

pulley times ils rpm. For 6-in. (0.5-ft) pulley. circumference is 3.14 x 0.5 = 1.57 ft. Belt speed is 400 x 1 57 = 628 ft 'mal

of pulleys or gears. rpm 01 times rabos for all the pairs gives tunal speed. Here. d comes to 900 x (20'15) x 18 12) - 800 rpm FOR TRAINS lir SI driver

LEVERS, LINKAGES, INSTANTANEOUS CENTERS Equal anca».

insfanianeous for en rawdS on

E3C

V, -u 't._ Fixed DIVOI 1 points

bar

D

VA

LE VER LAWS: Distante moved is directly, and force is Inversely. proporlional lo dis-

lance from the fulcrum, as per aboye

Al any given Instant. all points in any moving body are rotating about a single polo! called Inc mstantanoous center. In coree-bar linkage (aboye). center of BC is where extensions of levers AE CD intersect INSTANTANEOUS CENTER:

A

Distante moved= length o! arm x 2 x 3.1416x anglo (degrees)/360

of any point in rotating body is proportional to distante from axis MOVEMENT

Power Handbook

Thrcc-bar linkagc. at right aboye, shows application of laves of levers. rotating bodies, and instantaneous centers. AB and CD are rigid levers with fixed pivots. Bar Be connects thcm, jointcd at B and C. If point A moves with velocity V,. what is vclocity of Velocities at A and B are proporcional to distantes from thcir pivots. Triangles give us V,. To find velocity of C. locate the instantaneous cerner of BC first.

Point B moves al right ingles lo AB, so one instantaneous radius of Be must lie along AB extended. Extending De gives anothcr such radius. Intersection of thc extended lines provides the instantaneous center. Velocities al 8 and C are in proportion to distantes from the instantaneous center. Ilaving, found Va can be obtained by proportion, as 1', veas obtained from VA.

EFFICIENCY

OVERALL EFFICIENCY CT motor yner0tor set IS 791103 = 0.767. whiCh iS 76.7%

kfficieney of any machine is its °input divided by its input, measured in the same units of work or powcr. Sketch shows a simple case. Input providcd by thc motor is measured at 103 kW, whilc output from the gencrator is 79 kW. Dividing 79 by 103 gives 0.767. Thus. elliciency in percem is 76.7%. Note that this is the overall ciliciency of the motor-gencrator (m-g) set. To

lind thc ellicicncy of the motor alone, connect it lo an decirle dynamometer, a tactic normally undertaken by thc manufacturer to rafe his machine. This gives the output in kW. Mcasure the input with a watt:ni:ter to obtain power in the same units. A simple división provides thc efficiency. Gcncrator cfficieney would qua' m-g cllicicncy divided by motor eiliciency.

STRENGTH OF MATERIALS TENSION, COMPRESSION, SHEAR C_TENSIO-C/1 t

cc» On. 13001e Ductile

Brick Concrete Wood Load Sheet sachan nength x tnickness

SHEAR

Single shear

Double sheet

Force applied te materials Cali be (1) push, that is compression, (2) pulí, tension, (3) scissor-like action, shear. Alt materials givc with force: roo much force and thcy break. Engineers oftcn work with stress in psi rather than force in lb. Stress is force in lb dividcd by section arca in sq in. Curve at right shows how tension aireeis ductil° materials. Below thc elastic Emir, material stretches in proportion lo stress. If stress is removed, material returns to original size and shapo. Beyond a material's clastic limit, stress gives permanem set. Al yield point, material begins lo Ilow with no Mercase in unit stress (actual force over original section arca). Ultimare strength is force jost before breaking divided by original

section arca. Yield strength is stress producing 0.2% permanent set. Factor of safety is ultimato strength dividcd by allowablc or working stress. Annealed carbón steel has yield strength of 40,000 psi, ultímate strength of 60,000 psi. )4.1— af poi eankting tflumate strength =-- bold pavo

— riastr

Slrelch

DUCTILE MATERIALS stretch under ten sion. brittle materials give little before breaking at some predictable point

STRESS IN SHELLS

o Pu gane

ALLOWABLE PRESSURE = safe stress times Iwice watt micznesslinside diameter

In a Sealalt,N or weldcd tubo or drum under interna] pressure, the tendency to split lengthwise ("unroll") is twice the tendency to split around. For that reason, stress is always figured for a lengthwise instead of a circumferential split. Sketch shows how to translate interna] pressure into stress. Huid pressure tends to split tubo in hall. Force on "pistón arca" equals pressure P, pwig, times diameter D, inchcs. This force is resista'

by thc two metal sections A. The resisting force equals two times the thickness t times the unit stress s. EXAMPLE. What is the allowable pressurc for scamiass shell having a 30-in. incide diameter, 0.5-in.-thick wall, and a 15,000-psi sale stress? SOLUTION: Safe pressure is 2 x 0.5 X 15.000/30 = 500 psi. With 80% elli• cien) longitudinal riveted sco u t, the rafe pressure is 0.8 X 500 = 400 psi.

STRETCHING OF MATERIALS F I-0—:00 fi

Sta& —Ildr rad

ff —44

_

a

4-177-71ell

HEATING to explane) 1 in., then gripping rod will set up 25,000 -psi stress 10

When a material is stretched or compressed within its clastic liad, thc ratio of a pplicd stress (psi) over unit chango in length (in./in.) is a constant called modulas of elarririty. Flongation (in.) = length (in.) x stress (psi) divided by modulus. Seo pago 16 for thcrmal expansión, pago 23 for expansión of pipas in Piping. Flow & Tanks.

ELASTICITY AND STRETCH

FOR COMMON METALS Street. psi. eleeticity, to sketch psi X 101 1 in./100 10 8.300 16 13.000 18 15.000 30 25.000

Modulus of

Material Alunnurn Come r fron

Sleel Wood (aPP105 )

1.5

1.250 Power Handbook



BEAM SIZES* Standard "1" beams

Wide-nango -1" bearna Depila

'Height

Modulas

lartuting length

17 20 24 28 31 35

14 1 17.0 20.8 24.3 27.4 31.1

10.1 12 2 16 3 18 8 21 6 24.4

21 25 29 33 39 45

21.5 26.4 30.8 35.0 42 2 49 1 546

9.9 12.3 14 2 17 6 21.2 24.5 27.9

27 31 36 40 45 50

341 39 4 45 9 51 9 58.2 64.7

10.9 12.5 14.5 17.3 19.2 21.2

30 34 38 43 48

418

9.3

48 4 54.6 62.7 70. 2

10.9

49

1

1 'For sea« - r Pum,

FACTS ABOUT BEAMS Limibeg length

Mode 3 3

Warghl 5.7 7.5

Modulus 1.7 1.9

4 4

7.7 9.5

3.0 3.3

9.7 10.2

5 5

10.9 14.7

4.8 6.0

9.8 107

6 6

12.5 17.2

7.3 8 7

10.0 10.7

7 7

15.3 20.0

10.4 120

102 108

8 8

18.4 23.0

14.2 16.0

10 6 11 1

10 lo

25.4 35.0

24.4 29.2

11.4 12.1

12 12 12 12 15 15

31.8 35.0

36.0 37.8

11.3 11.5

40.8

44.8

14.4

50.0 42.9 500

503 58.9

15.4 11.4

64.2

11.7

12.3 15.4 17 2

10 1 10.8

When beam is bent. convex fubers are stretched under tension: (concavel libers are shor t ened by compression

(AreiNGTH

TENSIOF.

:.;TRENGI •

STRENGTH 3

Strength of beam is durectly proporcional to Three planks side by sude talco triple load

STRENGTH 1

STIR NGTH

as width.

UNGIUED 3 (UNGLUED 2 STRENGT14{ GLUCD 9 CLUB) 4

Strength ts also roportional to square of depth. load Three planks stacked olued tal e 1U110

SECTION MODULUS (MOMENT = MODULUS

moment onuals ~Pa times 20.030 re

BEAM FORMULAS

x

STRESS)

RECTANGLE: Modulus equals 6 width x depth squared

L is coches W is total load in pounds

SOLID CIRCLE equals il p :9et pan?

W

(rod): Modulus diameter cubed

HOLLOW CIRCLE (sube): Modu!Nal: thieknen x las equals average diameter squared

MOMENT = L x W

MOMENT 1 /. xL x W

To !muro sale momenl el 1 bearns lortger Iban lima mg length. reduce stress of 20.000 psi by fraction of limiting length over total 'engin

W Danger pon?

!langur

?"4MOMENT - 1 /. x L x W

FIGURING LONG BEAMS

pon;

L -111 MOMENT = '1. x L Y W

SAFE STRESS 20.000 psi

Sal n Ctlárn SIPCI

BEAM PROBLEMS EXAMPLE 1. What size " 1 - btain .pan a 10-ft (120-in.) gap to supporl an evenly distributed 50.000-16 load? SOLUTION. From "'kan) Formula` abo y e. inoment = 4 X 120 X 50,000 = 750,000 in.-lb. With sale stress of 20,000 psi, minimurn beam modulus 750,000/20.000 = 37.5. From "Rcani Sizes." use either a 12-in.. 31-1b/ft l'unge bearn, ora 12-in.. 35-lb/ft standard beam to support the load. EXAMPLE 2. How much wcighl will 6-in., 12.5-1b/ft standard "1" bcam support on one cnd If it is 15-ft long and Powet Handbook

UNIX' m a cantilever arrangcment? SOLUTION. From table, limiting length is only 10.0 ft. so reduce the allowable stress of 20.000 psi by multiplying by 10.0/15 to set a 13.300-psi safe stress. Modulus is 7.3, so sale inoment is 13,300 X 7.3 = 97,000 = 8080 ft-lb. Safe load is then 8080/15 e 540 lb. CAUTION. Use the bcnding formulas with caution, particularly ir the beam shortcr than Relee its depth or is slender, wobbly, and unbraccd. Cal' in a structural enginecr to &sign any permancnt bejín installations.

Any common structural tumber

600 to 1400 tasa

RIVETED JOINTS Slrene,111 ul a jUltll IN teman) ligurctl as lis efficiency- the strength of its weakesl

point divided by the strength of the solid ;Mate. To figure efficiency for a singleriveted lap joint, for example, lind thc weakest point first, which may be between boles, in front of thc riector the rivct itsclf. This point, expressed in psi. divided by the [date strength, in psi. gives Ihe elliciency. Welded joints have largely displaced riected joints in construction today. 11

PSYCHROMETRICS 1 0 6000 30°0

1

0

0. 030 /00o

r t • 04/4‘. nea,.. net ,,,

toCiel

—

0.028

‘c. 9 -Ir

0 026 0.024 0.022

• 0.020 0018 n O 016 O 014 0.012

2 •

0.010 o 1.75 0 008 ,ve 0.006 0.004 0.002 35

40

45

50

55



60

65

70

75

80

85

90

95

100

105

110

115

120

Dry . tx.ob temperature, F

PSYCHROMETRIC CHART, at barometdc pressure of 14.696 psia (sea levet). Chart relates seven key properties of moist air. Dry-bulla • wet-bulb temperaturas are most readily measured and usually serve to determine the remaining unknowns Atmosphcric air is a mixture of dry and a variable amount of water vapor) Atmosphcric pressurc is thc sum of the partid pressures of the dry air and the water, cach acting as if it alone occupied the volumc occupied by the mixture, Sinec thc wcight of moisture per pound of dry air is small. the partial pressurc of thc moisturc is vcry low --scldom ex. ceeding 0.5 psia. WATER VAPOR always behavcs as if no air was presenl. Al a givcn pressurc, water vaporizes or condenses al a tixed temperature. known as the saturarlos temperature. Put another way, saturated vapor at any given temperature has a tixed pressurc and dcnsity, as listed in the steam cables un pase 21. in whia specific volume is the reciprocal of dcnsi. ty. To apply the labios, assumc the vapot in an air/vapor mixture at 70F is satur Tablc 2 shows thc absolute pres sute of the vapor is 0.3631 mi. RELATIVE HUMIDITY by definition equal to the actual pressure of vapor comparcd to the vapor pressure salur 12

ated al the same dry-bulb temperature. If some of the moisturc was condensad unlil the vapor pressure was 0.1781 psi abs. the relativo humidity would be 0.1781/0.3631 = 0.49, or 49%. In othcr words, a pound of dry air containing vapor al 0.1781 psia, contains 49% as much vapor as it could if the vapor were saturated al 70F. DEWPOINT. Reference tu Table 2 shows that vapor with a pressure of 0.1781 psia has a saturation temperature of 50F. This rcpresents thc dcwpoint or saturation temperature of the vapor. Temperature of the air and vapor are always thc same. Thus. if the air/vapor mixture is cooled to 501 : by passing it over a cooling coil, che vapor would be saturated. If the air/vapor mixture werc cooled further, moisturc would be condensad. When the actual temperature of the moisture is grcater than its saturation or dewpoint temperature. the moisturc is superhcatcd. LATENT, SENSIBLE HEAT. To vaporizo water. hcat is added; lo condense it. heat

is takcn away. Al atrnospheric tcmperatures it takcs about 1050 Illu/lb. Thc exact amount of this laten: heat for changing state varios with the dewpoint or saturation temperature. Sensible hen, is what changes temperature. The amount dcpcnds on temperature and spccific hcat (0.24 btu/lb for dry air). TOTAL HEAT. For dry air, sensible hcat = total heat. For moist air, sensible hcat + latent hcat = total heat. The total hcat or enthalpy of an air/vapor mixture = dry-air sensible hcat + amount of vapor per pound of dry air x cnthalpy of the saturated vapor. WET-BULB TEMPERATURE can be determine(' by using an ordinary thertnometer with wctted gauze or a sleeve over the bulb. If air is circulated over the weited wick, some of the moisturc cvaporates, cooling the remaining water in the wick and lowering bulb temperature. Ilow much the wat-bulb is lowcrcd depends on thc air dryness and temperature. The wat-bulb temperature in combination with dry-bulb temperature gives Iwo valPOV/Of Handbook

coordinatcs for cnthalpy and humidity ratio (moisture cometo), and finas of constan( dry- and wat-bulb temperaturas, relativa humidity, and volume of mixture pa pound of dry air. The saturation inc l is also the 100% relativa humidity line. EXAMPLE 1. What are the properties of moist air at 8OF dry-bulb and 65F wetbulb temperaturas'? SOLUTION. From the chau, relativo humidity = 45%. humidity ratio = 0.01

ucs that can be 'llenad on a psychromctric chart to determine many other psychrometric properties. PSYCHROMETRIC CHARTS are graphic representations of the properties of air/vapor mixtures over a wide ranga of conditions. Knowing any two properties, typically dry- and wet-bulb temperaturas, all the othcrs may be found. The chart on the facing page pico highlights of various properties that are commonly shown on thesc charts. These include

lb of moisture per lb of dry air. volunte of the mixture = 13.8 cu ft/lb, enthalpy = 30 Btu/Ib dry air. saturation temperature (or dcwpoint) = 57F. EXAMPLE 2. A moisture content of 0.006 lb moisturc per lb is dcsired at a rclative humidity of 40%. What are thc other properties of Ibis moist air? SOLUTION. From thc chau, dry bulb = 69F, saturation temperature = 44F, volume of thc mixture = 13.45 cu ft/lb dry air. enthalpy = 23 131u/lb dry air.

