2017 ALIGNMENT CHARTS FOR THE ENGINEER PART I AIR AND STEAM BY STANLEY R. CUMMINGS, S. M. RESEARCH ENGINEER, THE HOOVER CO. FORMERLY, ASSISTANT PROFESSOR OF MECHANICAL ENGINEERING OREGON STATE AGRICULTURAL COLLEGE AND JOSEPH LIPKA, PH.D. LATE ASSOCIATE PROFESSOR OF MATHEMATICS, MASSACHUSETTS INSTITUTE OF TECHNOLOGY i JOHN WILEY & SONS, INC. 440 FOURTH AVENUE NEW YORK .. i 1 ALIGNMENT CHARTS FOR THE ENGINEER PART I AIR AND STEAM BY STANLEY R. CUMMINGS, S. M. RESEARCH ENGINEER, THE HOOVER CO. FORMERLY, ASSISTANT PROFESSOR OF MECHANICAL ENGINEERING OREGON STATE AGRICULTURAL COLLEGE AND JOSEPH LIPKA, PH.D. LATE ASSOCIATE PROFESSOR OF MATHEMATICS, MASSACHUSETTS INSTITUTE OF TECHNOLOGY NEW YORK JOHN WILEY & SONS, INC. LONDON: CHAPMAN & HALL, LIMITED 1924 Engin. Library TJ 151 .097 COPYRIGHT, 1924, BY STANLEY R. CUMMINGS AND JESAMINE MCILVAINE Stanbope Press TECHNICAL COMPOSITION COMPANY F. H. GILSON COMPANY BOSTON, U. S. A. Rulasz 12-25-40 mgu 11-13-29 *.R.=. - 2 5 -- 267. Engin. Want 4-16-25 11481 PREFACE Among engineers, there has always been a demand for graphical charts which would take the place of numerical computations. For such a chart to be valuable, it must be simple in construction and easy to read; it must give the desired degree of accuracy, and it must accomplish a saving of time and labor. These qualities are all embodied in the type of chart known as alignment or nomographic chart. Such charts have appeared more and more often in the engineering literature of the last few years and have been favorably received. It is the authors' purpose to issue a series of such charts covering a large variety of en- gineering subjects, and including such formulas as occur most frequently in practice. The subjects include perfect gases, flow of air and steam, air-compression, power, power trans- mission, coiled springs, steel and concrete beams, hydraulics, etc. The present edition is Part I of this work, containing 20 charts on "Air and Steam." Other parts are to follow from time to time. It is hoped that these charts will prove useful both to the engineer and the student of engineering. The authors wish to acknowledge their great indebtedness to Professor G. B. Haven, of the Department of Mechanical Engineering, Massachusetts Institute of Technology, for his helpful suggestions while the work was in progress. They are also deeply indebted to Professor T. H. Taft and Mr. W. H. Jones of the same department for their assistance with Part I. They are grateful to Mr. H. V. Sturtevant for his untiring efforts in drawing many of the charts. CAMBRIDGE, MASS., Dec. 22, 1923. STANLEY R. CUMMINGS JOSEPH LIPKA PRINCIPLES OF CONSTRUCTION AND GENERAL DIRECTIONS The following principles govern the construction of alignment charts. If an equa- tion contains three variables, a scale is constructed for each variable; the construction and arrangement of the scales are such that any straight line (called index line) drawn across these scales will cut them in points corresponding to values of the variables which satisfy the equation. If an equation contains four variables, a scale is constructed for each variable and an additional line (called turning line) is drawn; the four scales, grouped in two pairs, and the turning line are so arranged that two index lines drawn through the same point on the turning line will cut the pairs of scales in points corresponding to values of the variables which satisfy the equation. If an equation contains more than four varia- bles, this construction is extended, so that, in general, we have an additional index line and an additional turning line for each additional variable. For a detailed analysis of the principles and methods underlying the construction of alignment charts, see "Graphical and Mechanical Computation; Part I, Alignment Charts," by Joseph Lipka (published by John Wiley & Sons). Opposite each chart is its accompanying text. This contains:— Number: thus, II-3 indicates that the chart is the third chart in Section II, " Flow of Air and Steam." Type-name: thus, Flow of Steam (Rankine, unrestricted). π Formula: thus, w=280 d²P1. Notation: a specific description with proper units of each variable occurring in the formula. Directions: specific directions as to the method of reading results from the chart; thus, Alignment scales. P₁ W d. This means that values of the variables P₁, w, and d, which satisfy the equation, must align, i.e., their corresponding scale-points must lie on a straight line crossing the three scales. Thus, if the values of P₁ and d are known, the celluloid strip (accompanying the charts) is placed on the chart and maneuvered until the black line or index line which it carries cuts the P₁- and d-scales in points corre- sponding to the known values; it will then cut the w-scale in a point from which the re- quired value of w may be read. The value of any one of the three variables may be found when the values of the other two are known. As another example, let us consider chart NO. I-5, one turning line. PV = R. Alignment scales with T P T- q V T— q R This means that the index line joining the values of P and V and the index line joining the values of T and R must intersect on the turning line q. Thus, if the values of P, V, and R are known, the celluloid strip is placed on the chart and maneuvered until the index line cuts the P- and V-scales in points corresponding to the known values; at the intersection of this index line with the turning line q it is found convenient to hold a pencil or other point, and to revolve the celluloid strip into a new position, maintaining this intersection until the index line cuts the R-scale in the point corresponding to its known value; it will then cut the T-scale in a point from which the required value of T may be read. The value of any one of the four variables may be found when the values of the other three are known. Examples: Specific numerical examples are worked by means of the chart (generally corresponding to the broken index lines drawn). Remarks: these contain any necessary tables of values of constants appearing in the formula, statements as to limitations (if any) of the