PLACE ll RETURN BOX to remove this chockwt from your town]. TO AVOID FINES Mum on or baton dd. duo. DATE DUE DATE DUE DATE DUE I L:e -———I ___g ___-:J MSU I: An Afflrmdlvo Action/Emu! Oppommlly Institution ENERGY SAVINGS AND IRRIGATION PERFORMANCE OF A MODIFIED CENTER PIVOT IRRIGATION SYSTEH BY Sagar Raj Pandey THESIS Submitted to Michigan State University in partial fulfillment of the requirements for the degree of MASTER OF SCIENCE Agricultural Engineering Department of Agricultural Engineering 1989 ABSTRACT ENERGY SAVINGS AND IRRIGATION PERFORMANCE OF A MODIFIED CENTER PIVOT IRRIGATION SYSTEM BY Sagar Raj Pandey The goal of any modern irrgation system should be to apply water with the maximum uniformity and minimal labor and energy costs, while increasing both the quality and quantity of food production for maximum economic benefit. Proper irrigation system design is important to achieve maximum possible application uniformity and energy savings. The demand for energy in center pivot irrigation system changes with the change in the operating point of the pump curve because of the opening and closing of the end gun sprinkler. The primary objective of this research was to determine the possible energy savings of a modified center pivot irrigation system in which the operating point of the pump curve is fixed. Computer programs were written to simulate the design modifications that included auxiliary sprinklers in the system which open with the closing of the end gun sprinkler. The results showed that there are time and energy savings after the modification in the design of the system. The amount of energy savings I438 the function of the steepness of the pump curve, with high values for systems with steep pump curve. In addition, it was found that the capacity of the pump could also be reduced with the modified design. The application uniformity of irrigation improved significanlty with the modified system. The increase in application uniformity was between 71 and 161 in systems with steep pump curve as compared with the values between 3; and 8% in system with flat pump curves. This increase in application uniformity increased the potential yield of crop by 3: to 17$. Approved Major Professor / I Approved (v/é/WQ QAQVKL 0.0. ’ Depirjnajt Clairpersofi Dedicated to the Memory of my Grandfather Sardar Rudra Raj Pandey iv ACKNOHLEDGEHENTS Hy special thanks to my major professor, Dr. Vincent F. Bralts. I greatly appreciate his support, guidance, and patience in helping me to complete this research. I have very high esteem for him as a teacher, advisor, friend, and individual. I thank the other members of my committee, Dr. L. J. Segerlind from Agricultural Engineering Department, Dr. David Niggert from the Civil Engineering Department, and Mr. Austin Miller, 9.3., for their helpful comments. Special thanks to Mr. Miller for his thoughts, advise, and encouragement during the analysis of the results. In addition, thanks to the graduate students of the Agricultural Engineering Department for their help and support when things did not look encouraging in those frantic hours. My special appreciation to Mr. Jerold Bement for allowing me to use his center pivot irrigation system in Dowagaic, MI. Special thanks also to Mr. John Smith, MECP Irrigation Scheduling Technician in Cassapolis, HI, for helping to collect field data. Special thanks to my wife, Shashi, for her understanding, encouragement, patience, and support because of which this work became possible. rm orcou'rm LISTOF TABLSOOOO OOOOOOOOOOOOOOOOOOOOOOOOO ......OOOOOOOOOOOOOOOOO V111 LIST OF FIGURES.. ........... . ............... ....................... 1! I. INTRODUCTION ..... ...... ............ . ............ ........... 1 A. Scope and objectives... ............ ..................... A II. REVIEW OF LITERATURE AND THEORY................ ....... ..... 7 A. HydraulicSOIOOOOOOOOOOOO00......0.0.0....0......00...... 7 B. Uniformity and Application Efficiency................... 19 1. HydraUIic uniformity.’......OOOOOOOOOOO.......OOOOOOC 19 2. Application Efficiency............................... 2” 3. Application Uniformity....... ......... . ..... ......... 30 E. Hydraulic Network Analysis.............................. 33 F. Energy use............. ......................... . ..... .. “2 6. Summary and Discussion.... ........................ ...... 52 III. METHOWLWYOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO00.00.0000... 56 A. Research Approach ...... . ................... .. ........ ... 56 B. Theoritical Development ................................ . 58 IV. RESULT AND ANALYSIS.. .................................... .. 63 A. Hydraulic Analysis ................................. .... 68 vi 8. Field Performance Evaluation C. Energy and time savings ................... . ........... . D. Summary ........ ................ . ........... ............ V. CONCLUSION AND RECOMMENDATION ............................ .. APPENDIX A. Center pivot hydraulic network analysis model "SYSTEM” APPENDIX 8. Simulation output from network analysis model "NETWORK" APPENDIX c. System curve simulation model "SYSTEM" ................ APPENDIX D. Auxiliary sprinklers simulation model "AUXISPR" ....... APPENDIX E. Field performance evaluation model "FIELD" .... ..... ... APPENDIX E. Simulation output from field performance model "FIELD" APPENDIX 6. Irrigation system layout plan ........ ...... ........ ... LISTOF REFERENCE.........OOOOOOOOOOOO...O vii 107 115 117 119 136 1"5 1"? 139 155 167 168 10. LIST OF TABLES Page Example sprinkler discharge-head relationships as given by the manuracturer 00.00000 00000 OOOOOOOOOOIOOOOOOOIOOOOO00. 13 Component loss coefficient for a few components ............ 19 Numerical methods and the number of iterations required for convergence 0.0.00.0...OOOOOOOOOOOOOOOOOOOOOOOO 38 Discharge from different pumps with the end gun turned off .. 80 Change in the yield of corn with the improvement in the irrigation application uniformity after the modification of the design of the center pivot system .................... 106 . Electrical power required to operate different pumps for the end gun on, end gun off, and modified end gun Off condition .....OOOOOOOOOOOOOOOO0.0.0.0000...0.0.0.....00.108 Power and time required for pumps to apply varying amount of water in the field without modification in the design of the center pivot system L.................................... 109 . Power and time required for pumps to apply varying amount of water in the field with modification in the design of the center pivot system ..................................... 110 . Change in energy and time requirements with the modification in the design of the center pivot irrigation system 00.... ...... 0....OOOOOOOOOOOOOOOOOCO0.0000.111 Change in energy and time requirements for different pumps .. 113 . The total energy and time changes to irrigate three .crops for the uncle season .00... ..... OOOOOOOOIOIOOOOOOOOOOO11“ viii LIST OF FIGURES Figure 1. Example layout of a center pivot irrigation system . ....... . 2. Discharge in a conduit with varying x-sectional area ....... 3O DisCharge thruaDOZZIe .......... .....OOOOOOOOO0.0.0.000... A. Moody's Diagram showing the relationship between friction 10. 11. 12. factor f, relative roughness of the pipe e/D, and Reynolds number Re based on the Coolbrook equation for comerCial pipe ...OOIIOOOOOOO.........OOOOOOOOOOOOOOOO. Distribution pattern from the center pivot irrigation system with increasing discharge away from the pivot point ........ Distribution pattern in the center pivot irrigation system with uniform discharge from all sprinklers ....... . (a)Distribution of discharge along the lateral line of the center pivot system 0.0.0.0......OOOOOOOOOOOOOOOOOOOC (b)Distribution factor of pressure head loss along the center pivot system 0.0.0.0000.........OOOOOIOOOOOOOO0.00 The relationship between the area of the field and the lateral line distance that irrigates it for a 125 acres field. A typical application pattern from a single sprinkler with varying pressure .............. .......... .............. Surface storage curve for 0.3 intake family soil ......... .. Surface storage curve for 0.5 intake family soil and Pullamn soil ...... . ........... ......... ................. ... Illustration of the solution using the Newton Raphson method . ix 10 16 21 21 23 23 26 27 31, 31 36 Figure 13. 1“. 150 16. 17. . 18. 19. 20. 21. 22. 23. an. 25. 26. 27. 28. 29. 30. Elements considered in the Finite Element method ............ Sequence of the nodes and elements of a riser element ....... Example power nomograph to be used to determine the horsepower required to pump water when the total head, discharge, and pump efficiency are known ............................... Example pump characteristic curve ............. ....... ....... The plot of the operating point of the pump ................. The change in the operating point of the pump when the end 8"" 18°“ IO....00.0000000COOOOOOOOOO0.00000000000000 A network consisting of a pipe, a component, and a sprinkler . (a) Sequence of nodes and elements for the center pivot 81mu13t10n madel 0.0.0.0....00.0.0.0........OOOOOOOOCOOOO (b) The experimental network of a center pivot irrigation system .................................................. Experimental pump curve with constant head .................. Experimental pump curve with constant discharge ............. Experimental pump curve of a pump used in the field ......... Data for the hydraulic network analysis of the center pivot irrigation system 0.0.0.00I0.0I0.0.0.00.0000000000000000 Results obtained from the computer simulation model 'NETHORK' for the end gun on, end gun off, and modified end gun off condition ....................................... Graph showing the discharge-pressure relationships at the center pivot point for the end gun on condition ............. Graph showing the discharge-pressure relationships at the center pivot point for the end gun off condition ............ Discharge-pressure relationships of the center pivot system at the pump for the end gun on, end gun off, and modified end gun off condition ....... ................. ...... System curves of the center pivot irrigation system for the end gun on and end gun off condition .................... Pump curve for pump 'V' plotted with the system curve to determine the operating point of the pump ................ Page H0 R3 R6 ”7 H9 51 S9 65 65 66 66 67 69 71 72 7n 75 76 77 Figure . Page 31. Pump curve for pump 'H' plotted with the system curve to determine the operating point of the pump ................ 78 32. Pump curve for pump 'J' plotted with the system curve ‘ to determine the operating point of the pump ................ 79 33. Auxiliary sprinklers models and nozzle sizes used for the modification of the center pivot irrigation system .......... 82 3A. Sprinkler characteristics showing the pressure (P), radius of coverage (R), and discharge (Q) for the regular and aUXIllary sprinklers 0.......0.......0..00.0.00......00...... 8" 35. Data for the hydraulic analysis of the center pivot system after the auxiliary sprinklers are included in the design ... 85 36. Discharge-pressure relationship at the center pivot point for the modified end gun off condition ...................... 86 37. System curve for the modified end gun off condition with respect to the end gun on and end gun off conditions ........ 87 38. System curves and pump curves for the example operating conditions and pmps 0.00.00.00.000000000000...0000000.000.00 88 39. Field performance evaluation data obtained from the simulation model 'NETHORK' for the end gun on conditions .... 90 A0. Field performance evaluation data obtained from the simulation model 'NETHORK' for the end gun off with pmchonditlons 0.0.0000000000000000.0000000000000000000000 91 A1. Field performance evaluation data obtained from the simulation model 'NETHORK' for the end gun off with pumpvconditiona 0.0.0000.00000000000000000000.000000000.... 92 A2. Field performance evaluation data obtained from the simulation model 'NETNORK' for the end gun off with pumpHconditions 00.000.000.000...0.000000000000000. 00000000 93 A3. Field performance evaluation data obtained from the simulation model 'NETHORK' for the end gun off with the modification in the design ........ . ........... .... ........ 9A AA. Graph showing the relationship of the depth of application to the percentage of timer setting .......................... 96 A5. Graph showing the application depth along the lateral . line in the field for 501 timer setting ......... ............ 97 xi Figure Page A6. Example average application depth in the field for the end gun on and end gun off condition ....................... 98 A7. Example pattern of the portion of the field irrigated with the end gun on and end gun off ........................ 99 A8. Application uniformity and time required to irrigate the field using pump V without the modification in the deSIEn or the system .0...OOO0............IOOOOOOOOOOO......0 ‘00 A0. Application uniformity and time required to irrigate the field using pump J without the modification in the de818n Of the system ......0............O............000.0...101 50. Application uniformity and time required to irrigate the field using pump H without the modification in the design of the system ........................................ 102 51. Application uniformity and time required to irrigate the field with the modification in the design of the system 0000000..0000-0000..0.0.00.0..0000000000000.0000...103 xii I INTRODUCTION Irrigated agriculture has been practiced in various locations_ around the world for over AOOO years in the world. Most great civilizations started in river valleys of famous rivers such as the Tigris and Euphrates (Middle East), Nile (Egypt), Indus and Ganges (India), Hyuang-Ho (China). Over the years, great wars have been fought for the privilege of using water for irrigation. Worldwide in 1985, over one-third of the world food was grown on only 181 of cropland which was irrigated ( Postel, 1985). The most common form of irrigation worldwide has always been surface irrigation in its various forms (furrow, border, basin). The average irrigation efficiency of the surface (gravity) irrigation system is usually low. Worldwide average irrigation efficiency of surface irrigation is less than 371 (Postel, 1985). Sprinkler irrigation and drip irrigation were developed in the more industrialized nations with the advent of plastic and aluminum pipes after World War 11. These system of irrigation are capital intensive, less labor intensive, and have a greater energy demand than surface irrigation. Sprinkler and drip irrigation systems, however, allow for ,greater control of water delivery and thus have a greater water use efficiencies than comparable surface irrigation systems. Sprinkler and drip irrigation systems also tend to irrigate more uniformly than gravity systems. Water efficiencies typically average 70% or greater (Pastel, 1985). Irrigation in the United States has experienced unprecedent growth in the last few decades. In most cases the land was brought into irrigation using a variety of high-pressure sprinklers designs. in some areas, farmers have used these systems to irrigate hilly and marginal lands unsuitable for gravity methods. Overall, sprinkler irrigation accounts for virtually all of the net increase in irrigated area in the United States between 1960 and early eighties. In 1980, 321 of irrigated land was under sprinkler irrigation (Jensen, 1982). Today sprinkler irrigation are used on about 35-90 percent of 0.8. irrigated land with drip irrigation on only 2-3 percent. One of most common sprinkler irrigation system is the center pivot system. A single center pivot system irrigates about 150 acres, and is now used in much of the 0.5. High Plains. In addition, over 12000 center pivots have been installed in the desert nation of Saudi Arabia over the last several years ( Postel, 1985). The center pivot irrigation system , figure 1, consists of a lateral arm which rotates around the central pivot point. The water is supplied to the lateral under high pressure through the central pivot point. The lateral line sections are usually suspended on towers with the help of simple truss structures and consists of openings for sprinklers, through which water is discharged to the field. The sprinklers are spaced and designed to supply uniform depth of water in Field to be irrigated Center lateral line with 91 .t sorinklert 1"” pical sprinkl ' ‘ pattern "a ‘< \1 underground mainl ine .. or well at the center direction of rotation end n of the lateral 1i. 9" irrigated OTC! ' "'83: 3 f:8M‘1fl-‘I.. ”It-1"” I". 