COMBUSTION THEORY is a spccial form of oxidation. Oxygen combines rapidly with certain types of fuels, such as coal. oil, gas. or wood, and substantial heat is liberated. Under some conditions. combustion may be self-starting. For example. coa!

Evcrywhere, at all times, oxygen combines with olhcr elements. 'ron and oxygen combine to form the oxide known as rust. Silver tarnishes, copper takes on a soft green coas. This general process is callad oxidation. Burning, or combustion.

pilad outdeors combines slowiy with oxygen in the air, giving off heat. If the hcat doesn't dissipatc fas( cnough. temperature vises, and the reaction %peak up until it eventually bccomes rapid enough to be callad burning.

HOW BURNING STARTS Combustion usually begins when heat from an outsidc source is applicd to a fuel. Wood laid in the hearth will not ignitc by itsclf. The flauta from a match is nccded to light kindling. and the kindling heat to start the logs. Some matcrials, obviously, burn more rcadily than othcrs. In ganara', the degree of flan:mabita>, depcnds en how casy it is to turn the particular substancc into a gas, becausc nothing truly burns until it is a gas. This, in turn. dcpends nn

the ual urc and quant ny ot the su bsta ncc, comparad with the amount of hcat available to star( combustion. A candle illustrates the point well. To make wax burn, it is turnad into a gas with thc hcIp of a wick. The wick draws up, by capillary action, an amount of matad wax so small in relation to thc (lame from a match that enough heat is available to raise thc wax to the temperature nadad for vaporization. Once the v,ay 1. vaporizad (transformad into a

is relatively casy to star(. Burning thc wax gives off enough additional heat to continua the process of melting, vaporizing, and igniting. Although thc wick eventually is consumad. it rcally contributes nothing te the burning proccss. I I is menely a mechanical devicc to creatc conditions that are needed to staff and maintain combustion. In the powcrplant, thc wick Iinds its counterpart in burners, stokers, and furnaces. gas). burning

COMBUSTION OF GASES Two clements are basic to most fuelshydrogen and carbon. Sulfur and soma other elements burn and give off heat, but common practica considcrs thesc rcactions as nagligible, and the elements as impuritics. Ilydrogcn normally is a gas. and can be 'igualad only at an extremely low temperature, below -400F. On the other hand, carbon is a solid that does not vaporiza completely until the temperature machas 6300F. Hcating values are high: 62,000 Btu/lb for hydrogen and 14,100 Btu/lb for carbon. Carbon and hydrogen cxist in almost endlcss combinations callad hydrocarboas. Many of thesc compounds, such as methane (C11 4 ), normally ara gaseous, and (hese form a major par( of most ' important fucl gases. AIR REOUIREMENTS. Each clamen' or connx)tind in a fuel demands a certain a mount of oxygen for complete combusPower Handbook

tion. Thus, it is casy lo figure exactly how much oxygen is nadad te burn a given amount of gas and. further, just how much air is required. If exactly this much air is mixed with exactly the given amount of gas, a perfect or J' Inflan' mixture mutis. The relationship betwecn gas and air is callad the fuel/air ratio. usually expressed as a weight ratio. IGNITION TEMPERATURE. Still anothcr ingrcdicnt besides air is nceded - haat. If such perfect mixtures are heated graduany. the rale of chemical combination Mercases until a point is reached where the reaction no longer dcpends on heat from an outsidc sourcc, and practically instantancous combustion occurs. Thc lowest point al which this happens callad the igualan temperatura defined as that temperature at which more heat is generated by combustion than is los( te the surroundings. Below this point. thc gas/air mixture will not be self-sustain-

ing unless heat is supplicd lo it. Thesc observations apply cqually lo mixtures that are not perfect -that is, to mixtures a titile leaner ora hule Ochar iban perfect. Por each gas, however, a point is reachcd at which the mixture bccomes too lean en too rich for ignition and continuous burning al any temperature. In other words, a ranga of mixtures exists for cach gas within which ignition will occur and sustain itsclf. Limits of Chis ranga are known as the limits of flammability.

M111011 ISM u'

..0 melbane

17.0 ar

COMPLETE COMBUSTION calls for 17 lb ol an Ion each

of methane

1 lb of gas 13

PRACTICAL MIXING In actual practico, even adequate mixi g and burning happen only rarely. closest lo it is in the cylindcr of 11 Iras AA carbon bumed al edge ot llamo

•

L UIAVVIVS llame el incrindeSCern carnet)

-Cracking" by haat terciases ClirtiOn Air amas nene no? ye? heated -

No ar antas abone/ of bufar

Fish . M :r.pner

VurnInating gas THERMAL DECOMPOSITION of gas to a fishtail burner yields yellow flame

mix helor° reaching flame in Bunsen burner. produCing a blue color

GAS AND AIR

POOR MIXING (left) leads to incomplete combustion. Good mixing is at right

enginc. In bofia liring, and othcr furnace work, there is no single mass of rcady-made gas/air mixture, but rather air and gas strcams that must be brought togethcr and haced in some manner to star) and sustain ignition. The Bunsen burner found in laboratories and the fishtail burncr once used in gas-lighting systems illustrate good and poor mixing. The first display% a retativcly small, blue, nonluminous llame; the fishtail burncr shows a largor, ycllow. luminous llame. With the same gas, why the difference? VELLOW-FLAME BURNING. Thc answer lies in the way air and gas mix. and shows clearly how the time and place of Chis mixing affect the process by which Ihe fuel elements particularly hydrocarbons burn. In the lishtail burilar, gas fiows through a slotted opening in a (hin fiat strearn. The arca near the burner tip is cold gas in the process of picking up and mixing slightly with air. Suddcnly the mixture rcaches the correal proportion and starts to burn rapidly, only a few hundreihs of an inch from the relatively cold, loo-rich gas mixture. Itere a phenonmenon occurs which is conunonly known as cracking, or thcrmal decomposition. It happens Chis way: When hydrocarbons are subjected to high temperatures, thcy crack or break down chcmically into other carbonhydrogen combinations, and eventually into the basic ingredients--carbon and hydrogen. The cracking action sets free (1) hydrogen, which burns in the flame, and (2) carbon particles, which glow to incandescente and make the flame ycllow and highly visible. Normally. carbon burnout occurs by

the time Chis element reaches the Ilame's edgc. But if any cold object is placed in the (lame, the carbon particles are chillcd and burning stops. The resulting dcposit of lampblack or mor is identical with similar dcposits in boilers. The microscopic particles of carbon that stay suspended in the products of combustion are called, collectively, smoke. BLUE-FLAME BURNING. In the Bunsen burner, gas issucs from a small jet at the base of a mixing sube, and air is drawn in through a sido opening. Gas and air mix before reaching the llame arca, and the result—if the air opening is properly adjusted is a ncarly colorless blue flame. The reason Chis /lame differs from that of the fishtail burner is that gas and air mix before rcaching the combustion zonc and are heated more slowly. Oxygen has a chance to pcnctrate the hydrocarbon molecules (a process callcd hydroxylarion) and form oxygenated compounds. These compounds burn with an entirely different llame. There is no cracking, and no carbon is formed. Thus, a change in the point of mixing, and in the rato of applying hcat, produces an eniirely different burning process. GOOD MIXING VITAL. This simple example dentonstrates the importancc of thorough mixing. The example also shows what happens when carbon breaks away from its compounds. Microscopic partieles formcd are carried along in the stream with littic or no motion rclative to the air. This makes them extremely difficult to burn oren if the process is no( stopped, as by chilling. Finally, and perhaps most important. onc learns that combustion is not (irrite so simple as it seems.

BURNING HYDROCARBONS Complete analysis of flue gas (ron. and oil-fired furnaces often reveals the presence of chemical compounds known

Haat

nyckocartonsi 1

•

1 0x1rPer, I

Hydroxyfa red compouncks Ainehydes

Hear CO2 M'e?



CO Oxygen

Oxygen Ware,'

IN GAS FIRING oxygen may unite with hydrocarbons to form unstable products 14

as aldehydes. (Formaldehyde is a familiar mcmber of this group.) llow did thcy gel there? In the gas-burner example, oxygen associates with the hydrocarbon molecoles, yiclding hydroxylated compounds that are unstable. These form aldehydes which, in turn, break down or oxidize until formaldehyde is produced. What happens ncxt depends on how much oxygen is present. Formaldchyde may break down with hcat to form carbon monoxidc and hydrogcn, or it may burn to water and carbon dioxidc or carbon monoxide. Any CO and H, thus formed burn in the normal way. So it is evident that severa] changes in fuel composition occur between the beginning and the end of the burning process, and that many intermediatc

products form and disappcar. If combustion is incomplete, some of these compounds remain.and can be identificd in the Iba gas. CRACKING PROCESS. Thc gas-burncr example also showed that, under some conditions, hydrocarbons are crackcd instead of hydroxylated. Such cracking

HYDROCARBON FUELS that are cracked by heat dissociate to carbon. hydrogen POwer Handbook

hrcaks thcm down into carbon and hydrogen. which burn lo carbon dioxidc and water. Thc iwo ways hydrocarbons can burn are not separare processes. In virtually

al' practica) burning, both go un simultaneously, and the general character of the llame dependas on which predominares. This. in toro, depends un surrounding conditions. Early mixing and preheating

ot hydrocarbons and air. plus time tor the oxygcn molecules lo cnter the hydrocamben molecules. favor hydroxylation. Rapid heating, and lack uf mixing time, favor cracking.

ELEMENTARY FUELS For powerplant purposes, all cumples Focls can be considere(' as combinations ri ihree simple or elemcntary fuels: (I) gascous hydrocarbons, (2) solid carbon. and (3) mixtures of CO and H,. Each of (hese general groups coniains a varicty of fuel substances which differ in minor respects, but each group displays a generally similar pattcrn of combustion. Some commercial fuels are simple mixtures of (hese clementary fuels. For example. coal gas is essentially a mixture of gascous hydrocarbons and CO, plus II,. Fuel oil volatizes into gascous hydrocarbons before combustion: some cracking may occur, yiclding C and H.. To know what aetually happens when real burns, the individual burnim • pro•

ccsses of the elementary fuels must be undcrstood. Carbon burns ibis way: Oxygen penetratcs the carbon surface to break away atoms• which hook up with the oxygcn in a loose type of carbonfoxy. gen compound that is unstable. Depending on temperature and other conditions. Chis quasi physiochcmical compound brcaks up into CO, and CO. If excess oxygcn prcvails, CO is oxidizcd co CO,. If carbon is in excess, dioxide is recluya( to monoxide. Hydrogen and CO are grouped together as an clemcntary fuel because they commonly occur together. and their burning processes are alike. In ordinary combustion, conditions are such that

and simple chemical reactions: 2H. + 0, --•• 21-1,0 203, 2CO + O, The burning of gaseous hydrocarbons. the third elementary fuel, has already bccn comed.

both hura according jo I hese

tour un c.tabi •

Outet layer oí OxYgef carbon atetas

r

• 4‘ • •

~abad

• • niaead°5 oxygcn

• */

OXYGEN PENETRATES carbon surtace to

carbnn/oxygen compound

FUNDAMENTALS OF HEAT TEMPERATURE Hcat is onc common form of cnergy. Hcat addcd lo a body makes it honor: removing heat cools the body. I leat also melts solids into liquids and convcrts liquids into vapors or gases. Expansion is anothcr result of heating, which also triggers certain chemical reactions and welding of certain materials. Han cnergy can bc comed into mcchanical energy to do work. typically stearn moved through a turbine. But practically all mecha nica I, electrica I. and chemical energy used to do work eventually cnds up as heat through friction, change . of-state, and other tomes. HOY AND COLD. Temperature is a measurc of how much hotter or coldcr one body is (han another. Vade temperature is mí a direct ~asure of heat. it measures a resuli sometime.s callcd sensible heat or heat leve). The hotter a body (that is, the highcr jis temperature). thc highcr the heat levcl. MEASURING TEMPERATURE. In most engincering work. temperature is measured on the Fahrenheit scale, on which 32F and 212 I' are the freczing and boiling points of water. respectively, at sea leve). On the Celsius scale. 0°C is frecting and I 00"C boiling. Thc Fahrenheit absolute scale (somctimes called Rankinc. in dcg R) uses Fahrenheit Power )(anobook

dcgrces starting from absolute nro. The Celsius absoluto scalc (Kelvin scale, in kelvins) uses Celsius degrees from absoluic zero. Absoluto zcro is the temperature where molecules stop moving. a t — 459.69E or — 273.16'C. INSTRUMENTATION. Temperature is measured by :1 rango of deviccs— (ha. mocouplcs. RTDs (resistance temperature detectors), radiation and optical pyrometry, bimetallic thermometers, filled syslems such as Bourdon gages. In thermometry. a wide array of digital dcvices are used. Some are linkcd with thermocouples, olhers with RTDs and thermistors: most rcly on solid-state plug-in circuitry. Conversion of analog inputs to digital signals within thcrmal instrumenis is achieved by special analog-to-digital circuitry. Electrical transmission signals may be classilied as voltage, current. position, frcquency. or pulse. MI hvc can transmit analog signals, but only thc pulse type can be used for digital transmission. Most measurement signals in powcr applications are voltages (mV) and current,. (usually 4-20 mA. de). When a coded pulse is transmitted, the signal can be fed into a digital computer.

