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I ,~ 3 " o J . .1 5” I": . . . ‘ ". .'. 1' g _ . .3 ‘ ~ .: . . . -' 4 ' .'{Jdll.‘U-g.:.’,i.f':.‘.;."-' 0 .'~. 3 Au,- _;;.:'.'_"‘J|l|.0'.fl' .‘ “ A. .3 o «.0 n .J.’ ‘0 2 v .' .:‘9. . — :-':~ 3‘ h 304‘ "‘v. .6.qu ('fifdbw-fl*"" , “I- I .- 'I-:_'.' IIHINH CISHGAN STA | "HI HUN!"lHlllHl!INHIHUIHHHIWI 293 00628 6037 -_> _.A kaA‘ L“ .J i " . .. r 5 . w «grin... - .w. . l Haida" '0' \ .L .- I. l a. 4 ‘W $.12: . Vulo‘u’firhflu . N d] u‘ol cyl‘ flTL . l‘ W- ‘Itl' III: V.t|4o" THE APPLICATION OF NOMOGRAPHIC CHARTS FOR THE SOLUTION OF VARIOUS ENGINEERING PROBLEMS A THESIS SUBMITTED TO THE FACULTY (I . MICE IGAB STATE COLLEGE OF AGRICULTURE AND APPLED SCIENCE BY. Banana 0 hoBERTs Candida“ For The Professional Degree or Electrical Enema: ‘ June 1934 TH 5513 ._ , A .d~.-.fi HISTORICAL DATA 10343!) The history of the theory of equations takes us back to the year 300A.D. when Diaphantos, who has been called the "rather of ngebra", developed the use of abbreviations and symbols for.11gebra. However the history of graphical representation of functions goes back six hundred years earlier to the time of Euclid, Who represented certain functions by constructing graphs in the form of rectangles. Later in history, about 1500 A.D., in the writings of Bhaskara, a Hindoo scholar, we find the first steps taken to- ward constructing graphs of equations when positive and negative numbers were represented by opposite segments of a straight line. From.this conception, it was but a step more to graph the relationship of one variable to another by measurements along two lines. .This was done by Nicholas Horem, a French teacher, in the fourteenth century. At that time negative numbers were considered fictitious, so the graphs of Horem were constructed in one quadrant only. At the beginning of the seventeenth century, Rene' Descartes laid the foundation for graphical representation of functions in all quadrants. Descartes was a French phil— osopher and mathematician, born at La Rays in 1596. He adapted the system of locating points in a plane by means of their distances from.two fixed lines, and introduced the terms "ordinate" and "coordinate". The term "abscissa" was first used by an Italian writer about 1659. The Nemegraph, or alignment chart as it is sometimes called, is a special type of graph generally used to solve certain equations which would require considerable time if treated by mathematical analysis. This type of graph also originated with the French mathemticians. A most compre- hensive treatise on the subject was published by Professor Maurice d' Ocagne of Paris. In the field of Engineering a limited number of ~ formulae are used so repeatedly that any mans of conserv- ing time and effort in their application is desirable. The nomegraph is readily adaptable to this purpose since its scales are so easily constructed and read. or course, the nomograph is not to be used as an exact solution, since the accuracy of the graphical solution depends upon the precision of the diagram; however, the results so found may be regarded as a very close approximation and generally are within the limits of accuracy required in practical engineering. GENERAL APPLICATION OF GRAPES Graphs generally are constructed and used to present a picture of the variations of a given function, or the relation- ship of one function to that of another. Innctions of two variables are generally plotted as in Fig. l and functions of three variables, as in Fig. 2. star (I) 1‘3“” f JIP L111 rvrrtvfirr-vrrvvrivvv‘r x - VYTTT'TT"‘T"""7"I‘IT X Fig. 1 Fig. 2 To read the values of a point on these curves requires interpolation in two directions i.e. along the coordinate ales. This is generally mentally fatiguing especially if several readings are to be made and the divisions on the axes are small, in which.case, errors are easily made. A.function of two variables may be graphed in scale form. where interpolation is accomplished easily. As an example, we will construct a small scale showing the relationship of the radius of a circle and the area of the circle: Radius of Circle 0L t 21 =2 4 5 L . A 1 i ' I‘PI'I‘I'W'WI * ‘5 ‘ 1 T1, ' I ' r , :41 O 5 10 2O 30 4O 50 80 Area of circle "‘ ’ .t‘ It.‘ll Such scales are quickly made, easily read, and where used often, are decidedly advantageous over the type of graphs represented by figures 1 and 2 (page 5). SCALES Throughout this thesis the term, SCALE, will? be frequently used, and in each case it refers to a graphical I scale. Lipka defines a graphical scale as "a curve or axis on which are marked a series of points or strokes correspond- ing in order to a set of numbers arranged in order of magnitude." In the construction of nomegraphs, the scales are I generally non-uniform, depending upon the type of equation and its degree. In the following work equations will be developed for plotting tlm scales and will be expressed in terms similar to a Cartesian coordinate eqmtion with reference. to a given origin. A A factor of major importance in constructing the scale is the scale modulus to be used for each variable. We my define the SCALE MODULUS as a number (or fraction), which, when applied to the 1(1) we wish to plot, will determine the distance along the ordinate or abscissa for plotting the various values of x. In the subsequent work this dietance will be in inches. for instance suppose: rm - a x2 + 10 Assume X varies from O to 5, and that the " scale is to be 10 inches long. The scale modulus would be the scale length divided by the largest value of fix), which in this case would be: 10 - 1 IE- If we plot this f(I) along the "Y" ordinate, the plotting equation is: 2 y - (Modulus) fix) - 1 ( or + 10 ) inches. 1'6" Substituting values for I in fix), gives x - c 1 2 s 4 5 y - 5/8” 1" 2-1" 4" c-sn 10» '5 S Plotting these values produces a scale like the following: Values of I t ? 4 p I I I I 1 * f(X) 2"- 3" 4" 5" 6" 7" 8" 9" 10" Values of y ‘ -‘ Origin The scale is calibrated much the same as a slide rule, i.e. instead of marking the value of the f(X), the value of I is marked off at the point corresponding to tlm valm of NM. 101'! that the origin of the scale does not always coincide with the zero value of the variable I, however, in most cases it will or can be made to correspond by rearranging the equation unless the scale is logarithmic. THEORY OF THE NOMOGRAPHIC CHART .LU 11 at the present thee there are several good text books published which treat of the construction of nomegraphs. Each text adapts a different geometrical solution for the plotting equations; however, the final results by each method are identical. Rather than attempt a new geometrical solution for these equations we shall apply a mathematical interpretat- ion which.has the virtue of flexibility in arranging the variables. In an attempt to compare a graph constructed with Cartesian coordinates and a nomograph for the solution of the same equation, let us assume: Y r,cx) + f£(Y) + f3(zi) + C - 0 (where C is a constant) A, ’9 \.\ tfi’ Kf)‘ i.“ Z N- ._ ,4 _ s >4 .--- o : m ~ “'3 O : 3w - s : > ,1,” ‘ H e t a Fig. 1 Fig. 2 It appears that in Fig. 2 the scale of I and Y’in Fig. l are rearranged and that the curves zl , 22 and 23 became points on a third scale f4(zi) located somewhere between f1(x) and fz(Y). The plotting equations are developed to locate the position of these scales and to give their calibration. The scales 'f1(x) and fth) are so calibrated that when a given value of I and Y’ on.their respective scales are Joined with a straight edge, the correct value of 2 on the f4(§) scale may be read. 12 Let us consider the solution of these plotting equations nor the different scales by means of determinates. Fig. 3 Referring to the above Fig. 3; if a, b and c are in a straight line, then 1- ‘8' I‘1--"u"‘s ’2 - ’1 13 - 32 Or clearing fractions, 2. -‘ -x +x + -x -0 I[1‘72 1[1’3 2’1 aye 1:5’1 3’2 Iqmticn z may be set up in determine/xteforn, as 1 x1 1'1 l x y 13 The solution of an equation by a nomograph requires that three specified points that satisfy tlm eqmtion must be in a straight line. Examining this determinate (eqmtion 3) we see that x1 and y1 refer to scalar distances. In a preceding; part of this work, covering scales, it was shown how to construct a scale for the fix) expressed in scalar distances. So referring to the determinate (equation 5) we 1 and y1 by the functions they represent. In other words, 11 - m1 fix) and y1- M1 P(X) may replace x where: X is any variable. 1a1 is the scale modulus along the x axis. I1 is the scale modulus along the y axis. Referring to Fig. 3, we may consider "a" as a point in one scale, "b" a point in a second scale, and "c" a point in a third scale. Hence, if we can write the equation in determinate form of the third order so“. that only one of its fariables will be in each row, a nomograph can be constructed for the solution of tin equation, providing the value of the determinate is equal to zero. Throughout this discussion, only third order determinates are considered. Assn. tlm variables to be U ',- V and I. We may then write equation 3 in this manner: (I) (y) 1 nuns) nurun d. l mvffi) 19F”) ' 0 4) 1 mwfll) I'M!) F 14 This is a general type of determinate which would produce plotting equations for curved scales. These equations would be: “gr-Infill) Viv-15,317) H~Y=M,I(W) x - mu rm) ‘- x - mv f(V) L, = m, f(W) Ihere: x is the distance from the origin along the x axis. y is the distance from the origin along the y axis. If a determinate is eqml to zero, we may perform the following operations without changing the value of the determinate. 