0 Cheek valve water $OU'CQ Figure 1. Layout of the Center pivot irriga.ion system. the field being irrigated. The most important advantage of the center pivot system when compared with other forms of sprinkler irrigation is that it requires minimal labor. Energy costs, however, are also an important parameter that could-determine the viability of a center pivot irrigation system. Rising energy prices, especially when combined with falling water tables, can increase irrigation cost to prohibitive levels. Therefore the need to operate the center pivot system with maximum application and energy efficiency is becoming more and more important for the success of the irrigation systems. A. SCOPE AND OBJECTIVES The important factors that determine whether a center pivot irrigation system is properly designed are (1) Energy use, (2) Application rate, (3) Uniformity of application, (A).Instantaneous application rate. Any improvement that can be made in the design of the center pivot irrigation system will require the improvement in one or more of the above mentioned factors, but not at the cost of compromising the other remaining factor(s). Many researchers (Bermuth, 1982; Solomon and Kodoma, 1978; Heermann and Hein, 1968) studied and developed methods that improved the performance of the center pivot irrigation systems with regard to energy use, application rate, and/or application uniformity. Bermuth (1982), Solomon and Kodoma (1978) pointed out that the operating point of the pump shifts when the end sprinkler of the lateral (end gun) is switched on and off. The switching on of the end gun decreases the ’0 m A." b. .4 ea.‘ . : -u \., A! discharge and increases the pressure of the system. This in turn can change the application rate, application uniformity, and energy efficiency in the system. Therefore it is best to keep the discharge and pressure in the system constant such that the system will perform with maximum efficiency throughout the period of irrigation. Some researchers have suggested using the booster pump which is operated when the end gun is on. Pressure compensating sprinklers can also be used to keep the application rate and uniformity constant. But this does not help in operating the pump with maximum efficiency when the end gun is off. The possibility of using auxiliary sprinklers to keep the pumping rate and pressure constant throughout the period of irrigation has not been studied by any researcher. The auxiliary sprinklers would operate along the lateral when the end gun is switched off. As soon as the end gun is switched on, the auxiliary sprinklers would stop operating. This method would help to operate the pump with maximum efficiency throughout the irrigation period, thus saving energy. The overall goal of this research is to investigate the possible conservation of energy by modifying the design of a center pivot irrigation system. The effect on the application rate and application uniformity as a result of the proposed modification will be studied. The specific objectives are 1) Determine the relationship between energy saving and addition of auxiliary sprinklers in the center pivot irrigation system. 2) Make a comparative study of the application uniformity with and without the introduction of auxiliary sprinklers in the center pivot irrigation system. 3) Study the effect of the introduction of auxiliary sprinklers on the water application rate on the field and the saving of thme that is possible. II. REVIEH OF LITERATURE AND THEORY Any modification in the design of the center pivot irrigation system such as the introduction of auxillary sprinklers will call for a thorough look at the hydraulics, field uniformity, network analysis, and energy use in the system. Therefore, a literature review of the above aspects of center pivot irrigation system was undertaken. The following is a review of the literature related to the design aspects of a center pivot irrigation system : A. GENERAL HYDRAULICS 1. General The hydraulic principles of fluid mechanics are based on the classical equations of continuity and energy. A theoritical development on fluid hydraulics has been given by Garde and MiraJoakar (1977) and others. The continuity equation is based on the conservation of‘mass principle. Refering to Figure 2, the flow in the pipe is Q = V1 A1 = V2 A2 = constant [1] where Q = discharge, V = velocity of flow, A}: '211' DATUH Figure 2. The discharge in the conduit with varying x-sectional area. A = cross sectional area of the pipe, and the subscript 1 and 2 are the positions where 0, V, and A are measured. The energy equation in fluid mechanics is commonly known as Burnoulli's equation. Again refering to Figure 2, the energy balance equation can be written as mm + z, + (V1)2/Zg = 92/7 + 22 + (V2)2/2g + AH [2] OP [( 92-91 )/Y I + ( 22-21 ) = [( (V1)2-(V2)2 )/ 23 l + AR [3] where P a pressure in the pipe, 7.: specific density of water, V : velocity of flow, AH: hydraulic head loss between positios l and 2, and the subscripts again represent the positions where the values are evaluated. The hydraulic design of sprinkler irrigation systems based upon the above two equations has been presented by several researchers including James (1988), Wood and Charles (1972), Perold (1977). 2. Center pivot system hydraulics Hvdraulics of a Sprinkler: When a fluid is discharged from a conduit into the athmosphere through an opening of any form, the pressure intensity along the issuing Jet surface is athmospheric (Garde et al, 1977). In sprinkler irrigation, the opening is a nozzle. Nozzles are employed when a high velocity Jet is desired from a pipe. Considering a nozzle as shown in Figure 3 (Garde et al, 1977): lO vena contracta Figure 3. Discharge through the nozzle. 11 If section 1 is located in the undisturbed flow, while section 2 is located at the vena contracta, Bernoulli's equation gives 91/7 + (V1)2/2g = 92/7 + (ma/23 [u] The potential head AZ has been neglected in the above equation. Solving the above equation with the continuity equation gives Q = [cc/.41 - (cc)2(d/D)")] [(nazxm/z'éll HEP/"Ty [5] OI" Q = cd < naZ/u ) (23)” 50-5 [61 where Cd = coefficient of discharge, H = static head,. Cc = coefficient of contraction, d = diameter at the mouth of the outlet, and D a diameter at section 1. Perrold (1977), Karmeli(1977). and Bralts(1983) have written the equation of flow thru the sprinkler or emitter in the form of q = khx [71 where q sprinkler or emitter discharge, 9? u constant of proportionality, 3' II pressure head at the sprinkler or emitter, and 3: 0| sprinkler or emitter discharge exponent. It can be seen by comparing [6] and [7] that the constant of proportionality, K, contains variables such as the coefficient of 12 contraction, geometry of the nozzle, and gravitaional acceleration. The exponent x may be assumed to be 0.5 for fully turbulent flow from nozzle or orifice (Perrold, 1977). The nozzle with an x value of less than 0.5 would be pressure compensating in nature. A zero value for x would make the sprinkler fully pressure compensating (Bralts, 1983). All irrigation sprinkler manufacturers give the relation between discharge, q, and head, h, in tabuler form (Table 1). Lateral line hydraulics: Water flow in a center pivot lateral line is considered to be hydraulically steady, spatially varied pipe flow with the flow being along the streamline. The total flow thru the lateral line decreases every time it passes a sprinkler nozzle. Taking into account the conservation of mass principle, Q:VA and the fact that A will remain constant, the velocity then decreases along the lateral line. A large share of the energy loss is due to the friction in the pipe. In general, head loss due to friction can be written as hf : Rd" [8] where hf’: head loss due to friction, R = constant depending upon the relation used, Q = flow thru the pipe, and m : discharge exponent. Christensian (1992) used a generalized formula to calculate the 13 Table 1. Example sprinkler discharge-head relationships given by the manufacturers (Rain Bird; 1986). O a“ \‘I ~ \. *5 A". 1‘ x e“ 1" 0" o'“ 10400/10500 23‘ Traiectory Ring Name 870" 60 264 110 70 275 118 80 288 127 295 138 ’00 305 1.12 110 315 150 120 322 157 990 60 234 142 7 295 134 30 308 184 90 315 175 100 32 185 110 338 195 ‘2 :44 :8 l 100" '5 330 185 «‘0 3:4 200 80 328 213 337 227 100 348 2" 110 35! 250 120 368 269 1201" 60 318 228 70 332 243 80 345 283 90 358 278 100 371 290 110 382 308 120 392 323 1.293” 60 338 27s 70 350 295 80 364 313 90 378 338 1m 390 382 111) 402 372 120 412 392 1380" 60 352 1 324 70 257 383 80 383 374 90 398 400 100 409 422 110 421 441 120 431 488 1.450" 60 365 385 70 383 418 80 398 447 90 414 475 100 425 sec 110 433 525 120 4:0 550 O to. . ed, .“ +99 '9‘ 096‘ 0“" 10408/ 1 0505 23' Traieciory St. Bore Nozzle 0513" so :45 as 70 255 93 80 265 99 90 275 105 100 288 111 110 295 117 '20 305 122 0690" (:0. 269 110 70 231 118 80 291 127 . 300 138 1: :3 137 c 790 so :92 142 to a 154 so 31:1 164 328 199 80 336 218 90 347 227 100 357 2 0.9”" 60 329 226 70 3‘7 245 357 2 90 367 278 100 37? 10W“ 60 348 275 70 363 80 375 315 90 390 100 110 410 372 120 420 392 1 190 ' 60 366 331 70 381 80 395 374 90 410 400 100 420 422 1290" 60 385 390 70 4m 418 80 818 “7 90 427 475 1} \ I 0‘ +° 1" Q9 a 0 ¢ 0° 9“ 0e ‘8 ‘ - 9 0 6‘ o 6.0“. A“ (‘9 10508-00143“ Traieauy St. Bore Nozzte 0.513" 60114 45 70 06%" GO 119 47 0.790" 60 127 49 7155 e 6' 9‘ 19 head loss due to friction. This relation is commonly used in hydraulic design and analysis (Anon 2,1983; James, 1988 ) hr = KLo“ / ozm+n [9] where K =coefficient depending upon the formulation and unit used, m,n = exponents depending upon the formulation used, length of the pipe (m, ft), diameter of the pipe (mm,in), and L D 0 flow rate (liters/min, gpm). Comparing [8] and [9] gives a = KL / 02”*“ [10) Many semi-empirical equations are in use to calculate friction loss (hf) in the pipe. Some of the more common equations are Darcy- Weisbach, Hazen-William, and Scobey. The Darcy-Weisbach's equation for friction loss is h, = fLV2/2gD - [111 and z s R = 8fL / gn d where f = friction factor, L = length of the pipe, V : velocity of flow, and D : diameter of the pipe. The friction factor, f, in [11] depends upon the Reynold's number, Re, and the relative roughness of the pipe, K = e/D (Garde et al, 1977; ‘1 ‘0 II_ e., . ._. ‘v. I ' Q ‘i .‘I . ‘ I ~ I \ I '. ;., el 0 I ..c 'I 15 Jeppson,1982; Swamee and Jain,1978) For commercial pipe, the Coolbrook equation can be used to determine the value of f (Garde et al, 1977). This equation is 1/«T"- 2 log1o Re/K = 1.7V - Zlog1o[1+18.7(Re/K)/(Re/TD] OI‘ 1/77'2 1.11 — 2 10310 (e/D + 9.35/(R9773' [12] Equation [12] has to be solved implicitely to determine f. Moody's diagram (Figure A) is a plot of Equation 12. The drawback of the Moody‘s diagram is the difficulty of using in computer programming. Swamee and Jain (1978) developed equations which could be solved for f explicitely. One of their equations is r = 1.325 [ln(0.27(e/D) + 5.72 (1/11,,)°°9]"2 [131 .which is valid for a e/D range of 0.01 > e/D ) 10 and a Reynolds number range of a 10 > Re > 5000 Equation 13 has been shown to accurately represent the Coolbrook formula (Swamee et al.1978). The other popular expressions to determine pipe frictional loss .are the Hazen-Williams and Scobey formulas. For the Razen-Williams formula, K =(o.285 c1’1-352, m = 1.85, and n :1.17 in eq.15. In conventional units, Figure A. 16 we .. - wan-uh“ ' oer—o, ‘ Y I W ' H «r “on 4? ‘1 an. “30"“. menmvw l v , ‘— ”,__l 5. \ ”bee new \_.~ I M '““~ —_1 m———-— _ k 1“‘—M. “- “~‘—-—_- '5— ' ‘. M “4.%, v ' na. - .2 were“; - ': ———.‘-—‘ ...—a LIA“\ -‘——-—-' -.._.__ 3- __—T. ' __._... .—_3— fix '- ..- ‘ a "— \ " i 12“!““_ I .-'\A ‘_ ~. L m g e' ‘ -k _ _ ‘— 0 v .g 1 E W_ W I MMQ‘ o m : “— v z ‘ . ‘- \_~ 2 v Y 3 our ' .,:c.\~., _ .___- _ 1 __ J v - ...... “XV - ’7'“- ; __._ ‘ t“ .‘ “WW.M ' ‘ e 1*![91’ A ‘\r --.... V: v .1 ‘ VA ._ a. “A m \ a .v *— . I? L . ~_‘ I 2 v“ ‘ ' 41 L i Al : ‘2 - 1 -' \ ' I'f'f. ‘-i ‘ ' " " ' :- _.. \ ' [1'11 [111“ 1'1". 111 ' 1111' " J 1'111- 1~;_'\1 1‘” ~1- \1 111118;. w. . ,v 1. . . 1'17?) WW ,fr‘W—i. 75.".1 “ .fi ”W- M. _. E. —, _ .~ ‘ m ‘ ease- ‘ A“ * 1 ~ H ‘ - . Hi .; ¥\ . I... “HA..- 1 A ' L Yi V \ od‘ 1 a asann' a a oune‘ : aouuo' hut-“0.1;,- ' - \ - a a 4 "no" \441010' ‘ teens \m koooy's ciagram showing the relationship between friction factor f,relative roughness e/D, and Reynold's number Re based on the Coolbrook equation for commercial pipes (Garde et al; 1977). 17 hf = L Qm / (0.285 c)‘-85202mm [11] where C : Hazen-William‘s constant. Similarly for Scobey's equation, n=1.10, m=1.90, and K=K3/3A8. Scobey's equation was given in conventional unit in the following form by Anon 2 (1983) hf = Ks v‘-9/ D‘-‘ [15] where Ks = Scobey's constant, and hf head loss per 1000 ft of pipe. Jeppson (1983) wrote that the values of Hazen—William's C and Scobey's K3 values should clearly be depended upon Reynold number. The limitation of these two eQuations is that their coefficients are weakly related to the diameter of the pipe. Besides this, they are not functions of the Reynolds number (Wiggert and Potter, 1989). Since the flow in the lateral line is spatially varied with decreasing discharge, the energy grade line is an exponential curve rather than a straight line (Bralts,1983). Component loss: Besides friction, there is energy loss due to the presence of components in the system. This can include tees, connectors, elbows, valves, contraction and expansion joints. Component loss is generally denoted as minor loss and accounts for less than 101 of the total friction loss (James, 1988). It is a general practice to assume that minor losses are 101 of the friction loss in the network ( James, 1988; Anon 2, 1983) 18 Though component losses are generally considered as minor losses, they can have a value more than what would be comfortably called a minor loss. The many outlets on the lateral line can increase the component loss in the system until it is substantial ( Villemonte, 1977; Haghighi and Bralts, 1987; Saldivia, Bralts, and Segerlind, 1987) Component head loss in the lateral line can be given by a generalized equation hm = x (Va/2g) [16] where K : loss coefficient, and hm = component head loss. The value of loss coefficient for some components are given in Table 2 as given by James (1988), and Jeppson (1983): Table 2. Component loss coefficient for a few components. Standard Components K Standard tee; Entrance to minor line 1.8 Gradual contraction 0.0” Abrupt Contraction 0.50 Kincaid and Heermann (1970) reported that with a diameter ratio «dp/D < 0.2 and q / 0 <0.3, the head loss in feet for the flow into the riser is given approximately by hr : V2 / 23 e(9-ZQ/Q) [17] 19 where hr = riser entrance head loss, q : branch discharge, 0 = discharge in upstream mainline section, dp = diameter of riser, and D = diameter of main line. B. UNIFORMITY AND APPLICATION EFFICIEMCY 1.Hydraulic Uniformity To obtain desired discharge and application patterns from center pivot irrigation, an adequate pressure must be maintained throughout the lateral of the center pivot system. In the center pivot system, the water is introduced at the pivot point and flows outward thru the line, supplying each of the individual sprinkler heads. Since the water flows outward from the pivot, the pressure at the pivot point is higher than at the outer end of the lateral line. This is not the desired pressure distribution since the larger sprinklers at the outer end normally require higher operating pressures than the smaller sprinkler near the pivot point (Kincaid et al, 1970). To maintain adequate working pressure in the larger sprinkler, the pivot pressure must exceed the outer-end pressure by the amount of pressure drop in the lateral line ( Kincaid et al, 1970). Heermann and Hein (1968) reported that for practical purpose, the uniform distribution of depth of application is achieved by limiting the pressure drop on the lateral line to 20% of the higher pressure. The American Society of Agricultural Engineers (ASAE) also 20 recommends the pressure drop to be no more than 201 of the higher pressure in the system (Anon 1, 1986). Because of the design of a center pivot system, the discharge of individual sprinkler heads must be increased in proportion to the area each sprinkler irrigates in order to obtain a uniform depth of water distribution over the entire field (Figure 5). If the sizes of the sprinklers are the same, the spacing of the sprinklers farther away from the pivot must be closer than the sprinklers near the pivot(Figure 6). With approximately constant spacing of the sprinklers, the discharge of the sprinklers farther away from the pivot should be greater than that closer to it. Chu and Moe (1972) showed that 0 = Q0 (1-r2/RZ) [18] where Q = discharge from the sprinkler located at a distance of r from the pivot, 00 = total discharge excluding that from the end gun, r = distance from the pivot to the sprinkler which is under consideration, and R = distance from the pivot to the end of the system. Chu et al (1972) showed that the distribution of discharge in the lateral line according to the above equation closely resembles experimental field discharge in the center pivot system (Figure 7a). They also analytically derived an equation to determine the total pressure head loss in the system as no - h, = hm B(m+1,0.5)/2 -(v°)2/2g [191 where ho : pressure head at the pivot point, 21 - Sprinkler Pipe Line 1 ?atterns / O 1 4 ‘V”""’ Pivotv ‘ - / - Point ‘ i\ ' ‘ 1 Sprinkler Sozzles End gun Sprinkler pattern Figure 5. Distribution pattern from the center pivot irrigation system with increasing discharge away from the center pivot. r- Sprinkler Estterns F-Pipe Line ., -- .. Pivot- I r / \1/ q//./ 'vfi . ‘ :ind gun Point Figure 6. Distribution pattern in the center pivot irrigation system with uniform discharge from all sprinklers. 