F (F abs)

t CC 01,51 100

13731 T

Ty. 212 (672)1 BoAng

- (F - 32)/

F 1.8t + 32

E O (273) Freenng''

1

32 (492)

- O (4601

e 15



▪

SPECIFIC HEATS

HEAT AND SPECIFIC HEAT The standard unit of heat is the Bri ish thcrmal unit, or Btu. II is dcfined as 1/180 of the heat supplicd to raisc he temperature of 1 lb of water from 32 to 2I2F. For all practical purposes is mcans that I Btu is the heat supplie (o raise the temperature of 1 lb of wat 1 deg F. SPECIFIC HEAT. If objects of the sane weight but ot .lilTerent matcrials reccive the same anlount of hcat, they will h al up to diffcrent temperatures. How m ch cach object's temperature will r se depends un the specific han of the ma erial. Specific heat is defined as the h t t'expire(' lo raise the temneralure of I lb

of a substance I deg E By definition of the Btu, thc specific hcat of water is 1. All other %peak heats are relative to this. Ileat supplicd = lb of substance x sp ht x lemp risc. For example, warming 100 lb of fuel oil from 40F to 100F takes about 100 x 0.5 (see table) x 60 = 3000 Btu. Spccific hcats change with temperature, but often can be assumcd constant within the practical rangc of temperatures. Two spccific hcats are needed for gases, ene for hcating al constant pressurc, another for hcating at constant volumc (a closcd vessel). In most cases, volume and pressure change: see -Gas Laws." nexl page.

EXPANSION OF SOLIDS, LIQUIDS AddIng heat to substances malees Mem expand: removing hcat makes them contract. Within normal temperature rangos, solids or liquids expand or contrae( in proportion to the temperature change. To figure expansion or contraetion. multiply the original length of a material X cxpansion coefficient K temperature change. Heating a 30-i3. steel bar 40 dcg F, cxpands it 30 x X 0.0000067 (see table) = 0.008 in. Heating or cooling a substance witlout letting it expand or contrae( sets up stresses. II a rigidly held steel bar were coolcd enough that it would shrink 1 in./100 ft. if allowed, the sanee tensi stress would be set up as if the bar we stretched I in. (see "Stretching of Mate-

MaININ

e•

Acemite

0.514

Alcohol. ethyl Asbeslos

0.68

Asaban Bakelile Benzene BeCkerOfk Careen lel

Comen/ Coal Colee Concreta Key Air

Material

e• 0.53 0.16-0 2 0.12 0.5 0.217 0.21 ± 0.5 0.191 0.12 0.2 0.42 0.45-0.65

Gasolone Glass Iron Kerosene Lime_slone Marbie Petrck-um Sana Steel Stone Turpontine

0.25 0.22 0.35 0.412 0.2

0.201 0 16 '03 0.36

0.156

Wood

gasea

e IV

A' O 240

0.172 0.160 2.44 0.47

020 3.40

Carbon 010Xide Hydrogen

Melhans

0.52

'c - lobee nem. 'A - consient pressen, re = constant volume

F levg)

AVERAGE EXPANSION COEFFICIENTS Material Alammurn 13rass. casi Brome Cerner° neat

Concrete Copper Glass tubo Glass. plato Giass. Pyrea Ice' Pon, casi Iron. wrought Marble Masonry Mcnel StICOn

StainWss steel

Linea.'

Cubdcal1

Material

Linear'

Cubicar

13 3 10.4 10.2 6.0 80 92 4.6 4.9 1.8 28.3 5.9 67 2 lo 9 2.510 5.0 7.8 4.2 9.4

399 312 30.6 18.0 24 0 27 6 138 15.6 5.4 84.9 17.7 20 1 6 to 27 7.5 to 15 23.4 12.6 28.2

Suyo. case alloy

6.7

20 1

Steel. SAE 1020 Wood. Oak• Wood. oak' Wood. pino' Wood. pino'

6.7 2.7 30 3.0 189

20.1 81 90 9.0 56 7

e, pe, dg F

•eu

Acetrine Alcohol, ethyl Benzene Carbon tetrachlonde Mercury Petroksan. Penn Petra-le-pm. Casi Patri:Muna Tes Tachonarle Water

826 610 770 687 101

500 430 420 541 115

dep r x lo • 'Aloe. 59P - 4F to • 30F

I P ardid lo 9rale 'Access

ola"`

HEAT TRANSFER Ilcat Ilows Iron one reglen to anoth by thrce diffcrent methods: (1) by co duaion via transfer of kinelic energ betwccn particles or groups of particl al thc atomic leve(: (2) by radiation vi the emission of encrgy in the forro clectromagnetic waves: and (3) by vation via encrgy transfer by eddy mi ing and dilTusion. In practical enginecring work, hc transfer is usually by two or all thrcc thc abo y e methods. The net eltcct i often hard to predict and depends on way Ouids flow along the transfer sur' faces and on thc chape and material the transfer surface. Rapid flow in creases hcat transfer. U FACTOR. For practical purposcs, eng neer: &ten lump these factors into a 16

STEAM COIL PARAMETERS 13 Actual tube darnetev. in 3



A Length ol cod. 1!

1 0.6 0 3 0.2

2

10 20

40 60 100 X10

500 1000

-itrit9M1/411.4éi+114-rn-r- i 1 mi t (mit HM V II I t 1/11111411 1 2'3 4 6 O 10 20 30 50 100 700 300 \ 100 -60 40 20 ..-- -s. / \ E Overol' coeMiclent. U I.....1 DIAean temperature ditlerence

X

1 / ""(-....

2 \,r

Mea/ transmated.13tuihr

e 10

3 2

Y / ....... 4/

7"N

+-INDO( F 1

-....

4 5 0.5

)13

Answer -„ /0000 `50.000 10 20 30 50\ 100 1 200 500 1000 1 2 3 4 6 8 /0 20 30 50 100 200 500 1000 2000 5000

C Heating surface. sn II

FIGURING HEAT TRANSFER. Follow thesesteps to find heat transfer. connecting (1) Index. (2) 8 through X to C. (3) C to D. (4) E through Y to answer F

A

to

Power lisedtx•ok



• •

civerall "coefficient of heat transfer' adlcJ the U factor. in Biu/hr-sq fi-deg temperature difference. Each particular design of heat exchanger has a certain U factor for a particular service. As thc tablc shows. (hese facturs vary widcly. Heat transfer (in fitu/hr) = U factor X transfer arca (in sq ft) x 'man temperature difference. EXAMPLE. Use thc chart (bottorn. p 16) to lin() heat transfer through the wall of a 40-ft coil of 0.825-in. tubing if the mean temperature dilTerence is 14 dog F and the U factor is 160. SOLUTION. Follow steps and dashed line. In chart. The answcr. as plotted. is 9,500 Blu/hr.

RANGE OF U-FACTORS U-lar I o, a

Tipa a res ~unce, LIPA l0 load

laq ue lo howd Lepe lo oas !aun pros-esti temo to batel/ aquel 1.4540 tO ming halad Gas i aun pressuret to lepe Gas ¡atm prenote, to gas

Free

rapte

Tiptcal Dual

25.60 5.10 1.3 20.60 5-20 1-3

150-300 20-50 2-10

Water

c.;5: 1.Trei c,da.tan

—

50.150 25.60 2.10 24

Water

Hoi-water rap e Bone coceos

06.2

Gas (atm prensa/O/0 TOPO hAud Coodeneng rape to acopo Conoensng yapa to eteart Conoenswg vapor lo Imitad Condenen° vapor to Irgue ConOwrang y apar 10 gas ;Pm persone Condensan vapor lo Pealo 1.0p 0 Condenan; vapor lo bolina »ad CowrIoneng vapor topo:don apeo

1 -3 50-200 10.30 40 - 80 — 1-2 40100 3601300 50-150

al

Od

ex caten fp:nom:en Sleam supereeetws

240

150.800 20.60 60-150 15-300 2-10

Stearnarmer Deam.ca

Sleam 110kat Legua noten and ccnaensers

Orgarat ca001-waTer SlatAaal Matute Steam mas

ir

ara

heatetS

Scaiwfwnong evaporaron 51nm-water Steampel

13iu • hr•An callo F lot free and lorZed cnennolon

GAS LAWS -xcepl atar their condensing points, ost gases follow (hese roles: For a given pressurc, gas expands in xact proportion to absolute temperature V,/V, = TIT 2 ). For example, tripling bsolute temperature triples volume. For a given volume, the absolute ressure of a gas varíes in exact proponion w the absolutc temperature (13,/13, TIT,). For example. raising absolute emperature 30% raises absolute prest ure 30.11. For a given temperature, volume of a given wcight uf gas varíes inversely as Lbsolute pressurc (P, x V, = PI X VIL Wr example. doubling absolute pressurc ha 'ves thc voluIllt. These airee roles can be summed up in »le formula: Pxyzoky, be supplicd for complete combustion. Then the products of combustión will includc excess oxygcn (unchanged) plus nitrogcn oxides. The large amount of nitrogen in the air supplied performs no useful duty in the burning process. In fact, it can cause an air-pollution problcm: when combustion is not controlled properly. nitrogen combines with oxygen to form largo amounts of nitrogen oxides (NO,). In stoichiometric combustion, fuel reacts with the cxact amount of oxygen required to burn all carbon. hydrogen,

BURNING HYDROGEN. llydrogcn burns with oxygen te tocan %a ter. 11,0: 1120 2 H l + O, 4 parts + 32 parts -e° 36 parts + 8 lb 9 lb I lb + 61,000 Btu BURNING SULFUR. Although mainly a corrosion nuisancc. sulfur does contribute heat: S + 0, -w SO, 32 parts + 32 parts -w 64 parts +1 lb -lo 2 1b +3983 Btu 1 lb Of course, SO, is a pollutant which must be controlled if conccntrations exceed EPA-prescribed limits. In practice, combustión equipment is designed and operated to provide complete but not stoichiomctric combustión. and certainly, not incompleto combustion. Complete combustion assures that fuel isn't wastcd and permits flexibility in equipment operation. Converscly, incompleta combustion means inefficient fuel use. possibly hazardous operation. and addcd air pollution.

This 2.33 lb of CO can later be burncd to give up the rest of the 14,093 Btu originally in the 1 lb of carbon: +0,-2CO3 2 CO 56 parts + 32 parts -•• 88 parts 2.33 lb + 1.33 lb --•• 3.66 lb + 10.143 Btu

COMBUSTION REACTIONS Equations explained under "Chemistry of Combustion" were only the major reactions on a lb air/Ib fuel oasis. Very orlen in combustion calculations it's necessary to figure fuel, air, or combustion products on thc basis of volume, in cubic feet, or amount of substance, in mol es . Thcse have been calculated in the table bclow for theoretical air. How lo allow for the excess air that's needed for d'II-

cient combustion is cxplained in a following scction. FIGURING BY MOLES. A mole of any substance is dctined as a wcight in pounds cqual to the molecular wcight. Thc bcauty of using moles is that for mast gases the volume represented by a mole is pretty ncarly the same at a given temperature and pressure. (At 601: and 14.7 Asia, the volume of a mole of "per-

fect" gas is 379 cu ft.) This means that 1 mol of carbon is 12 lb. 1 mol of oxygen is 32 lb. and 1 mol of carbon dioxidc is 44 Ib. Thc number of molecules of a substance in an equation can be taken to mean the number of moles of that substance. So thc equation C + 0, = CO, actually mcans 1 mol C + 1 mol O, = I mol CO2 . It also mcans that 12 lb C + 32 lb 0, - 44 Ib CO:.

REQUIRED COMBUSTION AIR AND PRODUCTS FOR COMMON COMBUSTIBLES BURNED WITH THEORETICAL AIR REQUIREMENTS FON ONIN

Ab Finé

C

#1,*

S CO

Unto Mete Cu ft Lee Met a6 Lbs Oda Cu fl Lbs Mea Cu II Lbs Uds Cu ft Les Mote

0 1 11,

Q15

ellis

n

041 ,

Lb. MS Lbs Mole Co It Lea

O.

110112 C4 , FM OS - (5

N.

10 3.711 379 1425 32.0 105 1.16 05 119.5 712 16.0 26.6 1.0 3.76 1425 3,11 32.0 106 0.5 1.111 1MS 712 160 52.6 2.0 7.52 2093 758 64.0 210 2.6 9 40 947 5500 000 263 1129 3.0 1137 4260 96.0 316 3.5 13.17 1325 4960 112.0 369

N0,

HP

1.0 379 440 - 10 - - 1.0

NJ NO.