1. 2. 3. 4. 5. Interchange any two columns. Interchange any two rows. Add any two columns or any two rows. Multiply any column, or row, by a number other than zero. Divide any column, or row, by a number other than zero. Generally an eqmtion of three variables, f(U,V,W), can be expressed in one of the following forms: 5. 6. 1'1“!) + f2W) + f3”) + C - 0 Where 0 =- a constant. f1(U)‘fz(V) + ta") + C - 0 If an equation contains more variables than three, it can be reduced to three variables by the introduction of a quantity that represents the sum of the excess variables. For instance 15 assume the eqtmticn to be: f1(U) + raw) + f3(l) + r4(x) + c - 0 Let: f5(l) + r4(x) - r5(r) Ie ny then construct a nomograph for the solution of this eqmtion, and a second nomograph for the solution of the equation: f1(U) + f2(V) + f5(Y) + c - 0 These nomographs generally can be combined '30” that :5”) scale is used as a TURNING SCALE and does not require calibrating. In like nenner, equations containing several variables can be solved by ming turning scales and combining the nomographs . We will now proceed to develops the plotting eqmtions for the scales of each variable. First consider the type equation: 5. . f1(U) + faW) + to”) + c - 0 This can be set up in determinate form, as (a) (22:1 1 O‘ - 11(0) 1 1’3(W) + C Let: tlm modulus of the scale for f1(U) - Ma :0 w w w w w f2(V) .. MY " abscissa distance between the seros of tin above two scales - D inches. In determinate '7 : 8. 1/uu O ingl llultiply column (a) by l/Mu -Mu flu?) -Mv fan!) f5”) + c ' w (1:) ~ 1/14,r we then have: (by wing the operations stated on page 14) IL/MT -fz(7) 8 0 1 1/11' :30!) +0) w, 1/111u l/MY (b) O - Mu f1(U) l - Mv f2(V) "u Mu My r w + 01 I, + M. --—— [3‘ ’ J by D, and we have: 9. (I) c D .5133... In”. Mu+MY (‘7) - Mu f1(U) - Iv f2(V) JAE. [ram +0] Mun. Bow multiply column (b) of the above deteminate :- O .L'/ The plotting eqmtions are: y- -M f (U) inches. For U { u 1 x- 0 inches. y ' - Mv fan?) inches. For Y { x - D inches. v - Mu M. For '1 W [fSUH + C] A inches. I - M11 D inches. + 7 This will produce a nomograph of three parallel scales, such as the following figure: c a r E I A L I I 1 Values of W Values of Y rfi 1.6 Is will now consider the type equation, as 6. 11m - ram + ram + c -= o Iqmticn 6 can be set up in the following determinate form: (a) (b) c 1 - f1(U) 1o, 1 c - r30!) 0 1 f2(V) C Let: the modulus of f1(U) - Mu w w w f3(w) . MT " abscissa distance between tlm zeros of scales f1(U) and f3”) - D inches. In determinate lO: Divide column (a) by M. w n (b) n “u Ia then have: 0 1m - f1(U) F 0 l - Mu f1(U) 11. 1/1' c - f3(w) :- 1 o «- u. to") - 1m“I 1 raw) 0 1 1 raw) 0 (a) . (b) .1 1 ‘- Mu f1(U) 1 o - M' to”) - c 1 ll. 1'2”) Mu a, 0 Mn + I. 1am Na + “- fem Now multiply column (b) of the proceeding determinate (11) .by D, and we have: (I) (I) l D - Mu fl(U) “3+:12‘” “11"wa2‘“ The plotting equation: are: Y " - f (U) inches. ror u{ M“ 1 x - D inches. 1 - n“ I" 0 inches. m vf n, + u. ram 1 " “I D :2”) inches. "a + llw 13(7) y - - I. fsm) inches. 1.. .( x :- 0 inches. 19 20 If C - O, the plotting eqtmtions will produce a nomograph commonly known as a "Z" chart, such as the following figure: 3 aw; —a—_— J 1 l 1 Values of I l Valms of U l Origin j If 6 f 0, the plotting equations will produce a Fig. 5 nomograph of general configuration, such as the following figure: I 1‘3 '3? is -. E O c. m _ o ,8;- .,. a n3 ‘ ’> J Origin/”O snx‘ {— ‘rm .- 1 (J H'X/S As a special case of equation 4, let us consider the equation: (a) (b) (O) 1 1 O 'W 1 1 14o o 1 -T-%VT = O 2 1 1 “WW1 3 Let the modulus of 'f1(U) - mu m n w w fz(v) . M7 ~ In determinate ld: multiply column (a) by uh ~ - m an, .. ~ (a) -‘(-1) and we have: (7) (I) ° .2— r (U c "a 13(0) . “a 1 ,’ 15 o “7 1 - 0 H' f8(V) 1at" lln "v m1 “a to”) “v to") s 1 1 21 Hence the plotting eqmtions are: ,ry- I! f (U) inches. rcr Ué ‘1 1 I I 0 IDODOS e . y '3 0 11101108. For V f x -‘ IL, f3(V) inches. /y ‘ ll“ f5(W) 1116138. For I I - Iv 133(W) mmle Ihieh would produce a nomograph like the following figure: Fig. '7 22 Another method of plotting the f3(W), is to determine tlm modulus, 11', for this scale. We proceed by locating any point, P , on the equations deve10ped on the preceding page. \ “L \9 - \Q \ 9 (\[JW 85' \7 /\ f“ 0 I, T‘\ 7‘ N; w Q“ (h! 3 I \a?’ 9441),, /\ .-» ’ " .. Jive- , r ‘2 eat/5 so ,’ / Y‘ o( :6 01/” fi(WP)// Fig. 8 Let the coordinates of P be: y - 11' fswp ) Let the distance from the origin to P be: M' fswp) From trigonometry: 16 sin 0 - Mu fswp) Mu 3 n a, ramp) 1:" 17. sin 180 - (O + (I!) _ M, ramp ) - ll. - sin O “u ramp) M u Or 13, 3111 :19 + i) - sin (0 + C) "1"11 My ‘TV": Having determined tlm scale moduli, M and My, fan?) scale by using the plotting and x - It“ f3(lp) - sin (0 + 95) sine ORB angle (say 0 ) my be assumed and the other angle W) . may be determined from equation 16. Then the modulus hi. my be determined from equation ’18. 24 As a special case of an eqmtion containing four variables, let us assume the equation: 19. flu!) +12”) - gut) + 1'5”” . rsw) + c a 0 Setting up this eqmtion in determinate form, gives: (a) (b) ( 1 o - raw) so. 0 1 - fa”) ' .. o L gm ‘5") 1’1”” + G Let the modulus of the scale for f2(V) 8 MY :0 w w - n w w n f3(W) - u. Let the abscissa distance between the acres of the above scales - D inches. In determinate 20: Divide column (a) by "v n to ‘ (b) n u' and we have: fi- 0 - ram V 1 I”) 1 f”) f(U)+O — 4 s 1 u' it: 50 By performing sons of tlm operations as stated on page 1d upon determinate 21, we arrive at tlm following determinatd: (I) (Y) 1 O - Mv f2(V) 28. 1 D - M. fan!) :- 0 1 , I. ram D n. H. [mm +c] 1:. mm + n, rsm H, mm + u, rsm Tb plotting eqmtions are: I " - ll, fa(V) inches '{ for x - 0 inches. 3‘ n - " {8(W) indies. For I x - D4 inches. y_ with”) +0] for U A: fl { I" a”) + “v :5”) 16,0 r5m x. mm + u, ram To graph the seals for f1(U) and I, assign desirable values for I and plot the scales of f1(U) for each of the given values of i . h 876 These equations will generally produce a nomograph having a network of scales, as in the following figure. Values of ¢ ¢ A sé 9%,, b F p: b ‘54 _ ta S ‘ g L g 31 E E E .. _Z)' ___-_,,-1 _, _ ____ r Fig. 9 '1 SOLUTION The solution of equations by leans of a nomograph is accomplished by utilizing the min statennt of the theory, i.e., that three specified points that satisfy the eqmtion must be in a straight line. Hence by Joining the value of U on the. f1(U) scale with a straight edge to the value of V on the f3(V) scale, we read the valm of I on the fa") scale that satisfies the eqmtion. 27. THE SOLUTION OF VARIOUS ENGINEERING PROBLEMS BY THE APPLICATION OF NOMOGRAPHIC CHARTS SECTION A ELECTRICAL PROBLBIIS 28 29. RELATIONSHIP OF VOLTAGE - KIA. - CURRENT 3U Practically all computations for the solution of electrical circuits involve either the load in Kva., or the load current in ampercs. A simple form of nomograph can be constructed to express the relationship of lead in Kva., Line current, and phase voltage. 1. 2. 3. 4. 5. 6. The equations relating these quantities, are I - EiL. for single pinse circuits. I - I—g‘n- for balanced three phase circuits. to may express the above cqmtions in logarithmic form log I - log Kva. + 1:; 3 = O for single phase current. log I - log Eva. + log E + log 1.732 :- 0 for three phase current. Setting up equation 3 in determinate form, we have: I 0 log Eve. 0 l - log 3 - 0 (single phase circuit) 1 l log I Setting up equation A in determinate form, gives: 1 0 log Us. _ O l - log L - 0 (three phase circuit) 1 1 log I + .23856 31 For limitixx values, assum: Eva. varies from 10 to 100. I " " l. " 1000. r w w 100 a 10,000. ror scale lengths, let: Kva. scale be 6 inches long. I n n g n n . Hence the scale modulus for log Kva. - 6 and " " " " log E - 3. Assume an abscissa length equal to 5 inches. Applying these moduli and abscissa value to dcternimtes 5 and 6, and reducing tb determinate so tint it contains a column of units, gives Iron determinate 5: '7. l 0 6 log m. l 5 - 3 log E - 0 (single phase) 1 10 8 leg I 3' Prom determinate 6: 8. l 0 6 Log live. 1 5 - 5 log I! - 0 (three phase) 1 10 2(log I + .25856) ‘5' 32 The plotting equations from determinate 7, are: y - 6 log Era. inches m- m.{ x - 0 inches y - - 5 log 1: inches. For I { (single phase) 1 - 5 Inches. y - a log I inches. For I { x - 5.555 inches. Tl: plotting equations from determinate 8, are: y .