applicability of the chart, directions by which the usefulness of the chart may be extended, and references to preceding or fol- lowing charts where allied formulas are covered. vi CONTENTS PART I. AIR AND STEAM I. PERFECT GASES I-1. PRESSURE-VOLUME-TEMPERATURE OF GAS. I-2. PRESSURE-VOLUME OF GAS. I-3. TEMPERATURE-VOLUME OF GAS. I-4. PRESSURET-EMPERATURE OF GAS (Below atmospheric pressure). I-5. PRESSURE-TEMPERATURE OF GAS (Above atmospheric pressure). II. FLOW OF AIR AND STEAM II-1. WEIGHT OF AIR DISCHARGED THROUGH AN ORIFICE (Fliegner). II-2. WEIGHT OF AIR DISCHARGED THROUGH AN ORIFICE (Fliegner). II-3. FLOW OF STEAM (Rankine, unrestricted). II-4. FLOW OF STEAM (Grashof). II-5. FLOW OF STEAM (Rankine, restricted). II-6. WEIGHT OF STEAM DISCHARGED BY A NOZZLE OR ORIFICE (Thermody- namic Method). II-7. VELOCITY OF STEAM FLOWING THROUGH PIPES (Babcock). II-8. WEIGHT OF STEAM FLOWING THROUGH PIPES (Babcock). III. AIR COMPRESSION III-1. INTERMEDIATE PRESSURES - TWO AND THREE STAGE AIR COMPRESSORS. III-2. AIR COMPRESSION - SINGLE STAGE (Preliminary to chart No. III-3). III-3. AIR COMPRESSION SINGLE STAGE. III-4. AIR COMPRESSION — TWO STAGE (Preliminary to chart No. III-5). III-5. AIR COMPRESSION Two STAGE. III-6. AIR COMPRESSION THREE STAGE (Preliminary to chart No. III-7). III-7. AIR COMPRESSION — THREE STAGE. vii I-1 NO. I-1 FORMULA: PRESSURE-VOLUME-TEMPERATURE OF GAS PV R. T NOTATION: Pressure P in pounds per square inch absolute or volume V in cubic feet per pound of a given weight of gas of gas constant R and absolute temperature T. DIRECTIONS: Alignment scales with one turning line. P- T- T— q q - V R If value of P is read from the Pa-, Pb-, Pc-, or Pa-scales, the corresponding value of V must be read from the Va-, Vь-, Vc-, or Va-scales respectively. The T-scale is graduated in degrees Fahrenheit (thermometer temperature) for con- venience. EXAMPLE: (1) Given P = 100 lbs. per sq. in. absolute, V = 1.93 cu. ft. per lb., gas is air; to find T. Align P₁ = 100 and V₁ = 1.93, turn on intersection with q; align q and R for air, cutting the T-scale in the required value T = 61° F. (2) Given P = 1.0 lbs. per sq. in. absolute, T = 61° F., gas is air; to find V. Align T = 61 and R for air, turn on intersection with q; align q and Pa = 1.0, cutting the V-scale in the required value Va 193 cu. ft. per lb. REMARKS: The chart may be used to find the pressure P, volume V, or temperature T of a given weight of gas which at volume V₁ and temperature T₁ produces a pressure P₁; the formula is PV/T P₁V1/T1. Example: - 1 1 Given P₁ = 100 lbs. per sq. in. absolute, V₁ = 1.93 cu. ft. per lb., T₁ = 61° F., V = 60 cu. ft. per lb., T-50° F.; to find P. Align Pb = 100 and Vь align q and T = - 1.93, turn on intersection with q; 61°, turn on intersection with R; align R and T 50, turn on intersection with q; align q and V. = 60, cutting the Pe-scale in the required value P = 2.55 lbs. per sq. in. Absolute temperature in deg. F. = (thermometer temperature in deg. F.) + 460. I-1 2000 (Pa) 300 (PG) 20 (Pc) 4.0 (Pa) 2600 (7) 1900 19 2400 1800 250 18 3.5 1700 2200 17 1600 16 2000 1500 15 3.0 1800 200 1400 14. 190 1600 1300 13 180 2.5 1200 170 12 160 1100 150 1000 140 10 2.0 1.9 130 900 9.0 1.8 120 1.7. 800 8.0 1.8 110 1.5 (Four scales Pa, Po, Pc, P¿) 1400 1200 1000 900 800 700 100 700 7.0 1.4 600 1.3 90 600 6.0 1.2 80 500 Fahrenheit 500 70 5.0 1.0 450 4.5 0.90 80 400 4.0 0.80 50 350 3.5 0.70 Pressure (P) in Lb. per Sq. In. Absolute 400 300 200 100 Temperature (T) in Degrees Turning Line (2) 1 (R) 250 (V₁) لسل سلسا 30 2000 400 (VC) (Y6) 4.0 (Va) 350 25 3.5 1500 300 3.0 -20 19 250 94 76 18 سلس 2.5 17 200 سلسل سلس 16 15 1000 200 14 2.0 190 19 900 180 13 1.8 170 12 1.7 150 800 ·160 1.6 150 1.5 700 140 10 14 130 1.3 100 90 80 70 40 30 25 20 Volume (V) in Cu. Ft. per Lb. (Four scales Va, V6, Vc, Vd) 600 120 110 88 9.0 1.2 8.0 2 1.1 500 100 70 7.0 1.0 450 90 0.90 69 6.0 400 80 0.80 سلسسلسلسلي 350 70 5.0 0.70 4.5 300 60 0.60 4.0 250 50 3.5 0.50 45 0.45 3.0 200 40 0.40 190 -180 35 2.5 0.35 170 -180 750 30 0.30 140 2.0 1.9 -130 2 7 25 1.8 0.25 17 7 60 Air I lov N2 50 Constant (R) Value of 45 300 3.0 0.60 40 -50 250 35 2.5 0.50 -100 0.45 16 70 14 13 30 60 12 200 2.0 040 -150 " 190 -120 1.6 110 1.5 100 20 14 0.20 90 15 80 8 18 • 4 7 6 19 0.19 1.3 0.18 1.2 037 Q/G = q q j v v v v 0.15 1.0 014 0.13 0.90 0.12 0.80 Q11 15 1.9 10 180- 1.8 50 10 25 0.35 170 47 (R) -200 45 180- 40 سلسل سلسالمسلسل 0.70 0.10 9.0 0.09 0.60 8.0 0.08 1.6 150 1.5 0.30 20 140 1.4 19 130 1.3 18 0.25 17. 120 1.2 16 110 1.1 100 (Pa) 15 (%) 14 1.0 (P) -250 0.20 (Pd) -300 (7) 35 7.0 0.50 0.07 045 30 6.0 0.06 0.40 25 5.0 0.05 0.35 |(9) 4.5 0.045 (va) (VC) (16) 0.30 20 4.0 (Va) E 0.04 T-2 NO. I-2 FORMULA: PRESSURE-VOLUME OF GAS P₁V₁ = P₂V₂ or P₂ 2 P₁ Ala NOTATION: Pressure ratio P2/P₁ of the final and initial pressures (in same units) of a given weight of gas of exponent n, whose volume ratio of the initial and final volumes (in same units) is V1/V2. DIRECTIONS: Alignment scales P2/P1 n V1/V2 The ratios P2/P₁ and V₁/V₂ may be found from Chart NO. I-1. 2 Use low (high) range scale of P2/P₁ with low (high) range scale of V1/V2. EXAMPLES: (1) Given P2/P₁ = 2.61, V₁/V2 1 2.61, V₁/V₂ = 2.06; to find n. Align P2/P₁ = 2.61 (low range) and V1/V2 = 2.06 (low range) cutting the n-scale in the required value n = 1.325. (2) Given P₂/P₁ = 6.8, V1/V2 = 4.25; to find n. Align P2/P₁ = 6.8 (high range) and V₁/V₂ = 2 4.25 (high range), cutting the n-scale in the required value n = 1.325. (3) Given P₁ = 30 lbs. per sq. in. absolute, P₂ = 78.3 lbs. per sq. in. absolute, V₁ = 51.5 cu. ft., n = 1.325; to find V2. 1 Align P2/P₁ = 2.61 and n = 1.325, cutting the V1/V2-scale in V₁/V₂ = 2.06; then V₂ = V₁ ÷ 2.06 = 51.5 ÷ 2.06 - 25 cu. ft. I-2 4.6- 2.5 2.0 1.0- 48 1.7 50 7.0 8.5 40 6.0 35 5.5 30 5.0 25 4.5 20 19 18 17 4.0 ·16 15 3.5 - 12 3.0 10 8 1.5 -Low Range High Range→→ a 14 Pressure Ratio (P2) LO (n) L 1.2 (u) TTTT 91 91 Adiabatic (1-1405) for Air (၁ (2nd) Value of Exponent (n) (pv". c) (/) Volume Ratio Volume Ratio (1) Low Range - High Range (~) 借 ​-1.2 1.5 -1.3 2.0- 1.5 2.5- ·1.6 -1.7 3.0 -1.8 3.5 -1.9 4.0 2.0 4.5- 5.0 6 2,5 9.0 10 12 3.5 19 2 14 15 16 4.0 17 18 19 20 4.5 25 5.0 30 5.5 35 6.0 40 6.5 V2 (√₁) 15 45 ·7.0 50 I-3 NO. I-3 FORMULA: NOTATION: TEMPERATURE-VOLUME OF GAS n T₁V¸ª−1 = T¿V₂¤−1. 