0,8

-

- - -

0.00132 0.5

0.0t22

POP OPE POMO OF FIEL PIM ME Met 0007 Of FUEL 05 preM (ea N.1_ Ca~ 411._ Al Al CO, 140 10. N. CO. /40 60. 0. 11. - - 0.01133 0313 0.013 11101 31.6 31.6 - - 2.67 - 367 05 0.00496 - 000264 - 0.250 0.940 - - 31501 1.88 - - 94.6 -

0.130

10 379 64 1.0 379 440 1.0 379 44.0 2.0 758 86.0 2.0 758 86.0 2.0 756 M.O

-

-

2.0 36.0 10 16.0 2.0 36.0

540

- 0.00132 0.5 0.0422 0.00529 2.0 03 60 O.0006 2.5 0.211 000719 3.0 0253 0.0093 3.5 0296

-

0.0475

-

-

-

0.03496 1.80 0139 O 01913 7.52 0.556 0.02411 940 0.004 0.0296 11.29 0.824 0.0347 13.17 0.972

0.10064 1.0 0.116 0.00264 1.0 0.116 000526 2.0 0232 0.00526 2.0 0.232 000628 2.0 0.232

-

000526 00950 0.0064

-

8.0

26.3

- -

0.0312 1114

0.1176

-

00475 0.00526 0.0950 0. 0079

-

0.1425

MI

3.29

0.067 25.4 114 0 470 116 13.17 0362 137 10.13 0.403 153 1129 0.439 1863 12.29

0.090 36.4 303

•

0.1071

-

40.6 1.43 0.1167 022 3.73

9.0

0.312 11.84 20

44.6

0.179 677 0.571 0.125 47.4 -

-

-

0.0367 13.53 1.57 0.0025

0125 2.75 225 0.070 0.0365 79.15 3.33 0.601 0.0714 0.0714 27.1 1266 3.14 0.0647 0.10 25.3 1.8 2.93 23.7

-

v dee ol e t. and combeehon troducta atan me toming el estropeo. The «Ve molecaer Ver)Mg assurnamon• for ~mi& w•Ight wooduce • 019111 inconekthincy In the ~phi ce hydrOgen te 202. bel the aPIXO XIC/14 1* yak.* 04 2 is usad in IleutIng W and combushon products

Power Handbook

51

EVALUATING FUELS HEATING VALUES. Combuslion results in release of thermal energy or hcat. The amount of heat generated by compldte combustion of a specific fuel is constan'. and is known as the heating value. It may be determined directly by measuring thc hcat generated during combustion of a known quantity of che fuel in a calorimeter. Or it may be estimated from chcmical analysis of the fuel and the heating valuc of the severa' chcmical elements in the fuel. lligher (or gross) heating value (HHV) is found when water vapor in thc products of fuel combustion is condensed, and thc latent heat of vaporization of water is included in the fuel's heating valuc. Lower (or net) heating value (1.HV) is obtained when latent heat of vaporization is not included. Assume HHV if not otherwise indicated; LHV is mainly used

for internal-combustion enginc fuels. Heating values are always given in relation to a certain reference temperature, usually 60, 68, or 77F, depending on practico in a particular industry. Set table at bottom of next page for heating values of key substancias in common fuels. Not all heat released during combustion can be used effectively. Thc largest hcat loss is in che form of incrcased temperature (that is, thermal cncrgy) of hot exhaust gases aboye the temperature of incoming air and fuel. Othcr losses includc radiation and convection heat transfer from outer walls of combustion equipment to thc cnvironment. AS-FIRED VS DflY. Heating valuc and analysis of fuel as-fired varíes with the amount of moisture associated with (he fuel. Thc moisture, in lurn, constantly

changes in amount during recovery, shipping, and storage. Thus, chemists and producers sometimes prefer to quote the analysis of dry fuel, and givc the moisture scparately. Thc calculations in Chis scction, however, are based on fuel as-fired. Data given on a dry basis musa be converted to the as-tired basis. For example, assume that coal on a dry basis has 12,700 Btu/lb and 12.4% ash. If the as-fircd coal has a moisture content of 4.5%, the conversion constant is 1.00 — 0.045 = 0.955. So multiply thc dry-basis figure by 0.955 to obtain the as-fired figure, or 12,700 by 0.955 = 12.130 Btu/lb of as-fircd coal. This shortcut method is not intendcd for largo powerplants, which have the testing equipment and personnel to makc complete analyses based on chcmical'inputs.

CLASSIFYING FUELS Key fossil fuels usad in powerplants today are shown in the cables -- coal, oil. and natural gas. Dependence on thc last two is decreasing as coal usage increas. Other fuels finding application in powerplants are liquefied petroleum (19) gases, wood products. process wastcs such as bagasse, blast-furnace gas, and black liquOr—and, of coursc, nuclear fuel in utility stations.

COAL. The complcx composition of coal makes classification into clear-cut types difficult. Chemically, it consists of carbon, hydrogen, oxygen, nitrogcn, sulfur, and a mineral residuo called ash. A chemical analysis provides some indication of coal quality, but does not define its burning characteristics enough. The coal usar is interested not only in availablc Btu/lb of fuel and the amount of

ash and dust produccd, but also in handling and storing properties and firing capability. A detail description of coal qualities and their characteristics can be obtained from LIS Burcau of Mines publications. FUEL OILS are broadly classified as dis'Nate (lighter oils) or residual (hcavier oils). As the cable on next pago shows, ASTM has cstablished specifications for

ASTM SPECIFICATIONS CLASSIFY COALS ACCORDING TO RANK

Class

Group

I Anthrechic

2 Anthradte 3 Sernianthracile2

1 Meta-anthraCite

II

Beuminous

1 Low-volatile bituminous coal 2 Medturn-volatile biluminous coal 3 HIgh-voladle A bituminous coal 3 Hige-volatile B letunenous coal 5 HIgh-volatee C 8/tumbaos Wel

Fiad ~Pon hmlla, % (Dry, mineral. metter-Pse bola) Equal or grial., Lea. Iban Iban 98 92 86

982 928

78

as

69

78 69

I Subblturninous A c0111

2 SubterumInous e coal 3 Subbltumlnous C 1 Lignito A — IV Upnllic — 2 lignito B The ~recae Socia" for Teming a Masera cusallication inca set toren in Standard III SubbItuminous

ASTM D-388. "Spectandlon for claaaliketiOn ol COM* by nra." dote nd inducie ter COela principal/y nontanded vsMUts. valla neve nauta physical and chaman Prepone*. and *Mal come *Minn the ImIte ol luxad Carbon or caldtnc value of the high-volatlie laturranorn and subteturelnotta nana* All 01 thoso coas *the oonteln lema unan 48% doy. mineral-manes-Me luxad cubas or ha y a more than 15.500 Btuilb on e mona. minerel-mano-free basa

52

Volad* ~bar limita, % (Dry, mineralmalter-lree basis) Equal Greater or Si. Plan teso

CalorNIC nave 1en14% Btu/lb (810808.1 rnéremel-reetter free barde)

Equal /peritar Iban

Len Iban

2 8 14 14 22 31

Agg lomera ling CbaraCter

NonagglonwatIng 22 31 14.0003 13.0005 11,500 { 10,500

14,000

Cornmonry aggiomeraeng•

13.000 11,500

AgglomeratIng

10.500

11,500

9.500 8,300

10,500 9,500

NonagglornsratIng

6,300

8.300 6.300 • 4011.1 Mere to tos otelatning Its natund inheent montuna bol net ~ling viable water on the audacia ol the co.' aggiorrerating, classity si the inv-volatile caoup oí the bitumnout Crea. `Cona hanng 69% or more naco carbonan the dry. mimwel-nuttter-free basa sha,' be clualhed accotding to n'ad cerbon, regardlen 01 ~ea vana is recognereo that theta may be nonagglarmatng venenos in th"O gratos ol the Monino°. dan. and there are notable exceptlona in Ir* NO volase c biturninOtia group Poner Handbook

ASTM SPECIFICATIONS CLASSIFY OILS ACCORDING TO GRADE Flash pan! F

Pour poinl F

Water and sedimen1 vol, %

Mas ash el. %

Saybolt vitcoeity, lee

Carbon residuo cm 10% troncos. %

Universal at 100F Grade Min Mas 1 100 0' 0.05 0 15 0 1$ 2 100' 20' 0 05 .... 0.35 132.81 137-91 4 130' 20' 0.5 0.1 ... 45 12$ 5 light 130' 1 0.1 300 >125 5 heavy 1 130' 01 >300 900 140 6 2' 19001 (9000, pala minen n me mem aterecone 'ron me Montan Scans kv benne a IMWeati' 51ended 0.3%, "Standard s peaker-en loe IuM oils " When reaing no Utile. nem n mes Mal the unaa men ry mounerwel ol • gomn grade das n01 au100na1beany place in oil un the n'id Ice« grade Leían d meen dt reauremenis el the knor gime Or bytai ,viscosay aves e paren:reme are la nmennunn °ni> sed mi recoser-Ny inane Iones or newr par peno: may le yese:sed -anee.« roya ed tg condices o/ molard use naco poto pool cm Man erro totalice /e ent•Mum nsoeuiy reo Grade No 2 Quo be 18 cSt 13703 $4,444) unieran and me mínimum %% pode rol be »oboe "Vheef emaela loe 01 it metwed. Grade

Mln

Sáneme* vieceeity.ests 1226 itu

ale:0F Min Mas

al MY Iw atm

Mm glantY. deg AM Sullw, %

...

1.4 .... 22 0.5' 15 20' 36 0.5• 30 15.81 126.41' — -1-898l ".. — .... ... >264 .... Legal /5Y .... (61) (23) >65 (1948 142) 1401 ---. 1.6981 45 . 303 Legal (92) (Saar "" be asno un ges al las and fres are meted . in ~vos cunde me II. eme mis ines may appry %hen lomeolle AM el e remeted. fuel cd loling e /No ynconly nroe ola lane.nunnbet.ed grado. aten m and .ntmsng NO 1, may be Stebbled / nOtOtrOrt orare9 m>9°9-16' are s79181/ the nrcoury ruge ve me num arme mai to 3dentellect and advance renco roa be manee aten char9ng from une vscosely toree lo ~met Die nate. shas in in sr-nr:ren: hne to penad in< mor te trake Ira r«ossary admenetreS 'The arome, ol soler by maniaten plus me ~men' by

....

mechen snáll nol ~veo 2% ore reune of rebose by ant a/ion ~1 net /COM 0.5%. A ~Gol n guasee mai be nade lo, Mi mane and 'cernera In t'ea./ Ot 1%

No 6 han ol me be CleSfreed as 10/poto 16C8 may ! ce Ngb- pour (no man Low.pow loe el toso

GROUP CLASSIFICATION OF NATURAL GASES* Nitrogen, Group 1 High mere type II Hign meMane type NI Hlgh Btu type

%

6.3-16.20 0.1-2.39 1.2-7.50

Specilic gravily 0.660-0.708 0.590.0.614 0.620-0.719

Methwe, %

71.9-832 87 6.95.7 850.90 1

Blu/cu It dry 958-1051 1008-1071 1071-1124

Ftem Gas Engmeees Handbook. 1985

fuel-oil properties which subdivide the oils into various grades. The specifications are based on required characteristics of fuel oils for use in different types of burncrs. Characteristics which determine grade classification and suitability for givcn applications are viscosity. flash point. pour point, water and scdiment contcnt.

carbon residue, ash, distillation qualitics. specific gravity, sulfur, heating value, and carbon-hydrogen content. l.ow-sulfur residual oils are being marketed in many arcas to permit users to mee! SO, emission regulations; here, the key classification is percent sulfur contera. NATURAL GAS. Typical compositions of natural gas distributed for use as fuel

includc: methane, 70-96%; ethane, 1-14%; propane, 0-4%; butane, 0-2%; pentane, 0-0.5%; hexane, 0-2%; carbon dioxide, 0-2%; oxygen, 0-1.2%; nitrogen, 0.4-17%. Some constitucnts, notably water vapor, hydrogcn sultide. helium. LP gases, and gasoline are removed prior to distribution. Natural gases are usually classificd as shown in the Cable aboye.

FUELS AND ANALYSES PROPERTIES OF SUBSTANCES IN COMMON FUELS MMeMer Combustible eutretences Carbon (lo CO) Carbon Do CO2 Sultur (lo SO2 Sulfur (lo 30:1) Hydrogon Carbon monoxide Methane

Acetylene Ethylene Ethane Butano Propone Hydrogen aullido

Othor *obstantes Nitrogen Oxygen Carbón dioxide Adr Water vapor

Molecular symbol C C S S 147 CO CH. CM, C21-le C2/46 C4/1(v C,H8 HoS

Molecutor symbol Oz CO2 H2O

weight (approximele) 12 12 32 32 2 28 16 26 28 30 58 44 34 Molecular weight (epproximate) 28 32 44 29' 18

Density, , IIIIV," Btu/lb lb/cu tt 3.950 4,093 3.980 5.940 0.00562 61,095 4.347 O 0780 0.0448 23.875 0.0732 21.508 0.0783 21.636 0.0801 22.323 0.168 21.321 0.126 21.669 7.097 0.0961

Dembly,' lb roto tt

0.123 0.0373

Al 14 7 pela ano 32F. or the sate/anon temperature II highor Iban 32F. toptior Hoaling Value. vilote my-luden IMMO he& 00 vapenealion 01 sale/. ~recen to 60F *Id 30 in. Hg. LOW Egenneont vese Hat1tIng Value detIS not include 151001 nem or veponzation

Power Handbook

MI fuels are madc up, in varying proportions, of the substances listed in the cable "Properties of substances occurring in common fucls." Not listed is ash mineral impuritics that don't burn but do present a disposal problem. Fuel oropel-tics are reponed in two ways —by ultimare analysis or by proximate analysis. ULTIMATE ANALYSIS lists the vario°. chenucal constitucnts—carbon, hydrogen, oxygen, etc—plus ash as percents by weight. Any moisture present, however, is broken down into hydrogcn and oxygen. PROXIMATE ANALYSIS. This is the more common foro of analysis for coal. It tisis fuel constitucnts as percents by weight of moisture, volatiles, fixed carbon, and ash. Volatiles (tarry substanccs) arc driven off when coal is hcated in an air-tight retort to makc cake; thcy consist mainly of hydrocarbons. When cake. in turn, is burned in air, the lixed carbon is driven off, leaving behind only ash. Sulfur is usually reponed separately, but its wcight is included in the wcight of ash for the proximatc analysis. 53