- 6 log Kva. inches. For KYa. { x - 0 inches. {y =- - 5 leg 1!: inches. For I (three phase) 1 a 5 inches. For I { x- 5.555 inches. Ie my cmbine the two nomographs produced by these plotting equations. In doing this re calibrate the current scale on one side for single phse current and on the othar side for balanced three pllse current. SOLUTION Join the value of Eva. and B on their respective scales with a straight edge, and read the value of I on the current scale. (See the tollosing sketch) EXAMPLE Single phase : 3 - 8000 volts Kva. - 50 I - 25 sips. (from current scale) Three phase: E - 2000 volts (phase voltage) no - 50 I - 14.4 amps. (from cun'ent scale) .q C O3 0 fl Km 0 , O 1'0”» (>1 0 l I l l I l I l I r111 rtrT1II1[111vlrvivf1ullrinrlrrvfil— KVA. N O 55 v .v v~~r.« flwL'FTT srldm - -AQ‘L‘) - I‘v‘adlgt‘t I - "J's."ba(.ali. b.11t—‘s- for jalancec A.C. Circuits 60' 1000 — 100 50 : 400 . 300 500 F 400 i 300 )5 200 L 300 40 is; 600 100 ' E E L-4OO S" s: 100 1“: "’" 500 g . U €« . _. In C) L. 50 50 1— s00 V IL: ;".' 5-1 ( f“ 3.4. so .2 T 1000 +41. +3. _ p so 51' * P ,7- ‘ .' L. . ,4 “r: 10 1 ' ;-— [I] .17 , 5' \’ *- 09) CH C "1. :7‘ b— ” ‘ :1... .1”: 10 V3 : 5.000 04 f.“ fl. P (‘3 5 ‘5'; b 52+: {—4 " r" 4 0'? —- 5000 3 a UFO! “’4000 ~5000 , _ L— 0000 1 - 7000 “-8000 l t:lOOOO (:3 m ngCTuICnL yysiu: PLANNIKG DIVISIpP HCR - 3 10-95-33 INDUCTIVE REACTANCE 34 .35 There have been several formulae developed for the computation of the inductance and inductive reactance of conductors. Such authors as LB. Rosa and 3.1.. Curtis of the 0.8. Bureau of Standards treat the subject very thoroughly, taking into account skin effect of the current md change in resistance ofnthe conductor. However, for ordinary distribution work, such refinements are not required. In this particular case, we shall develop a modified formula from the basic relationships presented by E.B. Rose in his paper # s - 80 of the U. 3. Bureau of Standards. INTERIM. SELF INDUCTANCE OF A CONDUCTOR 45‘ '\ \ / ’.1 .. f If I is the direct current flowing in the conductor of radius a, and uniformally distributed throughout, then the flux linkages may be expressed as: a. l. I I - .E (rt/a3) I (2 u I r/az) dr :- u I3 where u - the permeability of the ”ndm‘we The internal self inductance then may be expressed as: I. L - W I " 2.2.. per centimeter length of conductor. IITEBNEL SELF INDUCTANCE or A CONDUCTOR lhere the current is distributed uniformally over the cross section of the conductor. I l : ; IIIII X 4 The total flux , I , outside the conductor may be expressed as: '5/N - .. hm I'm...” o-..~ 1- 212(1ogfl-l) r The external self inductance may be expressed as: 4. Lac/1&21(1ogg;_-1) , 1‘ Hence the total inductance of the conductor of length 2. centimeters is the sun of the internal and external inductances which.nay be expressed as: 5. L 5' 8'L( log.§JL - l + u ) r '1 . 37 mun. INDUUTANCE OF TWO PARALLEL CONDIETORS The mutual inductance of two parallel conductors of length 7. , radius r, and distance apart d, will be the number of lines of force due to unit current in one which will cut the other when tte current goes to zero. This value may be determined by integrating equation 5 between the limits of e and infinity instead of between r and 1nr1n1ty, and dividing the result by the current I. Performing this operation we have , the mutual inductance M : 6. 11 1‘ 22(log 22 -l) whercdissmell compared toZ. TEE SELF INDUCTANCE OF A RETURN CIRCUIT Ie may consider a return circuit formed by two parallel conductors similar to the case of tw0 coils (of one turn each) connected in series so that their individual effects counter- act the flow of magnetic flux. The effective inductance for such a condition my be expressed as the following equation: 7a L. - 2L " 2 M Substituting equation 5 for the value of L, and equation 6 for the value of II, we have: 8. L I'- 4 7.( log d + u ) ccntimtcrs. ' '5 I -9 .'. 41(1032 +%)x10 henrys. r _ It is assumed that the permeability, u , of copper is unity. Hence for the inductance to neutral for one wire of length 2 centimeters, we lave: -9 9. L.*2Z(10g° g +i»)x10 henrys. r Changing the Naperian logarithm to the loglo , the inductance per centimeter length is: , . -9 10. L./cm. - 2 ( 2.50259 log10 g + it ) x 10 henrys. r d + .1085? ) x 10-9 henrys. 9- z x 2.30259 (10310 1‘ £- 2 x 2.50259 ( log10 1.284 g) x 10.9 henrys. - r The inductance per 1000 feet of conductor to neutral is: 11. I../1000' .3 .1404 x 10-3 1310 1.284 g henrys.. r reactance The inductive/per 1000 feet of conductor to neutral for 60 cycle frequency is: 12. I 1‘- .052916 logl.,284 5; ohms. (logarithm to the base 10) - r To solve this equation requires the use of a log table. For general computations we may construct a nomograph to save tin, and the accuracy of which is within the limits required for distribution work. DU Setting up. equation 12 in determinate form, we have: 1 0 log d 0 1 " 108 1e284 a o 1‘ 1 1 1 Assume: d varies from 2 inches to 200 inches. r " " .10 fl " 1.0 " . a scale length of approximately ‘8 inches. an abscissa of 5 inches. Tl: scale modulus for: d - 4 and for: r a 8. Setting up the determinate to accomodate these scale moduli, gives: (from 15) 14. 1/4 0 log d O 1/8 - log 1.284 :- r 1/4 1/s x 1 0 4 log d l 5 - 8 log 1.284 - O . r '1 5 4 x 5 .0525 From.determinate 15, the plotting equations are: y - 4 log d inches. For: d{' x - 0 inches. y - - 8 log 1.284 inches. For: r f r x - 5 inches. For: a y = 4 I inches. I § x - 1.666 inches. The above formula applies to solid conductor. For stranded conductor the inductance at 60 cycle frequency will be approximately 15 x 10.4 ohms less than for solid conductor of the same size. Hence on the reactance scale we may plot the inductance nor stranded conductor by proportional readings. On the radius scale, we may plot the wire size corresponding to the radius on one side of the scale and on the opposite side plot the radius in inches. SOLUTION Join the value of the spacing in inches in the d scale with a straight edge to the value of the radius (or wire size) on the radius scale, and.read the value of the reactance on the I scale. (See the following sketch) Solid conductor: Conductor spacing - 50 inches Conductor radius Reactance /1000' Stranded conducta': Conductor spacing Conductor radius Hcactance /1000' .50 inches .1115 ohms. 50 inches .50 inches 0 1102 Oh” e 41 H 42 am mammam monommmoo f‘v‘vllv‘w‘c '.J...‘Lrt I IVD”””‘ mwdom O 0 Nb 0 O O... ..>. A... ...,.. 6 O O O O O . . p P — b L p p p - p - p—pC-bpbp.Ppppp—bpp.bppp.—_:-<—P>:—~p.0 _ _ __._ ‘__Mm__r_r_4_l.dl: no no no u. "a u. "a "mu. / / I/ 2 3 4 0 0 O 0 m 0 0 m0 r0 0 O o 0 0 O 2 5 A. 5 f0 7 8 90. 1.. fig mg: moeobeaoo 950m I: 9.35 OOOH mam mommaaomlmm 3B0 4 2 . 9 8 7 6 5 a m 1 m m. ‘ o. o. o o 0. ere—PerP—banhp-p pPpehrbe-nbbh—bbt—bb Dupe—pfhpppE—bpb.—ebDP—>ePe—heennsh-ebrp erp—thb—e- quad-ud—d.41‘1naufiddddddddi‘dd4q-1‘dd,41¢ dd<-14411ddd_1adJ—4444—4‘d‘—¢¢dfifidauddalnfla‘d‘dqd‘ddd—d: A. 3 2 1 0 9 8 7 .b 5 1 1 1.. .1 1 0 0 0 0 0 e c e e a a a e a a pm?“ 003 mmm @939ng ammo 0—1 I O 0 r0 ... v A. 7Q mum; 02H ______: _ @0113; I _ _. I 9 7 0 no 9... 1. I 2H mmHoamm momobpmoo . L ‘. \J.4 .. 5...! . .’\ I“. \ , . V - 4— ma (:9 n SVI'Tt TV’ '.L'E ICA. -— .30 Lam-I130 13 ILIJJ V.” JJ 1)... 7. 2000 M."— .70 r; J ”D 1 L4 IMPEDANCE CALCULLT IONS 44 lhen solving electrical networks it is often desirable to express the impedance in polar coordinates rather than in rectangular coordimtes to facilitate nethemtical operations. Hcscvc, it sat be barre in mind that the impedance is a complex number and not a vector and due care must be exercised when .ing this expression in predate containing vectors. In a circuit under consideration: Let: B - the resistance I - the reactance z - the impedance. It can be shown by analysis that the followixg relations exist: 2 2 23-3 +x ‘7» - X C‘Mlx/R 4/79; _ R The express ion for the impedance in rectangular cocdinates is generally given as: And in polar coordinates as: 2. Z Z O The value of tie impedance in polar coordinates is: s. 210 -(na+xz)*/tcn‘lx/R work and a table of trigonometric functions. Finding these values requires considerable slide rule The solution of this equation is accomplished with ease and quickly by reams of a neuograph. The nomograph may be constructed from the relationships stated on page 44, and my be expressed as: 4. 5. 7. will produce a nomograph like Fig. 2 2 3 +1 ~Zz-O 2 R tanO -X .0 Setting up equation 4 in determinate form, we have: 1 0 22 1 - n 1 :2 Setting up equation 5 in determimte form, we have: 0 1 -a 1 0 12 1 tunes 0 Ixaaining tlm above determinates we note that determinate 6 will produce a nomograph like fig. 1 below, and 3 42:15 Fig. 1 2 below. .7 y‘ 1’5 u U \ (II--...I._._+.._1_._--I_____. - - - tlat dc terminate '7 If the correct abscissa lengths are chosen, the two nomographs (Pig. 1 and Fig. 2) may be combined. This may be accomplished by making the lengths of the abscissa between scales I and H in Fig. 1 equal to that between I and R in Fig. 2. Assam limiting values of 10 ohms for the scales I, B, and Z. And assume scale moduli of T1555 for tlm scales of H and 2 which will make these scales approximately 8.5 inches long. Assume an abscissa length eqml to 10.2 inches. These. odd figures were chosen in order to construct a suitable. nomograph on 8*" x 11" ”1301'- Applying these scale moduli and abscissa length to determinate 6 , gives: (1) (If) 2 ' 2 11.75 0 z 1 0 z IIOES 2 s. 0 11.75 -a - 1 10.2 ~22 -0 II.55 11.75 11.75 12 1 5.1 12 25755- Hence the plottixg equations are: y - 22/11.” inches 4 ror: x - 0 inches 1 - - Ila/11.75 inches For: B { x - 10.2 inches 2 y - I /25.50 inches For: I{ x - 5.1 inches 47 Using a scale modulus of l/ll.'75 for R, and a scale modulus of l/25.5O for I, and applying these moduli to dctcrnimte 7 , with an abscissa length eqml to 5.10 inches, we lave: (I) (I) 2 ' 2 = 0 11.75 - n _1 5.1 - a 11055 9. 