22 hr = pressure head ata distance r from the pivot, the friction head loss of the main line operating as a supply line, hm B(n+1,0.5) = beta function, and m,n : exponents from the Christensian general formula for head loss. The above equation is also written as (ho - hp) / hm = B1m+1,o.51/2 [20] The above relation [20] does not include the potential head in the system. Taking into account the potential head in the system, [18] can be written as ho ' hr 3 ( hm B(n+1, 0.5)/2 ) " AZ [21] where A2 = potential head. Using [20], Chu et al (1972) showed that there is approximately a 501 pressure drop in the first 251 of the lateral line (Figure 7b). There is a limiting factor as to the acreage that can be covered by a single center pivot system because of the the hydraulic uniformity. The total acreage covered by a system can be increased substantially by increasing the length of the pipe line and the total system discharge to maintain a constant discharge per acre. However, the total head loss will be larger because the extra discharge must be pumped thru the entire lateral. Kincaid et al (1970) showed that the 301 increase in area (from 1A0 acres to 180 acres) and discharge increased losses nearly 1001. This relatively high head loss may cause higher than necessary pumping costs and it also provides higher pressures than required for the 23 __ Theory (Eq.18) 9 System 32 (Heermann,1968) ° University of Wyoming data Figure 7a. Distribution of discharge in the lateral line. ___ Theory (eq. 20) . U of Wyoming data ° System A (Kincaid,1970) r/It Figure 7b. Distribution factor of pressure head loss. 29 smaller sprinklers and lower than required for larger sprinklers. Besides this, the application uniformity can be drastically effected because of the high pressure loss in the lateral line (Kincaid et al,1970) . 2. APPLICATION EFFICIENCY The water application rate to a point on the soil surface varies continuously with time during application by any moving sprinkler system (Kincaid, Heermann, and Kruse; 1969). In the center pivot system, the application rate must vary along the lateral length from a low value near the pivot to higher values at the outer end. Figure 8 illustrates why the application rate must increase toward the outer end of the lateral. The discharge and linear rate of travel of the sprinkler heads must increase in proportion to the area irrigated. The water is thought to be applied in an elliptical or triangular pattern from the sprinkler to the field. Bittenger and Longenbaugh (1962) found that the triangular pattern with a spacing equal to the radius of the pattern itself will produce a more uniform distribution of water than the elliptical pattern when moved at a constant speed. But Kincaid et al (1969) conducted experiments and found that the assumption of elliptical rather than triangular-pattern sprinklers produced the accumulated application curve that agreed more closely with the experimentally determined curve. This means that the elliptical pattern has to be considered in obtaining the application rate and application depth in a center pivot system. Bittenger et al (1962) found that the most even distribution for the elliptical pattern exists at a spacing of about 1.A times the radius of the pattern (r). 25 '0 WA ERA 1- 25% oi .... PLlCATlON ALONG LATERAL 28 % oi Men 23 'l. of Area 28% oi Areo PER CENT OF AREA LATERAL LENGTH ACRES 100 1320' 125.66 75 l 143' 94.24 50 933' 62.83 25 660' 31.42 Figure 8. The relationship between the area of the field and the lateral line distance that irrigates it for a 125 acres field. 131 ' \ 26 The water distribution pattern from individual sprinklers greatly depends upon the pressure available at the nozzle. The distribution of water can vary a lot when the pressure is greater or less than the design pressure (Anon 2, 1983). When the end gun is switched off, the pressure in the lateral rises as a result of which the water distribution from a sprinkler can change to give a different application pattern in the field (Figure 9). Heermann et al (1968) showed that the total depth D3 of water at a distance 3 from the pivot for one pass of the system for the elliptical pattern is Where D3 N wT1/2 w)Z h1(1-m1)1/2 I [ 1- [lini(n1+m1) sinz¢ / (1-m1 z)]].5 d0 [22] i- 0 total depth of application from a sprinkler at a distance 3 from the center of rotation, angular velocity of sprinkler lateral, rate of application at center of sprinkler pattern, ratio of distance from sprinkler to point P, measured along sprinkler lineto pattern radius (s-R)/r, time required for one-half a single sprinkler pattern to pass point P, ratio of radius of rotation to pattern radius, R/r, angle of integration, equal to (n/2 - a/2), angle of rotation about pivot, distance from pivot to point P, distance from pivot to sprinkler, radius of sprinkler pattern, and subscript referring to the ith sprinkler on the system. numerical solution of [22] requires a considerable amount of A- Pressure Too Low SO 20 10 O 10 20 3O 8— Pressure Soiisfocwry 30 20 10 O 10 20 3O C-Pressure Too High Figure 9. A typical application pattern from a single sprinkler with the change in pressure. 28 computer time. Heermann et al (1968) and Bittenger et al (1972) reported that the skewness for the circular path was less than 0.01 inches of water from the inner to the outer radius for a sprinkler at a radius of more than 650 feet. Therefore, when the radius of circular sprinkler motion is greater than 650 feet, the application pattern are essentially equivalent to those of linear motion and the depth of application can be found from N 08 = n / 2w 2 (h, ri)/Ri * (1-m1) [231 1:1 The maximum application rate which occurs at the center of the pattern was given by Dillon,Hilel,and Vittetoe (1972) as h1 = 2 ql / nrl [2"] where h1 = peak application rate at the center of the elliptical path, . Q1 = flow per one foot wide band in gpm, and r1 = distance along the center line of the band from the center of the pattern. Dillon, et al (1972) also gave the relation between the distribution of water by a sprinkler along the lateral and the percent of area covered. The water applied to a one foot band of land which encircles the pivot is proportional to the area of the band divided by the total area irrigated by the system. This was expressed as ql:2LQ/R2 [25] where L : distance from the pivot to the middle of the foot wide band in feet, 0 11 flow at the pivot in gpm, and 29 R : radius of coverage of the system in feet By substituting [2A] into [25] and correcting for units, the following expression was obtained AI = 122.5 QL / Rzrl ' [261 Since the maximum application rate occurs at the end of the system, Dillon, et al (1972) approximated the maximum application rate in the system as a whole from the following relation 11 = 122.5 Q / R r [27] where h = maximum application rate at the last few sprinklers, and radius of coverage of the last few sprinklers on the system excluding the end gun-type sprinklers. P The intake rate and the storage capacity of the soil are important parameters that determines the maximum application rate that is possible without runoff from the soil. It is possible to use one center-pivot system to irrigate two fields if the intake rate of the soil is relatively high, provided the discharge and the pressure in the lateral line is sufficient. Potential runoff begins when the surface storage of the soil is satisfied. The surface storage is equal to the amount of water applied faster then the soil intake rate which eventually infiltrates into the soil. If surface storage is not considered, the feasibility of center pivot systems is limited (Dillon et al (1972). The minimum speed of travel of the end tower v ,such that there is no runoff from the field , in feet per minute, is determined from the 30 expression v = r /30t 128] where v = minimum speed of travel of the end gun, . the radius of coverage of the last few sprinklers on the 7‘: system excluding the radius of the end gun-type sprinklers, and t : maximum time in hours for an elliptical pattern to pass a point before the surface storage is exceeded. Dillon, et al (1972) determined the surface storage curves fer 0.5 and 0.3 intake family soils (Figures 10 and 11). The value of t in [28] can be determined from these curves based on the surface storage and intake characteristic of the soil. 3. FIELD UNIFORHITY: Karmeli (1978) estimated the sprinkler distribution pattern using linear regression. He noted that the use of linear regression, based upon the dimensionless cumulative frequency curve of the infiltration depth Y and the fraction of area (X) represented by Y: A + Bx, is an accurate method for describing sprinkler distribution pattern. This approximation proved to produce good estimates for both high and low quality distributions . Christensian was the first to introduce a uniformity coefficient (Uc) to a sprinkler system (Karmeli, 1978). 0c .-.[1 - (1: | Y; -71) ”171100 [291 or Uc:[1-AY/Y]100 31 as o «.09 W ffi unmanned ”'CICAHON Olen, an (out . 9_U_ M o a 3 4 3 C T0900 g ‘ 1 ‘71" Itoullta row 7'1: DATTEIN 70 P335 8 Ocuf. m Figure 10. Surface storage curve for 0.3 intake family soil. 0 O QCOC‘ “IN” V . O. u o -oo manna-4 Ant (elm n4" . \ - )6'090 J 41 - 1 4 (- 1mg orgy-ego row 11 rant-n 1: Pass 4 rent. wove! Figure 11. Surface storage curve for 0.5 intake family soil and Pullman soil. 32 where 8'? = mean deviation about the mean. Hart(1961) and Hart and Reynold (1965) developed a uniformity coefficient (Uh) with the assumption that water distribution from commonly used sprinklers, under regular Spacing conditions, may be described and approximated by the Gaussian distribution 0,, = {1 - (s .57 / 71100 [30] where s = standard deviation, and 7 mean of the sample. Karmeli (1978) pointed out that, with a normal distribution, the mean of the absolute values of deviation equal 0.7893, and the uniformity coefficient can be written as 0,, =11- 0.7893 / 71100 (31] Wilcox and Swailes (1997) first presented the statistical concept for the evaluation of sprinkler irrigation systems (Bralts, 1983). This statistical uniformity coefficient was based on the coefficient of variation, Vy, and is defined by the equation UH = 1-Vy or 11,, = [143 / 711100 [321 where U“ = coefficient of uniformity, Vy = coefficient of variation, 8 = standard deviation, and '7 : mean of the sample. 33 Heermann and Hein (1968) have developed a method to calculate the uniformity coefficient for center pivot system based on the "coefficient of uniformity " concept proposed by Christensian. This uniformity concept considers the ratio of a summation of the absolute areas to Uc Hhere Uc D3 3 The deviation of the mean volume from observed volumes for sub the total volume applied and is expressed as its, [0, - (1:0338 / zssi|1 100 [ 1 - 1 [33] 1203531) = Heermann and Hein uniformity coefficient, = Total depth at a distance S from the center, = Distance from the pivot to the collector, subscript denoting a point at a distance S from the center, and = summation of the total number of catch containers. ASAE standard (Anon 1, 1986) ) on the test procedure for determining the uniformity of water distribution of center pivot uses the above Heermann and Hein uniformity equation. The above equation equation. is more commonly known as the Heermann and Hein modified E. NETWORK ANALYSIS: The solution of the energy gradient in the irrigation lateral can be determined by approximation method or by iterative procedures. 1. Hardy;Cross Method: One of the first and probably the most widely used method of 34 analysis is the Hardy-Cross technique (Wood and Charles, 1972). This method was popular in the pre-computer days as a hand worked solution. This method has been incorporated into numerous computer algorithms for the solution of hydraulic network problem (Wood, 1972). The method analyses the network by using the (1) head balance technique in the loops and (ii)flow balance technique at the nodes. The energy equation of flow is used to write the head balance equation in each loop of the network, and it is written as th = 1:001n = 0 [311 The continuity equation is used in the flow balance technique. In equation form, the relation at each node can be written as 201" 3 20°“: =0 [35] The Hardy Cross method uses the combination of an assumed flow (00) and a corrective flow (A0) to solve the loop equations. The head loss equation around a loop that includes corrective flow is thy = as ( 01 +00)" [361 Once the corrective flow is established, a new assumed flow is determined by using 0121 = 01 + 001 [37] where i : iteration number. The use of the Hardy Cross method was also described by Chenoweth and Crawford (197A) and Jeppson (1982). Using (8] to write discharge 35 in terms of head loss. [35] is written as mar/111"” 1,, - twp/111"“ out = o [38] Jeppson (1978) reported that the number of simultaneous equation that must be solved is greatly reduced by using [38] instead of [35]. After substituting the Junction pressure, 2[(HJ-1 - HJ) / R I”m in : 21(HJ - HJ+1) / R I”Ill out [39] where HJ-1 = Junction pressure at the upstream node, HJ : Junction pressure as the present node, and HJ,1 = Junction pressure as the downstream node. The Hardy Cross solution for the corrective pressure AH at a node can be determined by using [8] and [36] to yield A11: 1 101: I 111W} x 1 (1/m)(hf/R)1/m ' ‘11 11101 Equation AD is non-linear and iterative. As the calculation of AH proceed over the network, the value of AH should be calculated at each node and correction should then be applied to the energy and hydraulic grade line elevetion (Chenoweth et al , 197A). The number of iterations required in the Hardy Cross method is «dependent upon the accuracy of the initial guess. In some cases, Wood (1972) found that convergence was very slow and not at all. 2 . Newton-Raphson Method: This method overcomes the drawback of the Hardy Cross method by (quadratic convergence. The development of the equation below is based 36 f(x) xl Figure 12. Illustration of solutions using the Newton Raphson Method 37 upon the method given by Shamir and Howard (1968). Referring to Figure 12, the value X0 is sought for the solution such that “110) =0 At the kth iteration, the approximation for X0 is denoted by Xk. The next approximation is given by xk+1 : xk + AXk = xx - f(Xk) / [ df(Xk) / dx] [Al] The equation for the Kth improvement axk can be written as AXk = rm + (ar/aXMx = o [1121 in which both f(X) and 3f/8X are evaluated using the present value of X. Because this method adJusts the flow rates in all the loops simultaneously, convergence using the Newton-Raphson approach is much quicker than using the Hardy Cross approach. However, both methods of analysis require initial guesses for the flow distribution and a very bad estimate of these values can lead to slow convergence or no convergence at all. 3. The Linear Theory Method: This method was first proposed by Wood and Charles in 1972. Some (of the special features of this method are that it is not necessary to «estimate initial flow rates, convergence is extremely fast, programming is easier, and it can be used for optimization analysis. The basic principle of this method is to transfer the loop 38 equations into linear equations by approximating head loss