HOW TO USE ULTIMATE ANALYSIS llore is a typical ultimate analysis coal reponed by test:

f

% by ISIPM

Ash Sulfur Carbon Hydrogcn Nitrogcn Oxygen

10.49 1.20 71.98 6.47 1.16

8.70 100.00 MOISTURE, FREE HYDROGEN. I t is com.

monly assuttled that all the oxygen in thc fuel is present as moisture. Since 1 lb oí hydrogcn exista for cvery 8 lb of oxygén in thc fuel, the amount of hydrogen tied up as moisture is 0.087/8 0.0109 lb/Ib of fuel. Total moisture is 0.0109 + 0.087 = 0.0979 lb. or 9.79% moisture. Free hydrogen is 0.0647 - 0.0109 z• 0.0538 lb, or 5.38%. HEATING VALUE. Multiply weights of individual constitucnts by their heati0g valucs and add: C: 0.7198 x 14,093 = 10,144 H: 0.0538 x 61.000 = 3,287 5:0.012 x 3.983 = 48 Heating value. Btu = 13.479

This method is only approximate. also, heating value is usually mcasured when the ultimare analysis is made. THEORETICAL AIR. Multiply the wcight per pound of each combustible by the wcight of oxygcn each combines with: C: 0.7198 x 2.67 = 1.922 FI: 0.0538 x 8 = 0.430 5:0:012 X I = 0.012 Thcorctical Q = 2.3641b Since air is 23% oxygen by weight, the theorctical air required per pound is 2.36/0.23 = 10.3 Ib of air. EXCESS AIR. More oxygcn is required for efficient combustion than the theo. retical amount. In this example, assume coal is burncd with 20% excess air. F.xcess 0, will thcn be 0.2 X 2.364 = 0.473. and total O, will be 2.364 + 0.473 = 2.84 lb. Actual air supplicd is 2.84/023 = 124. Or. 1.2 X 10.3 = 12.4 lb. COMBUSTION PRODUCTS. Since matter cannot be destroycd in combustion, the products must be I lb coal + 12.4 lb air = 13.4 lb total products/lb of fuel. However, 0.1 lb ash (in Chis example)

will normally go finto the ashpit. so stack gases will be about 13.4 - 0.1 = 113 lb. The wcight of dry gas will [hen be total flue gas leas moisture content. N4oisture brought in by the air supply is usually ncglected, so moisture content is nine times the total hydrogen weight, or 9 X 0.0647 = 0.58 Ib. So dry-gas wcight is 13.3 - 0.58 = 12.7 Ib. made up of: Lb 0.2 SO2 0.024 20.8 CO2 2.64 3.7 O, 0.473 75.3 Ni 9.57 Total 12.707 100.0 The oxygen is the excess value alrcady figured. Nitrogcn is lb in fuel (0.01) plus lb in air supply (12.4 lb air - 2.84 lb 02) = 9.57 lb. SO2 and CO: are figurad as: 0.012 x 2 = 0.024 and 0.7198 x 3.67 = 2.64. CARBON MONOXIDE. Despite excess air, CO is almost always present in flue gas. First problcm. sce below, shows how to figure combustion for Chis sanee fuel, using flue-gas analysis with CO.

EXAMPLES COMBUSTION OF COAL

H25 10 Ory flue gas per ID COELI - coal + sir - ash - moisture = 1+ 16.32-0.1049 - 0.58 - 16.64 lb Glyen complete analysis of coal and nue gas. 11. Figurtng CO loas: determine (per lb of coal) theOrelical mr, actual sir, dry flue gas. moisture in flue gas: siso per Fraction of carbon burning to CO cent incas, air and pareen loas from incomplate = 0.6 + (10.2 + 0.8) - 0.0555 combustion Weighl of carbOn to CO 1 Ultimate analysis of coal (as flred).% by = 0.0555 x 0.7198 - 0.040 weight: Blu ioss from CO = 0 040 x 10143 = 406

Ash Sulfur HYdrollen Carbon Nitrogen Oxygen

10.49 1.20 8.47 71.98 1.18

8.70

100.00% 131u/lb

Flue-gas *a:SME'. % by volutas: CO CO 0, Ne

13.800

10.2 0.8 9.2

80.0 101«. Flgurtng moles of carbon and nItrogen In flue gas' nue gas % Mas gas Mols C Mola N, 10.2 10.2 10.2 CO, 0.6 CO 0.6 0.6 0, 9.2 9.2 N, 80 0 80.0 800 Total mola 100.0 10.8 80.0 Pounds carbon Manad - 10.8 x 12 - 129.6 Pounds coal burned = 129.6/0.7198 180.0 Moles air supplied •• 80.0/0.79 )4 101.3 Pounds air supplied - 101.3 x 29 2938 Rallo aa to coal = 2938/180.0 - 16.32.1b air/lb coal MOMIure produced per lb coal = 0.0647 x 9 = 0.0 lb

64

CO2 N2

Total

2. Oxygen, combuslion products for 100 mofa of fuel Mole 02

Pareen! peal loss from CO 100 x 406/13.800 - 2.94% 12. Theoretical alr per pound coal Lb 0, reclutad Lb of com- per Ib Lb O,. COmb required bustible Carbon 0.7198 2.67 1.922 Hydrogen 0.0647 8.00 0 518 Sulfur 0.012 0.012 1.00 Total



Loss oxygsn In the coal

NATURAL GAS G,ven voiurnetric analysis of a natural gas. determine the volumetrm analysis of the Ruegas II the fuel M Manad with 40% excess air (by volume)

1. Gas analysis by volume. % CO H2 CH4 C2H4

Gas CO H2 CH4 C2144 )123 CO2

0.60 1.62 94.30 0.15

Mols 0.60 1.82 94.30 0.15 0.25 0.85

nooded 0.30 O 81 188 60 0.45 0.37 -

Mole produced



190.53

0.40 190.13

96.05

TONO

CO2 0 60 94.30 0.30 0.85 96.05

H2O

302

1.62 188.60 0.30 0.25

0.25

190.77

__111 0.25

190.77

0.25

MIAMI 02

in fuel

2.452 0.087

Nel oxygen needed 2.365 lb Theoreacal air per lb coal 10.28 lb air - 2.365/0.23 13 Excess air - 16.32 - 10.28 - 6.04 lb per lb coal 14 Pe r cent e ! xco8s air = 6 04 !O 28 - 58.8V.

0.25 0.40 0.85 1.83 100.00

02

3.

-

Figuring nitrogen in flue gas N2 regulad 1c< 190 13 mola 02

- 1901383.76 714.9 mol 40% excess N2 (by volume) -040x714.9= 286.0 N2 in fuel 1.8 142 in ffue gas = 1002.7 mola 4. Unburned 0 2 ' n II" gas - 0.40 x 190 13 -76 0 mola 5. Volumetric gas analysls (H20 disappears) % by vol Gas Mols 96.1 8.2 CO2 76.0 6.5 02 0.0 302 0.2 1002.7 85.3 N; To 1175.0 l Total Power HandbOOk



COMBUSTION SHORT CUTS Many plants, cspccially smaller oncs, are not equipped to obtain complete fuel. fluc-gas, and rcfusc analyses. So chart and scales werc designad for use whcn only fíat-gas analysis and fucl's heating value are known. Numerous checks have shown this approximate method (example workcd out next pagc) to givc results within 2% of thosc obtaincd by figuring from ultimate analysis. CORRECTINO HEATING VALUE. Ileating value used should be the as-fired value. If it is reported for dry or moisture-free fuel, multiply heating value by one minus the moisture content to correct for as-fired

O

1 u.

value.

EXAMPLE. Coal with a dry-basis hcating value of 12,700 Btu/lb contains 4.5% moisture. As-fired heat value is 12,700 (1.00 — 0.045) = 12,130 Btu/lb.

1.5

0.5

1.0

0

Pe-cerl CO

of carbon INCOMPLETE COMBUSTION LOSS can be oshmated directly from the CO and CO2 reacia-1gs of the flue-gas analysis. Dashed line shows solution for problem on next pape

FUEL OIL Theoret cal 14

13

ANTHRACITE COAL

ar. lb per lb fuel 16

15

17 1.0 1. 1 1.2 Water produced lb por Ib fuel (CO2 + • % 15 14 13 12 9 11 10

8

lo 20 30 40 50 60 70 80 90 100 Excess afr, % I1iptylsity ‘Inly i t1 /1 11 1 1,11 , 9 1, t 1 sil 1 6

7 8 02, %

THEORETICAL AIR REOUIRED

9

10

water produced from free hydrogen

16



in fuel can be estimated directly from fuel's heating value

BITUMINOUS, LIGNITE, WOOD Theorelical



6

17

aro, Ib per b fuel 8 10

12

24 Heahng vales, 1000 81u/lb

19 18

9 10 11 12 13 14 15 Hentina Ireenidierneinnipidriempletuttnep dtrinill 1000 Btu/lb 0.15 0.20 0.24 Water produced, lb per lb fuel (CO7 + '/:CO) % 19 18 17 16 15 14 13 12 11 10 1//10/11/151;ililt/ii?,n\11111/RSI,,/,¡?...ILI/i 0 10 20 30 40 50 60 70 80 90 100 Excess ale, % I t y t; s l yyslyyyl i i \ 11 5 1..vlirl t wit u 0 1 2 3 4 5 6 7 8 9 10

8

Thuoeffical a f. Ib per Ib fuel

15

11

8

07. % and

NATURAL GAS 14

TI/ce/inca] en, !b per lb kie1 9 10

7

Hoating 22 1000 Btu/lb

1 2 3 4 5

1 2 3 4 5 6 7 8 9 10 11 12 Portan! fuere hoal wasted by incompleto comtfusbon

Hoebng value. 1000 Btu/lb



1.9 2.0 2.1 2.2 Water produced, lb per b fuel CO2 + '/CO) , % 12 11 10 9 8 7 6 ITT11111111111111%11 r y r I riply\ O 10 20 30 40 50 60 70 80 90 100 Excess air. % 1 2 3 4 5 6

7

8

9

10



02. %

EXCESS AIR is estimated from CO, + 1/2 CO,

0.15

0.20 0.25 0.30 0.35 Water produced, lb per lb fud

(CO2 + 1/:00) .% 18 17 16 15 14 13 12

O 10 0

20 30

40

1 2 3 4 5 6

0.40

11

10

50 60 70

7

8

80 90 100

9

Excess as, %

10

02% in %. If flue-gas analysis is accurate, O, should line up with CO, + 1/2 CO

PROCESS WASTES

MUNICIPAL REFUSE

Process wastes consist of gases (coke-oven, blast-furnace, CO, refinery, etc), liquids (black liquor, sludge, dirty solvents, paints, oily wastes, etc), and solids (bagasse, wood byproducts, rice hulls, corn cobs, etc). Average hcating valucs rangc from 575 (CO) to 21,800 (rcfincry) Rtu/lb.

Municipal refuse c,onsists mainly of paper (50.7%), food waste (19.1%), metal (10%), glass (9.7%), wood (2.9%), textiles (2.6%), other (5%). Average hcating valucs rangc between 5000 and 6000 Btu/lb. Kcy elements of rcfusc prcparation: shredding, separation, decomposition, anaerobic digestion.

Power

Handbook

55



SHORT CUT EXAMPLE QUICK FIGURING OF BITUMINOUS-COAL COMBUSTION aituMinOv$ COMM 13.800 81u•lb

Now of fuel 2 Meeting value as.brou 3 Room temperature 4 flue-geS temperature 5 Five-gas enraye»:

CO, CO O, .. 6 Balar ellkienCyllrOM tan

SOF 520E .

10 2% 0 6% 9.2% 74.4%

COMPUTATIONS: 10 2 lb..11, fuel Theoreticai ea. from atare . 0.35 Ity lb fuel 8 Water produced by burning *vedad* hYd r alien 1 • 0.40 10.1b fuel 9 Total watt» from bufona 04 coal - 0.35 • 005 = 17.9 lb/lb fuel 10 Actual ale - 175% of theOletiCal - 1.75 x 10.2 18.7 lb/lb luto 17 9 + 1 0.2 I I Tono fiue gas - ata • coa; - ashes 18.3 Ib i lb fue 12 Ory nue gas - total nue gas - water produced = 8.7 - 40 13 Sensible Real feas ,ndry gas dry-gas weighl x temperature rue x x 0.24 • 1930 Bluélb fuel spocir•c heat 18.3 a (520

HOW CALCULATIONS WERE MADE: kens / ro 6. This data must be give or obtained. ¡tem 7. Read from iheurelical-air scal bituminous coal. !tem 8. Read scale markcd water p ¡fuetd lb/1b fuel. This is water produc by burning free hydrogen in fuel: it d not includc moisture originally in fuel air. Air's moisturc can be neglected. Dem 9. The 0.05 lb of original moi cure/Ib coal is assumcd. not measur Amount is too small for crrors to important. Item 10. First add CO 2 and onc-ha f Ihe CO to gel 10.5. Entcr scale markc CO, + 1/2C0 to get 75% excess air. No that the same scale corrcsponds w 9 oxygen showing there is no serious crr in original gas analysis. tren 11. All rnight fed to a furna goes up the stack or into the ashpit. Mo4t of it goes up thc stack. Stack dischar per lb of coal cquals 1 lb plus air su plicd per lb of coal minus refuse per lb f coal. Rcfusc contains most of thc orig nal ash in the coal and somc unburn carbon. Some ash goes up the stac

14 Mofa!~ Reat loas 'lb water - 1000 • .5

lemplialw0 -

1000 t. 260 - 1260 Btu water lb waler•lb coal 500 Btu/Ib fuel 1260 x 0.4 3.2% 16 Incompleto-combatían Ion. from chut 131unb fuel 17 Heat onfance 14.0 1.930 a Dry-gas losa 3.8 b. ~Mute losa 3.2 440 C. Incomplete-oombustion loes 4.8 660 tinaco:agitad-10f loases 3.530 256 Total loases 74.4 11270 1, Uselul haat ....... 100 1000 fuel g. Heahng value. 18 Calcula!~ ol Real balance fleme: Dry-gas losa - 1930 13.800 - 14.0% Moesture loss - 500 . 13 800 - 3.6% 440 Sto lb fuel Incom pleta oombushon loss - 13.800x 0.032 Unaocounted toscos - 3550-440 -500- 1930 - 660 Bto , 11) fuel 8tu.lb 13.80050.744 10.270 USeful haat -

15 Montura Mar los, -

particularly with pulverized coal, but it is only a small percent of the total wcight of stock dischargc. Assume (for any bituminous or anthracitc coal) 0.2 lb of refuse per pound of coal, for Chis calculation. Ncglect refuse for gas or oil. (tem 12. Dry gas and water vapor in fluc gas must be separated to calculatc thc hcat carricd off by each. ¡tem 13. This is the largest single loss product of the weight of dry gas times thc temperature difference between stack and room times 0.24. the specific hcat of the gas. ¡tem 14. This formula is based on room temperature of 80E, but is close enough for 60 lo 100F. ¡tem 15. Product of the results of 'tenis 9 and 14. liem 16. Follow dotted fines as shown en largo chart, page 55. ^ tren /7. From previous computations and givcn data, 1111 in the underlincd figures. Use 13,800 Btu as 100% to fill in thc peleen, column for a and b and 8tu per lb fuel column for e. Use given boilcr efficiency (tul Line e is g minus f d is e minus a. b. and e.