25.50 0 x“3 - 1 0 x3 =- 0 '25'3'5'0" 25.50 11.75 tan20 0 1 5 1 tan2 a 0 2 + tan2 0 From the above determimte, the plotting equations are: y - - Ila/11.75 inches. 1% For: x :- 5.10 inches. 2 y - I /23.50 inch“. For I { x - 0 inches. y - 0 inches. tor: . g z I - 5c]. tan 9 1n“..e 2 + tan 0 Plotting these eqmt 10m produces a monograph as shown on page 48. SOLUTION Join the values of R an I on their respective scales with a straight edge and read the value of 0 on the angle scale and the valm of 2 on the impedance scale. EXAMPLE 0 R - 6 I - '7 From nomograph z - 9.22 and 0 - 49.4 48 CHART ILL); hDAN CL .01.. 3 au 4. Cu rpp r:-—:pp—-:~—-b.—bbrrr.—.PP-Pbbbppk bkaFh Pb r- mofifiomflm O .9 8 an! 6 1 VA 1 .0 a 7 4°. m NoZdBm H9. ME moving 1 8 . .Htkpprerbrp_PLr_p__Lr»PFLb.IL =R+jX 0 Z u) .U .1. ‘.LiCTRICAfiu SYS'T-..T. PLANNING DIVISION .4 J _|fi..s.—+.a..Q—q+adfi_ 0 1.. —Jfiq.—-#——dutqu—dd—ddfiq—fi—jdq—«du 8 N. n! 5 «My. _ _: 144—1211411— 4. A «u l 310 l \ b .iC'r lU-lB-b' r L} VOLTAGE RMUIJTION II SHOR '1‘ LINE S 50 The distribution engineer of a Public Utility supplying electrical power is very much interested in voltage regulation. It is of great concern to him to keep voltage fluctmtion within a very snll limit, but while designing a distribution system to aeeemodate this limit, he must also be conservative in the investnnt and must anticipate the future growth of load. The years 1927, '28 and '29 rather upset his ability to prognosticate load growth, while 1930, $1332 and '53 left him with an excess of capacity. Voltage fluctuation limits are generally between 2% and 5% for the general distribution system supplying municipalities. Ihere separate power lines are built for individual customers, this limit is much higher. However, for some types of loads that fluctuate so rapidly that the regulator at the substation cannot respond, the voltage regulation is generally limited to Iii at the substation bus. A typical load of this classification is the starting current of a large motor which lasts only a few seconds, but which generally cames flickering lights on the system. It is for this type of load that the engineer must compute the voltage drop on the system. The calculations are Imde seeming a constant voltage on the transmission system supplying the substation. Tb results are gemrally determined as a number of volts drop on son base voltage, rather than the percent regulation. In calculations of this type, the length of circuit is short 9110qu that the conductance and susceptance effects of the line my be neglected. METHOD I FOR LL66 ING POWER FACTOR 52 Theory \ \\ \ \ ,/—m. F? / I 3\ P» "' if I '1 ,—"' If I f 4’“. | “ ,-—/’/ _ ___ --_.-1 L____ __d_‘,,_'"r ,p_,__ , ,1 . L l V /’ Jfi"_/I ‘f In the above figure: 1. TC- 35 + 55 - tlm voltage drop From the gecmtry of the above figure: 3. III-mess! +nsinfi 3. EU!- IIcos «IRsinlla s Where: Is :- sending voltage til I receiver voltage - load current in amperew - resistance of line reactance of line uNwH II a power factor at receiver end of line. 53 4. K - E - voltage drop per ampere. cacos¢+xsin¢+flIIXco 2¢+IRsin2 ‘IRIsinz {33' £3 17' ¢ ‘3; 95) Generally tl's length, F5, is extremely smll compared to E and my be neglected; in which case: 5. x:-: R.cos ¢ +=x sin ¢ However, if it is desired to use the complete equation (4) ' for K, It will simplify the construction of the nomograph to limit the working range of the nomograph within a given relation- ship of the resistance and reactance drops to the sending voltage. Assume the IR and II drops to be 5 9501‘ 3', the sendirg voltage. Substituting these values in equation 4, gives: 6. K-Rcoafl! +Xsin¢ +.0251ccs2¢ +.025Rsin2¢ - .05 1 sin 0 cos I I 2 2 2¢ - R (cos ¢ + .025 sin 0) +I (sin e + .025 cos 0 -.025 sin Setting up equation 6 in determinate form, gives: 1 o " ' ' -R 7e 0 1 “I w (cos I + .025 sinzfi) (sin! +.025 oaszo ¢.025 nines) -K 54 $8. + as a: mac. .. n 80 + u 5: n 355 . a a... 5. 3o. .. ems. moo. + 5. n .. use , 1; .u «on e305 . M a I b. oedema n I H _,.. memos." o I H , m .x H “Oh m . m "Oh 355 N m I h, . memos.“ m a a he "one 33:23 $3»...on on» .m enemas—“Been flesh mac... em a? mac... a mac +k 33» 3o... no 3. one... u 8o + e n M... a... 5.. use... nouns 39+ u on. n H h... o n _ M m m a .o I»: m m o a .5 :3 ”serum 3. 332533. on mamas." eeeuoens one :52. cases 323 935.34 3305 n 3 Heads memos." seasons as as: .3» 3 3:. as: H es. m you 838.. 38. any .355 33 no news: 38. a e5 .H on. m use can 3 so 333 832.3 ass-3 55 SOLUTION Join the valms of R and I on their respective scales with a straight edge, and read the valm of K on the desired power factor scale. See sketch on folloiing page. For single please circuits: Voltage drop - s I K - z .5?- x 103 for three phse circuits: Voltage drop - 1.758 I K .- i? x 103 It my be scan from the above that the single phase voltage drop is twice the three phase voltage drop for the sale In. load, providing the single piece and three phase circuit characteristics are to sum. EXAMPLE B - 4.5 ohms I - 4.0 ohm Power factor - .70 From nomograph K - 6.0 (a) Single phase current - 5 amps. Single pulse voltage drop - a x 5 x 6.0 - 60 volts. (b) Three phase current - 5 amps. Three phase voltage drop - 1.732. x 5 x 6.0 - 51.9 volts. : .(‘ for “WW A.C. ACAW a4 ? M.....Hfi.a...a op m..o..-....wo..x.j u .... 0 0... av, .3 . 7+.- 7 iv .-a 4 7. n}. l O —:C—C:—::_:p.—::T:._::_:r.—:»p_....—_Fs.T.:PrZL...__::_...._.:Lsr:T:.:E._ .. ohm.....&u Mom assume... or .95 398/ "M :. 9m cl AU 9. no 7 .; ‘ a 5 a I i. m 0*. rp-Lrsann.-»_Idperrad.prp|rpbrfhpbpPbb... we. Peppq pr. be. P.~P .>P~Aw O I . x s L I z z z x xx 1 I I I; s, x x, s M. u. .. . I IV I I I I I o L o o. . a .3 on 11 .4 d . n (I. 1 in h . A 4/ / A; / 7/ dz; 4/ d», 7, F i J—fl F d —. — H .— . a z z , , . . . . . r / l / I o‘ o o y c o M B W» H G i , m. . . ._ 00 «.0... 0 MW Ob. .— F w I _ _ m s 1m. M... Au _ . aw _ emu _ AH _ J1, |F — _ — a _ . . . _ . . . . .R. u . . . i . H _ u h . . . . om. _ t k b h. p m P e. p p. . _. r a p b. b F k n . a L _. F m r _ a . a a q A a O K x a a. as s‘ .. .M .. .. _. u w . ”no1.x_1:.\_4s.\..i_.._. .4.._r_._.m.m P \\ \ \\ \s X a‘ \ as x .. .. .. . cm. 1x. 1 .. . x. i x. i \_ . x. 1 a i an . a . H». H 1 T i h ..J . \ \\\ \\ \\ \\ X \a a \ a. i .. .. O 5 . \ ~ . . . . -. . . . L no —‘4 < ¢\.\+4 d -\.\1‘# < <\q\— 4 d . \¢\— 4 14 “x— 4 q a H\— 4 a q «a— d “A 1; _ q q .u _ - .J — d a 4 s d . u a 4 __ 4 u d 4‘ n \ b x . ._ x r . i u... r\ u\ m x 9 x8 x7 \6 i. E... I. 1...... “z t 0 \ \ \ \ \ \ x \ \\ xx \\ x \x M x x \a .i l. \\ \ \ \ \\ x T x i ‘ _ \ .\\ x xx x X xi 1. . i . O x \\ \ \x x xx. \ X i i \ . . _ H. a. a. a. pie 9 a a m. 8e H: «Cb—pr..—.n.:—:pr—sst_:: ::_..: ::r:. a._Ls.: .ZLL. ._:_:: bb:—..: spirsp i 1144‘saus_...444..4—..qs_:ds_~d.daus..fiiq.~_.44~d4q««_1.4._..4.4¢_.4—444q..q«_—-.41—.4¢<—...._m.. ‘ 0 no on. n! 6 :1. AI, 7) . 2 W... 1 mapw30n ow warranwm m n m 0 \\.~ 1" HCR- H PLANNING DIVIsI ELECTRICAL I K phase I 1.73” I K D 6. Single phase a Three Voltage drop for if! ., ..- v. r... «I: METHOD II FOR LIGGING POWER FACTORS 58 lhere one works mostly with a given voltage, it is often desirable to compute the voltage drop using this voltage as the base voltage. Hence the valms of resistance and reactance for transmission line, substation, distribution circuit, distribution transformer, and secondary circuit are expressed in relation to this voltage in terms of an equivalent circuit. A nomograph may be constructed with scales calibrated for resistance, reactance, power factor, length of circuit or a multiplying constant, current, percent voltage drop, and volts drop on some other base voltage upon which voltage flicker limits are based. from Method I, we will use the formula 1. E-E'-Er-Incos¢+IIsin¢-r IIcos -IR81n¢)2 s Since the last term of this equation has so little effewt upon the result (except at unity power factor or where B is very large compared with I and the power factor is very poor) we say dispense with it for this nomograph. Generally the current and power factor are estimated in practice so that the inclusion of this term might assume accuracy which is not warranted. Hence we will use the abbreviated equation: a. B.-lr-IRcos¢+IIsin¢ Where: I - line current R - line resistance I - line reactance 0 - cos-1(Power Factor) 59 Let: I. - sending voltage to neutral. Br - receiver voltage to neutral. r - resistance to neutral per unit length of line. x - reactance " " " " " " " . L - the number of units of length of line. Substituting these elements in equation 2, gives: 5. E-Er-LI(rooa¢+xsinf) Tb. percent voltage drOp may be expressed as: 4. D- 100 E8"Er n100LI(rcos¢+xsin¢) “‘3. rs Let eqmtion 4 be broken up into parts, so that: 5. 1-100(rcosfl+xsin¢) Is 6. B-Ld Then: 7a D'IB Equations 5, 6, and 7 will form three determimtes for plotting three nomographs, and if we choose the scales correctly they my be combined into one nomograph having two turning scales which will not require calibrating. 