by using the following equation: hli = RiQim = RiQiOm-lQi = R1'01 [A3] where 010 = approximate discharge in line i, and R1' modified pipeline constant. Reasonably accurate initial flow rate is calculated by assuming that the modified pipe line constant is independent of flow rate and, as a first approximation, is given by R1' = R1 1 [44] The solution of nodal equations obtained by applying this theory was highly accurate with fast convergence (Wood and Hayes, 1981). But in some cases convergence was never obtained. The linear theory method was used by Wood et al (1972) to solve network consisting of 58 pipes. The comparision of this method with Hardy Cross and Newton-Raphson methods showed that the number of iterations in this method is the least (Table 3). Table 3. Numerical methods and the number of iterations required for convergence. Method No of iterations Hardy-Cross 635 Newton-Raphson 2V Linear Theory A 39 H. Finite Element Method: ‘ Bralts and Segerlind (1985) reported that the finite element method could be used for hydraulic network analysis if the head loss equation can be written in a linear form. Considering the pipe segment shown in Figure 13, 21 + H1 = ZJ + ”J r Ron [”5] where Z = elevation, H = static pressure head, R0m = head loss due to friction, and i and J are subscripts denoting upstream and downstream end of the pipe element. Equation A5 can be rearranged to give RVm : [(21 + H1) - (21 + HJ)]1/u [”5] or Q = Cp [(H; - H5) + (21 - 21)] [171 where cp = [(21 + H1) - (zJ + H111 / 8"” is known as the pipe coefficient. For the simple network in Figure 13, the nodal equations for element (e) can be written as ’Qs(e'1) 7' 03(8) 3 0 [”8] and -Qg(e) + Qt(e+1) :0 [‘19] where 03 = flow to node 3, and Qt flow to node t. 90 (..11 (e) (.411 11 0011 a. Straight pipe element b. Simple one branch element Figure 13. Element considered in the finite element method. A A1 It A1 Based on the definition given by Segerlind (198A), the contribution of element (e) to equations to the global stiffness matrix are simply On and Q; where Qt(e) 3 ‘Cp(P3 ’ Pt) ' cp(zs ' 2t) [50] where H3 static pressure head at node 3, static pressure head at node t, :1: (T u 23 : elevation at node 3, and elevation at node t. N ('7' ll Equations V9 and 50 can be written in the finite element form as 03(e) r0,, -cp1 r9, cp A2 = 1 + ‘ [511 Qc(e) b-Cp Cp .1 Lpt -Cp AZ where A2 a 2g - 2,. Equation 51 has the standard finite element form {3(a)} g 1 K(e)] {9(a)} _ (f(9)] [52] where [K(e)] element stiffness matrix, [f(°)] = element force vector, {3(9)} : residual vector, and [H(e)] = pressure head vector. Haghighi and Bralts (1987) extended the above concept to include the component loss as a result of an elbow, a knee Joint, a valve Joint, an expansion/contraction Joint, and a booster pump in the network. Saldivia, Bralts, and Segerlind (1987) showed that the hydraulic design #2 of the sprinkler irrigation system could be done using the finite element approach. The sprinkler element and a pump was also considered in their analysis. The sequence of nodes and elements at the riser was considered as shown in Figure 1A. The maJor advantage of the finite element approach for the analysis of pipe network is. that the matrix is handed and symetric which minimizes the computer storage requirement (Segerlind, 198A). 2. ENERGY use Irrigation is a maJor energy user in production agricultural, accounting fo 13% of all energy used on farms ( USDA-ERS, 1977). In the Southern Plains of the U.S., irrigation pumping account for about 501 of the energy used on irrigated farm (Clark and Schneider, 1980) I Energy used in irrigation depends upon the method of irrigation. Keller and Bliesner (1983) have shown that center pivot irrigation system consumes less energy than the handmove sprinkler and traditional surface irrigation system when the water is pumped from a well of 500 ft deep. The main factor that makes the center pivot system compete with other methods of irrigation is that it requires minimal labour energy. The demand for the center pivot system increases with the increase in the cost of labor energy (Heermann and Hein, 1968). Massey, Skaggs, and Sneed (1983) suggested that the energy requirements for irrigation may be reduced by (i) improving pumping plant efficiency, (ii) increasing irrigation efficiency so that less water is required, and (iii) lowering the pressure head of the system. "3 (e) (e-l) 1e+11 Figure 19. Sequence of the nodes and elements of a riser element. 11.. ”a I'- '1 ._. ’v' l'l '§. “A Energy in irrigation is used to lift water from one elevation to a higher level, to overcome friction loss during conveyance, and to provide pressure for sprinkler operation. Energy requirement in center pivot irrigation is normally supplied by a pump that either runs with electricity, diesel, or gas. Capacity, head, power, efficiency, required net positive suction head, and specific speed are parameters that describe the performance of a pump. The power imparted to the water by the pump is called water power. Water power is (James, 1988) war = OH / x [53] where WHP : water horse power (hp, KW ), a discharge (gpm, m lacs), :1: O u 11 head (ft, m), and constant depending upon the unit used (K = 3960 for conventional unit; K = 0.102 for SI unit). 1% 11 Pump efficiency (up) is the ratio of the energy delivered by the pump to the energy supplied to the pump shaft; that is, the ratio of the water horsepower to the brake horsepower(BHP), or, np g was / BHP [SA] Overall efficiency (no) is the ratio of the energy delivered by the pump to the energy supplied to the input side of the pump driver; for ' example, the overall efficiency of a motor driven pump is the ratio of the liquid horsepower to the electrical horsepower (EHP) (Anon 3.1987). or, ' no a war / EH? 1551 A5 Anon 2 (1983) gave the nomograph developed by Kenneth Frost that shows power requierments with other factors known (Figure 15). The required net positive suction head (NPSHa) is the amount of energy required to prevent the formation of vapor-filled cavities of fluid within the eye of the single-and-first stage impellers. Continued cavition can severely damage pumps. James (1988) wrote that the net positive suction head required to prevent cavitation is a function of pump design and is usually determined experimentally for each type of pump. Cavitation is prevented when heads within the eye of single-and-first stage impellers wxceeds the required NPSH (NPSHr) values published by manufacturers. A well designed irrigation system have pumps that operate with maximum efficiency for the design discharge and head. This pump ' efficiency is constant when the discharge and head are constant. Curves relating head, efficiency, power, required net positive suction head to pump capacity are utilized to describe the operating properties (characteristic) of the pump. This set of four curves is known as the pump characteristic curves (Figure 16). An operating irrigation system has water flowing thru the pipes. ‘The head under which the system operates is dynamic. The dynamic head is 1nade up of several heads and is given by James (1988) as H3 : SL + DL + DD + H1 + H1 + H0 + VH [55] where H, a dynamic head, SL a suction side lift, 1'1 81 1 I- t (he Total Pumping Head In rent 116 19001;” P F’ IO. 1... F - 1.0 000 . 1.0 -100 ~10 F 400-1 -1.5 1.5 1- 20 r" .9 0 P109 3““ _,0 3.9.1.. -. _ 1.54. I ’o ‘03. i0 5 ”2.3 .3 o m- . . ”0" 2 .04. .5 3.0‘ ‘.°-1m - ‘\ mg" 71 1 .. 5 ’° In : 3 - , b . 140 \ 5 us 3 " ° 4.0» S. . ,0 s \ 150T i e J S! \ ~200 g ”I" -a.0 : - ”0 . \ : ~4.0 10 0 3 : cm ’0 7 30° . \ , 0 0 88-1 9 ...;oo \‘ 2 71“" 12.0 3 __ __... ,_ -- 60 .. . .. 'l "9.9 NO" 0 _.-—--"" e 10 . soo _ . . __ _ 0. '- P‘QO = 9.01' ... a— “’8... a 95“ m 10 8 ,0 I 0009”” 5 20.0" e - " = ' - )1-3‘1’ '25 T. 5‘ .- 5. 40-1 7: u”115011 3,- ~13 ”-9" I} .3 5 2000? 6 nan '; g .. 45 1b,. ? -’m g 104-11- s°.°‘- é a .5 ‘°°°"' 2 - u " 0° 1:. “g- b 40 I“ - .5909 30- ‘0.°dp a. 30004. F I- l" " ’3 d-I‘ rIOOOO ~40 1201- 13000“ J '14. _‘ 50 - 1601:". " 30 N 'l ~00 200-- z 70"- -l 0 - 8 "8 60000“ 1.39 ’ ~00000 ’°1’1oo 1001~ -20 I. '- b ‘ “°‘f 1°° ~20 «L4 .. :1” Figure 15 A power nomograph to be used to determine the horsepower required to pump water when the total head, discharge, and pump efficiency are known (Anon 2, 1983). Tolal pumping head (1111 120 110 130 90 80 7O 60 I17 Puma 01531132101315) "cm 220 '- 0 0.01 0.03 0.03 0.04 9.05 1 i 1 1 1 F’ 102 x_ 102 mm 11 in. 1: 4111.1 ‘ - Impede: o1ameier amp 3450 "m : _ 23$ mm (93- 1n.1 “1' to Ellie... - - U \ 2° )0 ’9 p27 :Cy ’2’ u 1- :25 mm \ \ / \. {1&3 -. ' E20173!- 111.) \ ‘ >2 ’ \ \ \ "\"va '- " “501303. "‘m "1" ~ .. .. " (Big-in.) ‘1' .. 52w-- \ @ W I I «1 .. 2-10 - I l \’ P r: - a i “- 1 1“ 1 1 1' 1 F' I1 J 1 1 I 1 1 0 100 200 300 000 $00 600 700 800 Pump dustharge 1n gprn Figure 16. Pump characteristic curve. N O 8 NPSH 1111 “8 DL = discharge side lift, DD = water source draudoun, H1 = Head loss due to pipe friction, H1 : component head loss, Ho VH operating head, and velocity head. Except for the suction side lift and the dynamic side lift, all the other heads are dependent upon the Q-H value in the system (James, 1988). Therefore any change in Q-H value will change the prevented when heads _uithin the eye of single-and first stage impellers exceeds the required NPSH (NPSHr) values published by manufacturers. A well designed irrigation system will have pumps that operate at the optimum point of the pump curve. An example of the optimum operating point of the pump is shown in Figure 17. Changing the diameter and/or speed of an impeller alters its characteristic curve (Anon 3, 1987). The changes in impeller performance resulting from changes in pump speed can be estimated using the following equations 02 = Q1 (NZ/N1) , [57] H; = a, (Na/N1)2 [58] 8P2 = 391 (NZ/N1)3 [59] where 1 and 2 are the subscripts denoting two different pump conditions. There are two center pivot irrigation design philosophies (a) Head 119 Pump urv Ooefating H93 1 Operaimg '“"""""""' "" - -l.\ Pom: 1 1 v I : Fk) ate .Delwered System Curve ' by Pump Discharge (Caoacny) Efficiency curve Figure 17. The plot of the operating point of the pump. I;¢ Ivc 1.- .1 ... 1|“ - 1": v.1. 5A.. \- I. v I” 1.. it) ... ‘ 50 design assuming that the end gun is in operation and (b) design assuming that the end gun is not in operation. Bermuth (1982) used a field verified computer computational technique and studied the effect of cycling an end gun on and off for a typical center pivot condition for consideration in selecting sprinkler nozzles for flow uniformity. Based on the flow uniformity concept, he found that the superior nozzling philosophy was to nozzle interior sprinklers assuming that the big gun is not in operation. But the difference between the flow uniformities for the two philosophies was extremely small. Besides this, the error induced as a result of the second philosophy was toward under irrigation. Bermuth did not consider the change in the performance characteristic of the pump in his study. Most design engineers use the second design philosophy when considering the pump that is to be installed. The main reason for using the second philosophy is to make sure that there is no cavitation when the converse fashion of operation is used. The end gun in the center pivot system is turned on when the corners of the field are irrigated. This will increase the discharge and decrease the head in the pump according to the performance curve (characteristics) of the pump. This means that the operating point of the pump will change. The change in operating point means a possible change in the efficiency of the pump (Figure 18). Many researchers (Bermuth, 1983; Solomon and Kodoma, 1978) have written about the change in operating point in the pump curve as.a result of turning the end gun on and off during irrigation. But no one has looked at the possibility of using auxillary sprinklers that 51 2- Pump Curve I Efficiency curve "(ft) end gun off system curve ‘ o-end gun on system curve 0 (919m) Figure 18. The change in the operating point of the pump with the change in the operating head. 52 turn on when the end gun is off end gun discharge. The sizing of these auxillary sprinklers can be such that their total discharge is equal to the end gun discharge. The use of auxillary sprinklers can keep the operating point of the pump constant so as to operate it with maximum efficiency throughout the irrigation period. SUMMARY AND DISCUSSIOU: The review of the literature related to hydraulics indicates that the Hazeneuilliam and Scobey equations are widely used to calculate the friction head loss in a pipe even though it is known that the Hazen- Hilliam constant C and Scobey constant K, does not accurately represent the friction factor in the pipe. The main reason for using these equations instead of the Darcy-Heisbach equation is because of the requirement to solve friction factor f implicitely for the later equation. This difficulty in solving for f can be overcome by using the Swamee-Jain equation that can be solved for f explicitely. The Swamee-Jain equation also contains the relative roughness, e/D, and the Reynolds number terms. The review of the literature on hydraulics also provides the theoritical framework for the analysis of the center pivot system hydraulics. The need to calculate the component head loss instead of assuming its value to be 10% of the total friction head loss is reflected in this section of the review. The review of literature related to hydraulic uniformity indicates 53 the difficulty and importance of achieving uniformity of pressure in the center pivot lateral line. Though the best pressure distribution is to have higher pressure at the end of the lateral line, this can not be achieved. Therefore a standard has been set by the American Society of Agricultural Engineers (ASAE) limit the pressure drop in the lateral line to no more than 201 of the pressure at the pivot. This limitation in the pressure drop will allow a better water uniformity in the field. It will also limit the pumping cost per field area to a reasonable value. The main focus of any irrigation system is the application of water to the field. Since the elliptical application pattern from the sprinkler produced the accumulated application curve that agreed more closely with experimental values, it is relevent to consider elliptical application pattern from the sprinkler. The intake rate and the storage capacity of the soil determine the maximum application rate of water in the field. It is possible to use one center pivot system to irrigate two fields if the intake rate of the soil is relatively high. The addition of auxillary sprinklers, which operate when the end gun is shutoff, and the increase in the angular speed of the lateral line can increase the application rate as a result of which the total time required to irrigate the field is reduced. The literature review of field uniformity was motivated by the need to know which field uniformity coefficient equation was the most valid for the center pivot irrigation system. SN Heerzann and Hein (1968) uniformity coefficient concept was developed from the basis Christensian uniformity concept. It takes into account the fact that the application rate in the center pivot [irrigation system increases farther away from the pivot. Any change in the system design concept should not reduce the field uniformity significantly or below the minimum appraved level. The literature review on the network analysis was required to incorporate the most efficient and accurate network analysis method in the development of the model. The linear theory model is an excellent tool for the analysis of pressure drop in a loop. But this method does not always give convergence in the flow calculation at a node. The Newton Raphson method is also a good technique to analysis networks. The Hardy Cross method requires the most iterations to solve the equations, though this does not necessary mean that it requires maximum computer time for the analysis. The finite element method for network analysis is gaining in popularity. Basic research for its use in sprinkler irrigation has been completed by researchers. The major advantage of this method is in computer network analysis. This method cannot be used easily for analysis with the hand calculator. The literature review on energy use is an important aspect in the development of the objectives that is to follow. Though many researchers have noted the fact that there is a change in the operating point of the pump with the switching of the end gun on and off, no one has looked