-

UNACCOUNTED LOSSES include (1) loss of combustible to ashpit. (2) radiation from boiler and furnace. (3) accumulated crrors of measurements and computation. If unaccounted losses run over 5%. or if they vary much from test to test. look for carbon in the refuse, mistakes in data or computations, or unusual conditions in steam generator. CARBON LOSS. The short-cut method en chis pase assumes that no analyses other than fluc-gas are available, and so lumps carbon loss in refuse into Ihe unaccounted losses. If coal's ash content is known, carbon loss can be calculatcd from measurements of coal fired and refuse produced (assuming negligible carbon loss up thc stack). EXAMPLE. Ash in 1 lb of coal as-fired is 0.074 lb, refuse produced per Ib of coal is 0.096 lb. The refuse must contain 0.096-0.074=0.022 Ib of combustible per lb of coal tired. Since thc volatile has probably been "cooked" off, this can be assumcd to be pure carbon with a heating value of 14,090 Btu/Ib. so carbon loss in refuse is 0.022 X 14.090=310 Btu/lb coal.

STEAM GENERATION GADD," CONDENSER PERFORMANCE Condenser-performance cal culations a !nade up of two parís: (1) the healttransfer function co be performed, a ' (2) the physical characteristics of tte condenser necessary to perform th t function. Assumc a hcat load of 2 Iltuar is to be condenscd at an absolu e pressure of P in. Hg. using cooling water at a temperature T, (all temperaturas are in degrees F). Then considcr e basic temperature rclations in a surta 56

condensa by referring te thc figure on the next pago. In a saturated steam mixture al absoluto pressure itereis a corrcsponding The difference steam temperature benveen 7', and thc cooling water temperature (T1) is defined as the initial temperature difference. 1TD. As the cooling water progroses clown the length of condenser tubas. its temperature risas continuously until it exits at

temperature T2. The difference between exit and inlet cooling-water temperatures is called the temperature risc. TR. The ratio of TR to lTD is defined as R. The difference between 7', and TI is known as the terminal temperature difference (17'D) or thc approach (A). T/I)should not bc less than 5 deg E From fundamental hcat-transfer relations, the log mean temperature difference (1.MTD) is defined as: Power Handbook

LMTD = (TM- TTD)/In(ITD/TTD) 11 = velocity of cooling water with= TR/In(1/1 - R) (1) in tubes, in ft/set Ry definition. the denominator of Eq 1 is For typical surface-condenser calcula- 1Th E, so that R = 1 - (1/e)". Valucs of R lions. thc following is a summary of Che are found from 'rabie 6. foregoing in normal order of use: The rcquired cooling-water flow (in P corresponds to T, (Table 1) gpm) can be calculated from the heat T, - 1TD balance: Eq 3 GPM = Q/(500 X IR) (2) K determines R (Tabla 6) sincc 500 lb/hr = I GPM. TR = R x ITD From a study of the heat-transfer Eq 2 characteristics of condensar tubes, the (4) IV, (GPM x 344)/(g X V) following rclationship is Irue: where N, = number of tubes (total), and K=(L)N,,XaXbXfXF,)/1," g = GPM/tubc at 1.0 ft/sea (Tabla 2). (3) Finally, where S = (N,)(L) X ni (5) L = effective sube length, in ft where S = sq ft of surface arca (total). IV, = number of cooling-water and m = sq ft/lineal ft of tubo (Tabla passcs 2). a = constant that varios with sube Note that the ovcrall cocfficient of diameter and wall thickness heat transfer (U) is nol apparcnt in the (Tabla 3) foregoing equations. Actually, thc (da= tube-material corrección fac- tion = (C X V" x xfx fe)/100 tor (rabie 4) has been absorbed into thc form of Eq 3. J = cooling-water inlet-tcmp Values of U for any condition can be corrcction factor (Tabla 5) calculated from = tuba o; aran cleanlines. factor = (K x riPtt x 500)/S

TI

1

• ea.

1-TEMPERATURE OF

SATURATED STEAM Absalule pressure. in. Hg

1. 1 . ind condensar cooling "atar and surface requirements. know. 1,Ig ¡bese conditions: Q = 1000 X 10' 411/hr, P = 2 in. Hg abs, T, = 75F 1Ict cooling water, and V = 7 ft/sec. .sume the condensar to be single-pass, Y:1 alean, with 1-in.-0D. 18 BWG tubes having a 50-ft cffcctive cngth. SOLUTION. The calculation follows Chis

T, =101.1F (Table I) lTD =101.1 - 75 = 26.1F K =150 x I X 0.069 X 1.0 x 1.025 (by interpolating) 0.851/7 0 ' = 114 R =0.680 (Tabla 6. by interpolating) TR =0.680 X 26.1 = 17.8 GPM = 1000 x 10. /(500 x 17.8) = 112.400

sequenic:

2 -VALUES OF CONSTANTS 'm' AND 'g' Tubo

Y. Ve 1 116 154

Aro/•84 lagth (m), W1 0 164 0.196 0.229 0.262 0.295 0.327

C(V") 267 767 263 263 259 259

11 0.88

1.04 148 1.99 2.58 3.25

58.8 640 68.4 72.3 75.9 79.0 82.0 84.7 87 2 89.5 91.7 93.8 95 8 97.7 99.4

1.0 1.1 1.2 1.3 1.4 1.5 1.6

1.7 1.8 1.9 20 2.5 3.0 3.5 4.0 4.5

101.1 108.7 115.1 120.6 125.4 129.8 1318

5.0

= 112,400/(1.99 X 7) x I 8068 S =8068 x 50 x 0.262 105,700 sq ft Note: ITD - TR = 'ITD = 26.1 17.8 = 8.3 deg F, which is greater iban 5 dog. as specified.

3-VALUE OF CONSTANT 'a'

FOR CONDENSER TUBING

G014 per lube al 1.011 sic velocity (g) for well BWG

In U .4

empereiure.

0.5 0.6 07 O8 09

PROBLEMS AND SOLUTIONS EXAMPLE

T2

LMTO

Tubo

19

20

21

0.72 1.09 1.53 2.05 2.63 3.33

0.75 1.13 1.59 2.12 2.73 3.41

0.77

0 79

115 1.61

164

215 2 76 3.44

2.18 2.80 3.49

TuS well gano. BWG

OD, In.

118 Y.

'A 1 1'4 n.

12

14

16

18

20

22

24

0.215 0.151 0.114 0.092 0076 0 065

0 169 0 127 0 098

0 146 0 111 O 089 074 O 063 0 055

0 116 O 093 0 076 0 065 0 056

O 110 0 089 0 073 0 063 0 055 0 049

0.106 0 085

0.081

O 129 0 101 0 082 0 069 0 059 O 052

-

0 068 O 059

o050

0071 0.062 0053 0.048

4-TUBE-MATERIAL CORRECTION

FACTOR 'b'

5-INLET-WATER CORRECTION FACTOR 'f'

Tul» wall gauge. BWG Tubo material Adrriralty metal Arsenital copper Aiummum bronco AIummurn Draw 90/10 Cu/Ni 70/30 Cu r éte low-carbco stee1 Staenless Med. l'ype 304.316 1 dannon

POIVIW Handbook

12 0.87 0.87 0 84 0.14 0.74 0.64 0.75

14 0.92 0.92 090

0.84

14 0.96 0.96 0 94 0 94 085 0 77 0.86

0.49 -

0.56

0.63

-

-

0.90 0.80 0 71

141 1.00 1.00

Intel komp, F

20 1.02

22 1.04

24 1.06

0.97 0.97 0.90 0.82 0.91

102 1.00 100 0.94 0.87 0.95

1.04 102 1.02 0.97 090 0.96

1.06 1 03 1.03 0.99 0 93 1.00

0.69 0.71

0.75 0.77

0.79 0.81

0.83

40 42 44

0.85

46

30 32 34 36 38

1 0.550 0.574 0.601 0.628 0 655 0.683 0.707 0.733 0.760

Intel terne. F 48 50 52 54 56 58 60 62 64

1 0 785 0 810 0 833 0.855 0 875 0 895 0.915 0.934 0.951

In14I lamia. F 66 68 70 72 74 76 78

80 82

11941 lemp, 1

F

I

0.970 0 986 1.000 1.010 1.020 1.029 1.037

ea

1 057 I 063 1.069 1.075

1045 1051

86 88 90 92 94 96 98 100

7080 1 085 1.090 1.095 1.100

57

EXAMPLE 2. Find tubo Icngth, condense" surface, and GPM for a fixcd TR, know ing that Q = 1000 x 106 Btu/hr. P

= twice N, for single-pass, or

6-VALUES OF R VS K

11,468

x 30.9 x 0.262 = 92,800 sq ft 1.50 in. Hg abs, T1 = 60F, V = 8 ft/sed, TR = 20 deg E (or 7 -2 shall not naced EXAMPLE 3. Find pressure for a givcn 80F). Assume the condensar to have condcnscr surfacc, knowing thesc condi100% alean, 1-in.-0D, 22 BWG, Typo tions: GPM = 175,000, S = 300.000 sq 304 stainlcss-stcel tubos. ft, with %-in.-OD, 18 BWG aluminum brass tubos, 36-ft effectivc Icngth, twoSOLUTION. Follow this sequence: pass dcsign. The heat load = 2500 X (Table I) T, =91.71" 106 Btu/hr for 8OF cooling-water temlTD =91.7 - 60 = 31.7 deg E peratura, with 85% cicanliness. 1,2 =TR/ITD = 20/31.7 = 0.630 SOLUTION. Starting with Eq 5: K =0.995 (Table 6, by interpolalN, = 300,000/(0.229 x 36) = ing) 36,400 Rcwriting Eq 3 as follows: V = (175,000 x 2)1(36,390 X LXN,=(Kxl,")/(aX6X.NF) 1.48) = 6.50 ft/seo = (0.995 X 800)/(0.063 K= (36 X 2 x 0.082 X 0.97 X 0.79 x 0.915 x 1.00) 1.045 x 0.85)/6.50" = 1.99 = 61.8 ft R = 0.863 (Tabla 6, by intcrpolatThus, with the fixcd temperatura risa, ing) only one total Icngth of cooling-water TR = 2500 x 106/(175,000 x 500) travel is possiblc when the vclocity and = 28.57 deg F diameter are fixcd. Thc condensar can be TR/R = 28.57/0.863 = 33.1 lTD = two-pass with 30.9-ft cffcctivc tubo deg Icngth, or single-pass with 6I.8-ft effecT, = lTD + T, = 33.1 + 80 = t'ya tubo Icngth. Thcn, for single-pass, 113.1 F 61.8-ft tubas: P = 2.84 in. Hg (Tabla 1, by interpolating) GPM = 1000 x 104/(500 X 20) 100,000 Howcvcr, lTD - TR = 33.1 - 28.57 N', = 100,000/(2.18 x 8) X 1 = 4.53 deg E, giving a TTD of Icss than 5734 5 deg F. Continuing the example, lTD = S = 5734 x 61.8 x 0.262 28.57 + 5 = 33.57F and 92.800 sq ft 7', = ITD + 7', = 33.57 + 80 = I I 3.57F. For doublc-pass, 30.9-ft tubos,

X

S = 11,468

025 026 0.27 028 029

0 221 0229 0237 0244 0252

055 056 057 058 0.59

0 423 0 429 0 435 0 440 0446

085 066 0.87 088 089

0.519

0.30 0.32 0.33 034

0 259 0.267 0274 0.281 0 288

060 0 61 062 0.63 0.64

O 451 0457 0 462 0.467 0.473

090 091 092 0.93 094

0.593 0.598 0.602 0 605 0 609

035 036 037 038 0.39

0.295 0 332 0309 0316 0.323

0.65

0.478 0.483 0.488 0493 0 448

0.95 096 097 098 099

0613

0.66 0.67 0.68 069

0.40 0 41 042 0.43 0 44

0 330 0.336 0.3.43 0.350 0 351

070 0.71 0.72 0.73 0.74

0.503 O 508 0 513 0 518 0.523

10 11 1.2 1.4

0.632 0.667 0.696 0728 0 753

0.45 0.46 0.47 0.41 0.49

0 362 0 369 0.375 0.381 0.387

0.75 0.78 077 078 0.79

0 528 0.532 0 537 0.542 0.546

1.5 16 1.7 1,8 1.9

0 777 0 798 0.817 0 135 0850

0.50 0 51 042 0.53 0.54

0.394 0.400 0 406 0.411 0.417

0.80 0.81 0.82 0.83 0.84

0.561 0.555 0.560 0.564

2.0 2.1 22 23 2.4

0 865 0 878 0 889 0 900 0.909

031

0566

1.3

0573 0.577 0.551

0.565

0617 0.831 0.625 0.628

Thus, P =

2.88 in. Hg (Tabla 1, by interpolating)

From 'Surface Condenses Calculations.Ecolaire Condense. Inc. Calcula:ion constatas are in accord Irish 'Standards for &rant Surface Condensers, - Heat Exchange Institute, 7111 edition.