10. UU Expressing equatisn 5 in determinate form, we have: 1 0 0 1 sin I - 100 I - A =‘ O - 100 r cos ¢ '1?“' IIpressing equation 6 in determinate form, gives: 1 O L O - A l 1 Expressing equation 7 in determinate form, gives: O B 1 - D l O To plot the nomograph, let us choose a length of lO indies for the scales of x, A, B, and D. Assume the following conditions: l.=- 24000/(3)i volts to neutral, the base voltage. 1‘ :4 r! t- H varie s from varies from varies from varies from varies from 0 0000 to 10 ohms. to 10 ohms (scale modulus - 120). to .087 (scale modulus - 120). to 100 to 10 61 Assuming an abscissa length of 10 inches and applying the given scale moduli to determinate 8, we have: 11. For: For tor of 2 (I) (y) l 0 twfix l 10 - 120 A I O 1 1o - H305 r cos p l +,sin I l + sin 0 From the above determinate, the plotting equations are: y - .866 x inches. =={ x - 0 inches y - - 120 A inches (do not calibrate this scale) AS x - 10 inches. {x . 10 inches. 1 + sin 5 Using the sane scale modulus for A and a scale modulus for B, and applying these scale moduli to determinate 9 with an abscissa length of 10 inches, we have: 0'.) <3) 1 O - 180 A 1 10 2 B I O 1 600 o L+60 Prom determimte 12, the plotting equations are: y - -1so A inches (do not calibrate this scale). For: A g x - 0 inches. y - 2 B inches (do not calibrate this scale). For B { x - 10 inches. (y - 0 inches. tor: I. x .. 500 inches. 1. + 60 Using the scale modulus of 2 for B and a scale modulus of unity for n, and applying these scale moduli to detrminate 10 with an abscissa length of 10 inches, gives: (I) (v) 1 o as 15. 1 10 -n - o 1 so 0 I + s From determimte 15, the plotting equations are: y - 2 B inches (do not calibrate this scale) For: 8 { x - 0 inches y - - D inches. For: I) { . x - 10 inches. - 0 inches. For: I g y “x - JL— inches. I "l' 2 63 The plotting equations from determinates ll, 12, and 15, will produce three individual nomographs. Since we have chosen the/53h moduli for the variables comon to each individual nomograph, we my combine these three nomographs into one nomograph and use the scales A and B as turning scales. This is similar to oomtructing the three nomographs and than superimposing them so that the scales common to each coincide. On page 60, we assumed the value of “GOO/(3)1} for Es’ which is the voltage to neutral of a 24 Kv. three phase transmission system. Hence the current, I, in determinate 15 is to be considered the three phase line current. We could construct another nomograph for single phase voltage drop; however, we may accomplish the same thing by constructing a second ---scale‘.. for single phase current on me some nomograph. tor a single phase system: (24 RV.) I. - 24000/2 to neutral Hence for single phase line current, the plotting equations for the current I , from determinate 15, may be expressed as: y- 0 inches. 1 { 20 '5 inches. 2 I/(S) + 2 1- See Sketch )9 HCR - 4 in rear of thesis. 64 SOLUTION (l).Toin the value of I with a straight edge to that of R on the desired power factor scale and locate the intersection with turning scale A . (2) Join this point with a straight edge to L on the length scale and locate the intersection with turning scale B. (5) Join this point with a straight edge to the value of I on the current scale and read the percent voltage drap. EXAMPLE x - 4 r - '7 Power factor - .80 L- 40units I=2amps., three phase From nomograph: Porcent voltage drop a 4.65 ( See Sketch # HCR - 4 ) MEIHOD III FOR LAGGIN} POWER FACTORS It is often desirable to compute the voltage drop on the overhead circuits only, where the impedance of the transmission line and substation may be neglected. Far cases of small loads, or where a quick estimate of wire size is desired for an individual power customer, this calculation is sufficient. Hence, we will attempt to construct a nomograph for close approximation using scales calibrated for conductor spacing, wire size, power factor, and volts drop per ampere to neutral. Using the approximate formula for voltage "droy- from Mehtod II, page 59, we have: 1. Ia-Er-Llucosfl +‘xsin¢_) For this nomograph we will. let: L - length of line in 1000 ft. units. I - line current in amperes. ‘ r - resistance to neutral per 1000 ft. x - reactance to neutral per 1000 ft. for 60 cycle frequency. From a proceeding part of this thesds (page 58), the inductive reactance per 1000 ft. of‘conductor to neutral for 60 cycle frequency was developed, as: 2. I - .0529 1% 1.884 .d. . a where: d - ccns’ucta: spacing in inches. a - conductor radius in inches. 67. Let equation 1 be broken up into two parts, so that: 3. E.-Er-LIK and: 4. K-rcos¢+xsin¢ Substituting equation 2 into equation 4, gives: 5. K = r cos ¢ + .055 sin i! log $1.284) ' a - rooso + .055 Sin ¢,(1og o + log _1__._s_a_4__) a - volts drop to neutral per amp. per 1000 ft. of conductor. lc will construct the nomograph for the solution of equation 5 only, since the solution of tlm voltage drop is (from.equation 5) Just a.matter of multiplication after the ”value of K is determined. Examining equation 5, we see that the rcsistencc,r, and the radius, a, of the conductor are directly related. Hence we will not require a scale for conductor radius if we construct a scale calibrated for wire sise. Setting up equation 5 in determinate fom, gives: 1 O - .055 log 6 60 o 1 K -0 sin I l r cos 5 + .055 sin 5 log 1.284 a 68 Assume a 3 inch scale for d and a 6 inch scale for K. It I varies from 1 to 100, its scale modulus - 30. If I varies from 0 to .60, its scale modulus = 10 issue the abscissa length - 10 inches. Applying trn above moduli and abscissa length to the determinate 6, we have: 7. For: For : For: (I) (y) ° - 1.59 log a 1° 10 K a 0 so 30 r cos 0 + 1.59 sin fl 108 l.i84 + . n 3 + sin 0 from determinate 7, the plotting equations are: y .- - 1.59 log .d inches .{ x - 0 inches. {y 10 K inches. K x - 10 inches. so r cos i + 1.59 sin d log Luz-2'1 y a ins. Iii-e sise and 3 + 8111 i Power Factor I - 3° inches. 3 + sin 0 69 . SOLUTION Join the conductor spacing,cn the spacing scale, with a straight edge to the wire size on the desired power factor scale, and read the value or K on the K scale. Then the: Three phase line drop = (5)13 L I K a KVa. I. K. Single phase line drop = 2 L I K a 2 Km. L K. ( See sketch on next page '70 ) EXAMPLE Conductor spacing -= 40 inches. lire size - #2 copper. Power factor = .90 From nomograph, K - .21 I - 10 amps. line current. Three phase line drop - (Ni x 4 x 10 x .21 - 14.6 volts Single phase line drop - 2 x 4 x 10 x .21 - 16.8 volts ( See sketch on next page) ”a ~ " U .J '37: 9.0-53 JO 1:! 1 ‘1 :11 OJ. com 23:11:31 ' 7 '.j) 1‘) :7) 3‘ . I Q - ‘- . .‘ " q 7-] O O O O . [#injllill[JIILI111114111111111 1111111111ll1111l1111 (.EO I ~ . § w Q 'a 0‘ Q 01: ’ I i . \ \ X //// ‘3; 0’” s. 0*" L: m * 2:» c3 0 .3 H 1 sh m at m~w-\\\\‘\- c3 c3 *4? fi‘C) O m (I: «It: I") O ' 4 ~' 1‘5 1* 1:; H \‘4. I I n- ‘w hdcu 2:213 5.51». H :2 O N F1 I I O. 3 O 0 a 2 n .1: 3 Do a. D. O h o '0 H o 0 2° " on H E . I! (0 E I: a ' '3 p. ‘.‘1 l‘ 3') 1 C) C) C t 1 3 C) OQC) C) Q '” C .. . .43, a-) N no V'H”; H C-3 “3 9,". Wlhwww I a»: WIDE: -l e._£iJS rename“ :I;,r3 'IU ‘5‘ [(11 SECTION B MECHANICAL PROBIEMS STRESSES IN THE STRAND OF ANCHOR GUYS '72 HEATIONSHIP BETWEEN THE ANGLE AND THE EIEMENTS OF ITS TANGENT In rough field work, or where instrument accuracy is not required, angles in the lead are generally measured by the tangent method. This is done by pacing off the legs of a triangle; one side (b) of which.is in.the direction of the lead and.the other side (a) in the direction of the offset. This method is also used to determine the steepness of pole guys, or. anchor guys. _-.b,——_,o o o n 9, a: ,h,/// lJffllD I / f .41.. a __4 From the above figures, the following relation may be $0) POLE expressed: 1e tan 9 - Q/b or in logarithmic form: 2. logtanO -10ga+1ogb-0 qution 2 may be set 'up- in determinate form, as: 1 o #1031: s. o 1 loge -o l 1 log tan 9 gives: 4. For: For: For: 74 Let: The scale modulus of a = 5 The scale modulus of’b - 6 The abscissa length .-4,5 inches. Applying these moduli and abscissa length to determinate I, (x) m l 0 - 610g b 1 4e5 3 1% a - 0 1 3 2‘ log tan 0 From the above determinate the plotting equations are: y'a - 6 log b inches. .{ x - 0 inches. y - 5 log a inches. a {I ' 4e5 inches. y - 2 log tan 0 inches. ratio a/b ‘ or ten 0 x a 5 inches. SOLUTION Join the values of a and b on their respedtive scales with a straight edge, and rdad.the value of a/b or 9 on the tangent scale. all-16 See shstoh on next page (75). EXAMPLE b =- 40 From nomograph, a/b - .40 e - 22° 1‘ 75 '1 ' - .0:‘ Veda-e : ” s U .U 1:. .u .32. U U U U U WU av ..v 9. HO 7 ..0 r3 4 may w...» l / r, 90 . 1 ..J U ml. at .1. J1 n0 .6 l a .M . thp—n—pprrF h Pup—p—_-.—.-PpF.-p F—Php—n—ph - 7b — — —:~h—:>p_-Pp—Pb-L)L O A. : .2... .K e Sagan e . m P» .3 . 7.. -- .iJ / n Wm . _ r.“ a _.. _ m1 1 T. X .3 O S _ r V _ Q OHM: a 5. .1 ,J 3 O. O O O .10 O O O I; o AW w W U W (W .., 0 0 Ru 1 C r. 5 a, I ,2 l .e h L P . — . P . — t r— - _..P P — —~ - p — p — b P «w.