2 Final volume V2 (any unit) and final absolute temperature T₂ of a given weight of gas of exponent n which at volume Vi (same unit as V2) has an absolute temperature T1. DIRECTIONS: 1 Alignment scales with two turning lines. √ 1 qv V₂ 2 αν n Чт T₁ Ят Ꭲ, The T1- and T₂-scales are graduated in degrees Fahrenheit (thermometer temperature) for convenience. EXAMPLES: 1 2 1 (1) Given V₁ = 50 cu. ft. per lb., V₂ = 10 cu. ft. per lb., Tı = 300° F., n = 1.4; to find T2. Align V₁ = 50 and V2 10, turn on intersection with qv; align qv and n = 1.4, turn on intersection with qr; align qr and T₁ = 300, cutting the T2-scale in the required value T₂ = 986° F. (2) Given Vi = 50 cu. ft. per lb., V2 = 10 cu. ft. per lb., T₁ = 300° F., T₂ = 986° F.; 1 to find n. Align V₁ = 50 and V2 align Ti 300 and T2 10, mark intersection with qv; 986, turn on intersection with qr; align qr and qv cutting the n-scale in the required value n = 1.4. REMARKS: Absolute temperature in deg. F. = (thermometer temperature in deg. F.) + 460. I-3 10 003 80 004 (V2) 3.0 70 0.95 0.08 60 0.07 0.08 60 0.09 0.10 السلسل سلسلسل سلسا 45 40 0.15 35 0.20 30 0.25 0.30 25 20 19 18 17 18 15 14 12 "1 13 Initial Volume (V₁) (Any Units) 0.40 0.50 0.80 0.70 0.80 0.90 1.00 1.5 20 2.5 30 Turning Line (9%) or Volume Ratio (V2/V,) (A) سلام 25 25 Final Volume (V2) (Same Units لبلالي 20 18 བ 17 14 15 16 19 12 10 1600 (TI) 1500 1400 3.5 1300 4.0 1200 1100 4.5 1000 5.0 900 800 700 7 600 (n) 17 16 15 14 Value of (n) in Exponent 500 400 300 in Degrees Fahrenheit 1.3 1.2 1./ 200 8 -50 4.0 30 100 Initial Temperature Turning Line (97) |(27) (T2) F -250 זייויזין Temperature (T2) in Degrees Fahrenheit Final سلسلا יזיין 100 200 300 400 500 600 700 5.0 35 -100 60 800 7.0 40 900 8.0 -150 45 6.0 9.0 1000 10.0 50 4.5 1100 1200 35 4.0 -200 15 60 20 (VD) S 30 25 30 (V2) 1300 1400 20 70 1500 (T) (T2) (97) 1600 -250- 80 יין יויויד -200 -150 -100 -50 I-4 NO. I-4 PRESSURE-TEMPERATURE OF GAS. (Below atmospheric pressure) FORMULA: NOTATION: 1 n TP₁a = T₂P₂ 2 Final pressure P2 in pounds per square inch absolute or final absolute temperature T₂ of a given weight of gas of exponent n which at pressure P₁ has a temperature T1. DIRECTIONS: Alignment scales with two turning lines. P₁ qp P₂ qp n Ят T2 Ят Ti The T1- and T2-scales are graduated in degrees Fahrenheit (thermometer temperature) for convenience. EXAMPLES: (1) Given Pi = 2.45 lbs. per sq. in. absolute, P2 = 14.7 lbs. per sq. in. absolute, T₁ = 50° F., n = 1.3; to find T₂. Align P₁ = 2.45 and P2 align qp and n = 14.7, turn on intersection with qp; 1.3, turn on intersection with qr; align qr and T₁ 50, cutting the T2-scale in the required value T₂ = 310° F. 2 (2) Given P₁ = 2.45 lbs. per sq. in. absolute, P₂ = P2 14.7 lbs. per sq. in. absolute, T₁ : 50° F., T₂ = 310° F.; T2 Align P₁ = 2.45 and P2 align T₁ = 50 and T₂ to find n. 14.7; mark intersection with qp; 240; turn on intersection with qr; align qr and qp, cutting the n-scale in the required value n = 1.3. REMARKS: Absolute temperature in deg. F. Pressure in lbs. per sq. in. absolute = (thermometer temperature in deg. F.) + 460. (pressure in lbs. per sq. in. gage) + 14.7. I-4 (P.) 10 * سلسل سلسسسل Initial Pressure (P) in Lb. per sq in. absolute 12 13 14 15 16 20 19 18 17 Turning Line (qp) (P2) in Lb. per sq. in. absolute ③ ليسليسلسسلسلاسار السلسلسله سلسسلسلسسلسلاسللللل السلسسلسا 20 Final Pressure 23 os ir t w 17 " |(2p) יויזין ~ 1600 1500 (T2) (2+) 1400 1300 1200 1100 600 600 700 800 Degrees Fahrenheit 1.7 1.6 1.5 1.4 1.3 1.2 (4) יויו Value of n in Exponent T 500 100 (TZ) 200 300 Temperature (T2) in 400 Final (2+) 1./ Turning Line (97) 500 500 Degrees Fahrenheit זירוזייייןיויןיוייןייויןייי 400 700 Ope 300 200 200 E Initial Temperature (T) in لسلاسللللللل י 1600 1400 1500 1300 1200 1100 800 7000 600 100 I-5 NO. I-5 PRESSURE-TEMPERATURE OF GAS. (Above atmospheric pressure) FORMULA: NOTATION: 1-0 1 T₁P₁n T₂P2 ¤ Final pressure P₂ in pounds per square inch absolute or final absolute temperature T₂ of a given weight of gas of exponent n which at pressure P₁ has a temperature T₁. DIRECTIONS: Alignment scales with two turning lines. P₁ qp P₂ др T2 n Чт qT T The T₁- and T₂-scales are graduated in degrees Fahrenheit (thermometer temperature) for convenience. The qp turning line is graduated in the pressure ratio P2/P₁ for convenience. EXAMPLES: (1) Given P₁ = 15 lbs. per sq. in. absolute, P₂ 75 lbs. per sq. in. absolute, T₁ = 240° F., n = 1.35; to find T2. Align P₁ = 15, and P2 align qp and n = align qr and T₁ = 75, turn on intersection with qp; 1.35, turn on intersection with qr; 240, cutting the T2-scale in the required value T₂ 600° F. (2) Given P₁ = 15 lbs. per sq. in. absolute, P₂ = 75 lbs. per sq. in. absolute, T₁ = 240° F., T2 600° F.; to find n. = Align P₁ = 15 and P2 1 align T₁ = 240 and T₂ 75, read intersection with qp; 600, turn on intersection with qr; align qr and qp, cutting the n-scale in the required value n = 1.35. REMARKS: Absolute temperature in deg. F. = (thermometer temperature in deg. F.) + 460. Pressure in lbs. per sq. in. absolute (pressure in lbs. per sq. in. gage) + 14.7. I-5 100 90 80 70 60 50 1000 900 800 (PJ) 6.010 (P2/p.) (P2) 10 0.015. 1600 (Tz) 1(97) 1500 (T) -260 1400 1300 700 0.020 1200 15 1100 600 0.030 -200 1000 500 0.040 20 900 0.050 -150 0.060 400 25 800 0.070 0.080 350 0.090 30 700 0.10 -100 300 250 200 150 Initial Pressure (p,) in Lb. per sq. in. absolute 0.15 0.20 0.30 0.40 0.50 0.60 0.70 0.90 1.0 ៖ ៖ 080 1.5 2.0 3.0 4.0 5.0 40 95 6.0 7.0 8/0 b.o 110 30 15 25 20 Turning Line (qp) or Pressure Ratio (R/P,) Final Pressure (P2) in Lb. per sq. in. absolute זיןיויויזין 35 600 40 500 50 400 60 70 80 90 100 150 (n) 300 200. 