FIGURING BOILER EFFICIENCY In these energy-critical times, knowing where w, = stcam flow, lb/hr how to find boiler efficiency is important. lt is a key indicator of unit per= reheated steam flow, lb/hr formance and tells both designer and operator if encrgy is bcing wastcd. Rabear«, Two mcthods of finding ovcrall cflisteam, ciency are included-using steam tablas, which is more accurate than applying a •• nomograph, the second method. The nomograph mcthod is recommended for quick, on-thc-spot answcrs of rcasonablc accuracy. 000

e

WITH STEAM TABLES This efficiency is figura' according Ito the ASME mcthod. Enthalpies can be found in the steam tables on pagas 21 and 22. Remember that hf stands for cnthalpy of liquid and hi for enthalpy of saturated vapor; also set the adjactnt drawing. Thc basic cquation is: OVERALL EFFICIENCY, steam generator = Output/Input, both in Btu/hr (1) in which Output = w,(11, - 114 + wÁho 2) + B(hfi - his) 58

= blowdown, lb/hr and in which Input = F X HHV

enz

(3)

Steam out, Ws

000 000 000 000 000 00 • 000 000 000 000 000

Fue'

74

Eitomlown

FesoWater

STEAM GENERATOR showing where to measure temperatures, determine steam and feedwater flows. and find enthalpies (from steam tablas) for figuring efficiency Pene Handbook



where

F = fuel input, lb/hr (as fircd) 1111V = fuel higher hcating valuc, Btu/lb (as fircd)

EXAMPLE 1. Find the overall efficiency of a steam generator (without reheatcr) producing 56,000 lb/hr of 600-psia, 8001 ; stcam. Continuous blowdown = 2500 lb/hr, and feedwater crucis the economizer at 300F. The furnacc burns 5960 lb/hr of coal, which has a IlliVof 13.106 Btu/lb. SOLUTION. Substituting in Eqs I, 2, 3: Efficicney = 156,000(1407.7 296.6) + 2500(471.6 - 269.6)1/(5960 x 13.100) = 0.823. or 82.3%

ECONOMIZER EFFICIENCY = Hcat absorbed/Hcat availablc. in Btu/hr (4) in whieh Ileat absorbed = wy(hfi hl)), where wr = fcedwater flow, lb/hr and in whieh Hcat available = II, x E where Hr = heat available in flue gas,

Btu/lb of fuel hcat available in dry gas and in fluc-gas vapor (1, - r40.24G + (1, -'A)0.46 (Al, + 8.94H2 + M„{G - 151, - 7.94(H, - 02/8)11 (5)

where G =

[1 ICO2+802+7(N2+CO)J/ [3(CO2 +CO)] x (Cb+S/2.67) + S/I.60

Al, = Ib moisture

(6)

per lb fuel burned

= lb moisturc per lb dry air te furnacc C, = lb carbon burned per Ib fuel burned = C - RC, C, = lb combustible per lb refuse R = lb refuse per Ib fuel Note: 112, Ar2, and 02 in FA 5 = lb of cach cicmcnt per Ib of fuel (as fircd); and CO2 . CO, 02 . and N: in Eq 6 = percentagc parts of volumetric analysis of dry combustion gas entering economizer. S in Eq 6 = lb sulfur per lb fuel (as fircd). EXAMPLE 2. From the previous example, assumc thc steam gcncrator burns coal with an ultimate analysis of 68.5% C, 5% 11 2 , 8.9% 0,, 1.2% ,V,, 12% S, 8.7% ash, and 4.5% moisture. Air enters the unir with dry-bulb/wct-bulb temperatures of 63F and 56F, respectively. The psychrometric chart on page 12 shows 56 grains of vapor heid bv each lb ( = 7000 gr) of dry air for Chis condigan. Carbon in refuse is 7% whilc rcfuse is 0.093 Ib

per lb of fuel. Feedwatcr 'caves the economiza at 370F and cnters at 300F, whilc %e gas enters at 850F with an analysis of 15.8% CO,, 2.8% Oh and 81.4% A', (CO

is negligible).

SOLUTION. Beginning with the numerater of Eq 4, Hcat absorbed = (56,000 + 2500)(342.8 - 269.6) = 4,282,000 Btu/hr. Note: 342.8 is found by interpolation. Next, carbon burned is round from G =C- RC Ce. = 0.685 - 0.093 X 0.07 = 0.678 lb/lb fuel Thcn, heat availablc in dry gas is found from Eq 6, G= Kit x 0.158 + 8 x 0.028 + 7 x 0.814)/(3 x 0.158)1 X (0.678 + 0.032/2.67) + 0.032/1.60 = 11.17 lb/lb fuel Solving for heat available in nue gas, Hr = (800 - 300)0.24 x 11.17 + (800 - 300)0.46(0.045 + 8.94 x 0.05 + (56/7000)(11.17 - 0.678 - 0.12 - 7.94(0.05 - 0.089/8))1 = 1472 Btu/lb fuel Thus, heat availablc = 1472 x 5960 = 8,773,000 Btu/hr and cconomizer efficiency = 4,282,000/8,773,000 = 48.8%

100 -

150 -

100

200 -O

go u. 250

C-1119

a E

ó 300

2

u-

so

E o

350

- 70

400 - _ - _

- €0 450 -•-

500 2,-50 NOMOGRAPH P CrlArer Handb004‘

helps find overall efficieney of steam generators

• -

E 59

AIR HEATER EFFICIENCY = Heat absorbed/Ileat available, both in Blu/Ib 1 fuel (7) in which Hcat absorbed = t,)(0.24 + 0.46/14) where = air flow through heatcr, lb/11X fuel = A A = total air to furnace, lb/lb fuel =G C.- N, - 7.94(14 02/8) G = similar to cconomizer buti based on gas at furnace exit = externa' air provided by fan or other source, lb/lb fuel and in which Hcat availablc = (15 - 11)0.24G + (/ - r,)0.46(Aff + 8.9411, + /144)1 where G and A are similar w G and A in the /wat absorbed equations, but are based on gas entcring the heater. Other factors are thc samc as for the economizer. EXAMPLE 3. Thc steam generator in the previous example has this gas analysis al the air-heater inlet: 15.0% CO,. 3.6% 81.4% N2 (CO is negligiblc). At the furnace exit the gas analysis is: 16.0% CO2, 2.6% 02, 81.4% N,. Air cntcrs the air hcatcr at 63F with 56 grains of vapor per

lb of dry air, and Icaves at 480F. Gas enters the air heater at 570F: 15% of air lo the furnace comes from a mili fan. SOLUTION. First, solving G for gas at furnace exit and using the result to find A to furnace, G= [( 1 1 X 0.16 + 8 x0.026+7 x 0.814)/(3 X 0.16)] X (0.678 + 0.032/2.67) + 0.032/1.60 = 11.02 lb/lb fuel A = 11.02 - 0.678 - 0.012 7.94(0.05 - 0.089/8) = 10.01 lb/lb fuel. Then Heat absorbed = 10.02(1 0.15)(480 - 63)[0.24 + 0.46 (56/ 7000)] = 867 Btu/lb fuel Solving G for gas entering heater and using the result to find A to heater, gives [(11 X 0.15 + 8 X G 0.036 + 7 X 0.814)/(3 X 0.15)] X (0.678 + 0.032/2.67) + 0.032/1.60 = 11.71 lb/lb fuel A = 11.71 - 0.678 - 0.012 7.94(0.05 - 0.089/8) = 10.7 lb/lb fuel. Then Hcat available = (570 - 63)0.24 X 11.71 + (570 - 63)0.46[0.045 + 8.94 x 0.05 + (56/7000) x 10.7] = 1560 Btu/lb fuel and Air-heater efficicncy = 867/1560 = 0.5 5 6. or 55.65

WITHOUT TABLES I iere's a nomograph that can be used to check overall boiler efliciency without the need to refer to steam tables. To apply the nomograph on page 59, operating pressures and temperatures must be known, plus fuel hcating value and evaporaban as pounds of steam per pound of fuel. EXAMPLE. In one week, a boilcr produced 22-million lb of 750-psia, 900F stcam while burning 1345 lons of 12,000-Btu/lb coal. Feedwater temperature averaged 260F. Thc average evaporation was 22,000,000/(1345 x 2000). or 8.18 lb steam/lb fuel. SOLUTION. On the nomograph, conncct 750 psia on the stcam-pressure scalc with 900F on the stcam-temperatura scale, and extend a straight fine to interscct pivot line no. 1 at A. Connect the intersection with 260F on the feedwatertemperature scale and mark thc intersection with pivot line no. 2 at B. Connect with 8.18 on the R scale and extend the line back to pivot lint no. I. intersecting at C. Connect C with 12,000 Btu/lb on the fuel-heating-value (H) scalc, and extend line to steam-generator-efficiency scale at right to obtain a value of 83.7% for overall efficicncy.

STEAM TURBINES FIGURING TURBINE PERFORMANCE Turbincs convert the energy in steam to shaft energy. How effectively this is done is always of interest in selecting and operating thcsc units. Turbine performance is measured in several ways: STEAM BATE is the amount of stcam requircd by a turbine to produce a givcn unit output, usually expressed in lb/kWh. It is of dire,ct practica] concern in figuring boiler steam needs and performance changes of individual turbincs. It has only limited value for comparing different turbines becausc it doesn't reflect changes in throttle pressure as does hcat ratc. HEAT BATE is the amount of energy nceded to produce a given unit output,

usually cxpressed as Btu/kWh. Because it dcals with basic heat units, it is better than stcam rate for comparing turbines. THERMAL EFFICIENCY is the ratio of energy output to energy input. It is dircctly rclated to hcat rate; that is, thermal efliciency = 3413/heat ratc, in which there are 3413 Btu in each kWh (see chart bclow). ENGINE EFFICIENCY is the measure of how well an actual turbine compares with an ideal one. Expanding from givcn throttle lo ex haust conditions, a pound of stcam does maximum work if expansion is at constan) entropy. This ideal condi. tion is nevar attaincd. When actual work done by a pound of cxpanding stcam is

divided by ideal expansion work, the resulting ratio is called the cngine efliciency. It serves as a measure of design effectiveness. How to find stcam ratc and thermal efficicncy for all types of turbincs is shown on the (swing page, How to find hcat rate and engine cfficiency for key types of turbincs is also shown. PERFORMANCE VARIATIONS. Stcam ratc. heat ratc. thermal cfficiency, and cngine efficicncy plotted against shaft or gcncrator output in kW' show turbineperformance variation with load. The common method for showing performance for autornatic-extraction and mixed-pressure turbines plots total hour17.Mr3P-4931 re N nit she

PIerrnai offiCCACy.

25

24

23

22

21

20

19

15.000 14.000 15.000 16.000 17,000 ia.000

le

11

6 15

10.000 20.000 21.000 22,000 73.000

sirkidrz,) CONVERSION SCALE pernuts quick transfer from thermal efliciency. Ve, lo heat rete, Btu/kWh, for usual ranga 60

Pont Handbook

BASIC CYCLES on which turbines operate affect performance factors. Text below shows calculation procedures

Sleetn w,

Sisan]

Mesen In A «11?

‘x

Sol%

• T.,/ Fxhausr num To process Use actual h„ To waste: use saluiabon h, cOodS,Oond3 g lo ~sume

NONCONDENSING OR BACKPRESSURE UNIT HEAT RATE, Btu• kW] • w.(b. - h')/ P

n

when steam exhausts lo process h' = actual exhaust enthalpy.13tu/lb

HEAT RATE, Btu/kWh = [(w, - wgXh, -

- 11,1 + wfh, - h.)) where

saturated-water enthalpy at exhaust pressure. Btu/Ib (see MolIler chart) h„ = vaporization enthalpy of leakoff steam al discharge pressure. Btu/lb ENGINE EFFICIENCY, % = (sama as aboye)

ALL TYPES OF TURBINES STEAM RATE, lb kWh -

w,/ P

total steam llow, lb total generator or shaft output.

steam leaving turbine system from glands and 'cake lb/hr = exhaust-steam enthalpy at entropy o/ initial steam, Btu/Ib h„, = leakoff-steam enthalpy at entropy of initial steam. Btuílb

THERMAL EFFICIENCY, % 3413/heat

ly steam flow against output for a range of extraction flows. OPERATING HINTS. In day-to-day operation, reading steam pressures at various turbine stages for known loads or flows, serves as a good indicator of intcrnal changas affecting operation. Whcn thc turbine is first installcd. usc properly calibrated measuring instrumcnts to takc data for curves like those shown on thc next pago. Thcn takc similar rcadings at intervals during the turbinc's operating schedule and compare them with thc initial curves. If throttle steam conditions and cxhaust pressure vary appreci•

ably, stage pressures must be corrected before comparing with thc curves. Turbine manufacturcrs supply such correction data. As long as steam conditions and internal parts stay the same, pressurc readings will fall on the curve linos. If pressures show increasing percentage deviations from these curves with time, &posits are plugging nozzle and blade passages. Sudden changcs in pressure readings my be caused by interna] damage or changas in steam extraclion. Check pressurc gaga and line before opening thc turbinc.