- 1_._._._._1 1 . _..:11451111 .1: . _.....1.._._._::. _ . _ ._ . 11:1... L MB? .85 4., no 9 19876 5 3 a 0.1.1.07 URU 4 0.0 .....~ ~ A e e 0 e e 0,; e s eflVwUfiUU O M ./ n% a.“ 1:. . .. . e e e e e e e e e p\a oapum ,,, MW WW WW we . J _.w . . .V J. 1.. U .M u a r ., l \w F .4 _J .8 7 ,3 Q. l l w... r p b . P u v ._p.H..LbP..PFJUp-.:..Pp«~: F 4 p — — — _ pp _ pF . p . —b pp 11am. Division p 8533 76 TENSION IN A POLE GUY AT A DEAD END b 5 - ._ _..____1_1 J— M L‘— q A From the above figure, the following relation may be expressed: . T 1e T8 '3 h/ 81D. 9 3 h sin (tan'l a/b) OR in logarithmic form: 2. log '1'8 - log Th + log sin (tanfl1 a/b) Expressing equation 2 in determinate form, gives: 1 0 log Th 3. o 1 - 103 sin (tan’l a/b) a o 1 l 1 T . °3 8 Let: The scale modulus nor Th -:6 The scale modulus for the ratio a/b = 6 The abscissa length - 5 inches. NOTE: In the above figure, ‘1‘h is assmned to be the equivalent - conductor tension acting at the point of guy attachment on the pole. Applying these scale moduli and abscissa length to determinate 5 , gives: (I) (v) 1 O 6 leg Th 4. 1 5 -610g sin (tan'la/b) -= 0 _l 2.5 5 log T 8 Iran the above determinate, the plotting equations are: y - - 6 log T inches. A For: '1' { h . h x - 0 inches. . -l y = - 6 log sin (tan a/b) inches. For: Ratio a/b { x - 5 inches. y - 5 log T inches. For: T { 5 8 I - 2.5 incheae SOLUTION Join with a straight edge the values of Th end a/b on their respective scales and read the value of T8 on th T8 8031‘s - ' Sec sketch on next page (78). EXAMPLE 1 - 20 b a 40 hence a/b - .50 Th - 4000 pounds From nomograph, T8 = 9000 pounds 1 See sketch on next page. Th Horizontal Tension L— LSQQJJ b - 15055 L )- F 10003 L- 3000 r. 00.20 ‘ 70.30 — BOQJ r *"' 5000 :=-~.,4’.J"O (T 20050 p r p C "' ZOO-U b r r r p b )- r rj 1.4 C L C '78! ¢ IP—’.T.f1 ‘3 1g J F's -4 ” :AUOJOJ (’-l r3UUJ¥ _ :BUOQU :70050' " bOUQUU E r- -- /\.u . -:;00'00 _ ~40000 ’ ‘ E cog-50000 - E-I : _..-g E t 5’ E—EOQOO : (3 h — ._ )- n : i: I 1 H _ a reef) a P r . :‘labud - 3“ ‘rz‘UOU‘ 34 : m -— * F I n 8000 a __ a m L533, ‘4OL EH6 C L L )- L .-, )- _‘f‘v '- FCCKJ H.000 L08 h ‘T :2 in A _— 955 Sin (tan-'ar'o, ficoerts ’- ' ‘a" r: UlJLle-«L 1510 c . ;;.'5 tan 1 111.11- )2. " " ‘J J '.T '_‘ AAUAA '79 THE HORIZONTAL COMPONENT OF THE TENSION IN A GUY WHERE THE GUY SPLITS THE ANGLE OF THE CORNER here there is a small cormr in a lead, it is often desirable to install the anchor guy so that it splits the angle of the corner and thus leaves the climbing space clear on the pole. If the total conductor tension on each side of the corner pole is the seen, the horizontal component of the tension in the anchor guy nay be expressed as: 1. Th - 2 T sin *0 - 2 T sin fltan-la/b) Or in logarithmic form: 2. logTh-logZT-uvlogsinie -0 Setting up equation 2 in determinate form, gives: *1 0 log 2T 5. 0 1 log sin 4&0 a 0 1 1 log rh Let: The scale modulus of 2T - 6 The scale modulus of 0 - 5 T1! abscissa length - 4.5 inches. 80 Applying these scale moduli and abscima length to' determinate 5 , gives: (I) (y) '1 0 6 lg 2T 4e 1 4.5 3 118 Bin *9 1 5 2 log Th From the above determinate the plotting equations are: y - 6 113 2T inches. For: 2 T { x a 0 inches For: Ratio a/b er angle 8 I - 4e5 1110138. == 210 T incles. For: T {y g h h x- 5 inches SOLUTION y - 5 log sin i-e - 5 lcg sin t(tan-1a/b) ins. lith a straight edge, Join the value of 2T and the ratio a/b or O on their respective scales and read the val!» of Th on the Th scale. See sketch on next page - 81. EXAMPLE 2T - 5500 pounds ratio a/b :- .5 From nomograph, T - 800 pounds. 11 See sketch on next page. 8T HeriZontal Conductor Tension -.. _. O— - .3 ., - . "r - -.. - _.-.fi '11... . .3 - - , 3 .7 4 - .. . —-—-‘~. GUY CHART Giving horizontal mansion in tu; Where Guy Splits Angle of Corner 20000 ;9 rTTIfl soar-cc 70_r3 eoifz 50f?19 — 6000 éodgzg "" e ( ~3--. a ié'.6 - 5000 ~0 3—.5 0 ‘L 4 '- ‘H'J "' e 4000 3.. fig - 20"- ? . 600 E g 500 o I '3 E a 400 (D ) ,Q t 5 30° '83: 10 E.“ F _— 5000 8' ‘2': E g 300 "* 3: 3 a: g P '3 «— l a: F S ’0 545.09 C- 2000 3 " '08 ’ 2;: 28*: :- «v : m 50 _‘_ eUO p 40 '- e05 L- 50 F— e04 * :2 .4 : 2° — .33 F L . . " 1000 10 .02 Th 3 4 T Sin 15 Harold L. noberts Elec. system :lan. givision A‘- - ACE; - 1L) 5 - l; - 31 82 'Often share there is a corner in the lead, it is not possible to install an anchor guy that splits the angle of the eermr. In such cases, two guys are installed: one in the direction of me lead and the other perpendicular to the lead. Especially does this condition exist at street intersections there the line crosses the street in a diagonal manner. HORIZONTAL COMPONENT OF THE TENSION IN A GUY IBERE THE GUY IS IN THE DIRECTION OF THE IEAD From the above figure, the following relation my be expressed: l. 1' - '1‘ - T cos (tan-1 a/b) Lh l 2 . . We shall break equation 1 into two parts, so that: 2.. K - T cos (tan-1 a/b) 2 and Setting up equation 2 in determinate form, gives: 1 0 K. e 1 " T I 4 0 2 O 1 cos (tan'la/b) 0 Let: The modulus of K a 1/2000 Tm modulus of T2 = 1/2000 The abscissa length a 3.33 inches. Applying these scale moduli end abscissa length to determinate 4, gives: (1) (y) 1 O K/ZOOO 5. 1 3.35 - Tz/ZOOO‘ B O l 3.33 ccs(tan"1a/b) 0 l + cos (tan'la/b) From the above determinate, the plotting equations are: y I: K/2000 inches. (do not plot this scale) Ior: K{ I B 0 incheSe y = - Tz/ZOOO inches For: T2 { x- 3.33 inches. 0 inches. y I For: Ratio a/b { 1 or angle 0 3.55 cos jtan- alb) inches I - -1 l + cos (tan a/b) Setting up equation 3 in determinate form, gives: 0 1 T1 60 0 1 " TLh - o l 1 K Let: The scale modulus of T1 = 1/1000 The scale modulus of TLh - l/ 1000 The abscissa length n 7.8 inches. Applying these scale moduli and abscissa length to determimte 6, gives: (1) (y) ' 1 0 Tl/iooo 7. 1 7.8 TLh/lOOO a o 1 5.9 K/ZOOO From the above determim to, the plotting equations are: 7 = Tl/lOOO inches. For: T1 { x a 0 inches. ~ y - - Lh/lOOO inches. For: TLh { x - 7.8 inches. y - K/2000 inches. (do not calibrate this scale) For: g x- 3.9 inches. We see that by choosing the correct scale moduli for T1 and 1' , the scale modulus for K from deteminate '7 is the sen as that round in determinate 5, so that we may use the K scale for the turning scale and combine the two nomographs. Hence the K scale willmt need to be calibrated. Tb solution for this nomograph will be given later. (See pus- 86) HORIZONTAL COMPONENT OF THE TENSION IN THE GUY where the guy is perpendicular to the lead From the figure on page 82, we derive the following relation: -1 8. Tph T2 sin (tan a/b) Or in logarithmic form: 9. chT - log T2 - log sin (tan-la/b) =0 ph Setting up equation 9 in determinate form, gives: l 0 log sin (“11-1 a/b) l 1 log Tph Let: The scale modulus for T2 - 6 The scale modulus for ratio a/b or 9 a 3 The abscissa length :- 4.5 inches. Applying these scale moduli and abscissa length to determinate 10 , gives: (I) (I) l O 3 log sin (tanu1 a/b) 11. 1 4.5 6 103 T2 - O l 1.5 2 log Tph 86 From determinate 11, the plotting equations are: ror ' Ratio a/b y a 3 lg sin (tan-l a/b) inches. or angle 9{ x a 0 inches. y - 6 lg T2 inches. For T2 { , x - 4.5 inches. For T h{ p P x - 1.5 inches. SOLUTIONS A. For the horizontal component of the tension in the guy, where the guy is in the direction of the lead. 1. Join the values of T2 and the ratio a/b on their respective scales with a straight edge and locate point K on the TURNING SCALE. 2. Join this point (K) with a straight edge to the valm of T1 on the T1 scale an read the value of TLh on the th scale. . See sketch on page 88. B. For tin horizontal component of the tension in the guy, where the guy is perpendicular to the lead. With a straight edge, Join the values of the ratio a/b, or angle 0, and T2 on their respective scales and read the valm of Tph on the Tph scale. See sketch on page 89. T1 - 4000 pounds. T2 II 2500 pounds. a/b =- .50 I+ From nomograph on page 88, T a 2170 pounds. Lh '+ From nomographs on page 89, T =- 1120 pounds. ph C ’.RT 5:- ['Y 88 " ‘ - - r V - - r ’ I 7,: I". V (-9 "' __ 3: ‘10: . .3 3 31mm rum-w - L O O '7) (D C) t") C} C) O '3 O '3 O (3 \¥) 1.) I4 C1 1‘) '4‘ .) LlLLlllllllllllllllfillllll1L1LLlLlLLllllhlAILAllLiJ F ' I ' I 3 1 1 1 3 1 7‘) Q (“3 :2 O "D 3) r3 .3 q 0 O -" ff) 0 Q 4 01 H 'd" .0 '3 UL versus: Jogonpuoo O 2‘ 5?, .9 o O 9 2 m ’20 I) C \.£\A I O : ‘.i . L14 '1 CE» 9.? b I I £31110 41111311 .—£ {:4 I! S Q it {2.4 3.4 f E‘ F—J \\ E“. \ ‘ \ L.-- ..... .h\ 00 O C O 8 8 o c o C‘ O Q Q r') O K) J. 3 '.n T J IlllllllhllllllllllLllllllllllllllhlhl1L 11111114 T 1: waisuei Joanna-110:) E3 $3 a.) o ~3-r-4 (n (0 C‘--r-i 7“. a, t r "H -'I‘J r413 '13 I I O b.‘ 'I“ C: (Er-4 ‘ 141-4 Ur‘i I U z: :1: ‘~ I) C‘. I Q) a) r-ir-i L“) (J) CA. 9 89' SKY CKART Giving Horizontal Tension in Guy Where Guy is Perpendicular to Lead Angle 9° ‘1: 3 3—5000 _.5030 50 *9 *‘ I " "4000 % 50 l .. fir _— 40 JE- 3 3000 3.4000 3*— 03 E800 E E 4— :600 m _ 1 3 - :500 g : .0 f—°c -—400 E4 . l - _ P. h L _300 3.5 _ 2000 B... -- ‘0 _ . “I \ +2 c: t m E'EOO g 3 0 d e]. : i3 }_ *4 5 E 09 ° : o _ .3 , d» e 1". .. Q» q ‘C".O'7 a: ’ D f _.06 E 80 H _ :94 r b s ‘— " :‘60 .- eOD b 50 g T [ g H04 F40 E ph :1000 '3 cu- — 8 H F— O t"'.O."5 _ a _b _ T _ 900 c.) P '- 20 ' 9 Lead 1 t-800 u a: - I“): _ '8 i T.