1.7 46 1.5 1.4 42 Value of n in Exponent 150 100 50 O 200 50 250 300 350 -100 in\ Degrees Fahrenheit Final Temperature (T2) Turning Line (97) ויויויו! Initial Temperature,(T) in. Degrees Fahrenheit لسار "דיזיין 50 100 150 200 300 400 500 600 700 400 800 - 150 20 20 500 30 600 -200 40 15 50 700 ៖ ៖ ៖ ៖ 80 70 10 ](P) 100 90 (P/P) (P3) المسلسل سلسله 800 900 1000 1100 1200 1300 1400 900 · 250 (T2) |(9T) 1500 1000 (Ti) 1600 50 II-1 NO. II-1 WEIGHT OF AIR DISCHARGED THROUGH AN ORIFICE. (Fliegner) FORMULA: w = 0.4163 d2P I1 where P₁ or 2 P2. = NOTATION: Weight w in pounds per second of air flowing through an orifice of diameter d inches from a reservoir at pressure P₁ pounds per square inch absolute and absolute tem- perature T₁, to a small straight pipe within which the pressure is P. pounds per square inch absolute, if P₁ = or > 2 P₂. DIRECTIONS: Alignment scales with one turning line. 2 Pi q T₁ q W d If value of d is read from the da- (dû-) scale, corresponding value of w must be read from the Wa- (Wb-) scale. The T₁ scale is graduated in degrees Fahrenheit (thermometer temperature). 1 EXAMPLES: (1) Given P₁ = 100 lbs. per sq. in. absolute, T₁ Align Pi = 100 and T₁ align q and d₂ = 0.15, per sec. (2) Given P₁ = 100 lbs. per Align P₁ = 100 and T₁ align q and dь per sec. REMARKS: = = 835° F., d = 0.15 in.; to find w. 835, turn on the intersection with q; cutting the wa-scale in the required value w = 0.026 lbs. sq. in. absolute, T₁ = 835° F., d = 1.00 in.; to find w. 835, turn on the intersection with q; = 1.00, cutting the wь-scale in the required value w = 1.16 lbs. If P₁ < 2 P2, use Chart NO. II-2. II-1 600 (P) ((9) 100 550 500 3.0- (W) (d. -90 (WG) 80 70 2.0 -60 450 -50 1.5 400 -40 0.55 (do) -3.0 0.50 ليل -0.45 350 1.0- 09- 0.8 300 250 200 190- 180 170 160 150 140 130- 120 110 100 90 80 Pressure fore Orifice (P) in Lb. per Sq. In. Abs. e Before 0.7 0.6- 0.5. 0.4 (Two Scales We, Ws) 30 20 ויויויזין 15 0.3 0.20 Turning Line (9) Q15- 0.10- 009- 0.08- 0.07 0.06 0.05 0.04 0.03 70- Flow through Orifice (w) in Lb. per Sec. ogon 65 10 9 Tuntuu tuntuu luua tuntuu 4 3 Orifice (d) in Inches (Two Scales de,do) יויוייזיויויויידי זיויוי 2.5 -100 0.40 T 0.35 السلسال 2.00 067- TTT -1.80 1.70 0.30 1.60 1.50 -1.40 Q25 -1.30 2.0 1.5 Diameter of Orifice (T) in Degrees Fahrenheit -200 דידי 300 400 ليد 500 -1.20 0.20 1.10 Before 600 -019 1.00 -1.0 0.18 -0.9 0.95 0.8 -0.17 ~0.7 0.90 0.020 -0.6 -016 0.85 олабшор mperature 700 800 0.5 0.15 60- 0.015 -900 -0.80 0.4 -0.14 -0.75 -1000 0.010 -0.3 50- 0.009 0.13 0.008 -0.70 -1100 0.007 -0.20 0.006 -0.12 -0.65 40 -1200 0.005 0.15 0.004 0.// 0.60 -1300 0.10 (T) 140 -1400 (P) |(9) 0.003 30 (W) 0.0025-(m) السلسة -0.09 oro (d) E010 (4) 0.55 વ -0.08 II-2 NO. II-2 WEIGHT OF AIR DISCHARGED THROUGH AN ORIFICE. (Fliegner) FORMULA: w = 0.834 d² P₂ (P1 P2) 2 where P₁ < 2 P₂ T₁ 1 NOTATION: Weight w in pounds per second of air flowing through an orifice of diameter d inches from a reservoir at pressure P₁ pounds per square inch absolute and absolute tempera- ture T₁, to a small straight pipe within which the pressure is P2 pounds per square inch absolute, if P₁ < 2 P2. DIRECTIONS: Alignment scales with two turning lines. P₂- qp qp- T₁ (P1 P2) Чт Ꮊ - W d If value of d is read from the da- (dь-scale), corresponding value of w must be read from the Wa- (Wb-) scale. The T₁-scale is graduated in degrees Fahrenheit (thermometer temperature). EXAMPLES: (1) Given P₁ = 181 lbs. per sq. in. absolute, P₂ = 100 lbs. per sq. in. absolute, T₁ = 1140° F., d = 0.20 ins.; to find w. = 81; turn on the intersection with qp; 1140, turn on the intersection with qr; Align P2 = 100 and P₁ - P₂ align qp and T = align qr and da per sec. = 0.20, cutting the wa-scale in the required value w 0.075 lbs. (2) Given P₁ = 181 lbs. per sq. in. absolute, P2 = 100 lbs. per sq. in. absolute, T₁ = 1140° F., d = 1.50 ins.; to find w. = P2 81, turn on the intersection with qp; 1140, turn on the intersection with qr; Align P2 - 100 and P₁ align qp and T = align qr and dû per sec. REMARKS: = 1.50, cutting the wь-scale in the required value w = 4.21 lbs. If P₁ = or > 2 P2, use Chart NO. II-1. II-2 (da) (do) 0.50 -2.50 (W.) (WG) 2.0- 50 200- 190- 180 (P₁) (P-P) 200 (97) -40 190 (T,) 1.5- 180 045 170 170- 30 160 160 -150 1.0- 150- -140 0.9 130 0.8- ·20 1.40 -0.402.00 -120 0.7 130 -110 0.6- 15 120- -100 יזין -1.90 -100 0.5- 90 110 0.4 80 100- 90 80 70 60 50 45 40- 35 30 25 Back Pressure at Orifice (P) in Lb. per Sq. In Abs. Turning Line (90) Turning Line (87) Pressure Drop through Orifice (P,-P₂) in L.6./per Sq. in. Abs. سلسل 10 9 8 70 0.3- -6 -60 0.2 5 50 4 0.15 45 40 3 (Two Scales We and wo) -1.80 035 1.70 بلب 0.10- Q.09- 0.08 -2.0 30 0.07- 0.06 -25 0.05 \0.04- 1.0 سل سلسال -12 °ཝཱ༤ཐུ ཝཱ ཨཱུ སཱུ – Ş༡ ཚྭ A 0.9 -0.8 003- -0.7 -0.6 0.02 -0.5 -0.4 0.015 0.3 0.010 Flow through Orifice (w) in Lbb per Sec. Before Orifice (T) in Degrees Fahrenheit (Two Scales 200 de and do) 1.60 -0.30 -0.30 1.50 -300 T T 1.40 1 L -0.25 -1.30 ·400 -1.20 500 -600 Temperature Orifice / (d) in Inches -1.10 -0.20 -1.00 -0.19 700 -0.18 0.90 of -0.17 800 -0.2 -0.16 0.80 0.009 0.008- 0.007 20 19 900 0.006 0.15 Diameter -0.15 18- المسلس 6 0.005 -1000 17 0.004 0.10 0.14 0.70 ·3.0 16 -0.09 15 4.5 0.08 -1100 0.003 -0.07 -0.13 14 ·4.0 0.06 1200 13 3.5 0.002 -0.05 -0.12 -0.60 12 -1300 3.0 0.04 0.0015- -1400 2.5 -0.03 0.11 10 (P) (90) 0.0010 -1500 (Wa) (2+) (P-P) -2.0 0.02 -1600 (WG) (T) -0.10 -0.50 (da) (do) II-3 NO. II-3 FORMULA: FLOW OF STEAM. (Rankine, unrestricted) W AP₁ П 70 280 d²P₁ NOTATION: Weight of steam w in pounds per second flowing from a reservoir at pressure P₁ pounds per square inch absolute through an orifice of area A square inches, or diameter d înches to a pressure of P₂ pounds per square inch absolute, when P₂ = or < 0.58 P1. DIRECTIONS: 2 Alignment scales. Pi W d If value of d is read from the da- (dь-) scale, corresponding value of w must be read from the wa- (Wb-) scale. EXAMPLES: (1) Given P₁ = 80 lbs. per sq. in. absolute, d = 0:20 inş.; to find w. Align P₁ = 80 and da lbs. per sec. 0.20, cutting the wa-scale in the required value w = 0.036 (2) Given P₁ = 80 lbs. per sq. in. absolute, d = 2.00 ins.; to find w. Align P₁ = 80 and dp lbs. per sec. REMARKS = 2.00, cutting the wь-scale in the required value w = 3.6 If P₂ > 0.58 P1, use Chart NO. II-5. 