Power Handbook

kWh

tate. Btu/kWh in which

- hj 4

w(h, - 11,31/P where

when

where w. P -

HEAT RATE, Btu kWh = (w.(h.

+ wat, - h„)]/ P

when steam exhausts to waste h' =saturated-water enthalpy

at exhaust pressure Blu/lb (see Mollier chart) ENGINE EFFICIENCY, % - 3413 P/((w,

REGENERATIVE FEEDHEATING UNIT

STRAIGHT-CONDENSING UNIT ti

Btu/lb (see Mollier chart, page 62) P - total generator or shaft output, kWh

Bakti lvedoanp r•eawaler lo bode/

where

w, = total steam flow, lb h, - steam enthalpy entering turbine,

Saltitation h, coowsponcling lo exhaust progsore

3413 = Btu in each kWh

enthalpy of feedwater leaving last heater. Btu/lb enthalpy of feedwater leaving boiler-leed pump. Btu/lb k = enthalpy of feedwater entertng boiler-leed pump, Btu/lb (see Mollier chart) ENGINE EFFICIENCY, %=3413 Pl[wg,(h, h

+ h„,) + wu(h, - h„) + wje(h, - h,,4 + wg(h, - h„) + wfh, 11w))

where 14. Me,

lb/hr

NI,

bleed steam flows.

= enthalpies of bleed steam at initial steam entropy. Btu/lb - exhaust steam flow leaving turbine. lb/hr exhaust-steam enthalpy at initial h,„ steam entropy, Btu/Ib

h„„ h„g.

EXAMPLE 1. A regenerative turbine produces 11,600 kW at its generator terminals when taking 100,000 lb/hr of steam at 800 psia and 900F, and holding condenscr pressurc at 1 in. Hg abs. Stcam flema and conditions at' the thrcc blecd points are: No. 1: 12,220 lb/hr at 230 psia and 1332 Btu/lb No. 2: 9470 lb/hr at 50 psia and 1210 Btu/lb No. 3: 7400 lb/hr at 1088 Btu/lb

7 Isla and 61

1.5 1650

1.6

Entropy. Mor os per 1.7 1.8

100

20

SiOaln

~Ye-.

75 1600 5017

17 So g 25o.2

1550

0

/..p Weed

0

100 25 /5 125 50 l 'e r ceni oi mleO papaciyo. ram1stear e 110w TURBINE CURVES, plotted when machine is new, are watched for

1500

deviations that indicate fouling, plugging, blade wear, etc 1450

1400

1350

1300

Y 10 20 40 60 100 200 400 1000 2000 4000 10.000 ii4/n steam presswe. IXS1a AVAILABLE ENERGY for condensing turbines depends on inlet steam pressure and temperature as well as backpressure.

1200

1150

12,000 1100

11,000

Y 10.000

1050

á 9000 1000 8000 950 7000

6000

1400 600 1000 1200 181ot stoem temperature, F THEORETICAL HEAT BATE of condensing turbines generally

900

200

400

600

improves with rising throttle-steam pressure and temperature 850 1.5

1.6

1.7

1.8

Entropy, B/u/18 per 0e9 F

1.9



850 2.0

IMOLLIER CHART is handy for estimating turbina performance.

Section shown is the most-used area of the complote chart

High-pressure gland-leakoff steam goes to No. 1 heater. Enthalpy of feedwater entering che feed pump is 245 Btu/lb: it 'caves at 247 Btu/lb. Feedwater temperature Icaving the last heater is 389F. Find hcat ratc. thermal efficiency, steam rate. and engine efliciency. SOLUTION. From the Mollicr chart on page 62. h, = 1455 Btu/lb. From thc steam cables. by interpolation, = 363 Btu. Thcn [100,000(1455 — 363) Heat rato + 100,000(247 — 245)1/ 11,600 = 9431 Btu/kWh Thcrmal efficiency = 3413/9431 = 0.362, or 36.2% Steam rate = 100.000/11,600 = 8.62 Ib/kWh

Locate intersection of 800 psia and 900F in chart at Icft. Entropy is 1.64. Drop straight down at chis cntropy to 230 psia. where hé, = 1300 Btu/lb. At 50 psia, haz = 1160 Btu/lb: at 7 psia, h,, •c 1025 Btu/lb. At 1 in. Hg abs. h, 882 Btu/lb. Exhaust steam. w, = 100,000 — 12,220 — 9470 — 7400 = 70,910 lb/hr. Denominator of engine-efficiency equation = 12.220(1455 — 1300) + 9470(1455 — 1160) + 7400(1455 1025) + 70,910(1455 — 882) 48,500,000 Btu/hr. and Engine efficiency = 3413 x 11,600/48,500,000 = 0.816.81.6% EXAMPLE 2. Steam is supplicd to a 5000kW t u rbine at 1000 psia and 800F: it

exhausts at 3-in. Hg abs backpressure. 1f the cngine aiciency is 74%, what are the heat rate, thermal efficiency, and steam rate? SOLUTION. From bottom drawing, facing page, the thcoretical hcat ratc 9000 Btu/kWh. The actual hcat mit = 9000/0.74 = 12,160 Btu/kWh. cfficicncy = 3413/12,160 = 0.281, or 28.1%. From middle drawing, facing page, availablc cncrgy = 490 Btu/lb. Energy relcased = 490 x 0.74 = 363 Btu/lb. output = 5000 X 3413 = 17,060,000 Btu/hr. Steam flow = 17,060,000/363 = 47,000 lb/hr. Thus. steam rate of turbine at full load = 47.000/5000 = 9.4 Ib/kWh.

HYDRAULIC TURBINES TYPES OF TURBINES A hydraulic turbine converts the energy in falling water, which is created by the combination of hcad and flow, inca rotating mechanical energy. A generator, in turn, converts chis mcchanical cncrgy into clectric energy.

Gate,,N

Turbines are usually designed for spcciñe applications and outputs, because of a site's operating characteristic.s. A good undcrstanding of both energy requircmcnts and characteristics of thc water resource is essential to proper turboma-

chincry sclection. Knowledge of turbine typcs is also important. IMPULSE VS REACTION. There are twu main typcs of hydraulic turbincs: impulse and reaction. Impulse turbines are usually the best choice for high

Runner

Dalt tubo linar

CROSSFLOW TURBINE

CROSSFLOW TURBINE Is an example of the impulse design — the best choice for high heads

Read cover (tixed) WiCket gafe

7 8 910

FRANCIS TURBINE is a reaction type of hydro unir Powor Handbook



15 20

30 40 50 BO 80 100 150 200

Effective head. ft SELECTION CHART plots operating head, power output. and flow range for vanous types o/ hydraulic turbines 63

individually. The units use pressure as wcil as vclocity lo turn the runncr. By using a gradually enlarging draft tube between runner dischargc and tailwater, reaction turbines take advantage of the total hcad availablc to thc tailwater. Because of (heir grcatcr flexibility, they are used in most hydro plants. Two types rxist — Francis and propeller. A FRANCIS TURBINE has a runner with fixed blades. Water enters the turbine in a radial direction with respect to the shaft, and is discharged axially. Principal components include the runner, a water-supply (spiral) case to guide the water to the runner, wicket gafes to control the quantity of water and distribute it equally te the runner, and a 90deg-elbow draft tubo to return thc water to the river (sketch. page 63). PROPELLER TURBINES are generan>, used for heads between 10 and 120 ft. They have a vertical shaft, a spiral case. an clbow-typc draft tube, and a runner

heads aboye 1000 ft —although clfi cient modem units exist for low-hca applications, down to about 20 ft. IMPULSE TURBINES have thc advan lagos of high reliability and low maint nance tests; moreover, their ellicienci are often over 90%. Thc units use a high-velocity stream of water. which strikcs buckcts mounted around thc rim of a rotating turbine wheel, or runne rotation of the runncr generales power much like the tradicional water whccl. THE CROSSFLOW TURBINE (sketch top) is une example of an impulse tu bine used in the hcad rangc between and 600 ft. Here, a mctering vano at the intakc maintains high efficiency (abot t 85%) over a wide rangc of flows. REACTION TURBINES, whilc doing l hc same job as impulse units, work en different principie. Ficre, thc runncr s submcrged in water at all times. Power s developed by water flowing over the blades, rather than striking cach orle

with fixed blades. When the blades are adjustablc. the machine is called a Kaplan turbine. Thc propcllcr &sign has good efficiency at the optimum flow point, but it tails off rapidly as flow changes. The Kaplan unit has a flat efficiency curve over a wide rangc of flows. AXIAL TURBINES are propcllcr dcsigns in which thc water flows to thc distributor coaxially with the shaft. They may have a horizontal or slightly inclined shaft, and cither fixed or adjustable runner blades. Three specilic types of axial turbines: Rim-generator type, in which the generator rotor is on thc periphery of the turbinc runncr. Tubc type, in which thc generator is located outsidc thc water passages. Bulb type, in which both generator and runncr are enclosed in the water passages. The steel capsule around the generator space gives Chis type its namc.

SELECTING HYDRO UNITS sclection will depend on severa] parame-

LOW - HEAD UNIT SELECTION can be

tcrs, including rated power output, efliciency, raied discharge. exeavation re-

mude by using the chart, page 63, which

shows the operating-head limits and flor rangc of differcnt turbines for small hydro installaiions (under 15 MW5. Ifere's how Lo use the chau: A poientibl sito with a hcad of 20 ft and a flow 41)f 1000 cu ft/sec (cfs) can devclop abo t 1.5 MW. From the chau. the preli nary turbine choice is between tu bulb. rim, and propcllcr types. Thc lin

quired. runner di:inicien and %pool.

EXAMPLE 1. What typc of hydro turbine

is indicated for a maximum hcad of not less than 200 ft, a flow of 1000 cfs, and a unit capacity of not less than 10 MW'? SOLUTION. From the charta only the Francis and crossflow dcsigns can operate at heads of 200 ft or more. However,

because of the rclativcly high flow rato and capacity, the Francis turbine is the betel- choice. EXAMPLE 2. A bulb turbine is selected to operatc : q its maximum cffective hcad. What capacity can be expected if the flow is 500 cps?

SOLUTION. From the chau, a bulb tur-

bine has a maximum effective hcad of 60 ft. Wherc the 500-cps and 60-ft-hcad fines intersect, gives 2-MW capacity.

GAS TURBINES KEY TURBINE TYPES Gas turbines are seeing incrcasing use ri s gencrator- and mechanical-drive uni s: applications include base-load, pcaking. and standby service. Kcy dcsigns a c: simple open cycle, regcncrative cy e, and regenerative cycle with intercooli g and/or reheating. See drawings, T-S d agrams, and equations on facing page. SIMPLE OPEN-CYCLE turbine has thrtc main scetions—rotary compressor, combustor, and gas turbine. Thc compres. r draws air in from the atmosphere a d pressurizes it adiabatically. The air t n flows into the combustor. Fuel injec d into the combustor's air burns to ra sc the temperature of both air and comb stion gases, al constan) pressure. The hcatcd, pressurized air/gas m xturc (hen flows into thc turbinc and d s mechanical work on its rotating sh ft. The air/gas mixture expands and c is 64

as it gives up cnergy to do ibis work and partly pressurized air back to atmosphcrthen exhausts to the atmosphere. The ic temperature bcforc compression is turbinc develops all the mechanical work complctcd in thc compressor's second of the cycic. parí bcing diverted to drive ME and the air is sent to the regenerathc compressor; the remainder is net tor. Intercooling lowers the average temperature of heat rejection and thus Q,. output of the cycle. REGENERATIVE CYCLE. Most practica' boosting Pi, the mechanical work availgas turbines exhaust their gas at temper- able, and raising the thermal cfficicncy atares higher than thc air discharged by of thc cycic. thc comprcssor before it enters the com- REGENERATIVE CYCLE WITH REHEATbustor. By passing the air circuit through ING. I lere, the turbine is split into two a heat exchanger called a regenerator, the parís, letting the hot gas expand partly in air can be heated to the exhaust temper- the first section. Thcn the partly ature before it enters the combustor. expanded gas is reheated in a sccond Both heat added to and rejected from thc combustor back to the initial temperacycic are reduced. improving the oyeran ture al the turbine's inIci. The lowpressurc gas then completes its expanthermal elliciency of the cycic. REGENERATIVE CYCLE WITH INTERCOOLING. Here, the compressor is split

into two parts and an intercooler is inserted between. This device cools thc

sion to atmospheric temperature at the turbine outlet. Rchcating raises the average temperature of heat addition, raising Lite overall cycle thermal efficieney. Pownr /landbook

SIMPLE OPEN-CYCLE UNIT

1111



REGENERATIVE-CYCLE UNIT



REGENERATIVE UNIT WITH INTERCOOLING AND REHEATING

4

L

v

,AN A

y

(

T

W- 0,,+ O, - 0.,) = (1-2-34-5-6-7-8-810)

6 5

3 1 „¿O I

SYMBOLS USED THROUGHOUT EXAMPLES

104.4, ; I

C -eomprossor F = combusto?' T8 = turbine R = regenerator intercoolor

Wc =compressor work {compresslon),

W RIOChartiCal work, Bici/lb sir )v,„ bobine work (oxpansion), Btu/Ib

alr

T, P, •

Wc = Ivo:,

71

1,*

Stuab air T - temperature. F P, prossure rallo 4. c, = SPecific heats at constant prossuro and volumo. respectively, Stu/lb k = ratio of Specific heats. Q. = heat added to cyclo, Btu/Ib sir

= 0,(T — T,)

= heat rolocted from cycle, Btu/ib air O, = haat clrculated within cyclo, Stuilb bit

TE = thermal efficiency

EQUATIONS, REGENERATIVE CYCLE

(S)

TE = W/ 0. (W,, - W