into the possibility of providing auxillary sprinklers in the lateral line such that thes sprinklers operate only when the end gun is off. 55 This can keep the operating point of the pump constant so that it is possible to have maximum pump efficiency throughout the irrigation period. The energy saving with this design concept of the center pivot system may be substantial, especially if the pump characteristic curve is steep. In. ‘0 '5'. ”a. 5 fl" _’ III. METHODOLOGY The review of the literature on the center pivot system in the previous chapter revealed that no one has looked into the possibility of energy conservation in the center pivot irrigation system by modifying its design to include auxillary sprinklers. There is a possibility of energy conservation by keeping the pump operating point constant throughout the period of irrigation. The application depth and application rate are also important parameters that are to be considered in any modification of the center pivot system. The following is a summary of the research approach to be followed in this study. A. Research approach The research approach that will be followed to achieve the major research objectives is proposed to be the following: Objective 1. To determine the relationship between energy savings and the addition of auxiliary sprinklers in the center pivot irrigation system. The approach to be followed to achieve the above objective will be the development of computer simulation models. The hydraulic network 56 57 analysis model will be based on the finite element approach. This model will be used to simulate the best positions and specifications of the auxillary sprinklers in the modified design of the center pivot system. This simulation will be done to get the constant discharge and pressure head throughout the period of irrigation so as to operate the pump with constant maximum efficiency. Simulation will also be done to determine the water power (UP) for varying amount of end gun discharge that will be passed thru the auxillary sprinklers when the end gun is off. Some model results will be compared to field data collected on existing modified center pivot system. Objective 2. To make a comparative study of the application unifbrmity and depth with and without the addition of auxillary sprinklers in the center pivot system. It is proposed that the above objective will be achieved by using the Heermann and Hein theoritical concept on the depth of application on the field. The application depth for the following conditions will be studied (i) the end gun is on (ii) the end gun is off, and (iii) the end gun is off with the modification in the design to include auxillary sprinklers. The study of the application uniformity will be based on this simulated total application depth of water. Objective 3. To study the effect of the introduction of auxillary sprinklers on the water application depth in the field and the speed of the lateral arm movement. The approach to be followed to achieve objective 3 will be to 58 the relationship between the addition of the sprinklers and the application rate for different speed of the lateral in the model. The model will simulate the condition when the application rate after the design modification will be the same as without the change in the design. The calculation of the irrigation time that can be saved with the modification of the design will be studied with the help of the model. B. Theoritical Development The theoritical development section covers the (a) hydraulic analysis and irrigation performance evaluation section, (b) energy evaluation section, and (c) an example implementation section. The hydraulic network analysis of the center pivot irrigation system network is based on the finite element approach. The pressure head difference between two sections of the network is the result of the friction head loss, the components head loss, and sprinkler head loss. Applying the energy balance between two points in the network, figure 19. gives (21+H1) - (ZJ+hJ) = AH [60] where the subscripts i and j denote the upstream and downstream conditions respectively and all the other variables are as previously defined. The head loss term, AH, of [60] can be further divided into three equations AH BO” pipe head loss [61] 59 sprinki Pipe element component 0' q i lateral length—'- Figure 19. Network cosisting of a pipe, a component, and a sprinkler. 60 AH KCQ; / ngz component head loss [62] AH Kaan sprinkler head loss [63] Equation [61] can also be written as Q = Cp (H; + 21) - cp (HJ + ZJ) [61] where Cp :1 [(H,.zi) - (HJ+ZJ)]1’m/m 1 / R"". Similary [62] and [63] can be rearranged to give Q = cpc1H, - HJ) + cpc121 - 23) [651 where Cpcs ungn / 8K130, D : diameter of the pipe, Kij‘ component loss for the ij element, and Q Csp (H1 - NJ) [56] Where C3p= (Hi-HJ)(a-1) ' Ksp. The elevation differences in the sprinkler element and the component element are~neglected in [65] and [66]. The calculation of the head loss due to friction will be based on the Darcy Heisbach equation. The friction factor f will be calculated using the Swamee-Jain equation, [13], which has been explained in the literature review section. The sprinkler head loss depends upon the sprinkler coefficient k and the exponent 'x' which is given by [7] as Qakhx 61 The values of q and h in [7] are given in the pressure-discharge chart provided by the sprinkler manufacturer (e.g. Table 1). The value of the exponent, x, is usually taken as 0.5 for a non pressure compensating sprinkler. The exponential value, x, and the coefficient, k, of [7] can be determined by linear regression of the known discharge-- pressure relationship between q and h. The pressure along the length of the pipe line of the center pivot irrigation system will be initialized using the model developed by Chu and Moe (1972). The initialized estimated pressure along the center pivot system pipe length using the Chu and Moe model is determined by using the beta function and is written as no - h, = (hmB(m+1,0.5)/2) - AZ [67] where the variables are as explained in the literature review section. The‘ distribution factor H is given by :1: I ' (hp'hfl)/(hO'HR) OP 3: I - 1-Bx(m+1,0.5)/B(m»1,0.5) (68] where x = r/R, and 83(m+1,0.05) = the incomplete gamma function. The value of m in [68] is 2 when using the Darcy Haisbach's equation to determine the head loss in the pipe line. Therefore the distribution factor H in [68] can be estimated using the following equation (Cho and Moe, 1972) H = 1 - (15/8)(x-2x3/3+x5/5) [69] 62 where x = r/R Considering [60], [61], and [62], the discharge thru any element is Q = (Cp+C3p+Ccp)[AH]-CPAZ [70] The global stiffness matrix can be built for the network of the center pivot system to solve the_pressure head loss at the nodes on the basis of [70). The Gaussian Elimination method will be used to determine the solution to the global stiffness matrix. The irrigation depth in the field will be determined using the Heermann et al (1968) equations [22] and [23]. The irrigation application uniformity is estimated using equation [33] developed by Heermann et al (1968). The American Society of Agricultural Engineers (A532) also recomends using equation (33] to estimate the irriagtion application uniformity for the center pivot irrigation system. A part or whole of the end gun discharge will pass thru the auxillary sprinklers when the end gun is turned off. This distributed discharge required at different segments of the lateral line will be calculated using the Cho et al (1972) relationship given as q = 0(1-(r2/321) [71] where q : extra discharge in the lateral line at a distance r from the pivot point, Q s discharge from the end gun that is to be distributed along the lateral line, and R : the distance from the pivot point to the radius of coverage of the system. 63 The auxillary sprinklers sizes and spacings will be determined based on the discharge required to pass thru it for the expected pressure in it. Once the auxillary sprinklers are included in the model, the pressure distribution along the pipe line will be calculated for the modified system. Field performance evaluation will be done for cases when 33%. SOS, and 66$ of the circle is covered with the end gun on. .The calculation of the power imparted to the water by the pump, i.e. water power (HP), is done using the following relation UP = QH/K where the variables are as defined in [52]. The total discharge, Q, and the total pressure head, H, in [52] are the values calculated by hydraulic analysis. The discharge and the pressure at the pump will be determined by plotting the system curve and the pump curve. The point of intersection of the system curve and pump curve is the operating point of the pump. The pump efficiency np is determined using the pump curve (e.g. Figure 20).The pump efficiency is determined by plotting the efficiency curve. The efficiency curve that passes thru the operating point of the pump gives the efficiency of the pump. It is possible to determine the brake power of the pump by using equation [53] once the pump efficiency and water power are determined. The total energy required will be given by Total energy required = (BF/n)'time of operation of the system an where n = electrical efficiency. The field data is obtained for the verification of the model from Mr. Jerry Bemeant's farm in Cass county of Michigan. This farm has the option of operating the system with the auxiliary sprinklers attached to the pipe line so a part of the the end gun discharge is passed thru these sprinklers when the end gun is turned off. This system is 787 feet long from the pivot point to the end gun sprinkler. It has 28 primary sprinklers manufactured by Nelson Irrigation Corporation. The pivot pressure and the spacing of the sprinklers are noted. The lateral speed of the end tower is also noted. The system is set up as shown in figure 20. The model will be run to determine energy savings fer three different pumps (i.e. three different types of pump curves). The two extreme cases will have horizontal and vertical pump curves as shown in figure 21 and figure 22. The pump with horizonatl pump curve as shown in figure 21 will have the operating pressure head always constant. The pump with vertical pump as shown in figure 22 will always give almost a constant discharge. The pump with the pump curve as shown in figure 23 was being used in Mr. Bemeant's farm in Cass county of Michigan. Pumps with horizontal and vertical pump curves are included in the study to show the maximum possible energy and time savings that is possible for different type of pumps. The comparision of the results obtained for these three different pumps will give the range of energy savings that may be possible. 3 5 53 55 57 no pun (z) (4) $2) 511 56) 1flvotlg) 13) jg '53) A ’55) - - l " ‘ “a 5* ‘5 e 727 f: d. Figure 20a. Sequence of nodes and elements for the center pivot simulated model. L_ ?87 feet ‘5 l terai-Line 'T"' 20 feet Pu- Sell Figure 20b. The experimental network of a center pivot irrigation system. head (ft) 66 pump curve V discharge (9pm) Figure 21. Pump curve to operate the pump at constant pressure. head (ft) pump curve discnarge(gpm) Figure 22. Pump curve to give a constant discharge for all pressures. 67 RCV SYil'Lfli Ill 5:!" UMV. 1 5' 1'" HOHOAI : 219 272 7971‘ Agrigulgur“ in”, :l ‘l ‘ 9 0011945148 #74 7J0 (In? 3g’r2’“ M“ 5%“ ire—Java! PER STAGE Figure 23. Pump curve for the pump used in the field. IV RESULTS AND DISCUSSION The results and discussion section consists of the analysis of the data generated by the four computer models, NETWORK, SYSTEM, AUXISPR, and FIELD. The results are presented for the hydraulic network analysis, field performance evaluation, and energy and time savings. The discussion follows the result of each section. A. Hydraulic Analysis 1. Center Pivot NETHORK Simulation As presented in the methodology section, a computer program NETHORK (Appendix A) was written based on the finite element method to simulate the discharge from the center pivot system when the end gun is on and when the end gun is off. The input data for the example system is given in Figure 29. In general, the center pivot system has a 787 feet long Lateral line on which there are 28 sprinklers. The distance from the center pivot point to the last tower (not the end of lateral line) is 765 feet (Figure 20). The discharge-head equations for the sprinklers were obtained by regression analysis of the discharge-head relationship given by the manufacturer (Anon A, 1986). Twenty five of the sprinklers have the exponent of pressure head between 0.19 and 0.51 which compares 68 69 000010030” EQUATION 0.271HOOO.SI’ 003' (H000. ", 0. ‘. (No.0. :0, 0. '8 114000. “0 1.201H000.“) 1.021H000.301 3.02‘H000.=°! 1.491H000.3°’ 1.02CH000.:°’ 1.301H0000308 1063(H0.°.0.8 1.02‘H00003OU 1o‘1‘H000o308 ‘0“ (10000.50. 1.771H9.003°’ 2.131H000.303 2003‘”..°o:°. 2.001wooo.a91 20‘21H00000.’ 2.001Heoo.301 2.201Hooo.:21 2.0.1H000.:°’ 2.001H000.30! :. :9 111000. 30’ 3.:01H..°.:¢’ 3.031wooo.311 80.001N0000598 0070 0N0 0£SUL70 «a so oomneim in can one-.- 20 Total length so lateral. 707 00 01mtmncm to last ton-o. 703 ft Hams-um 0110-0010 error in calculating pressure he... .3 90 0'0 0P021N0 DISTANCI F000 CALCULATE: N0 (Ge) P1V07 14c) K0 0 1 20.0 20.0 0.2002 0.3090 2 23.3 20.3 0.2003 0.4902 3 20.3 03.0 0.0303 0.3031 0 20.3 113.3 0.9100 0.0300 3 22.3 100.0 1.2000 0.0020 0 20.3 170.3 1.0171 0.3007 7 20.3 203.0 1.0171 .3007 -0 20.3 231.3 1.0033 0.0973 9 20.3 200.0 1.0193 0.3010 10 20.3 203.3 1.2012 0.3003 11 22.3 221.0 1.0310 0.0007 12 20.3 309.3 1.0193 0.3010 13 20.3 370.0 1.0193 0.3010 10 20.3 000.3 1.0300 0.3009 13 20.3 033.0 1.9721 0.3010 17 20.3 090.0 2.1201 0.3039 10 20.3 320.3 2.1201 0.3039 19 20.3 333.0 2.0029 0.0093 20 20.3 301.3 2.0199 0.0912 21 22.3 010.0 2.0733 0.0993 22 20.3 002.3 2.2232 0.3107 22 20.3 071.0 2.0733 0.0993 20 20.3 099.3 2.0733 0.0993 23 20.3 720.0 3.1900 0.0971 20 20.3 730.3 2.1900 0.0971 27 20.0 703.3- 3.0310 0.3000 20 2.7 707.0 13.9972 0.0990 01002700 010TAN02 0000 P1V071091 (inches: #000 70 3. 0 703.0 2.70 703.0 707.0 '1'! 0000HN030-0.00000010enem SLO?! CF The F12L3 13 N60L1010L0 R1523 oxanemta - 1.0000 1fl¢h00 2N0 Gun RISER DiantTER - 2.0000 inches x FOR TEE 0N0 EL2OU CCHPONINTS - 1.00 Figure 211. Data for the hydraulic network analysis of the center pivot irrigation system network. 69 000000010" 20007108 . :9 010.0. 51 9 0.39 ‘NflOg ", 0. ‘. ‘Ma :0. 0.911H000...3 00:0 119900. “’ 1.021H090.303 1.02‘H000.3°’ 1.091H000.3°' 1.021H000.303 0a3° ”90.0030, 10‘: ‘m. "’ 1.021H090.3°’ 1.020100%”) 0 0“ 00000.30. 3. '1 (“030’ 2.13‘H990.3°’ 2.131H900.309 20 :3 111000.301 2. 00 "9.”. "’ 20‘1119900. ‘9’ 2. 00111000.”! 20 2‘ (HO-00.323 2. 00110000. :0) 2. 00 00.00.30, 3.20 01000.30) 3. 20 "9.00.309 3.031H90°.317 8‘.“ 111000.301 0070 AND mu 9h 04 sprinkler-a 1n the enema- 20 Total length 09 lama-al- 707 00 010x015:- tm 1am: ton-or- 703 00 110111“ 0110-001. arr-ov- 1n calculating groom maaa- .3 00 0'0 firACINI 01370003 F003 CALCULATED 70 He) PIVOT 1901 KC 0 1 20.0 23.0 0.2002 .3090 2 23.3 20.3 0.3003 0.0902 3 23.3 03.0 0.0303 0.3031 0 20.3 113.3 0.9100 0.0300 3 32.3 100.0 1.2000 0.0020 0 23.3 170.3 1.0171 0.3007 7 23.3 203.0 1.0171 0.3007 .0 20.3 231.3 1.0033 0.0973 9 23.3 200.0 1.0193 0.3010 10 23.3 233.3 1.3012 0.3003 11 32.3 321.0 1.0310 0.0007 12 20.3 309.3 1.0193 0.3010 , 13 23.3 370.0 1.0193 0.3010 10 23.3 000.3 1.0300 0.3009 13 23.3 033.0 1.9721 0.3010 10 32.3 007.3 2.1201 0.3039- 17 23.3 090.0 2.1201 0.3039 10 23.3 320.3 2.1201 0.3039 19 23.3 333.0 2.0029 0.0093 20 20.3 301.3 2.0199 0.0912 21 32.3 010.0 2.0733 0.0993 22 23.3 002.3 2.2332 0.3107 23 23.3 071.0 2.0733 0.0993 20 20.3 099.3 2.0733 0.0993 23 23.3 720.0 3.1900 0.0971 20 20.3 730.3 3.1900 0.0971 27 20.0 703.3: 3.0310 0.3000 20 3. 707.0 13.9973 0.0990 010H3720 01370NC! 0000 01V071001 (inch-o: each 70 3. 0 703.0 2.70 703.0 707.0 01'! RCUBHN030-0.0000001nenaa SLO?! CF 7%: FIELD 15 N60L1010L0 01523 DZAnETEH - 1.0000 inches 2N3 EUR 01520 01000720 - 2.0000 inches < FOR 70: ans ELBOU CCHPONONTS - 1.00 Figure 211. Data for the hydraulic network analysis of the center pivot irrigation system network. 