2 The Grashof formula is given by Chart NO. II-4. II-3 300- 250- 200- 190 180- 170- 160- 150- 140- 130 120 100· 90- 80 70- 60- 50 45- (wa) (w₁) T 1.0- -30 0.9- (P.) -25 0.8- 0.7- -20 0.6 0.5- 15 Pressure before Orifice (P.) in Lb. per Sq. In. absolute 0.4- 0.3- 0.2- 0.15- 0.10- 0.09- 0.08- 0.07 0.06- 0.05 0.04- 0.030- 0.025 0.020- 0.015- بيلسلسلسيللي Two Scales We and wad ידיוידי Flow through Orifice (w) in Lb. per Sec. -10 9 8 7 6 L 4 3 2.0 -1.5 سلسسلسله -1.0 -0.9 -0.8 0.7 +0.6 (da) (da) 955-3.00 -0.50 0452.50 -0.40 -035 -0.30 Dic er of Orifice (d) in Inches (Two Scales da and do) T 2.00 1.90 1.80 · 1.70 160 -1.50 IAO -0.25 130 ·1.20 -0.20 -1.10 0.19 -0.18 -1.00 سلسل -0.5 -Q.17 -0.4 -0.90 -0.16 40 0.010 -0.3 -0.15 0.009 -0.80 0.008 35- -0.14 0.007 -0.20 0.006 -0.13 30- -0.70 0.005 -0.15 0.12 0.004 25- 0.10 0.003- Q11 0.60 -0.09 -0.08 (P) 20- 0.07 (wa) (ww) (da) 0.10 ± 0.55 0.10-0.55 (de) II-4 NO. II-4 FORMULA: FLOW OF STEAM. (Grashof) w = 0.0165 AP 0.97 0.0165 π 4 d²P,0.97 NOTATION: Weight of steam w in pounds per second flowing from a reservoir at pressure P₁ pounds per square inch absolute through an orifice of area A square inches, or diameter d inches, to a pressure of P₂ pounds per square inch absolute, when P₂ = or < 0.58 P1, DIRECTIONS: Alignment scales. d P₁ If value of d is read from the da- (dû-) scale, corresponding value of w must be read from the Wa- (Wb-) scale. EXAMPLES: (1) Given d = 0.25 ins., P₁ = 80 lbs. per sq. in. absolute; to find w. Align da = 0.25 and P₁ lbs. per sec. = 80, cutting the wa-scale in the required value w = 0.0568 (2) Given d = 0.90 ins., Pi 0.90 ins., P₁ = 80 lbs. per sq. in. absolute; to find w. == Align dь 0.90 and P₁ = 80, cutting the wь-scale in the required value w = lbs. per sec. REMARKS: If P2 > 0.58 P1, use Chart NO. II-5. The Rankine formula for P₂ = or < 0.58 P₁ is given by Chart NO. II-3. 0.737 II-4 0.55 (da) 0.50 0.45 0.40 3.0 **(do) 1.0 0.9- (Wa)30 (W6) 25 0.8 0.7 20 0.6 25 5 0.5 15 2.0 0.35 1.9 1.8 17 030 1.6 1.5 14 0.25 1.3 6.2 0.20 " 019- LO 018 (Two Scales) Diameter of Orifice (d) in Inches 04 10 0.30 9 0.25 8 020 015 7 6 5 0.10 3:0 0.09 25 2.0 0.05 1.5 དྷྱ ༔ ༔ སྦྱིg 8 0.08 0.04 1.0 0.030 0.9 0.8 0.025 07 0.020 0.6 - 0.5 a17 2015 0.90 04 Q16 Weight of Steam (w) in Pounds per Second (Two Scales) 300 (P) Pressure Before Orifice (PI) in Pounds per Square Inch. Abs. لسل ויזיויויויויד لسلسيللي 250 200 190 180 170 160 150 140 130 120 HO 100 90 80 70 60 50 45 40 0.010 0.30 0.15 0.009 0.80 025 35 0.008 0.14 0.007 020 0.008 30 0.13 0.70 0.005 015 012 0004 0,10 اله 0.60 0.0030 0.09- 0.08 0.0025 0.10 (da) 0.55 ](db) QOT 0.0020 (wa)aos. (wb) (P) 25 20 II-5 NO. II-5 FORMULA: NOTATION: FLOW OF STEAM. (Rankine, restricted) w = 0.0292 A [P₂ (P₁ — P2)]† 0.0292 T 4 2 d² [P₂ (P₁ — P₂)}} Weight of steam w in pounds per second flowing from a reservoir at pressure P₁ pounds per square inch absolute through an orifice of area A square inches or diameter d inches to a pressure P2 pounds per square inch absolute, when P₂ > 0.58 P1. DIRECTIONS: Alignment scales with one turning line. 2 P2 d др W (P₁- P₂) Яр If value of d is read from the da- (dь-) scale, corresponding value of w must be read from the Wa- (Wb-) scale. EXAMPLES: (1) Given P₁ = 100 lbs. per sq. in. absolute, P₂ = 80 lbs. per sq. in. absolute, d = 0.25 ins.; to find w. 2 P2 Align P₂ = 80 and P₁ - P₂ = 20, turn on intersection with qp; align qp and d₂ = P2 0.25, cutting the wa-scale in required value w = 0.057 lbs. per sec. (2) Given P₁ = = 100 lbs. per sq. in. absolute, P2 = 80 lbs. per sq. in. absolute, w = 2.05 lbs. per sec.; to find d. Align P2 - 80 and P₁ P₂ 20, turn on the intersection with qp; align qp and wb = 2.05, cutting the d-scale in the required value d = 1.5 ins. REMARKS: If P2 = or < 0.58 P1, use Chart NO. II-3 or Chart NO. II-4. II-5 0.55 3(da) 30-7 (db) 1.5 ; (W₁) (W₁E (wa) 40 250 (P2) (20) ) (PP) 0.50 1.0 0.90 0.70 2.5 0.45 0.60 30 25 25 200 190 0.80 180 20 170 20 160 150 0.50 15 140 040 130 0.40 0.25 0.35 1.9 a20 1.8 120 10 0.30 2.0 2 880 9.0 110 0.15 1.7 0.30 0.25 0.20 0.19 0.18 0.17 0.16 Diameter of Orifice (d) in Inches (Two Scales) 1.6 010 1.5 0.090 0.080 0.070 1.4 0.060 1.3 0.050 0.040 1:2 0.030 0.025 0.020 1.0 0.015 0.90 0.010 ȧ0090 Weight of Steam Discharged (w) in Lb per Sec. (Two Scales) z q ÿ ÿ ÿ ñ pmp لسلسال سلسلسلتي 8.0 100 7.0 69 6:0 90 5.0 80 4.0 70= 3.0 60 2.5 2.0 50 1.5 45 40 1.0 0.90 35: 0.80 0.70 80 0.60 Back Pressure" (P₂) in Lb.iper Sqing Abs. Turning Line (9%) Difference between Initial and Final Pressures (P.~ P₂) in Lb. per Sq. In. Abs. لسلسا 780 170 { 160 150 140 130 120 110 للللللللللل 100 90 80 70 60 50 45 40 35 30 25 0.50 25 0.40 0.30 20 Pressure Drop: 20 19 18 17 16 15 14 19 0.25 0.15 0.0080 18 13 0.80 0.0070 0.20 12. 0.14 16 0.0060 15 0.0050 0.15 14 0.13 10 0.7Q 0.0040 19 12. 0.12 0.10 0.0030 0.09 8 0.08 0.0025 (P) 10 0.11 0.07 0.60 0.0020 0.06 (2p) 0.10 (da) 0.55 (do) 0.0015 (Wa) (Wb) 0.05 (Pi-P₂) السلسسلسلس 7 6 II-6 NO. II-6 WEIGHT OF STEAM DISCHARGED BY A NOZZLE OR ORIFICE FORMULA: (Thermodynamic Method) 223.8 At √H₁ – H₂ H2 W Vt NOTATION: 223.8 π d² √H₁ — H₂ 4 Vt Weight of steam w in pounds per second flowing through a nozzle or orifice of minimum cross-sectional area At square inches or minimum diameter d inches, when the heat drop from entrance to minimum (throat) section is (H1 H2) in B.t.u. per pound. H₁ (H₁ is the total heat content of the steam as it approaches the nozzle or orifice, and H₂ is the total heat content at the throat section, assuming frictionless adiabatic expansion (see remarks below); vt is the specific volume of the steam at throat conditions in cubic feet per pound. DIRECTIONS: Alignment scales with one turning line. (H₁ — H₂) q q W Vt d If value of d is read from the da- (dь-) scale, corresponding value of w must be read from the wa- (Wb-) scale. EXAMPLES: = (1) Given (H₁ — H₂) = 100 B.t.u. per lb., vt 2.5 cu. ft. per lb.; to find the flow through an orifice of diameter d = 1 in. Align (H₁ — H₂) 100 and Vt 2.5, turn on intersection with q; align q and dp sec. = 1.00, cutting the wь-scale in the required value w = 4.88 lbs. per = (2) Given (H₁ — H₂) = 100 B.t.u. per lb., vt 2.5 cu. ft. per lb.; to find the flow through an orifice of diameter d= 0.40 ins. 1 2 Align H₁ H₂ = 100 and Vt align q and da per sec. REMARKS: 2.5, turn on intersection with q; 0.40, cutting the wa-scale in the required value w = 0.781 lbs. If the final pressure into which the nozzle or orifice is discharging is greater than the critical pressure, then the throat pressure is the same as the final pressure. If the final pressure is less than the critical pressure, then the throat pressure is equal to the critical pressure, which for initially dry steam is usually about 0.58 p₁. The value of (H1 H₂) for an actual case will, in general, be less than 60 B.t.u. per lb. If value of vt is less than 1.0 or greater than 10, use values on the chart and multiply or divide the answer by a power of 10; thus: if in example (1), vt = 25 instead of 2.5, read w = 0.488 instead of 4.88. II-6 500 (H-H₂) |(9) 450- 400- 350- 300- 250 200 190- 180- 170 160 150 140- 130 120 110 100 90- 80 70 Change in Total Energy or Total Heat Contents (H,- H₂) of Steam in Orifice, in B.T.U. per Pound Expanding Adiabatically through 60- 50- 45 40 35- 30- 25 20 19 18 17 16 15. 