— 3—700 ]_J a}; :/ d b __ ”-1.0 ) ’ ”—600 L—.01 L500 Electrical Systgm Planning Division HOB-8 5-11—31 SAG AND TENSION COMPUTATIONS FOR SPANS ASSUMING PARABOLIC FORM UNDER CHANGES IN LOADING AND TEMPERATURE 91 It is often required mat the planning engineer design special ecnstructicn to aoaomcdate certain field conditions. This occurs when the standard types of construction are; not suitable due to various restrictions. Such cases under this , classification are river crossings, lake crossings, and super highway constructions at complicated intersections. After pole locations lave been established, the next part of the design problem is the determination of conductor tensions and sags for stringing and loaded conditions so tilt the proper amount of guying my be obtained, and that proper clearances ny be maintained. for spans up to five hundred feet with the nornnl sags used, we will genrally be within construction accuracy to consider the span assumes parabolic form. With this assumption we will proceed to develop the relationships of sag and tension. SYMBOLS w - weight per foot of conductor. L - length of span in feet. .- sag in feet. - 3/1. - sag factor. - conductor tension in pounds (for which guying is computed) s d '1' E - modulus of elasticity of conductor in pounds. A - area of conductor in square inches. t - temperature in degrees Fahrenheit. 8" - coefficient of therml expansion of conductor per I? ~.,- sero - stress length factor. e - elongation per foot length of conductor. A - length of conductor in feet. THE PARABOLIC CABLE := L mm. _ x ,1 1 /i I \\ . fl ,/ 'T _‘ i T “‘ " L" w L i :4 L : I . ‘4'”"' : J ;. *-L - -3' Ii ‘5 From.the above figure the following relationships may be developed: a 'L 1' T n 2. 2L- L(1+ 802 - 32 d4 +eeeeeee) T T Since d is a very small number, all terms after the second in the above series (equation 2) may be disregarded. Hence we use the formula: 3. fl-LH'I-de) T— In this developmmnt we are mainly interested in the horizontal tension T, since this is the value for which the guying must be designed to meet, and the pole foundations are designed to support the vertical load. The following solution is based upon assumptions that are within the limits of accuracy required in practical overhead-lines construction, and is not to be considered a solution for precision work. 93. We nay define the length factor as the length of conductor divided by the length of span, or, the length of conductor per root length of span; and for a conductcr under tension T, my be expressed as , 4. T The length factor for zero stress in the conductor nay I - PL 8 l + 8 d2 (using equation 3, page 92) be expressed approximtely, as: 5. 20-1+8d2- T 5 LE The elongation of conductor per foot length of span may be expressed as: (at temperature t) 6. 7. gives: 9. et-Zo-l Hence using eqmtion 5 : e-Bd -'T "T E- Substituting the value of T from equation 1 into equation sag '1'" et+wL - -O Expressing equation 8 in determinate form, gives: 1 0 wL E - O O 1 et For a suitable nomograph we will use: Scale modulus for IL 2 4 x 104 IE’ Scale modulus for at = 2 x 10 Abscissa length a 10 inches. Applying tin above moduli and abscissa length to determinate 9, and rearranging the determinate, we have: (I) (y) 1 O 4 £1 1 10 AE 3 lo. 1 10 2 .t x lo a 3 4 l 1600 d 256 d x lo 1 + 160 d 3(1 + 160 d ) From determinate 10, the plotting equations are: l'cr: 1L 4 y:- 41L 110 inches. IE I!’ x - 0 inches. 3 y - 2 .t x 10 inches. For: a { t x =- 10 inches. 4 y - 256 as x 10 1mm... 3(1 + 160 d) For: d x - 1600 4 inches. 1 + 160 d See sketch # X — HCR - 11 in rear of thesis. The nomograph constructed from the plotting equations on page 94 will solve the condition of changes in loading providing the temperature remains constant. To make the solution applicable to changes in temperature, we use the following reasoning: The zero-stress length factor for temperature t is, l .' 2 11. g -' l + 8 d - w L _ 3 5 I AE Hence, the zero-stress length factor for temperature t2 is, 2"- I . 2, - la. 0 (Q +1 ,,( t2 t1) Since 22 is very nearly equal to unity and cr(t2 - t1) is very small, we may assume: . Z" I 2’ s": I- 13 a a + (t2 t1) Subtracting unity from.each side of equation 13 , gives: 14. (2.-1)-(8.-1) + 04152-131) Or, from equation 6, we have: 15. °tz a 't, + ”Y (tz - t1) ‘We shall now construct an auxiliary scale for c<(t2 - t1) so that it may be used in conjunction with.the et scale of the nomograph. It may be seen from.the plotting equation of et, that a length of two inches on the °t scale is equivalent to .001 foot change in.length. Hence we will use the following equations for the auxiliary scales: 3 -6 For copper: y - (2 x 10 )(9.6 x 10 )(t2 - t1) inches. -6 For acsa y - (2 x 105M104 x 10 )(t - t1) inches. 2 For Steel: y - (z x 103)(7.o x 10'6)(t2 - t1) inches. 96 In order to construct a nomograph to a large enough scale for desirable accuracy, and yet not exceedingly large in size, the scale w I. was folded on the value 5. Thus we AB have a Scale"A" for w L for values from O to 5, and All a lcale "B" for w L for values from 5 to 10. Likewise A3 the values of the sag factor "d" were plotted on two separate scales to accomodate Scales "A" and "B". Hence when solving a given condition of w I. on scale "A" use "d" to scale ”A", and for a given condition where w L is to scale "B", use "d" to scale "B". SOLUTION If the known conditions are: w , d , at temperature t1 1 l and we wish to find d2 under conditions of w2 and t2 , proceed as fo llows: 1. Join the point wlL/AE on the vL/u: scale with a straight edge to tln point ‘11 on the "d" scale and locate the point al on the .t scale. (For temperature t1) 2. To the point '1 add the length «(t2 - t1) from the auxiliary scale to locate the point e2 (For temp. t2). 3. Locate tm point sz/AE on the wL/AE scale. 4. Join the point a with a straight edge to the point 2 sz/AE on their respective scales and read the new value d2 on the ”d" scale. See Sketch X - HCR - 11 in rear. of thesis. EXAMPLE See Sketch # X - HCR - 11 Assume the following stringing conditions: Using #4 TBWP Medium Hard conductor,solid. L '- 150 feet. w1 - .164 pounds per foot (no wind and no ice)/ t - co° Fahr. - .053 square inches. 3 - 14:10‘5 s1 - 18 inches. (1.5 feet) ‘1 - all. -.010 1' - wlL/Bdl :- 307 pounds. rind tin value of the sag, s ahd the tension, T2, 2! for loaded conditions of an 8 pound wind, i inch‘of ice, and a temperature , t2, of 00 Fahr. This involves a change in temperature of 60 degrees and a change in weight of wire per foot from .164 pounds to 1.16 pounds. Proceed as follows: L x 10473.1: 2 .533 '1 - e010 1 e1 - -.0004 (not necessary to tabulate). (From nomograph) e2 - e1 - °<60 - -.00098 (not necessary to tabulate). 21. x 1047;]: - 3.77 d2 - .0215 (from nomograph) Hence: .2 - e0215 x 150 - 3.825 ft - 38s? 1110138 T2 - 1.16 x 150/(8 x- .0215) - 1010 pounds. ANALYSIS OF A FLEXIBLE DEAD-m POLE It occurs sometime, in subdivisions and on certain private preperty, that an anchor guy cannot be installed at - the end of the had. In such cases some means must be devised to hold the last pole, or as it is generally called, the "dead- and pole". However, the pole next to the dead-end pole in the lead generally can be anchored and treated as a rigid structure. lith such construction, the last pole is called a FIEXIBIE DEAD-END POLE. The amount of tension that may be held by this pole depends mostly upon the pole foundation. It has been found from practice that a pole may be loaded up to one-half its ultimate strength if a log heel and breaster are used, two- thirds its ultimate strength if crushed stone is need around the pole, and full ultimate strength if the pole is set in concrete. However, consideration must be given to the type of soil and- the grade of construction from which a suitable safety factor. may be determined. We will now analyze this problem with an attempt to arrive at its solution. :3 Had a" 6‘ 0' H- 0 d O worsen-mafia :» R ¢f’ 6,- ti Ih 0, 0* >1 0 100 SYMBOLS number of conductors. weight per foot of conductor uhder stringing conditions. I! I! fl '0 N I! ' loaded 0! . distance between poles at ground line in feet. span length in feet under stringing conditions. ' " ” " " loaded ” . conductor sag in feet under stringing conditions. ' " " " " loaded " . Si/L1 - sag factor. conductor tension in pounds under stringing conditions. ' " " " " loaded " . pole deflection constant in pounds per inch deflection. T/P - pole deflection in inches. conductor area in square inches.(cross-sectional area) modulus of elasticity of conductor in pounds. temperature in degrees Fahr. for stringing conditions. ' " " " " loaded " . coefficient of thermal expansion of conductor per degree Fahr. conductor length in feet under stringing conditions. 1' N 1! 1' R 10‘ dad H . ‘4 I _..—_..... .51 LSSUMPTION: The final tension in the wire will be equal to _..» _1 _. _. -......._ the resisting force offered by the pole, and that the conductor assume parabolic form. By analysis of conditions in tin above figure, we have: 1e £5 - n Tb feet. 15 P 2. 