70 well with the usual turbulent flow equation Q : kHO‘S I [7] As a preliminary test, the program NETWORK was run ten times each with given center pivot pressure. This pressure was increased from 80 psi to 125 psi in 5 psi increments with the end gun on, end gun off, and the modified center pivot system conditions. Figures 25a and 25b are the output from these runs for the end gun on and end gun off conditions. The reason for these preliminary runs was to establish the discharge-head relationship of the system at the center pivot point, making it possible to determine the pivot pressure for any given discharge thru the system. The allowable error for the runs were 0.5 feet of head. The pressure drop along the lateral line with the end gun off is about half of what it is when the end gun is on. These results indicate that the pressure along the line increases when the end gun is off. It can also be seen that regardless of the pressure at the pivot point, the percentage of total flow passing thru the end sprinkler (end gun) when the end gun is on is constant (22.92%) for all the runs. In addition, the discharge ,0 , and the pressure head , H, at the pivot point_ illustrated a very strong association (82:0.99999) as shown in Figure 26. Thus it can be shown that the total of all flows passing thru the sprinklers (Q) (discharge thru the pivot) is related to the pivot pressure (H) by O = 57.007 H0'502798 [721 The Q-H relationship equation for the end gun off condition was 71 090 0109 .9 Rue Pow-as ualvae Dena mama ”may m. 9019“ 0am. 3a. «0 10011 1071 19011 1971 19311 1991 10,911 can" u so 0awou l 90.00 100.91 09.0 100.3 10.30 :9.a2 700.7 170.3 22.02 2 93.00 190.07 73.0 170.3 11.22 .3.“ 010.9 191.9 22.02 3 90.00 200.03 70.2 190.7 11.90 27.37 030.0 107.2 22.02 0 93.00 219.30 92.3 190.0 12.00 29.01 937.0 192.0 22.92 3 990.00 23909‘ ..a. 3900’ 93.0. 300.3 '7’.’ 9,703 :30‘3 7 110.00 230.23 93.0 ...7‘0.9 10.01 33.32 923.2 237.0 22.02 9 113.00 203.01 100.0 231.1 13.03 30.73 900.0 211.7 22.02 9 120.00 277.37 100.3 201.2 13.00 30.15 900.3 210.2 22.02 10 123.00 230.92 100.7 231.3 10.27 37.01 900.0 220.7 22.02 7'9. 0 90 9| ”7909.991! 07 798 09907 10 01900 07 0- 37.007999000J027909 D. .09 '0 0191 Patna 9.1m Pena Inna PO. III". ham 0'. 90 19819 1!?) 12019 ('71 10019 1971 (0’09 10!!“ 19 09 R1909) — n “ — _ _ fl — _ 1 00.00 100.91 70.0 172.0 3.02 12.32 022.2 0.0 0.00 2 93.20 190.07 79.0 103.0 3.03 13.00 001.0 0.0 0.00 3 90.00 223.03 00.1 190.3 3.90 13.73 000.0 0.0 0.“ 0 911.00 219.30 03.0 -- 203.2 0.21 10.30 070.0 0.0 0.“ 3 100.03 2121.10 93.3 210.1 0.31 13.03 090.3 0.0 0.” a 103.00 292.70 90.2 220.9 0.00 13.70 713.7 0.0 0.00 7 110.00 3:10.23 102.9 237.0 7.12 10.03 730.0 0.0 0.” 9 113.00 213.01 10710 200.7 7.02 17.13 707.1 0.0 0.” 10 129.” 223.92 117.0 270.0 0.01 10.00 779.2 0.0 0.” 19.0900WP07M'9W7100W09 0000.090MM9 Modified end gun off 900 9191909 0.1900 90-0 lead Idea. Idea. 0919" 09.0 99-0 fl 19011 1m (9019 ('77 17911 “1771 (0709 (07197 13 01 0919009 .1. * * ~ “ * .— “ a... 1 19.99 111.91 71.0 111.1 1.11 11.19 191.1 9.9 0.90 1 99.99 199.07 79.9 177.9 9.09 19.51 799.9 9.0 9.99 1 99.90 109.93 91.1 197.9 9.99 19.99 991.7 9.9 9.99 1 199.90 131.10 99.2 199.9 9.79 11.91 999.9 9.0 3.90 9 111.90 299.91 191.9 199.9 11.19 11.99 919.7 9.0 9.09 9 119.99 177.17 199.0 299.9 11.91 19.97 909.9 9.9 9.90 19 223.99 199.92 112.9 191.9 11.97 17.99 999.1 9.9 3-30 1112 0 vs 11 02107101191119 1? 9112 71m 19 01920 99 OI 13.91919"9.1917991 Qifliifl'OOO Ila .IOOOOOUUOIIE 0 boogbuuwui'a EDigure 25a, b, and c. Results obtained from the computer simulation model 'NETHORX' for the end gun on, end gun off, and modified end gun off conditions. 71 010 0109 U * 0101 P011100 11011101 00110 110110 0000' 1101-00 0011101 0000 " 3110 10011 1": 10011 1971 10011 1071 100m 100m 10 01 0010011 3 .3000 170.97 73.. 170.3 11.22 03.99 910.9 191.9 22.02 3 90.00 200.03 72.2 100.7 11.00 27.37 030.0 197.2 22.02 C 73.00 219.8. 02.3 190.. 13.00 2°.I1 037.0 192.0 22.02 3 1&000 23009. “0. 200.7 03.0. ”0‘, M0, 1'70: 22.02 0 103000 293070 .10: 3000. 030“ ’00.. 9°00. ”303 :30" 7 110.00 230.23 93.0 23.9 10.01 33.32 923.2 207.0 22.02 9 113.00 203.01 100.0 231.1 13.03 20.73 900.9 211.7 22.02 9 120.00 277.37 100.3 201.2 18.03 30.10 900.8 210.2 22.02 10 18.00 280.92 100.7 231.3 10.27 37.01 900.8 220.7 $.02 11¢ O V. 99 110131109111- 07 79C PIWT 1. 019011 .7 0- 37.fl7999000.m no 0101 CV 0101 #011100 101901 00110 111110 00n- nnns 001901 0.10 001- 10 (FBI! 1!?) 1”" 1"" 1"” 1'71 10'”! 1M1 10 09 “19001 - _ ~ — — _ u — — 1 .9000 199091 ’90. 17:0. 2%.: 130:: .330: .0. 9b.. 2 03.03 190.07 79.0 103.0 3.03 13.” 001.0 0.0 0.” 3 90.00 253003 .900 19903 30'. 0:07: ..00‘ .0. 00.. 0 911.00 219.80 03.. . 203.2 0.21 10.30 072.. 0.0 0.” 3 103.03 2.21.14 93.8 210.1 0.81 13.08 090.. 0.0 0.” 0 1013.00 2032.70 99.2 220.9 0.02 13.70 713.7 0.0 0.” 7 110.00 230.23 102.9 237.. 7.12 10.03 730.0 0.0 0.” 0 113.00 2123-01 16700 200.7 7.02 17.18 797.1 0.0 0.” 9 120.00 277.37 112.3- 279.0 7.71 17.01 703.3 0.0 9.“ 10 120.00 223.92 117.0 270.0 0.01 10.00 779.2 0.0 0.” monumenmemusmu o-.1..090111uo.0031001 Modified end gun off 9110 901900 001900 9000 11000 90100 00100 001900 0000 0999 D 17011 1771 (7011 1777 17111 '1771 107191 107191 13 01 9919091 Co. * m _ “ --. ~ —. fl 1 11.11 111.11 72.1 111.1 1.11 11.11 111.: 1.1 1.99 1 15.02 120.07 71.0 177.0 0.00 11.11 790.0 0.9 00°. 1 90.00 100.93 91.1 117.1 9.19 10.19 121.7 0.0 9.00 0 91.00 211.19 91.9 110.0 9.19 11.93 100.0 9.0 0.00 1 100.90 131.10 10.2 291.1 9.71 12.92 199.9 0.0 309' 9 101.90 202.70 90.1 111.0 19.21 11.91 990.1 9.0 9.9. 1 111.90 295.11 101.9 200.0 11.19 21.00 929.7 0.0 9-99 10 123.00 219.12 102.1 291.0 12.07 17.10 999.1 9.9 3.39 7118 0 vs 11 Ill-07107131117 A7 7712 11m 18 017:: I! O- SS.91110"9.3937031 QflfiflQQOOOO l‘a ......ODUOI‘a 0-1 0 .....uuuu'li Figure 253, b, and c. Results obtained from the computer simulation model 'NETHORK' for the end gun on, end gun off, and modified end gun off conditions. 72 1000 Regression equation 0 - 57.007(H~0.502798) (1) R002 - 0.99999 3 Q) 0) CD .9. .E ”E“ 900-: Q . C” V an 0 (I: <1 I 0 £2 CD 800 1 200 300 HEAD (feet) in log scale Figure 26. Graph showing the discharge-pressure relationship at the center pivot point for the end gun on condition. 73 similaxiydetermined with a coefficient of determination (R2) of 0.99996 (Figure 27). The regression equation for this condition is Q = uu.89o H°-5°37°5 [731 2. System Curve Simulation Equations [72] and [73] were used to determine the system curve at the pump for the given elevation head (20 feet), lift from water level to the pump (90 feet), diameter of the main line (8 inches), and the length of the main line (3510 feet) (Figure 20). Program "SYSTEH"' (Appendix C) was used to determine the Q-H relationship at the pump for the given data. The system curve based on the output results from "SYSTEM" (Figures 28a and 28b) were used to determine the Ooh °relationship at the pump (Figure 29). These are the system curves for the system. The energy savings and field performance evaluation were done for three different pumps : pump 'H' with the horizontal pump curve, positive displacement pump 'V' with the vertical pump curve, and the pump used in the field pump 'J'. Pump curves for pumps H and V are hypothetical. These are the two extreme cases for pump curves. They are included in the study to show the range of energy and time savings depending upon the type of the pump used. Figures 30, 31, and 32 show the pump curves for the pumps 'V', 'H', and 'J' and the system curves for the end gun on, end gun off, and the modified end gun off conditions. The intersection of the pump curve and the system curve is the operating point of the pump. 7N 800 Regression equation 0 - 44.890(H«0.503705) Raz - 0.99998 .93. (D (J (I) Ch .9. .S ,_\ E 700 CL C” \_/ UJ 0 0: < I 0 L”. O 600 200 300 HEAD (feet) in ‘log scale Figure 27. Graph showing the discharge-pressure relationship at the center pivot point for the end gun off condition. 75 stun“ «99 um. 19.9 LIFT um ma. 99.9 lament” or 9'!!! "n ”Iain-9. 999999 stunt: or ru 1m quaint. 9.9999 91119-29 or 7": 99:7109 Dirt-19.9999 1.3.2?! or '99 onxvuv nu treat-1919.9 annual? or T119 "an 99 run Hm. 99.9999 09 m sum 19 n: 9199909. 9999 999' I199! 9 '99: 9999 9999 9919 999 9999 999 9999 99991'9999 999 '9 9811991 99:91'71 _ _, _ —-—- 999 119.99 19.9919 9.9999 9.1999 79.999 999.9999 999 119.99 19.9999 9.9919 9.1919 99.999 919.9999 999 119.99 99.9999 9.9999 9.9999 199.919 999.9999 999 119.99 99.9999 9.9999 9.9999 199.991 999.9999 999 119.99 99.9199 9.9999 9.9199 199.999 999.9999 999 119.99 91.9999 9.1999 9.9999 199.191 999. 99 999 119.99 99.9999 9.1919 9.9999 199.999 999.099“ 999 119.99 99.9999 9.1999 9.9991 919.991 999- 999 119.99 99.9999 9.1999 9.9199 991.999 999. 9 999 119.99 99.9999 9.1999 9.9919 999.199 999.999“ 1999 119.99 99.9999 9.1919 9.9999 999.999 999-99 9'9 99. 99' o“ cum: I!» "799 99.9 U"? um "719 99.9 mas: o: m: 9199 (199811109.999999 911‘: ‘ ~ : C! m: 9!!! 12992.91.“ 9.99999 91Lnt?:1 09 fit 'JC7ICI 9199.19.9999 or n: 92.1".“ 919: "tat-1919.9 when}?! 9! his turn: 19 7.1: nm. 99.9999 mm 9' in am 19 9:9 "999-9.9999 99" '98:. I '91: Ill- fISC 9999 VII III. 9'! III. 99999 III. 69' f9 9=L9919 99:99979 I! I? '7 _ _ _ —-—- _ “ 00—— 999 119.99 19.9999 9.9999 9.1999 119.999 999.9999 999 919.99 19.9999 9.9919 9.1919 199.999 999.9999 999 199.99 99.9999 9.9999 9.9999 191.979 999.9999 999 119.99 99.9999 9.9999 9.9999 991.999 999.9999 999 119.99 99.9199 9.9999 9.9199 999.999 991.9999 799 119.99 91.9999 9.1999 9.9999 999.999 999.9999 . 999 119.99 99.9999 9.1919 9.9999 999.991 999.9999 999 119.99 99.9999 9.1999 9.9999 999.999 999.9999 999 119.99 99.9999 9.1999 9.9199 999.999 999.9999 999 119.99 99.9999 9.1999 9.9999 999.999 999.9999 1999 119.99 99.9999 9.1919 9.9999 999.191 999.1199 II. 99. 9'9 CICIIIII 9999II9 9119'!!! 9999 99999 99.9 LH-‘T 99.991971- 99.9 999:2?939 99 911 9199 11l¢lll909.999999 911:3732 9! TR! 9199 119C39990 9.9999 911n77;1 9' 9!: 9097998 9199.19.9999 LlrzTfi 99 991 9311V299 9999 III€I99909919.9 :Sl'P1CXIIT c! ?H! 9997!! A? 9*! 'IVUIO 99.999. . 9999-!!! 9! 999 99999! 99 998 91999-9.9999 3 '982. I '99: Ill. 9993 I99. VII Ill. 9'. 9999 90791 '99. 99- 9981171 99:11?!) 9' If 9' 999 119.99 99.9919 9.9999 9.1999 ”.999 999.9999 999 ..9.99 19.9999 9.9919 9.1919 99.399 111.9999 999 919.99 29.9999 9.9991 9.9979 111.91? 999.999. 999 119.99 (9.9999 9 9999 :.:999 199.919 999.9999 7:9 .19.: :‘.9199 9 9999 9.9199 191.999 999.9999 ’99 119.99 11.1999 9.1999 2.9999 973.991 919.9999 999 119.: 19.9999 0 1919 9.9999 “'-3? 999.9139 999 1.9-: 99.9999 9 1999 9.9991 .19.:9’ 991.9999 '99 . 9.9 09.9999 9 1199 9 9199 999.999 999.9999 I 999 119.99 91.9999 9.1999 9.1719 19’ 99’ 099.9999 .999 1.9.99 99.9119 9.1919 9.9999 993.911 991.9199 Figure 28a. b, and c. Discharge pressure relationship of the center pivot system at the pump for the end gun on, end gun off, and modified end gun off condition. 19. Auntsv 41 fi-~/\ .hs s 76 I] II II T‘T H T‘T ': ”T w 1=__T_ J w m fl/L 0 an n 1 ....l IITIITI DIP- _ n n I....r............. .. I I .m .m. _ _ E a 11-9.1174... 450 --—.--——— ——— Comb O SAHEDIAIJ) THEN J 8 J + 1 I? J > 1 THEN PRINT§14,TAB(12) USINQ_ ' 51.5! lll‘.‘ I"‘."i_ SIZE(J),SAHEDIA(J-1),SAHEDIA(J) END I? END IF LATDIA(I) = SIZE(J)/12 IF I s 1 THEN n PRINTI1N,TAB(12) USING“ l'.ll 0 5.. SIZE(J),SAHEDIA(1) GOTO “5 END IF END IF "5 . NEXT Emu: 1r 122 REHIIIIIIIUIIIUIIIIIIIIIIIIIIIIIIII000.IIIIIIIII...UCICIISIIIIIIIIIlIII REH""""'*'|Reiative Roughness of all the elementseesoneness-essence REHOIIIIIIIIIIIIIIIIIIIIIIIIIIIIH0.000000000000009.CUIUIOIIIIIIIIUIIICC INPUT”Roughness of the pipe in inches (default is 0.0006) s ",8 IF E : 0 THEN E = 0.0000 PRINN1‘1,:FRINTJ111,TAB(12) USING"PIPE ROUGEu'ESS=l.HHHinches“;E FOR I = 1 TO ELEMENT - IF I/2 O IN'NI/Z) THEN EBYDU) : E / (LATDIA(I)'12) END IF ‘ NEXT REHIIIOIIIIIIIIIOIOII0000090000000...'IllIUDOIIIQCCICQIII000000000000... RD!’"'“"I)3CLUDE ms wzwmw AT DIFFEREITI' Pom-s w m FIELD-"u" REflQQOCIIIIIIIIIIDIIOOBBEOOOIIIEIIDCGQIDII9000000099000...0000000000.... 50 INPUT"Is the field assumed to be plain, i.e. negligible slope? YIN “,2: GUTO 7O 60 LOCATE 16,25:lNPUT'Please press 'Y' or 'H' ', 23 7o 117 z: = w on 2: s "y' mm: PRINTO1‘1,TAB(12)'SLOPE OF THE FIELD IS NEGLIGIBLE' FOR I = 1 TO NNODB 2(1) 2 0 NEXT I ELSSIF 28 : 'N' OR 23 = 'n" THEN can. wmmuuonesncmm ) .2( )) ELSE GOTO 60 END IF ' REHIUICIIIIIIIIIIIIIIIIIIIIIIQIII0000000000000IOIIIIIICIDIIIICIICISOIII mounuuueeumma mp AREA op THE RISER sueuuuueuume masseuse"use"useuneasueeeeueuuse"eeeueeeueuuusueeuse INPUf'Diameter of the risers (excluding end sprinkler) in inches: ',__ DIARISER FOR I = 1 TO ELEMENT IF I/2 : HRH/21m AREARISU) = (3.111/11)'((DIARISERI12)‘2) DIARISER( I ) :DIARISER/12 END IF NEXT I FRINTHRJAM 12) USING'RISER DIAMETER : NJ”! inches';DIARISER INPUT'Diameter of the end sprinkler riser: ”,DIARISE:CLS PRIN’II1‘1,TAB( 12) USING__ - "END GUN RISER DIAHE‘I'ER = "J!" inches";DIARISE maximum): (3.111/111'((DIARISE/12)“2) omnissmsuzmm = mums/12 123 RfinlllliillillIIIIIIOIIIOICIIIQIIIIIIIIIIIIIIIIIIIUICIIIIIICOCDIIIIIIIIIII REH"""""""" INITIALIZE THE 93355033 AT THE ”0955 senses-0000000000 I REHIOIIIIIIIIIIIIIIIIAND DISCHARGE FROH THE spaIHXLERSeeIIIsneeeeseneesee 353000Ilsesenasseseeeneeeeeeeeeneeeeeeeseeeeeeeeeeeeeeeeeessenseleseeeeeee 80 HEADLOSS: PSUH-OPPRES:sumlen=0 FOR I z 1 TO ELEMENT IF I/2 : INT(I/21THEN - SUHLEN = SPACIHG(I-1) +SUHLEN X=SURLENILATLEN man-1013111101 IS CALCULATED 1131110 1111-: BETA 01511113011011 rune-11011"!- 015111c1 = 1 - (15/81'11- (2'(X‘3)/3) + ((1‘51/511 9(1) : HEADLOSSGDISTFACT + 099323 If P(1-2)-P(l)=o THEN 9(1) = 9(1-2) - 0.0001 3(1): 9(1)-2.3111 P(I+1) = P(I)-0.5 : H(I+1)=P(I+1)'2.311l 0818(1) = 10(1)'a(1+1)‘2190u(1) END 11 ' was! 1 REHOUIICIDIIICIIIIIIOIII0.000.000IIIIOIICIIIICIIIIIIIOIIIUCDIOIOIIIIIIDII REH"’ ALCUL rLoa N THE LATERAL LINE 10 CALCULATE 131c11on Lossu REHIIIEGDG9GSIEII§§IIIIIlglIlillliliOllliiOOOOIIIOOOODIIIOCOIIIGOCOIIIOII QLAT‘ELEHERT) : 0 FOR J :1 TO ELEMENT IF I/2 <>INT(I/21 THEN IF 1:1 THEN LAT(ELEHEHT-J):QRIS(ELEHENT) ELSE OLAT(ELEHENT-J)=QRIS(ELEHENT-J+1) + OLAT(ELEHENT—J+2) END 1? END IF next J 11 coup = 1 THEN 1? ITER=0 THEN 931n1112,' INITIALISED DISCHARGE” PR1H1112,' ELEMENT DISCHARGE” PRINT, ‘2 , ' ...-D... ----0----“ 103 1:1 10 ELEMENT . IF I/2<>IN’1(I/21 111211 911N1112,usxw0' 111 111.11”;I,QLAT(I) ELSE 931u1112,05130' 111 111.11";1,ans(1) END IF nsxr END 1? END IF 1.. 1a 44’ 124 REHIIIQIIIUIIICIIII!IIIIIIIIIIIIIIIIIIIIII.IIIIIIII'II'IIIDIIIIIICCICIIII. . RD!""""VELOCITY.RSY}IOLD 110531011011 L055 111 THE LATERAL LINE""""" REKIIIODIIII!IIDIIIIIIIIIIIIOOIIIIIIIIDIIOIIIIIIIIIIIIIOIIIIOIIIIIIOIIIII. BER (11.311110 115003111 = 0.00001059 011 11' I sec AT 680st: FAR (20 um c1 mam s 0 90 FORII : 1 T0 element . 11‘ 1/2<>1H1(1/21111E11 - ' AREALAT(I) = (3.11/11-(L11011(11‘21 VELL11(1)=QL11(11/(11118.8IAREALA'1(111 RENO”) = (VELLATU)‘LATDIA(I))/(0.00001OS9) ram-1(1) =1.325 0 (1.0010271112311101 . _ 5.72'((1/RENO(I))‘O.9))‘(-21) 11m = 8'1-‘111c1(1)'srnc11:0(11 / 32.29(1118.8"'21'(3.11159‘21'HL1101M111‘511 ' 1110110n(1):n(1)'((0L1111)1‘21 E1111 1? 1011111c=111101103(1)+10m1¢ NEXT I REE!0.0000000000000000!!!IIIOIIIIII.0000000000IIIII0.0000000000000000... Rani-"11311211141115 cozrrxcmrr 10 BUILD THE GLOBAL 311111711133 11111111301.“ RExIIGOQGQIIDICICICOGIIQIIIIIQIIICOCDQIIIICIQIIQDIQIIIICIIIOII.DCI.I.... REN VALUE OF m FOR THE DARCY'S METHOD 3 2 N = 2 ° 333000000000eeseseeeeeesngrgaglug pxpg c0533Iglgyweeeseeeeeeeeeeeeeeenee 110 1011 1 s 1 1'0 Ema-:11: 11‘ I/2(>IHT(I/2) 1111.31 CP(I) = O 11 1 = 1 111121 was = (2111.110) - 2(2)-H(21)‘((1-11)/11) 0010 120 E1111 11 0EL11n=z(1-1)+a(1-1)-z(1+1)-fl(1+11 IF DELTAh :0 THEN DELTAh : 0.0000001 NUHER = DELTAh‘((1-H)/N) 120 60115? : (11(1))‘(1/11) CP(I) : NUHER/CONST END IF NEXT I 353300000000eeseeeeeDgTEgHINRE SPRINKLER COEFFICIENTeeIIeeneseesneeseeee FOR I :1 TO ELEKENT IF I/2 : INT(I/2) THEN CSF(I) : (Xc(I)) /((H(I+1))‘(1-EXPON(I))) 125 END IF NEXT I REH"""""CALCULATE THE TEE AND ELBow Loss COEFFICIENT KPC""""" KCF : 1 IF FLAGG:0 THEN PRINTI1N,TAB(12) USING”K FOR TEE AND ELBON COMPONENTS : ll.‘l';KCF FLAOG=1 END IF FOR I = 1 TO ELEMENT 11 1/2=1u1(1/21 THEN LS?(1)=(BBECP)/(32.2'(3.1112‘2)’(DIAEISER(1)‘n)-(118.8‘2)) CPC(I)=((H(1)-H(1+1))‘-0.5)/(KSP(1)‘0.51 END IF NEXT 1 REHQINBIIICIIIIICIIIIIIIIIIIIICICIIIIIIIIOICICIIIIIICI00000000000000.0000 REHQOIIIIICIIIIIIIIDEVELOP GLOBAL STIFFHESS HAIR]XIODIIIICIIIOIIIIIIIIII. “EH,00.0.0.0IIIIIIIIOCGIIIQIIBIIIIIIIIIICCIOIIIIIII00000000000000.0000... RONzNNODE COL=NNODE+1 LOCATE 6,23:PRINT'Building global stiffness matrix' LOCATE 7,30:PRINT""Flease Nait'FF' FOR R z 1 T0 RON 0(R) a 0 RC(R.COL)=G(R) OLDRCIR,COL1=H(R1 IF ITER:0 THEN FRSTRC(R,COL)=N(R) FOR C = 1 T0 001 “C(th) 3 0 “C(C1R) 3 0 NEXT 0 NEXT R FOR 1:1 TO ELEMENT IF 1/2 <> INT(I/2) THEN IF 1:1 THEN 30(1,1) = CF(1) RC(1,2) : ~CF(1) RC(2.1) = -CP(1) 86(292) 3 CP(') 6(1)=CP(1)'(Z(21-2(1)1 + 6(1) 0(2):-CP(1)-(2(2)—2(1)) . 