14 13 2 12 10 (Hi-H₂) Turning Line (9) (9) (Wa) (Wo) -300 (W₂) السلام (da) (do) 0.55 -3.00 055 9. (71) 8- 1.0 7 -200 6 -0.50 1.1 5 150 1.2 2.50 045 100 90 80 70 -0.40 རྗ 60 1.3 3 1.4 سلسلسال -1.5 2.0 -50 1.6 1.5 1.7 40 -1.8 1.0- 30 1.9 0.9- -2.0 0.8 0.73-20 10 0.6 0.5- 2.5 0.4- 0.3- Specific Volume of Steam at Throat Section (2) after Adiabatic Expansion through Orifice in Cu. Ft. per Lb. ·3.0-0.20・ سلسلاسل Flow through Orifice (w) in Lb. per Sec. (Two Scales We and wo) 0.15 -3.5 0.10 0.09- -4.0 0.08 0.07- 0.06- 4.5 0.05 15 10 9 8 7 -6 5 4 3 2.0 -1.5 -035 0.30 -2.00 - 1.90 1.80 -1.70 -1.60 1.50 -0.25 -1.40 Diameter of Orifice (d) in Inches (Two Scales do and do) T T T T 1.30 1.20 -0.20 1.10 -0.19 ·1.00 0.18 -0.17 5.0 0.04 -0.90 0.03 سلسل 1.0 0.16 0.9 0.8 0.7 -0.15 -6.0 0.02 0.6 -0.80 0.5 0.14 0.015 7.0 0.4 0.13 -0.70 0.010- 0.3 ·8.0 0.009- 0.008 0.007- -0.12 0.20 -9.0 0.006 0.11 0.60 1 0.005- -0.15 ∙10 0.004- (31) (Wa) (wb) -0.10 (da) -0.10 -0.55 (do) II-7 NO. II-7 VELOCITY OF STEAM FLOWING THROUGH PIPES. (Babcock) FORMULA: — V = 15,950 (P1 P2) d wil (1 + 3.6) NOTATION: Velocity V in feet per minute of flow of saturated steam through a pipe of diameter d inches, from a point at a pressure of p₁ pounds per square inch absolute to a point 1 feet distant at a pressure of p2 pounds per square inch absolute, if the weight of a cubic foot of steam at the pressure p₁ is w₁ pounds. DIRECTIONS: Alignment scales with one turning line. 1 Ap V- q q P d The variable w₁ is replaced by the variable P, where P is the mean pressure in the pipe in lbs. per sq. in. absolute. The variables P1, P2, and 1 are combined into the variable Ap, where Ap is the pressure drop in lbs. per sq. in. absolute per 100 ft. of length of pipe, or, Ap = 100 (pı — p½)/1. EXAMPLE: Given Ap = 1 lb. per sq. in. absolute, P = 100 lbs. per sq. in. absolute, d = 5 ins.; to find V. Align Ap = 1 and P = 100, turn on the intersection with q; align q and d = = 5, cutting the V-scale in the required value V = 5730 ft. per min. 95.5 ft. per sec. REMARKS: Babcock's formula for the weight of steam flowing through pipes is given by Chart NO. II-8. II-7 (AP) (V) 800 |(9) 20- 45,000- ·700 15 40,000 -600 35,000- 10- 9 30,000 500 8 7 6 25,000- -400 4 -350 20,000- 18,000. 3 -300 2.0 1.5 1.0 0.9 9990 0.8 0.7 0.6 0.5 0.4 0.3 0.20 0.15 18,000 17,000- 16,000- 15,000- -250 14,000- 13,000- 12,000- 200 -190 11,000 -180 -170 •10,000- 9000- -150 -140 8000- 130 Pressure Drop' (Ap) in Lb. per Sq. In. per /100 Feet of Length Velocity of Flow (V); Left Scale in Ft. per/Min. Right. Scale in Ft. per Sec. 120 7000- · 110 6000 ·100 -90 5000- -80 4500 70 4000 0.10- 0.09. 0.08 -60 3500 0.07 0.06 3000 -50 0.05 0.04 2500- 40 0.03 2000- 0.020 1900- 1800 -30 い ​1700- 0.015- 1800 1500- -25 0.010 1400 0.009- 0.008 (AP) 1300 (V) Turning Line (q) (d) (d) (9) Inside Diameter of Pipe (d) in Inches السلي זיז 5 6 7 8 -9 -10 -0.7 (P) -0.0 0.9 1.0 2 3 1.5 4 5 2.0 6 2.5 3.0 4 Mean Pressure in Pipe (P) in Lb. per Sq. In. Absolute 7 8 9 10 15 20 25 -30 -40 50 -60 70 80 15 90 -100 -20 -150 -30 40 (P) 200 250 300 II-8 NO. II-8 WEIGHT OF STEAM FLOWING THROUGH PIPES. (Babcock) FORMULA: NOTATION: W = 87 W1 (P1 — P2) d5 +3.6) 1 + 1 (1 Weight w in pounds per minute of saturated steam that will flow through a pipe of diameter d inches, from a point at a pressure of p₁ pounds per square inch absolute to a point 1 feet distant at a pressure of p2 pounds per square inch absolute, if the weight of a cubic foot of steam at a pressure p₁ is w₁ pounds. DIRECTIONS: Alignment scales with one turning line. Ap- q d P- q The variable w₁ is replaced by the variable P, where P is the mean pressure in the pipe in lbs. per sq. in. absolute. The variables P1, P2, and 1 are combined into the variable Ap, where Ap is the pressure drop in lbs. per sq. in. absolute per 100 ft. of length of pipe, or, Ap 100 (P₁ — P2)/1. EXAMPLE: = Given Ap to find w. = 1 lb. per sq. in. absolute, d = 5.0 ins., P = 100 lbs. per sq. in. absolute; Align Ap = 1 and d = 5.0, turn on the intersection with q; align q and P = 100, cutting the w-scale in the required value w = 175 lbs. per min. REMARKS: Babcock's formula for the velocity of steam flowing through pipes is given by Chart NO. II-7. II-8 5 (P) (^p) 2- 015 0.20- יויויןיוייזיוין 03 OA 0.6 05- 0.7- ៩ ៩៩ 1.5- 20- Pressure Drop (4p) in Lb. per Sq. In. per 100 Feet of Length 50 40 30. 20- 9 10- 8 오 ​Mean Pressure in Pipe (P) in Lb. per Sq. In. Absolute t 010- 0.09 60 0.08- 0.07 70- 0.06. 80- 0.05- 90- 100- 0.04 0.03- 150 0.020- 0.015- 200- ༣) ༠༥༩ Turning Line (9) 250- (P) 0.010 |(AP) (9) (11) ((2) (w) -40,000 -30,000 (d) 45 40 ·20,000 -35 35 ·10,000 30 -9000 ·8000 7000 ·6000 -25 ·5000 ·4000 -3000 -2000 سلسال -1000 900 ·800 700 -600 -500 -400 པེཝེཝེསྶ ·16 ཆེ ལ 60 السلسلسالسا 97 Weight of Steam Flowing through Pipe (w) in Lb. per Min. * 20 30 q mp | | | $ $ ༠ ༤Ëཎྞེ (d) in Inches Inside Diameter of Pipe (d) in Inches -200 300 ญ & 22286 ន ง -5.0 4.5 -4.0 -3.5 3.0 I-III NO. III-1 INTERMEDIATE PRESSURES - TWO AND THREE STAGE AIR COMPRESSORS FORMULA: P' P' = √/P₁²P₂; p" = (Two stage) 2 (Three stage) NOTATION: (Two stage.) Intermediate pressure P' in pounds per square inch of a two stage air compressor which compresses air from an initial pressure of P₁ pounds per square inch absolute to a final pressure of P2 pounds per square inch absolute. 