0C; - n TO feats Subtracting cé from --L gives, r‘ - ‘f - n _- 3. NC. “/3 m(T° Tb) Analyzing the elongation of the conductor due to changes in temperature and tension, we lave: - - 3 W n j. a 4. RC A6 (k, .. (to tb) + [E (T,3 Tb) Likewise: 5. 2,- - ,.‘.:,\'(t - t ) + Eur - T ) .. x h b O E'- b .0 102 Expressing and in terms of Lb and Lo respectively, we have: ° 2 e. 26 - L + 8 Sb b 2 '7. flc =- L + 8 Sc 0 c Substituting equation '7 in equation 4, we have: t ) 8 32 8 82 8. rid-n,- (L + c)o<(t -t)+(L + 0 MTG-Tb) c n; o b o rt; Substituting equation 6 in equation 5, we have: 2 2 _. g S 9. ,Zb-rz-(L +Bsb)x(t-t)-(L+ b 'r-r) - b b Subtracting equation 6 from equation '7, gives: 2 2 10a .2;- ‘1' - L "’ L + 8 ( so - Sb o c b 3 r; L?) Subtracting equation '7 from equation 6, gives: 2 2 lle R'JECUL-L +8(§_P_ -SO 5 b ° 3 Lb 12:? By examining tln figure on page 101, we see that: ,12. L -L -'-‘:’i'-*’,-.- n‘ .. b 0 ' ' Mme Tb) By ordinary comparison: 13. Tb/Tc - (wb dc)/('c db) Without introducing an appreciable degree of error, we may make the following assumption: . 2 . 2 I‘bI‘c - Lo" Lo “mtmw n . oaotqns S "anemone has on .H soapssuo msaeb 1 o .. o “ urmlmH 1 mumumw ”pH... adwfim m H+ an O O on a Moe- m. 9+ 9 .3 manom- n no posited- newsman o. "mo .na soapeowo macaque has or .eouoeneee hoonsooo poouoauuse mafia fin nE _.. m o ZNoW+d+ mania " .mdm+i + Iwmmwhw . o... +r evade- .. no- m moa+ on. .3 as «a «a H. aopuotoamomteiatnqmem a n “mopam .eano» wnfiwqoeueou can .ea oped ma soaposuo weapspapmpom o a o o e on a the E0 . ouameumflmi..patx.mow+iu zoom; at. o a- Toem+i+ n a 3 {Nu N " odom u. no massed" eon . a .3 season on» weavers noon «scram .ma soaposoo wnauspapmnsn use .HH seaweeds on dance a soapasuo moapeom o .n neonu one on one I .nk .n a m is none 0 33...... a 33 pa 3...... IImlmH o . on o I A I" A an "eschews has or .N noupesue moamb o o H+E I. E a as o _o H + fid a w as p o u o o I M a I o m. + a .on as «a we r .o» I on. n4 as no who a “on .ma ooaaeaue mnoenuo has or .eopoonnoe hooneooo esoaenmusa new: 6 o m I. {unifies .. 3.... m+:+E m I o o .. 3m. I m m” on + as .3 _H. O , n4 «A «e r .9» I avhhmcm m+ andI H4 mu 0_ m n "nobam .nanop woawesnnoon use .ma can“ ma nauseous wnaasaavensm o m. IIIMMII Emu N o I Ame-» oovm+aof osmofiem+duroom+i+ :79. 3e M+C+ ofllmH on. .3 _N .— N6 m o o o o ” ax m _I o weapooa one . a. he cannon on» masseuse noe- uoobaw .ma soaposde meanneauopoo and 0H qoawoauo o» assoc m seaweeds menupem sheen one a» use he .ou .oe .om .09 one: IIIIIII on case on sIIIII 105 ANALYSIS OF A FLEXIBLE DEAD-END POLE USING TWO DIFFERENT SIZES OF CONDUCTORS SYMBOLS All functions referring to the first conductor size have subscript "l", and all functions referring to the second conductor size have subscript "2". The additional subscrdpt "b" is med to denote stringing conditions at temperature t and the additional subscript "c" b, is med to denote loaded conditions at the temperature t c' Other than the above subscript notations, the symbols used are the same as those in the previous problem. 106 \— ' fl \ I {/\w L . \\_“ // / . l J n- ‘ I". ./ . 5'0 .. ' ‘7 "I" ,I' III 3Q ‘— .— 7 If)" i I ., (~ ['41 .~"(’I “1 "I \J [I ‘ .5” 1Q .‘ \ 0 l (I. n .‘ .l l 1:”: \f’ 5" I“. I"? Q I Q. I ,4." H..- ~ — L. .-_ .._. ASSUMPTION: That tle final tensions in the wires will be equal to the resisting force offered by the_pole, and that the conductors assume parabolic form. By analysts of conditions in the above figure, we have: 1. .1; _, n1 Tib + “2 Tab feet. ‘ 2 P 2. c... - “1 Tie + “a Tso feet. P Analyzing the elongation of: the conductors due to changes in temperature and tension, we have: J - g .. r'x‘ - R I': " 3, , _. L”; x. (t 13b) + I (T10 1311)) 1 ,-. 1,... =r t - t + 11.. a T 4. fl... / I, b) .. (T20 2b) 1mg . e 3.-.}.IE“- '- + l) .- 5 , 30 C ,..H (tb to) K/il(le T10) N. \ K: i 3 .r. - ..'_,.,'- .4. , ‘2' u + 1.3.: .- 6. . ”b to) w ”an :20) 2 _. mad one on + pHa Ho I .H ooHuosHoe 89E 1 D. I M H m «H . H pH.e «H a _+ H u + _..n maavmgaafidv one.“ pm H NAM: «n + SH: an n O I .H I «Amen we me o e a ..5 oH_m «H H no N .m a an as has .- .oHe -omeiHAHHV n + a “Halo r m m m 1 o H oH m «H m .IIIIILwLwLmrIII H m a H H 4 ".1 «lasagna? 3 £9 a: n + a: n I. ' m‘ I H n 0H _. .H vs. NA Ami .H e o o. H , m ..I N N l shoes. one on .ome .3: one .mfimfl .Htmfl .np .pme .nmm .pma .nHw .nHm .nHa . son: 0 IIIIIIIII «a one eoHo oe IIIIIIII- H _u on .o om. om . + «a as m ea ea . «a + no a "oneoHHou 3» as chunks as .._—3995 aspoeHn on» ea no! as even»!— oaea on» weHhHAmo hm m «A one me + oAH .« eoHposvo seem An I A I A O i. H oH m «H L H H + m a H M «H H , A+ ” _ a + Afidv e M. mafidvmn +AA§VHHHL nv»... #omn. I Aavo 9N nu u m o no: I _ HIEIII omen m «9+ on. a: «H o _ in a- on: :2? L: H. -.mzHlfiaHn + on E5 1 F N N N I o 1 oH m «H 1 A m «A m + T+ A A.# M A «Andy as + Handy mHHVNo + HHHV omen ‘oH pH pH ._.. .319- HIEIII easing HE: NH or _ no s. 03H 31.... Ho we ..oHimaimoo+He..H§1m o o on on .oH .oH .oH a green one 9» .nme .pAk one .mafidv .AAM4V . p . we . m . a I m H on! IIIIIII- one one oHa ooHe ca IIIItII- .NH .AA .OH 109 NOMOGRAPHIC SOLUI‘ ION OF THE CUBIC EQUATION Examining equation 16 on page 103, equation 20 on page 104, equations 7 and 8 on page 107, and equations 10 and 11 on page 108, for the solution of flexible dead-end pole problems, we note that they are of a cubic type. This is rather a difficult type of equation to solve; hence we will construct a nomograph for the solution of this special form of equation. We note that the above mentioned formulae are all of the type: 3 10 T +KT2 +C -0 Expressing equation 1 in nomographic form, gives: 1 o - K 2e 0 1 - C 3"" 0 T2 1 T3 Assume the scale moduli for K and C - unity and assume an abscissa length - 10 inches. Setting up determinate 2 to accomodate tle above scale moduli and abscissa length, gives: (1) (y) 1 0 - K 5. 1 10 " C - 0 3 1 10 T 1'2 + 1 T2 + 1 For For For 111 From determinate 3, the plotting equations are: y = - K inchOSe K1 x 8 0 inches y = - 0 inches. c{ x = 10 inches. 3 y - T 11101168. T i r2 + 1 x = —J-‘-9-—-- inches. 1" + 1 We will construct the nomograph for the following values: K varying from + 5 to - 10 C " " 0 to - l5 ‘1' " " 01'. to + 10 See Sketch # x - HCR - 6 in rear of thesis. If tln computed value of K and C do not lie within the plotted values on the chart, let: T - a T' where a is any desirable number. Then, substituting (aT') for T in equation 1, we have: 3 (T') + K(T')z + c - c a :3" Then: For value of K on chart use 10 = K (c'omputedl For value of 0 on chart use 0' =- c (computed) a3 Then value of T from nomograph - T' - T/a or T =- a T' SOLUTION With a straight edge, join the values of K and 0 on their respective scales and read the value of .T on the tension scale. See .Sketch #‘X - HCR - 6 ‘in the back of thesis. EXAMPLE Using a flexible dead—end pole, assume it is required to determine the stringing sag at 600 rahr. for a 40 foot, class 2 Western.cedar pole so that it will not be over stressed when holding three #0 bare copper conductors when subjected to a loading of an 8 pound wind and one-half inch of ice at 00 Fahr. Assume the soil conditions are such that a safety factor of 3 is chosen. For such a case : n a 3 (#0 bars capper conductors) w - .326 pounds (per foot, no wind and no ice) w a 1.262 pounds (per foot with an 8 # wind and i" ice) L - 100 feet (assumed span length) ' tb - 60o Fahr. (stringing temperature) t a 00 Fahr. (loaded temperature) AE = 1.1606 x 106 - 9.6 x 10'6 40 foot, class 2 western cedar pole P a 60 pounds (per inch deflection) Ultimate strength a 3700 pounds Safety factor - 3 From these conditions: 0 T - 3700 - 411 unds tension or conductor OF. 8wind "ice 0 ._.-fl 1 p0 p ( D # 3% ) L - 100 - 3 x 411 98.29 feet (formula 21, page 104) I2 x 60 3 S a 1.262 x 9.63 x 10 - 3.70 feet (formula 1, page 92) ° 8 x 411 d - 3.70 (sag factor) ° 95.29 Using . equation 19, page 104, to show how little effect the portion 3 d2 has in the term (1 +§ d2): Solving for the value of K from equation 19, page 104: a K -‘5 I 1,415 x 1.1606 x 103 + 1.1606 x 1.0038 x 9.6 x 60 _ 411 ‘6’ 3 xxl.1626 x 10 + 1.0038 - " 310e5 Solving for the value of c from'equation 19, page 104: c - - .106 x 9.63 x 1.1606 x 109 6 24( 3 xxl.l6269x 13 + 1.0038) 6 - " 0981. I 10 To solve this on the nomograph we will use the paviously mentioned transformtion (page 111) and let a - 100 Then: K' - - 310.5/100 - - 3.105 6 C' II - .981 I 10 /106 '- - .981 From the nomograph (Sketch # I - HCR - 6 in rear of thesis) T. - 302 Hence: Tb = 3.2 x 100 - 320 pounds (tension par conductor for 60° Fahr. no ice and no wind) L = 100 - 3 x 320 =- 98.67 feet (stringing condition) b ‘ IE x 60 3 Sb - .326 x 9.;11 x 10 a- 1.23 feet (string. sag) CONCL$ ION I have found in my work as a Planning Engineer for a utility company that these nomographs are practical, and sufficiently accurate, and that they provide economy of time and effort. Therefore with entire confidence in their value in the engineering field, I submit them to the members of the Graduate Council of Michigan State College. 116 BIBLIOGRAPHY "The Construction of Graphical Charts" -- Paddle. "Solid analytical Geomtry and Determinates" -- Dresden. "Bulletin of the Bureau of Standards, Volume 4" -- Dept. of Com. 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