0(1) ELSE 126 RC(I-1,I-1): 09(1) + 090(1) + RC(I-1,I-1) RC(I-1,I+1)=-CP(I) — 090(1) + 90(1-1,191) RC(I+1,I-1)=-CP(I) - CPC(I) + RC(I+1,I-1) RC(I+1,I+1): 09(1) 1 090(1) + RC(I+1,I+1) 0(1-1) = 09(:)¢(z(1+1)-2(1-1)) + 0(1-1) 0(1+1) :-CP(I)'(Z(I+1)-Z(I-1)) + 6(I+11 END 19 ELSE RC(I,I) = 090(1) + 90(1,1) 90(1,1+1) =.090(1) +‘RC(I,I+1) RC(I+1,I) =-090(1) + RC(I+1,I1 RC(I+1,I+1)= CPC(I) + CSF(I) +RC(I+1,I+1) END 19 NEXT 1 LOCATE 6,23:PRINT' LOCATE 7,30:PRINT' ' mFFFFFFF'F'F'INCLUDE THE BOUNDARY CONDITION IN THE NEWORK'F'F'F'F”.. 3011,1131 RC11,COL)=HSUN FOR I = 2 T0 Rm 35(1111 3 0 I REHIII0000.0IIIIIIIIIIIIIIIIIIIIIIIIIIIIIOI'IIUIIIIIIIIII.0000000000.... """"" ELININATION METHOD TO SOLVE 909 HEAD"'*""""" 2::IIIIOIIIII22N§§::NIOII!!!IIIIIilillllllilfllflfiillliIDIICIOQIIIIOODIICO 19 ITER = 0 THEN FOR 1:0 10 ITR . LOCATE 13+I,31:PRINT' 1 . . END IF LOCATE 11,11:PRIN'I"1!0 OF EQUATIONS ';RON LOCATE 11,117:PRIN'I"SOLVING ECUATIW " LOCATE 13+ITER,31:print'ITERATION N0 ';iter LOCATE 10,30:PRINT""Please Uait"" 19 ITER = 0 THEN I9 CHANGE = 1 THEN LOCATE 13.68:9RINT-END GUN 09:88“ LOCATE 1A 68:?RINT'RUN NO ; 19 CHANCE=1 AND SFLG<>1 THEN LOCATE 15,1:PRINT”RUN COMPLETE" ELSE LOCATE 13,1:PRINT”END GUN 011' LOCATE 11,1:931NT-AUN NO ”:NORUN END 19 END 19 127 FOR L = 1 TO RON LOCATE 11,6“:FRINT L FF:RC(L,L) FOR J a 1 TO COL ' RCIL,J)=RC(L,J)/FF NEXT J FOR K = 1 TO RON ’ FF:RC(N,L) IF L-K=0 OR PFC-0 THEN 0010 130 FOR J:1 T0 COL ' RC‘K,J) s RC(E,J) - FF'RCIL,J) NEXT J 130 NEXT X NEXT L ITR = ITER REHIIDDIII00000000000000.0000!IllllIDQIIDIIIDIDIIIDDDIIDI0000000000000... REM""""'1TERATION NO AND 19 CONDITION 909 NONE ITERATIONS”""""*' RE“.000090000!III!III...IIOICDIIIIIIIIDIIIDOID000000000000000000.00.00... ITERFLAG s 1 FOR I : 1 T0 RON IF AESIRC(I,COL)-0LDRC(I,COL))>ALLERR THEN ITERFLAG 8 0 END IF NEXT FOR I s 1 TO RUN IF ITERFLAG s 0 THEN OLDRC‘I,OOI) : RC(I,¢OI) END IF NEXT I REHIIDCIIIDII000IIIIIIIIIIIIDIIICIIIDDIIIIIIDIIIIDIIIIIII0000.00.00.00... REH"""""CALCULATION 09 DISCHARGE 990a THE SPRIHKLERS*"""""""' REHIIIIDIID.IDIIIIQIIDIIDOIIIIIIDIIIIIIO|III.I...DIIIIIIOIIOIIDIOIIIIIICI FOR 1:1 T0 RON HIII=RC(I.COL) F(I):H(I)/2.311l IF I/2=INT(I/2)THEN QRIS(I):KC(I)‘(H(I)‘EXPON(I)) END IF NEXT I REH""""’CALCULATION O9 DISCHARGE IN THE LATERAL ELEHENT""""'"" QLAT(RON)=O 909 1:1 TO ELEMENT x : ELEMENT-1+1 19 1/2 <> INT(K/2) THEN OLAT(I) : QRIS(K+1)+OLAT(K+2) END 19 NEXT 1 128 ITER=ITER+1 IF ITERFLAG = 0 THEN OOTO 90 FOR I = 1 TO RUN IF I/2:INT(I/2) THEN QI I):QRIS( I) ELSE GUI-TOUT“) END IF NEXT I REHIIDIIIDIIIOIDCDIDIOIOIII000000000000.0000.00.00.00...III000000.000... REH”“PRIHT PRESSURE AT THE HODES AND DISCHARGE 99011 THE S9NINxLEEs-"u REHQQIIOQGSIIIOOEOIIIDOIIIIOQGIillilllllbillBIO!IIIIIIDIIODIDOIIOIDIIIII 19 FIRST : 0 111921 9111191112,'TOTAL ITEAATION = ';ITER:Drlnt"2. 9111111112, chr$(15); ‘ 9111111112 - HEAD(9T) 9911:1112I'INODE INITIAL 91NAL D199EI1NECE 9111111112," NO (11:) (Psi) (ft) (031) (ft) 1 91111111123 - - -- —— FOR I: 1 T0 Rm D199:1OO9((9115T110(1,00L)-H(I11/993190(I.COL11 9111111112,US1NC __ 9111 111.11 111.111 111.111 111.11 11.11 11.1 1,99STEC(I,COL),rratrc(1,coi)/2.311A,H(1),b(1)/2.31111,DI99,_ (diff'100)!(frstrc(i,col)),I,0(I) NET: 1 REHIIQIDI.I'DDIICDIIIIIIIIIIIICIOOIIDIIIII000000000000000000000000.0000... """RR RESULTS IN THE DATA 91LE TO BE USED 9011 IRRIGATION""'"“ EESIIIIOIII:25!IDUOIIGDOIPERFORHARCE EVALUATIogOIOIIDOOICIIIOIIDIOIIIIIIOI 9011 1:1 10 ELEAENT - 9HINT11O,USINo-11,111.111,1111.111';1,9(1);N(I1 19 1/2:1111(1/2)THEN I9LO:19L0(1) :0:01113(11 CALL INTEHPOLATIOINQSPM ),RADCOV( 1.1,I9L0,0,SOLU) 991m:1O,USIN0'111.111,111.111';0111$(11,SOLU ELSE I 9HIN111O,USIN0'1111 .111,111.11';0LAT(1),S9A01N0(1) END 19 NEXT END 19 mauuneuuueuse THE SUMMARY op m mum-3000000000000"senescence IF COMP : 1 TI'EN FRINTF12 ”EL LATDIK AREA QLAT VELxITY REYNOLD FRIC(f) R H L053 TLQSS' PRIN'IF12,__ 11 9011 1:1 TO ELEMENT 9111NT112, USINO_ 129 ~11 11.11 111.11 1111.1 11.11 11111111 1.1111 11.1 11.11111 11.11 112. 1,LATDIMI),AHEALAT11),0LAT(1),VELLATU),HENOU),991011I).R(1),FRICTIm,- 10T9111c 119111- 9111111112, :911111'1112, :911111'1112, 9111191112," 11009 1191101111 1191101951) ELEV 11(1) LA'X'DIA" 9111111112;- ---- --. ,1- ...... 11 9011 1 :1 10 101009 «9111111112, 11311111 "111 111.1111'1'1 -111.11111 11.11 11.1111111 11.11';_ I.H(I),H(1)/2.3111,2(1),a(1),LATDIA11) 119.111 9111111112,:911111'1112,:911111'1'112, 9111111112,'91.EH 11009: 0130 91110 CSP 09c 09 .1 911111T112,- ...... --.. -... --- --- -..- 11 9011 1: 1 T0 1111009: 19 1:1 THEE 9111111110,usu10__ ~ 11 11 11 111.1 11.11 1111.111 1111.111 111.1 111'}. 1,1,2,0LAT(1),91110'11011(1).09c11),CP(1).11(1).H(2) 19 m THE! 19 112:111111/21 THEH 911111-1112,usx110_ ‘ " 11 11 11 111.1 1111.111 1111.111 111.1 111 . I,I,I+1,QRIS(I),CSP(I),CPC(I),H(I),H(I+1) ELSE 9111111112,1131110_ .. '1 11 11 11 111.1 11.11 1111.111 1111.111 111.1 111 ,~. 1-,1-1,1+1,01.1T(1),91110T101111),c9cu).09111,H11-1),H11+2) 1:110 19 E110 19 11m 1 E110 19 19 11111111:1 THEH 0010 110 19 0990mm 111101 9110910:0111$(ELE1m) IF CHANCE=1 THEN ENDFLO=0 ITERR(NORUH)=ITER : PSUER(2,NORUN)=PSUH : HSUHR(2,NORUH)=PSUH’2.311N PBJDR(L‘ORUEJ)=P(ELB€E2H‘) : 0E11D111110111111):Er1091.0 : 1193109111109.011):9E11D11(110111111_)92.31111 : 0311111112,H0111111)=0LAT(1) : 9D1109111110111111):9S11H-9(EL911E111) : 110110911111011m1):90110911(11011U11)'2.31111 0901091 110111111) : 1009(0911011111011011) IQSUHmz ,1101111111) RE”!!!IIIII!II.QIOIIDIiiI'lfiilCIIOIICIIIIIICCIIOCCIIIIDIIIII'IIIIIOCIIIII MIIICIIIIIIODIFFEW CONDITION OF 919555033 AND FLQQIOIICIIIIIIIIIIIDOI mliilltlilIlllillllllill FOR m RwllilllilllillllllliiICIIIIIIIIIIUI 1210 mung-Y- 19 11011011:11111111 A110 (CHANGE=1 011 CHANGE:2) THEN 110(10):"? 19 1111;111:011 then ENDS$(1)=”Y' 150 130 19 (NORUH=RURH AHD CHANCE:0) on CHANCE:1 THEN 9N05111)=“N' 19 CHANCE : 2 THEH EHDS$(1)=”Y' 19 CHANCE : 0 AND ENDS$(1)="N' THEN 0010 160 11011011 : 110111111 1 1 ITER30 9SUH1:9SUN+5 099HES:u+9(ELENEH1) 09HEAD:09PHES¢2.3111:91H00091:099HES:H(ELEHEN11:09HEAD 9SUH : 9SUH1 :HSUH: 9SUH92.3111 :9(1)=PSUH:H(1)=HSUH 19 CHANCE:1 OR 09900N : 1 THEN NHODE: NNOD£+2 ELEHEHT : ELEHEHT+2 LATLEB : LATLEH + SPACISG1ELEHEHT-1) END 19' 19 ENDS$(1):'Y' THEH 0010 80 LATLEH : LATLEH - S9ACIHC(ELEHEN1-1) 19 09900N:0 THEN HORUHz1 9SUH = 99 :HSUH: 9SUH°2.3111 :911)=9SUH:N111=HSUH 099HES:09:09HEA0:09992592.311u:H(ELEHEHT)=09HEAD 9(ELEHEH1):0990ES END 19 ELEHE31 : ELENENT - 2 NxonE : NN00E - 2 CHARGE 8 1 09900N : 1 0000 80 ELSE RE“.IIIIIQIOIICIOI!|IOIOIIUOIOQIIIIIIIIIRUIfllllflllillili...".Ililmu IF RUNH:1 THEN GOTO 170 160 19L0012):N0900:00HH112)=0:xccc12)=0:9199(2)=0 CALL S911HKLERC0999_ (9SUHH1).0$UNR(),DUNN1(1,2,19L00().xccc().EX99( 1.2) xccc:xCCC(2) :9199:9199(2) 931N1111,:9H1N1111,:911N1111. 19 CHAH : R CHANCE:2 THEN 91111111 _ GE 00 . ' 911001111011- 19 CHANGE : 1 THEN 901N1111 . ”' END GUN 099- 9HIN1111, . 91111111111,TAE(12)_ .. "111111 Pplvot Hp1vo: Pend flend‘ Pdrop H6909 09111011 06M 1- 91111111 11.11101 12)_ .. ”NO (951) (91) (931) (91) (931) (91) (0911) (6910-- 91111111111,1AE(12)_ ., I--_ ............ ---- ..... —— — ----.- --.-a- P 90111 1101101101 911N1111,1AE(12)US1N0 "Ill 111.11111.11 17.1 111.1 11.11 11.11 1111.1 111.1'}- 131 I.PSUHR(2,1),HSUHR(2,I),9EN0H(1),HENDH1I),90809911),HDHOPH11).QSUHR(2,1),_ QEHDR(I),QEBYQP(NORUH),ITERR(I) NEET I 901N111n,991N1111,1A0(12) ~THE 0 VE’H HELATIONSH19 AT THE 91101 IS GIVEN 01' 901N1111,TAE(25) USINO_ A ' 0:111.111(H1-1.111111)';xccc,EX99 170 19 CHANCE:0 OR CHAHCE:2 THEN LOCATE 15,1:PRIHT'RUN COMPLETE“ 19 CHAHCE:11HEN LOCATE 15,68:PRINT”RUN COMPLETE“ I? INK$(1) :"y' or IHK$(1):"Y” then 0070 150 STOP:END _ END 19 SUB DIFFSPRINXLER; (PLAT(2),RADCOV(2).QSPR(2),SPACIHG(1),CHEKSPAC,ELEHEHT,IFLG(1)) _ fifiulllliillIOIOIIIIIIIIIICQQIIOGICIIIIIIIIIIIOIICOIIIIOIOIIOIIIIIIIIIIII REHIOOCIIIIIIIOC SUB DIFFSPRIHKLER‘NfiODE) IOICQIIIIIIOBIIIOIIIIIIII Rfinilfllilll.IDIIIIIDOIIIIDIIIIIOIIBBOOIDIIIOIlllllIllilliiiliiillilllllfl DIN FLAG(30) COLUH=1£FLAG=2 REM...I'll...II.CIIIIIOOIIIIDIIIIICIIQIIIIII'IIIIDIIIIIIIIIIIIIIICIIOII 11E1-1 1119111 951, 0911, A110 RADIUS 09 COVERAGE 911011 1111: 91LE REHIIGOCOOOQIIIIQOllfliiilllllilliII...III!.IIIIII!GIICIOOIIIOIOOIIIIIII 901 (:1 TO ELENENT ' 19 K/2 = 101(112) THE! =FLflG SELECT CASE COLUN CASE 1 9031=2:COLUN=2 CASE 2 POSI:23:COLUH=3 CASE 3 < 90$1=56=COLUN=1 END SELECT LOCATE 9OSH,9001:L9HINT LOCATE POSH+1,POSI:LPRINT LOCATE POSN+2,POSI:LPRINT,’SPRINKLER N0 “:112 LOCATE POSN+3,POSI:LPRINT,'P(p31) R(ft) 0(gpn1' 180 901 1:1 10 100 INPUT 19,9LAT1K,I) 19 PLAT(K,I):999 THEN 19L0(x):1-1:1:100:0010 190 1N9UT 19,0ADCOV(K,I),OSPH(K,1) LOCATE POSH+1,POSI: 9HINT111,USIN0_ "111 111.1 111';PLAT(K,I),RADCOV(K,I),QSPR(K,I) 9HINT USINC "111 111.71 111~;9LAT(H,1),0399(1,1),HA0cov(x,11 POSN=POSN+1 190 NEXT I 132 END 19 19 COLUH:1 THEN FLAG=POSN+2 NEXT I EON 1:1 10 ELEMENT 19 1/2 <>INT(1/2)TNEN IN9UT 19,39AC1N011) END 19 NEXT 1 INPUT 19,0NEES9AC END SUD SUB mm3m161EXSpae,Lam) nggueeeeeeeeeaalaeeeea10110000010100000090100111101000000000000000000000! REM”"'*°'*'CHECK IF THE TOTAL LEKSTH IS EQUAL To SUM OF ELEHZHT LEBGTH" 35500001000011.1101101110.001011011110000110100111000aces-10001100000000. PRIHT'The total length Of the lateral does not match with the sumaatlon' PRIHT'Of the distances given above. Input the spacing Of the sprinklers” PRIHT“Calculated length Of-the lateral line = ';1atlen PRIHT”uaer input suzzatlon Of the lateral eleaenta = ';chekspac PRIHT'PROGRAH TERHIHATED':STOP:END END SUD SUB IHTERPOLATIGN(PLAT(2),VARI(2),I,IFLG,H,SOLD) REHDOOIICIIliilallieilllillllliOIIIODIICICDCIIIIIIIIIUCIDIICIIUIIIIOIIIII HEN-9'19!!!itINTEHPOLATION 09 0,H,AN0 RADIUS 09 CURVATURE'"""""""' REHOIIIIIICCIICIIIEUII’IIIIIIIIIIIIICICIIIOIICCOIIIIMIIIDIIIIIICIIIIINI. EXACT = O:LFLG=0:HFLG=O FOR J = 1 TO IFLG IF PLAT(I,J)O then SOLU : VARI(I,EXACT) ELSE SOLU =(((VARI(I,hflg)-VARI(I,lflg))'(H-plat(1,1f18)))[_ (plat1l,hflg)-plat(l,lflg))) + VARI(I,ltlg) END 19 END sun 133 R .. , 2 ,SPACINGU),ELEHBU,IFLG(1),KC(1),EXP'0N(1),FLA :gsfsfiyfifilgfifggzfizIggczuuuuuluuuuuunuuuuuuuuuue) REH'""""DEFINING 011 CALCULATIHC 591111111 THEN LATSUHso FOR I = 1 To stuzur IE I/2<>1NT(1/2)THEI 1E FLAG = 1 THEE CLS:PRIHT”D13tance from pivot Elevation“ PRINT” (feet) (feet)' . PRINT'- --------- FLAG = O :POSN : l ELSE LATSUH : SPACING(1) END IF LOCATE POSN,6:PRINT LATSUH:LOCATE POSH,30:INPUT,e1ev(I) 905» = POSN+I : LATSUH:LATSUM+SPACING(I) ELEV(1.11=ELEV(I) 1E 903» = 25 on P08" :50 THEN ELAC=1:COTD 200 END IF 200 NEXT I END SUB .135 SUB PipeDiameter(size(1),:anedia(1)) . REM-00000000000:-00000000000000:100aIDDOOIODICDDDOEIDDOIOIDI00000000000. mIIIQIlIIIIIIIIlIIINPU-r THE 01mg op THE p1pgiltlllllllililllllllll mllaiillllillliiIIIOIIIIIIIIIIIIIQEIIIIOIIIOICIIIIIIIIIIIIIIIIIIIIOIII CLS - INPUT'HOH many sizes of pipe are there in the lateral line? ”,aizefl PQSI = 6 - Print'Pipe NO. Size in inches Distance from Pivot in feet ' PRINT" .- START DID“ PRINT' -. _ PRINT:FOR 1 = 1 T0 sumo LOCATE 9031.1:911113 11031-5 LOCATE PCSI,17:INPUT,SIZE(I) IF 1:1 THE)! LOCATE POSI,36:PRIHT 0:1.OCATE 11051.52:1111>UT,3A11ED1A(1) ELSEIF IN THE" LOCATE POSI,36:PRIHT SMEDIMI-‘IthATE POSI,52 INPW,SA}EDII( I) END IF P031 = 9031 + 1 NEXT I SIZEHO=2 SIZE(1)=5.79:SAMZDIM1)=765 SIZE(2)=2.73:SAi-EEDIA(2):781 END SUB APPENDIX B OUTPUT RESULTS FROM THE SIMULATION MODEL ”NE'I'HORK" Case 1 : end gun on YCVQL 17:33710N 0 O ILLDUIILI ERIC. U .3 '7 “CA31'71 “DOC 3N1T1QL1333 ‘1NNL 31FFIRN‘C3 '0 Ht) 1.0.1 H11 10011 Ht) 3 1 303.10 33.000 203.603 33.00 0.00 0.0 3 301.53 37.330 303.03. 37.33 '0.33 -°.1 3 300.67 35.730 303.061 37.83 '0.7. -0.0 ‘ 1”.” 35.131 3%.?” “.33 $.17 -O.3 3 1”.b7 '33.”1 330.717 35.“ '1.” '0.3 O 1'3.°‘ €3.57. 1’Q.C$3 36.33 .0.70 -0.0 ’ 1".” 33.17. 1”.” “02. ‘10:, *0. 3 193.33 30.'3° 1.3.133 33.73 '0... -0.3 . 1.3.13 I0.020 1.7.... 33... '1." '0.. 13 193.33 3‘.°7. 1.6.... 33.09 .1.39 '0.. 13 1'30" '30:" 1...“ “0” -10’0 “0.. A: 1”.“ 83.5 1”.“ .‘08‘ -1.‘3 ..., 1: 1.3033 .205 1”. 1” .‘0‘3 -30.. -... 1‘ 3'10“ 33.537 1...1" ..001 ‘10“ .00. 1, 1".” .1033, 1'30'1’ 33.5 .10“ -101 1. 19033 .10”, 112.9c3 “O‘Q -10” -100 1’ 1‘3.” .30‘.” 113.3” 33.23 -2023 ‘10: 1. 1‘3003 .10”. 1.1.027 ‘20” -2002 .101 1. 1“.” .00“. 191.233 a.” -30: .10: 3° 13...3 80.73. 136.71. 33.31 03.1. 01.2 :3 1::0‘. 3002.. 1.9..” 33.13 -203, -103 23 1C3.16 83.0.. 1E’.303 31.9. .3.3b '1.3 33 1:1.” ”.3” 193.353 '30‘. '20:, .10: 2. 1n.” 0:” 11:30:32 .30: -20.. .10‘ a 1:20” ”.Gn 16703.3: .30” -10.: -10; 2. 162." ”.65, 1:10:37 e3. 1‘ -20” .10. :7 191.33 .537 161.37‘ €0.59 -3.33 '1.‘ 33 191.33 73.6.3 185.07. €0.36 -3... '1.3 3' 1:90.: ”03.. 1:30”. ”02° -20“ -10, 30 331903 ”.2“ 1‘30“” 93.50 -207. -10, :3 1”.” ”.7“ 163.973 ”.39 -203. .1.3 n 1”.” ”...: 1‘.o..7 ”.03 .20” .10. a 17'. fl "0 3‘3 1133.”. ”0.7 -10 33 .10 3 3‘ 1”0:3 77.333 13302" ”.73 -20“ .10. 33 173.0. 77.033 133.671 ".77 -3.33 01.3 3. 1’3.” ”.271 133.033 ”0" -20” .10. 37 1’70‘3 70.771 131.“ ’.O” -102: -10, 3. 173.0. ".030 133.“. ”02° -20” .10. 39 1".” ’3033. 1”.” 7.033 ‘30“ 01.3 ‘0 Afio.’ 7‘0“. 1.2091 "0” '20,: -10: ‘1 175.33 70.359 17'.633 77.74 01.79 '1.0 .2 1".” 3..’" 133.0113 "o’. -30.. .1.: ‘3 170.13 76.303 173.331 77.13 '1.33 ‘0.’ All 177.00 11.009 101.570 ".00 4.01 -1.! 0: 113.91 70.10: "In“ ".00 .1." «.0 .. 130.90 71.233 101.837 79.00 ~2.:2 -1.0 ‘7 173.7. ’b.033 177.0I7 I..33 -1.0. -0.0 a 1".” 75.60. 131.” a: .20 ‘: .10. 9, 173... 73.9.. 177.77. 7b.7‘ ~0." o... :0 170.7: 70.¢‘1 100.001 70.20 4.3: ...; n 1’..’1 ’30.” 1”."3 73.19 -2.” -1.’ 33 173.33 ’3.'33 17b.231 76.37 -0.01 -0.3 3‘ 175.71 75.090 17b..70 73.3. '0.13 ‘0.1 33 1’3.” 73.990 172.010 '7‘.“ 1.07 1.0 :0 :70. 71 ‘1. 030 170.020 70.01 o. o: o. o 37 173.33 73.330 171.30‘ 70.11 3.3. 1.! '36 gugun 311::azaazsazaéutuueasuafiuaun z ELI!!!” In 0. Q0130“...- 'LOU I33.3 3.0 333.3 3.7 313.3 618 313.0 7.3 .0... 3.3 7.3.3 ... ’3... 110’ 77b.’ 13.3 753.3 13.1 7‘I.1 13.0 730.1 31.3 70... Z3.‘ 0.... 33.3 ..‘0' 33.3 b‘1.’ 37.1 O1‘.. . .3 333.1 3... 333.3 39.6 330.3 11.3 0..., 13.3 ‘00.. 33.7 133.3 13.3 ’3... 13.3 330.. 13.. I11v' 13.. 13.. 337.1 ‘3.0 133.1 1.301 137 Case 1 : end gun on 0. U70“ 0.0 3.07 dun!" M 007671.! 0 .1 m0 707 .1 ...“ '7' 00 0'1 .... «10001 1.0 1.0-10"” 1"" '7‘ - _ .— ; 0.00 3.10 070.70 10.00 000000 0.010000 0.000000 1.88 1.73 7 0.00 3.10 072.10 10.0: 000707 0.010000 0.0000000 1.30 7.00 7 0.00 7.10 010.00 0.00 00477: 0.010037 0.0000000 1.80 0.07 7 3.00 7..0 013.00 0.0! 152200 0.1110007 0.0000000 :. 7.: 0 0.00 3.10 000.01 0.00 0001-00 0.010001 0.000003: 1.00 0.07 :1 0.00 3.10 700.10 0.7: 003030 0.1110111 0.0000010 1.30 0.11 1: 0.00 0.10 700.00 0.01 6:010 0.010723 0.0000090 1.: 0.:0 : 0.00 0.10 7'70.” 0.07 0315:: 0.910700 0.0003310 1.31 10.37 17 0.00 0.13 703.2 0.31 020.130 0.0107” 3.W.:0 1.17 11.70 10 0.00 0.10 700.10 0.13 0121.33 0.1110010 0.0001120 1.1: 17.07 31 0.03 0.13 730.70 0.00 002-337 0.010600 0.000002! 1.” 10.13 :7 0.00 3.10 700.03 0.00 3031770 0.010013 0.0000000 1.00 :0.“ a 0.00 0.10 .00.” 0:31 :01000 0.01007: 0.0000000 0.00 10.00 27 0.03 0.13 000.19 0.10 3.10:: 0.01233! 0.m 0.” 10.00 2' 0.03 0.13 001.70 7.” $37241 0.015103 0.0-6:071:21 0.6 17.33 :1 0.00 0.10 010.00 7.00 341210 0.010.100 0.0039020 0.00 10.77 8 0.03 0.13 33.13 7.10 23120 0.017."? 3.0063511 0.71 10.00 :0 0.00 0.10 $3.07 0.70 30:71.3 0.01610 0.003122: 0.00 70.00 7'7 0.00 0.10 325.31 ...: :03. 0.01: ‘11 0.01am 0.37 30.07 3' 3.03 0.13 000.73 0.03 ”0010 0.017505 0.00-2‘ 3.8 31.1. 03 3.03 0.13 122.21 3.13 330.40 0.016’1‘7 3.62.1212! 3.3 §.13 0.0 0.03 0.13 ”.01 0.73 710:2: 0.0164? 0.052223 3.33 8.00 a 0.00 0.10 $30.00 0.27 100700 3.31“?) 0.02:2: 0.20 3.77 0.00 0.10 311.70 3.00 173110 0.010770 0.03233 0.8 10.00 31 0.03 0.13 200.01 1.3. 100778 0.017127 0.0031102: 0.17 3.11 33 0.33 0.00 87.00 1:." 20.2037 0.01000: 0. ‘7 0.8 77.00 a 0.2: 0.00 1H. 10 0.“ 210370 0.017070 0.0000110 0.01 . .33 up .0 07‘ '07: e, 93 9 13.0001 Q 1. 3 a w“ 1' U -— - -— — — - .— 1 1 3 023.7 1.3 0%.“ 703.0 703.1 3 7 3 3.0 0.010 110.- 700.1 000.1 3 : O 08.3 1.. 011.-7 703.1 000.0 0 0 0 3.7 0.010 03.371 200.0 ”.7 0 0 0 070.0 1.00 010.000 100.0 100.0 3 O 3 .0. ‘0‘ .30“ 3'0. 3'0. 7 0 0 010.0 1.8 017.08 100.0 I'd 0 0 0 7.0 0.60 03.000 100.1 1‘0 0 0 10 00.0 1.00 003.0” 100.1 100.7 10 10 11 0.0 0.0“ 03.‘ 100.7 100.3 11 10 1: ”0.7 1.. 0..- 100.7 103.0 11 17 17 7.0 0.- 8.000 173.0 703.3 17 10 10 0 1.. 0..“ 100.0 100.3 10 10 II 11.7 0... 8.- 100.3 103.3 10 10 1. 770.7 1.31 000.000 100.0 103.0 10 10 17 13.0 0.070 I.” 10.0 107.3 17 10 13 703.3 1.17 030.307 173.3 101.3 10 10 10 10.1 0.070 8.870 101.0 101.3 10 10 70 700.1 1.10 000.100 101.0 100.7 70 70 31 10.0 0.” 71.3” 100.7 100.0 33 3. a ~03 1.” ”,0” 1‘0, 1.0. = :3 7: ’30, .033: 1.0”. 1'0. ...: :3 =1 20 7... 1.03 0..- 130.3 1‘3 70 70 a 3.0 0.710 17.311 100.0 107.2 33 3‘ 30 *0. 30.. '1'.” 1.0. 3.’0. '0 :0 :7 3. 0.100 17.83 107.0 100.8 17 20 70 000.: 0.00 700.000 107.0 100.7 a 8 7'0 3.0 0.181 17.70! 100.7 1&0 70 :0 30 001.7 0.. 700.53 100.7 16.3 70 30 71 17.1 0.100 10.08 1&0 100.0 31 13 33 010.0 3.. 000.” 