1 (Three stage.) First stage pressure P' in pounds per square inch absolute and second stage pressure P" in pounds per square inch absolute of a three stage air compressor which compresses air from an initial pressure of P₁ pounds per square inch absolute to a final pressure of P2 pounds per square inch absolute. DIRECTIONS: Alignment scales. (Two stage) 1 P₁- P' -P2 (Three stage) P₁ P'- — P' — P₂ - EXAMPLES: (1) (Two stage.) Given P₁ = 14.7 lbs. per sq. in. absolute, P₂ absolute; to find P'. P2 = 415 lbs. per sq. in. Align P₁ = 14.7 and P₂ = 415 (two stage compression scale), cutting the P'-scale in P' 78 lbs. per sq. in. absolute. (2) (Three stage.) Given P₁ = 14.7 lbs. per sq. in. absolute, P2 14.7 lbs. per sq. in. absolute, P₂ = 2400 lbs. per sq. in. absolute; to find P' and P". Align Pi = = 14.7 and P₂ = 2400 (three stage compression scale), cutting the P'- scale in the required value P' 80.3 lbs. per sq. in. absolute, and cutting the P''-scale in the required value P" = 439 lbs. per sq. in. absolute. REMARKS: Pressure in lbs. per sq. in. absolute = (pressure in lbs. per sq. in. gage) + 14.7. III-1 10 20 (P,) 19 السلس 18 17 16 15 14 Initial Pressure (P) in Lb. per Sq. In. Absolute (283 Stage Compression) 8 (P) 13 12 سلم مسلس (P2) BOD (P2) 4000 ................ 600 550 500 80 450 400 350 60 55 50 300 (p1) 25 30 35 40 45 50 55 60 70 80 90 First Stage Pressure (P') in Lb. per Sq In. Absolute (3 Stage Compression) Intermediate 35 1 100 110 Pressure (P) in Lb. per Sq. In. Absolute (2 Stage Compression) ………….........................Toolblodford § 10 56 70 120 السلس 130 (P') 120 110 سلسلالالالالا 00 45 250 י' Second Stage Pressure (P") in Lb. per Sq. In. Absolute (3 Stage Compression) 200 190 180- 30 170 180- 150 25 140 130- 120 20 110 100 (P") 1 1 1 1 250 (P₂) Pressure (P2) in Lb. per Sq. In Absolute (2 Stage Compression) Final 200 300 3500 450 450 3000 400 350 (3 Stage Compression) Lb. per Sq. In. Absolute 2500 لسل للسلسلات اسيا 888 40 30 800 15080 Pressure (P2) in 120 20 110 00 100 00 8 89 لسللللللل 70 Final 450 (P2) (P2) 60 400 600 III-2 NO. III-2 AIR COMPRESSION - SINGLE STAGE. (Preliminary to Chart NO. III-3) FORMULA: NOTATION: (See Chart NO. III-3.) DIRECTIONS: Alignment scales. M₁ P₂- 1 A P2/P1 n 1 - 1 M₁ The compression pressure ratio scale P2/P1 also carries a scale of final pressures P₂' in lbs. per sq. in. absolute, when the initial pressure is 14.7 lbs. per sq. in. absolute. The value of M₁ is to be used in transferring from Chart NO. III-2 to Chart NO. III-3. 1 III-2 95 Compression Pressure Ratio (PR) G 16 15- 14 10- -150 -140 -160 12- -170 13- -200 88 14.7 Lla per Sq. In. Abs. 50 -80 80 ∙110 -120 130 Pressure (P) in Lb. per Sq. In Abs. when Initial Pressure & Final 35 (H/P) ___ (P) 20- 19- 18 -----300 17-250 1.35 1.30 1.25 1.20 سلسل 1.70 1.60 (n) 1.50 1.40 (M₁) 1.5 -1.2 1.405 Adiabatic for Air 17- 1.10 1.15 Compression Exponent (n) Constant (M) for use ∙1.0 -0.9 -0.8 on Following Chart Na 3. -0.7 -0.6 1.05 (n) זזיןייייך ་་ है III-3 NO. III-3 FORMULA: AIR COMPRESSION SINGLE STAGE 144 n n 2 h.p. P₁ 330 -1} n NOTATION: 1 144 P. (1) M 330 Power h.p. in horse-power required to compress a given weight of air from an initial pressure of P₁ pounds per square inch absolute and a volume of 100 cubic feet to a final pressure of P2 pounds per square inch absolute, with a compression exponent n. DIRECTIONS: Alignment scales with one turning line. M₁ h.p.' h.p.' h.p. n P₁ The h.p.'-scale serves as a turning line and also gives the power required when the initial pressure is 14.7 lbs. per sq. in. absolute. The quantity M₁ EXAMPLES: ( P₂a n 1 n - 1 must first be found from Chart NO. III-2. (1) Given P₁ = 15 lbs. per sq. in. absolute, P₂ = 150 lbs. per sq. in. absolute, n = 1.25; to find the power h.p. Chart NO. III-2: 0.585. align P2/P₁ = 10 and n = 1.25, cutting the Mi-scale in M₁ 1 1 Chart NO. III-3: align M₁ align M₁ = 0.585 and n = 1.25, turn on intersection with h.p.'; align h.p.' and P₁ = 15, cutting the h.p.-scale in the required value h.p. horse-power per 100 cu. ft. of air. - 19.15 1.25; (2) Given P₁ = 14.7 lbs. per sq. in. absolute, P₂ = 147 lbs. per sq. in. absolute, n = to find the power h.p.'. Chart NO. III-2: align P₂ = 147 and n = 1.25, cutting the Mi-scale in M₁ = 0.585. Chart NO. III-3: align M₁ = 0.585 and n = 1 1.25, cutting the h.p.'-scale in the required value h.p.' = 18.75 horse-power per 100 cu. ft. of air. REMARKS: 1 To find the power required to compress an initial volume of V₁ cu. ft. of air, multiply the result obtained from this chart by V₁/100. III-3 0.9- 70- (M) 60- 0.8 50- 0.7- 06- 0.50 0.45 0.40 0.35- Q.30- 0.25 ارد ויויו זיויויויןיויוידי ידידי ין 1152 Value of Constant (M) See preceding Chart No. 45- 40 35- 30 ויזיויןיויויו וידיוידי 25- 20- 18 17 16 15- 14 13 12 0.20- 6 019- 0.18 5.0- 017- 4.5 910 015 4.0- Q14 3.5 013 012 اله oro יווי יוידיוידי سسة (M) 2.5 (h.p.) (h. p') (h.p.) 28 8 8 Turning Line (h.p!) or Power Required when Initial Pressure = 14.7 Lb. per Sq. In. Abs. Theoretical Power Required (h.p.) in Horse-power per Hundred Cu. Ft. of Air per Min. at Initial Conditions རྒྱང་བའདས།སྟེ་ཆུ་པ རྒྱུ ་ཆེ ཐ 10 હું જે છે એ -15 38 50 60 -80 70 (n) Compression Exponent (n) ·1.10 -1.15 ·1.20 ·1.25 -1.30 -1.35 흐 ​ Initial Pressure (P) in Lb. per Sq. In. Absolute لسلسسسسسسسساسسلسل سلسالسلسسلسالسلسلسة སྱེ་ (P) -12 13 Tr 14 -15 G -16 ·17 ·18 (n) [1.40 -1.405 Adiabatic for Air III-4 NO. III-4 AIR COMPRESSION - TWO STAGE. (Preliminary to Chart NO. III-5) FORMULA: NOTATION: (See Chart NO. III-5.) DIRECTIONS: Alignment scales. M₂ n 2 20 - 1 P₁/P₁n M₂ The compression pressure ratio scale P2/P₁ also carries a scale of final pressures P2' in lbs. per sq. in. absolute when the initial pressure is 14.7 lbs. per sq. in. absolute. The value of M₂ is to be used in transferring from Chart NO. III-4 to Chart NO. III-5. III-4 Compression Pressure Ratio (Pa/p) (P³/R) + 250 Final Pressure (P) in Lb. per Sq. In. Abs. when Initial Pressure = 14.7 Lb. per Sq. In. Abs. 15 14 -200 13 -190 ·180 Adiabatic for Air 1.4051.4 ∙1.60 -1.70 (n) ~1.50 Compression Exponent (n) 1.35 1.30 1.25 ·1.20 1.15 1.10 (n) 1.05 זי יזיויו דיויויויויויויויןיויויו דיר זיוייו -0.20 -0.25 0.30 0.35 Constant (M2) for use on Following Chart No. 5. -0.50 0.45 -0.40 ז'ןיויויזיויןיויויויויןיויזיזיזין יזיויויזיןיו זידידי ·0.65 -0.70 (M₂) -0.60 0.55 III-5 NO. III-5 FORMULA: AIR COMPRESSION - TWO STAGE n h.p. 330 n 1 134 P. (2011) (P)-1} n'- 2 144 330 P(2) M n 2 NOTATION: 1 Power h.p. in horse-power required to compress a given weight of air in a two stage air compressor from an initial pressure of P₁ pounds per square inch absolute and a volume of 100 cubic feet to a final pressure of P2 pounds per square inch absolute, with com- pression component n. DIRECTIONS: Alignment scales with one turning line. M2 h.p.' h.p.' h.p. n P₁ The h.p.'-scale serves as a turning line and also gives the power required when the initial pressure is 14.7 lbs. per sq. in. absolute. The quantity M2 = P₂1-1 n P₁ 2 n - 1 must first be found from Chart NO. III-4. EXAMPLES: (1) Given P₁ = 15 lbs. per sq. in. absolute, P₂ = 300 lbs. per sq. in. absolute, n = to find the power h.p. = 1.25; Chart NO. III-4: align P2/P1 20 and n = 1.25, cutting the M2-scale in M₂ 0.35. Chart NO. III-5: align M₂ = 0.35 and n = 1.25, turn on intersection with h.p.'; align h.p.' and P₁ = 15, cutting the h.p.-scale in the required value h.p. 22.9 horse-power per 100 cu. ft. of air. (2) Given P₁ = 14.7 lbs. per sq. in. absolute, P₂ = 294 lbs. per sq. in. absolute, n = 1.25; to find the power h.p.'. 2 Chart NO. III-4: align P₂' = 294 and n 1.25, cutting the M2-scale in M₂ 0.35. Chart NO. III-5: align M₂ = 0.35 and n = 1.25, cutting the h.p.'-scale in the required value h.p.' 22.9 horse-power per 100 cu. ft. of air. REMARKS: = To find the power required to compress an initial volume of V₁ cu. ft. of air, multiply the result obtained from this chart by V₁/100. For the intermediate pressure of a two stage air compressor see Chart NO. III-1. III-5 075- 0.70 0.65 0.60 Q55 0.50- 0.20- Q25- 0.30 0.35 山 ​0.40 0.45 Value of Constant (M2) See preceding Chart No. III-4 a/9- 018 017 016 0.15 Q/4 013 0/2 اله (M₂) ויזיזיזין. (M2) (h.p.) Turning Line (h.p!) or Power Required, when Initial Pressure 14.7 Lb. per Sq. in. Abs. ཕྱི (h.p.') لسل · 19 -18 -17 91. 20 -200 (h.p.) (n) 1.10 יון سلسسلسا Theoretical Power Required (hp.) in Horse-power per Hundred Cu. Ft. of Air per Min. at Initial Conditions -4.0 4.5 יד TTT 25 ייןייויןיונדיזין ויזידיו 30 35 35 -40 $ 55 ·60 88 8 100 8888 150 1 Compression Exponent (n) 1.30 ·35 1.35 3.0 1.25 1.20 (n) -1.40 1.405 Adiabatic for Air 1.15 Initial Pressure (P) in Lb. per Sq. In. Abs. 1 ་་9 -10 12 14 15 ·16 لسلسلسلسل مسلسل سلسان لسل سيلا (P,) , III-6 NO. III-6 AIR COMPRESSION - THREE STAGE. (Preliminary to Chart NO. III-7) FORMULA: NOTATION: (See Chart NO. III-7.) DIRECTIONS: Alignment scales. M3 11 P₂ д-1 3 д - 1 P₁ P2/P1 n M; The compression pressure ratio scale P2/P1 also carries a scale for final pressures P₂' in lbs. per sq. in. absolute when the initial pressure is 14.7 lbs. per sq. in. absolute. The value of Ms is to be used in transferring from Chart NO. III-6 to Chart NO. III-7. III-6 } Compression Pressure Ratio (P2/p) (P2/p,) 300- 250- 200- 190- 180- -3000 170- 160- -2500 150- 140- -2000 130- -1900 —1800 120- —1700 110-1600 ·1500 100 00 -1200 80- -1100 Final Pressure (P') in Lb. per Sq. in. Abs. when Initial Pressure = 14.7 Lb. per Sq. In. Abs. -900 -700 ·600 40 35 500 30. 450 (P') 400 Compression Exponent (n) 130 -1.20 Adiabatic for Air 1.405+1.40 1.70 (n) -1.60 ·1.50 (M3) ·1.15 ·1.10 Constant (Mg) for use on Following Chart No. II-7 זיןיויויויויןיויויויויויויויויויןיז זיויוי 0.50 -0.55 -0.60 -0.65 ויויויויויןיוידיויוין וייויין (n) (M3) 1.05 ·0.10 יויויויןידידי -0.15 -0.20 זיןיויויויוייז 山 ​-0.25 III-7 NO. III-7 FORMULA: AIR COMPRESSION-THREE STAGE NOTATION: 144 n h.p. 330 P₁(3)()-1) 2 n 144 n 330 P. (1 321) M Power h.p. in horse-power required to compress a given weight of air in a three stage air compressor from an initial pressure of P1 pounds per square inch absolute and a volume of 100 cu. ft. to a final pressure of P2 pounds per square inch absolute, with compression component n. DIRECTIONS: Alignment scales with one turning line. M3 h.p.' h.p.' h.p. n P₁ The h.p.'-scale serves as a turning line and also gives the power required when the initial pressure is 14.7 lbs. per sq. in. absolute. The quantity M3 EXAMPLES: (1) Given Pi = = (P₁) 2 A 3 д - 1 1 must first be found from Chart NO. III-6. 15 lbs. per sq. in. absolute, P₂ = 750 lbs. per sq. in. absolute, n = 1.25; to find the power h.p. Chart NO. III-6: align P2/P1 0.298. align P2/P1 = 50 and n = 1.25 cutting the Ma-scale in M; - Chart NO. III-7: align M3 = 0.298 and n = 1.25, turn on intersection with h.p.'; align h.p.' and P₁ = 15, cutting the h.p.-scale in the required value h.p. horse-power per 100 cu. ft. of air. 2 29.3 (2) Given P₁ = 14.7 lbs. per sq. in. absolute, P₂ = 735 lbs. per sq. in. absolute, n = 1.25; to find the power h.p.'. Chart NO. III-6: 0.298. align P₂' = 735 and n = 1.25, cutting the M-scale in M、 = Chart NO. III-7: align M, = 0.298 and n = 1.25, cutting the h.p.'-scale in the required value h.p.' 28.7 horse-power per 100 cu. ft. of air. REMARKS: = To find the power required to compress an initial volume of V₁ cu. ft. of air, multiply the result obtained from this chart by V₁/100. For the first and second intermediate pressures of a three stage air compressor see Chart NO. III-1. III-7 0.75- 0.70- 0.65- יויוין (M₂) (h.p.) 150- 140- (h.p.) -300 250 130 120 0.60- 110 0.55 100- 90 Q50- 045 040- 0.35- 0.30 Q25- 020- 4/9- 018 no Q17 Value of Constant (M) See Preceding Chart No. 16. 80 70 60 50 45- 40- 35- יו Turning Line (hp) or Power Required when Initial Pressure = 14.7 lb. per Sq. in. Abs. 25- 20- 19- 18- 17 16 15- M 016 13 0J5- 014 013 012- MC اله મા 12- 〃 "1 ເ لسا (M₂ (h.p.) Conditions سلسلس ཅེཥུ ༄ སྐྱེ དྨེ ལུཿ བཿ ཝཿ * ཙྩི Theoretical Power Required (h.p.) in Horse-power per Hundred Cu. Ft. of Air per Min. at Initial गगगगगगगग 3.5 " 2 سلسل -30 -35 -50 -45 -40 06- -100 Compression Exponent (n) (n) 1.10 (P) 20 1.25 (P.) 1.20 - 1.15 Initial Pressure (P) in Lb. per Sq. In. Abs. ཅེ་ -12 13 14 -15 91- 17 -1.30 1.35 3.0 1.40 (h.p.) -1.405 Adiabatic for Air (n)