1&3 I... a :3 a 79.0 0.101 13.80 185.0 1fl.0 :7 :2 :0 7a.! 0.71 ”.03 18.0 100.0 30 30 a 77.0 0.101 13.811 100.: 16.1 :3 70 30 ”.7 0.00 000.8 100.0 103.0 10 7.0 77 70.0 0.101 13.83 103.0 101.0 77 30 :0 010.: 0.00 011.370 16.0 103.1 30 :0 70 31.0 0.170 17.000 10.1 10.0 70 70 00 000.7 0.. 00.000 103.1 103.0 00 00 01 3.0 0.107 :1.003 '. .0 177.’ H 00 07 1.0.0 3.03 000.000 1&0 107.1 93 ‘7 07 ' .7 0.710 10.50 103.1 170.: 47 07 00 073.7 3.80 1107.!!! : .1 101.7 .. 00 0s 3. 0.100 11.07! 101.7 170.9 0! 00 00 :00.0 :4: 1171.003 101.7 101.0 00 00 0’ 30.0 0.31: 10.370 :01.0 177.0 0’ 00 00 230.0 0.20 107.077 :01.0 101.. 00 03 0' 70.0 a.::: “7.230 :01.: 37.0 .0 00 0‘7 711.0 0.3: 2033.103 101.1 :00.0 79 70 0! 07.0 :.:3.' 0.010 .00.0 170.0 71 7‘: 3 :0... 0.17 007.00? 100.0 100.7 1.1 72 2 07.0 3.37 0.000 100.7 170.8 0‘ 2: 70 777.1 0 :2 20.2” 100.7 :"'. 20 70 70 10.1 0.01 220.330 97.0 170.0 20 :0 2‘ :00.1 ..00: 30.007 1.70.0 171.0 27 01 :0 :0 3.00 0.30 170.0 171.0 Case a I. -0 138 end gun off, :umo 'J' ’07". {MAY’D‘ O S W 1.101101 0 .7 0'7 1000 ° 0100“ 400 170171“. #1100 11m w run 4 1111 10011 ‘ 1011 10011 1111 2 an 1.... — ‘ _ _ 1 30.20 110.000 200.200 110.00 0.00 0.0 1 7:9,; 2 :30 0. 109.” 50203 107.33 ”00‘ '00. z 3.. 3 30” 10.0“ 233.17: 107.3 *0“ *0: 3 m.‘ 7 2.50.73 1041.501 £22.07? 107.07 4.07 4.7 7 73.7 0 2.1.00 100.577 2.31.100 1%.“ 4.00 4.0 0 7.1 7 207.07 100.073 251.027 100.00 4.07 4.7 7 717.7 0 200.00 100.13: 23.007 100.10 4.¢ 4.0 0 7.0 0 210.73 107.037 207.073 10.1.11 4.07 4.7 7 710.3 10 217.77 107.020 218.717 107.07 4.00 4.0 10 7.3 11 217.00 107.120 207.030 107.77 4.07 4.7 11 7N.1 12 207.70 107.100 207.720 107.20 4.07 4.0 17 11.0 17 217.01 100.202 200.710 100.0: 4.87 4.1 17 070.0 10 200.03 100.330 200.050 100.0: 4.00 4.0 10 10.2 17 200.00 103.830 203.67 100.17 4.71 4.1 17 000.7 13 200.70 105.707 219.100 100.07 4.07 4.0 13 17.1 17 203.73 108.007 200.002 103.70 4.23 4.1 17 003.0 20 200.07 100.000 200.327 100.71 4.10 4.0 20 20.0 21 202.07 103.102 203.230 100.20 4.17 4.1 21 027.7 22 203. 10 103.211 203. 020 100.32 '4.10 4.0 20 3.7 27 202.03 100.711 201.702 100.03 0.07 0.0 27 777.7 20 202.00 100.372 202.071 100.00 4.10 4.0 20 3.0 27 211.27 104.372 241.000 100.27 0.10 0.0 a: 777.7 20 201.77 100.070 202.01! 100.70 4.10 4.0 20 3.0 27 200.01 100.000 200.77: 100.07 0.10 0.0 27 700.0 23 201.17 100.30 201.771 100.0 4.10 4.0 :0 8.0 27 227.77 103.030 237.720 107.71 0.11 0.0 27 ”.7 70 200.07 100.050 200.022 104.17 4.10 4.0 70 9.0 -31 237.07 102.787 222.077 107.13 0.00 0.7 31 077.0 32 200.02 103.007 200.207 . 103.70 4.07 4.0 :2 87.7 33 235.07 107.307 237.023 102.72 0.00 0.7 8 0..7 30 237.07 107.007 237.770 100.7! 4.00 4.0 30 3.0 10 230.“ 103.107 30.003 102.03 0.01 0.3 :7 120.1 :0 237.3 103.073 237.011 103.30 4.“ 4.0 30 3.0 :7 227.07 102.770 210.001 132.20 0.02 0.7 :7 371.7 :0 237.01 :03.300 230.000 :03.00 4.07 4.0 :7 3.7 37 2:7. 70 102.000 23.082 102.00 0.77 0.7 37 3.3 00 237.00 103.230 222.00: 103.32 4.00 4.0 00 33.0 ‘2 237.60 107.177 237.007 107.20 4.07 4.0 ‘2 “.3 «13 237.20 102.057 233.707 201.1: 1.07 0.0 0: 272.0 u 23.27 :03.000 30.377 107.13 4.00 4.0 00 10.7 03 237.10 102.777 220. 700 101.20 0.07 0.0 07 270.0 00 230.10 103.000 230.200 103.00 4.02 4.0 10 “-3 41 237.0: :cz.::0 203.000 1:11.02 1.00 0.0 07 190.1 4.0 7.10.1: 103.02: 230.101 113.30 4.0: -0.0 10 “.3 49 110.07 102.011 203.000 .00." 1.00 0.0 u 149-9 10 110.00 103.007 30.107 103.01 4.01 4.0 00 00.0 :1 110.93 132.007 2 ..130 100.:0 1.00 0.0 01 77.3 12 173.00 103.001 270.001 101.00 1... 0.0 12 10-1 01 170.12 102.001 112.013 1:0.0: 1.1a 0.0 13 . 17.7 :0 030.12 10:.000 :. .107 132.:0 4.3 4.1 '39 Case 2: end gun off, pump 'J' 'L 5".“ W0 1.” out" w fltfltfl . G an ’0? fl '0 4'7! 1. au- 0". "in. q lac-rue” In .90, n 3.“ s... '30... 0.9: «m 0.9"... mm he. I.“ Q a... a.“ '7.” 0.01 «um 0.0“... mm x.” ..u . a.” ,3.“ '3." Lll «nan 0.0“". mm 8.00 ..n 4 0.“ a." ".7.“ OJ: :vrrn man-m 0.000092. 3.“ «u I 0... :.II ‘9'... 0.» :nao 0.0“": mm hi, 2.» ;: a... a." '00.“ 0.30 am" 0.93.03. um t.“ o.» :3 a... a." on.“ I.“ mm. 0.93““ mm 0." 0.37 33 .0“ .0 .. On.” '0'. m‘” 0.0|2$OO .0m .0” .03. :7 0.00 a... no.” I... am: memo-u 0.0m ...! 0.20 I' 0.00 0.1. «3.0: 7.6 0.019000 0.000052: 0.. 30.0. : O.“ 0. II All.” 7.00 “G's! 0.0!!!“ mm 0.0! to... :3 o..- o.|I ”.31 7.8: =21“ mots?“ mm: 0.1! u.” a 0.00 0.10 :73.” too urn. 0.015329 0.00m: 0.00 £3.00 :7 a... 0.1. $00.00 ...9 ma 0.0130” mm: ...: 11.0! 29 .0“ ‘0 '. 223."! 0.3. m... 0.033383 .0M‘ .0. ‘30.‘ at o... a.“ on.» no 37209 0.032;! mm 0.” u.” n 0.40 0.10 0:3." 3.99 2333! 0.031317 mm 0.0 no.“ a 0.00 0.5. 013.07 3.30 nun «new: um I.” 13.00 37 0.0. O.“ ”I.“ 0.” ”nu 0.0“!” mm 0.8. 88.. :9 0.00 0.3. 333.9 0.3: true-o menus: 0.000%: 0.. u... 0| a.“ on. 315.0: 1.“ "was mourn mm 0.8. a.” I! o.“ 9.30 271.6 3.3: [31“. 0.01712: mm 0." so. u 0! 0.0 a.“ 23‘.“ a.» uni! menu): mm 0.18 no.3: 0’ a... 0.30 we.“ 3.: was. 0.01:»? mm 0.. “.83 00 o.“ o.“ um. MI ”ca 0.0"“? mama o... u.» 2: a... a.“ 0.8: 1." our.» 0.02-am: um 0.03 1.... 3.3: a.» to.” 3.8 um «am: 0.00». 0.8 to.“ on m on: "I: C a: 9 “on n u I III 0 0 fl — — _ - — - c- * 3 4 3 mo. .0. 003.133 ”‘03 at: t 2 a 3.0 0.0:: noun an.) 23.: I a o 117.: I.” on.“ 153.3 23.8 0 O I O.) 0.037 It.“ 82.8 33.! . . O 75.. ‘0“ “.83, 83.! ”‘0. o o 7 3.; mm 00.”! a... 233.0 7 0 I 7:7.9 1.. a..- 31.1 ”.1 O C 0 9.0 0.“ CI.” no.3 ”.9 . . t. ".o. ‘0', 0"..‘. ”0‘ 3‘0. I. I. H 9.. 0.... 37.377 100.. “.7 H :0 n um: I.‘ a..- 20.0 2.7.. 33 I! 38 ".0 0.“. 8.718 301.9 207.0 I! I! u -.I O.” ".0“ 307.0 8’.‘ u u I! “.8 0.‘ I.” 20.0 at... I! u u a... o.” ’3..- 2474 2.0.; u u n 19.3 0.03: 2...! 2.6-I 20.0 W I. II a...) 0.0 “3.8! 20.! 203.! 8. II I. 17.1 0.07. a.” fit! 20... H II a “3.0 0.. 700.63 ”3.3 2“.8 20 20 II 30.6 0... awn 3&8 3.3.: : a a ‘13.: a.” and“ 2a.: 208.! a :2 23 3.0 0.... “.896 208.4 3.3.. :3 :2 2| ”0.! 9.1! 00.7" and 202.7 2 20 a . .0 3.!” 28...! :ll.’ .NJ :2 :0 20 273.0 a... 00.... :fl.’ ”2.. :0 20 27 3.6 9. “I 5.4:: :CIJ 200.0 :7 a a $10.8 0.8 In.“ 10.0 20... a II I! .3» 0.80. $3.3” 100.0 23.9 3' a a 823.9 0.. 018.000 30.0 a... D so a: a... on. a.“ no... no 38 30 a 602.. 0.. 0.2.!” 300.. 300.8 x: a :3 3.7 0.“: u.” no.3 87.4 3 3'2 30 “.7 0.4! 10%.“? 20¢.) 23'.- ” 30 I! $3.. 0.“! H.” at. 37.0 :3 84 30 08.3 a.” 2.000.?“ 23".. a... 3A a 37 13.0 0.“! n... 230.0 a... n :. a ”‘0: ’0” ",.O... :3... m0. :0 SI :9 3.0 0.381 ".8” 23.: 23.! :V a to 33.8 0.1. :31.“ 23%| 78.. to a u z... 0.»; 19.0.. 22.0 83.1 n u u :1»! 3.3. ‘39.!” 5.. 33.. 03 a u “.8 3.8.7 0.200 238.! 29.. ‘3 t! u 33.. :di 01"..” 33.: 23.0 M u as n.) 3.»: 9.3: 33.0 134.0 0! u a 230.0 :.:I 0090.033 :3... ma .. C. .’ ..., :0 i" ’ ’3. :3.’ mo, 0' to a. 2.0.. 0.3. Iii-3.90 3...! 1': .3 '0 00 u «.3 : :00 1216 29.1 . .0 0' a 30 '08.. 0.35 ”03.02! 333.1 38.3 3: so an 00.. ‘ :00 1.0.0 39.! :32.- Ot e: a: I' u o." mace: mu 2:... 2: 2: :1 00.4 1.:00 0.“: 218.: 23:.0 1! 2: :0 00.? 3.” 120.3:0 218.: 237.9 :0 so as u.’ an: an. 0:0 237.0 287.: M :0 a. too 0.50 -..;oo run 28?.) 140 Case 3: ena gun off. puma 'V' '37“. IYWYIW I 0 WO‘ ‘0 0.00 1.01.011" OLLCHAILI ERROR - .3 '7 ~0001071 "0171* ”NIL 31"....“ 0‘21 0.11 a001 -0.11 .001 8 323.70 :00.070 323.73. 100.07 0.00 0.0 322.02 139.003 322.030 139.30 0.12 0.0 321.00 139.230 321.130 130.93 0.23 0.1 320.72 133.730 321.070 130.91 -O.11 -0.0 320.90 133.030 319.023 133.37 0.33 0.1 319.70 133.330 319.730 133.33 0.01 0.0 320.02 133.031 310.337 137.01 0.00 0.1 310.00 .137.931 310.323 137.72 0.17 0.1 310.99 130.007 317.009 137.19 0.39 0.2 317.03 137.307 310.773 137.03 0.33 0.1 310.11 137.020 313.070 130.00 0.71 0.2 30.... 3:70 02' 30:0“: 03‘.“ 00" 00‘ 110.11 130.702 310.100 133.09 0.00 7.2 310.03 130.910 313.229 133.00 0.92 . 0.3 13.32 130.010 312.739 13.3.31 0.01 0.3 313.0. :30.373 312.029 133.17 1.03 3.3 310.32 :' .073 311.003 130.73 0.97 0.3 310.90 130.237 311.331 130.72 1.13 0.0 313.79 133.737 310.020 130.13 1.20 0.0 310.10 133.917 310.237 130.23 1.20 0.0 313.00 133.017 300.002 133.03 1.07 0.3 313.32 133.001 309.301 133.03 1.33 0.0 312.30 133.101 307.200 132.92 1.00 0.3 312.93 133.330 300.097 133.07 1.02 0.3 311.70 130.080 300.001 132.30 1.72 0.0 312.39 133.133 307.722 133.13 1.30 0.3 311.20 130.033 303.393 132.21 1.01 0.0 311.91 130.903 307.013 132.03 1.37 0.3 310.73 130.003 300.030 131.30 2.10 0.7 311.02 130.731 300.271 132.31 1.00 0.3 310.20 130.231 302.032 130.90 2.03 0.0 311.03 130.370 303.730 132.27 1.71 0.3 309. 09 :30.070 302. 100 130. 70 2.31 0.0 310.73 130.032 303.233 132.00 1.70 0.0 309.37 133.932 301.020 130.30 2.37 0.0 310.00 :30.313 300.002 131.09 1.01 0.0 309.30 133.013 300.733 130.11 2.77 0.9 310.20 230.220 300.007 131.70 1.03 0.0 339.00 133.720 799.700 129.70 3.01 1.0 10.00 130.130 300.179 131.00 1.09 0.0 10...- :13..30 297. 99’ 120.72 0.33 1 . 1 109.92 ”0.001 303.907 131.31 1.92 0.0 000.70 133.301 299.370 129.0. 3.07 1.0 309.03 130.003 303.007 131.00 1.99 0.0 020.07 133.303 197..33 120.77 3.3. 1.! 139.77 110.019 103.09. 131.09 1... 0.0 330.02 113.319 397.320 123.72 3.39 1.1 109.70 110.000 303.031 131.30 1.97 0.0 300.39 133.300 790.230 120.17 0.00 1.3 339.?) 10.001 223.390 131.33 1.90 0.0 330.37 133.301 290.223 120.10 0.00 1.2 J39..'7 100.000 (03.739 131.11 1.07 9.7 0100H0003 ELIHINT fLOU NO 10.01 1 020.0 2 3.0 3 020.0 0 0.7 3 010. 1 0 3.7 7 010.3 0 0.9 9 001.0 10 11.1 11 790.3 12 12.0 13 777.9 10 10.9 13 703.0 10 7.2 :7 703.0 10 :9.0 19 720.0 2° ' 23.0 2 703.3 22 20.9 33 070.0 20 20. 23 007.9 20 20.7 27 019.2 20 20.9 29 390.3 30 30.2 31 330.1 32 30.0 33 310.0 30 30.0 33 000.0 30 30.0 37 602.0 30 00.3 39 001.0 00 83.3 01 330.1 02 30.0 03 300.1 {0 03.0 03 200.7 90 30.0 17 110.7 10 10-0 09 100. 7 20 30.0 31 109.0 32 20.0 33 32. 1 36 73.1 Case 3: end 141 gun off. pump 'V' 0- UNIO 0000 I" 10.3199 m 991071. 9 9 \fl0 ’07 I O r91 10 ”1 3,1 91m .0 1.473101 "1 0'71 - — 0.. C.“ ' 1.00 0.10 170.27 10.00 0,121 1.010.“ 0 0000000 ..32 1.3 1 0.00 0.10 970.77 10.01 19.000 0.01000: 0.0000070 1.30 :... 2 0.00 ' 1.10 110.07 1.93 03290 0.010002 0.00001000 1.: 0.01 ° 0.00 1.10 110.30 9.. 090213 0.010070 0.0000030 1.31 0.32 9 0.00 r. 10 001.01 9.77 000*» 0.010090 0.000033: 1.07 ..n 11 0.00 1.10 «0.30 9.00 039:3 0.1110710 0.000730 1.23 0.00 12 0.00 1.10 777.09 9.09 032191 0.010700 0.m 1.8 9.20 10 0.00 1.10 “03.03 9.30 0219:! 0.010701 0.000973 1.17 10.03 0’ .0“ ,0 0. ’“fl ... 010371 0.010811 0.0391135 00‘: 0‘0. 19 0.00 2.10 7'20.“ 0.8 000m 0.0101109 0.0000220 1.07 12.02 21 0.00 1. 10 103.01 0.. 000% 0.010079 0.000053 1.10 13.77 33 0.00 1.10 070.00 0.8 mm 0.01:»: 0.000030 0.90 10.70 :2 0.00 1.10 «7.07 '.90 :90: 0.0130111 0.000.341 0.00 10.27 37 0.00 1.10 019.19 9.! 300.131 0.013173 0.0.7::“1'1 0.79 10.20 :7 0.00 1. 10 390.31 7.20 3737? 0.013271 0.003.231 11.73 17.” 31 9.‘ 0.10 230.” 0.10 3057"! 0.0113313 0.0073370 0.70 17.” a 0.00 1. 10 210.03 0.8 337?.-.) 0.01: 3:3 0.00:;‘71 9.07 10.39 a 0.00 7.10 9N.” 1.00 it”. 3 0.011713 0.035;“! 0.09 10.” :7 .0“ .0 0. “'0“ 3.3. 3" '9 0 0.0121763 0.032213 .9“ 0.0“ 39 0.00 0.10 001.. 0.90 227100 0.010130 0.030373! 0.3 19.00 01 0.00 1.10 . .07 0.37 193100 0.010010 0.00001) 0.8 19.99 03 0.00 2.10 3”.“ 3.70 171109 0.010013 0. -..T’ 9.8 20.21 0‘ 0.00 0. 10 300.70 3.8 107017 0.017700 0.00:3:33 0.10 20.37 07 0.90 1.10 710.71 2. 11m 0.01m0 0.00:9:3 0.11 20.09 09 9.09 1. 10 100.70 2.01 91332 0.010750 0.0mm 0.07 20.00 :0 .0. .0 0. 1”.” '.0” .0.” ..m” .0m .0“ 0.... a 0.8 1.30 33.12 3. 03700 0.02“ 0.~1030 0.01 10.90 m “I 01: 9913 2' 90 O «1001 '0 U 2 ”I 9 1' C _ a— — 0— - — c- g 1 1 2 020.0 1.8 03.2 323.0 13.0 8 0 0 :0 .0.0. ”0' m9. m9. 7 2 0 020.0 1.30 017.7fl 22.0 81.1 0 0 0 0.7 0.010 70.200 321.1 321.1 , . O ...-0 1.: 0a... 9‘00 :0... 0 0 7 0.7 0.010 00.007 319.0 319.7 7 0 0 010.3 1.31 000.077 339.0 210.0 g a 0 0.9 0.- 01.101 310.0 010.0 9 0 10 901.0 1.0 33.012 310.0 317.1 10 10 11 21.1 0.0. 30.001 317.1 310.0 11 10 12 790.2 1.! “909 317.1 210.9 12 12 12 12.0 0.89 8.0. 310.9 210.0 12 12 10 777.9 1.8 0&010 310.9 210.7 10 10 10 10.9 0.007 8.090 210.7 210.1 10 10 10 703.0 1.17 , 009.80 310.7 210.0 10 10 17 17.2 0.- 8.299 313.0 312.0 17 10 10 700.0 1.12 072.070 313.0 212.0 19 10 20 720.0 1.07 ...007 312.0 211.0 2 70 21 73.0 0.070 10... 311.0 310.0 21 20 a 703.0 1.10 021.18 311.0 310.2 a a 23 10.9 1.07 10.070 310.3 ”.0 =2 20 070.0 0.90 “$.07! 310.2 2%.! 10 1 10 10.7 1.000 :1.012 309.8 307.0 1'! 20 20 007.9 1.. ‘11.027 309.3 100.0 20 20 27 10.7 0.09! 1.0.11 28.0 ”0.0 1, 0. 3. .0.0’ :0” no: 38.. ”'0’ 20 2'0 79 10.9 1.000 13.32 307.7 33.0 39 70 30 990.0 0.73 “.111 307.7 307.0 :0 3 31 10.2 1.110 11.090 107.9 700.1 31 30 32 330.1 3.70 702.101 307.0 700.0 2'3 3'3 33 ”0. .00: 1.0, o 03 303.7 3 32 30 310.0 1.07 972.371 300.1 300.9 30 10 3'! 10.0 1.1:: 0.399 3&0 302.1 a :‘ 20 0‘0. 3.0. "a,“ :30, m0: :0 30 3'7 10.0 1.120 10.007 1a.: 301.0 37 30 30 007.0 1.02 1001.200 303.3 300.0 30 3 39 00.0 2.120 9.7.9 3 .0 ”.7 3' 0. 0. “‘0. 0.” 3.0”.” ,“0. ”0.. 00 .0 01 13.: 0.100 1 :00 100.3 799.0 01 .0 09 ”0.1 3.32 1110. 091 130.0 330.2 02 02 0! 33.0 0.100 0.“ 100.2 ".0 03 02 00 100.1 1.32 1002. 902 330.2 300.0 10 00 09 01.0 °.100 0.;71 330.0 299.2 19 u 00 110.7 1 10 1119.771 1:0.0 3:2.0 00 00 07 00.7 1.100 1.009 103.0 797.0 1’ o. 09 710.? 3.11 991.799 503.0 233.7 09 19 O9 :1 0 0 100 1.010 103.7 17’ 1 .1 .0 00 1.13 g 07 1079 912 113.7 713.0 : 11 91 :0 0 1 193 3110 103.0 700.1 31 10 32 1'1 9 3 :3 [$0.320 103.0 103.0 11 1 3 '1 0 - 101 7.799 573.0 790.7 11 11 $0 .1.1 1 01 3199 002 333.0 701.2 '1 :0 33 ,1 1 - 10: 10.199 '03.) 202.0 :3 3.9 20 0 J 00 0 030 103.1 202.0 142 Case 4: end gun off, pumo'H' 'UTIL XTIl0713l 0 0 ALLnuAILl canon 0 .3 '7 HIIDC'T! N003 :NlTIAL ltNflL 90 «997 «00:! coca «00:: x 214.39 92.000 210.390 92.00 2 233.23 92.23: 213.072 92.00 3 232.07 9:.73: 213.04: 92.03 9 211.03 99.030 212.703 92.00 3 230.09 91.234 232.090 92.02 0 210.00 92.000 212.029 92.03 7 209.33 90.300 233.700 9:.02 0 209.:0 90.003 220.920 91.23 9 207.99 09.90: 210.700 91.39 :0 207.04 09.333 209.910 90.02 x: 200.09 09.33: 200.079 90.7: :2 200.37 09.230 209.032 90.00 :3 203.22 90.704 200.702 90.32 :0 223.30 00.730 209.213 90.00 :3 203.90 03.230 207.009 09.9: :0 203.93 90.230 207.000 09.73 :7 202.00 07.733 200.070 09.30 :0 202.32 37.730 200.033 02.00 :9 201.07 07.230 233.973 39.:: 20 202.73 07.200 203.099 09.00 23 200.00 00.780 200.93! 09.03 23 200.08 00.792 233.220 63.70 23 :99.00 00.292 203.000 33.20 20 199.0! 60.390 200.007 80.00 23 193.33 03.090 203.030 07.00 20 £90.02 30.019 203.370 33.20 27 197.07 03.329 202.09! 07.0: 23 £90.00 03.000 233.329 07.97 29 190.00 03.100 201.920 07.30 30 :97.33 03.373 202.03: 07.73 3: :90.:0 30.073 200.037 00.90 32 £90.02 03.003 202.329 97.33 33 193.00 00.303 199.990 00.32 30 :90.00 00.030 20:.930 37.30 :3 390.92 00.330 :99.003 20.30 :0 :93.0L 30.029 201.390 07.22 :7 n90.43 00.:29 :99.270 20.2: : 193.22 90.039 201.307 37.09 '9 n90.00 03.939 :90.039 03.93 ‘0 :74.90 00.320 201.000 30.99 0| I93.7l 23.020 I9I.007 33.00 .2 n90.0u 00.:90 200.002 30.09 .3 193.90 93.090 n90.903 33.2: ‘0 :90.93 00.x:9 230.090 30.03 .9 «93.20 07.039 107.737 03.30 ‘0 .90.): 00.003 220.302 00.70 «7 :93.:3 13.303 190.707 03.:0 .a 394.22 00.020 200.300 00.7: 49 a71.07 03.320 190.033 00.07 :3 |'9.:0 00.009 230.039 00.73 3 393.02 03.309 093.001 26.74 32 079.00 0'.00: 130.030 03.72 :3 "2.00 03.30: 073.339 00. 73 :3 193.:0 23. :99.922 00.09 OXICHIII' DIFFIINII‘ ILIHENT FLU! (0%) I O I”. 00” 00° ‘ .’°.: -0.20 90.3 2 3.8 -o. 7‘ -0.3 3 007.0 -0.‘2 ‘00: 6 3.. -O.’5 '0.8 5 0.30. 9000‘ .003 0 ‘07 -‘0:. ‘00. ’ “0. -O.I3 '00. C 7.3 -‘03‘ do. . “‘0, -00” *0: 0° .0. -‘03‘ 40’ “ “:0. -‘03‘ ‘00. ‘2 ‘00‘ -‘07: .00. z m.’ -‘030 do, 0‘ ‘30‘ -|.C. '90. 13 030.. -2... '0.. $6 I90. -200‘ -... 0, “0‘ -|.I. .00, 0. ‘30, -20 ‘3 -‘0‘ 0' 0, -1003 -100 20 I'o’ '2.8' 9!.8 28 37300 -101‘ “‘0‘ a a0. -2030 -.0‘ fl ”0. -20” -|.2 2‘ 2303 -2.2' 92.2 23 53.07 9205‘ 9|.3 2. 2303 -20“ -‘03 27 ”0‘ 92.07 -Io3 2' 23.. 02.3. 9|.3 3' ‘79., -1.” -‘0. a no: '20:, ".2 31 .3307 -2.” -..: n ”a, “3.3: 9|.2 33 42009 91.79 0|.3 3‘ 300. -20 9° '1.‘ I: m0° .30“ ‘.0‘ It ”a. 92.5. 9|.) ‘7 '39.: 93.12 '0.‘ I. .300 -1.37 9|.3 :9 0: .30 0. -‘0. 9° ”0. 02.23 'I.t ‘1 2’0.’ 93.39 ‘l.‘ 91 00.. ‘\.II .90. ‘3 m0. -’0a .‘0’ “ ”0° °l.33 0|.l ‘5 135.8 -’9fl ... 7 ‘. 0‘ 9|.IQ 90.0 ‘7 I7‘o3 "9 a ’I .’ ‘. .00. 7|.IS 7!.0 ‘9 l!‘.. ‘J.13 7|.) 5° 14.. -. ' .7 '00. 5: "0: '3-31 °l.7 32 94~0 .. 0 " ..0. s: .‘O D. -!0:’ ..0. 143 Case 4: end gun off, pump 'H' !. ..7333 “I 3." “:3” m. '.‘C73u ‘ ‘ 'n‘ '3 "‘ “I "I“C) Q 'fi."' 0"! O... 3. i. I’°.2° ..I? $77176 0.0350” °.m 3 ” .‘u‘. 1. I. “1.00 .. £3 =’«:. 0.0!:OJI Com :0.‘ . °-“ 3. 8. 0.3.3. .1.” 3“”. 0.0I30‘2 mm 3.. ° 3. u 3. .a.” I..:. 2‘37“ 9.0‘30’33 0.0000031 :- . O... ‘ :. .33.“ .... :."°, 0.0l5077 0.“: ’0” 2: a... '.|. “3.08 '.I1 :3“, 0.0|3!°‘ O.MI 9.” ;’ a... :0; .30": ..3 3““. 0.03:": 0.“! O.” a” a... a.“ on.“ “.3 sum 0.01:2" 0.0m: OJ. L. O... 3. t. 2”.” 7.” 3772-2: 0.013277 0.0m: 0.” :3 0.00 a.“ m.” an 317m 0.0mm 0.0m. 0.7! :3 ,0... a." an.” o.” m 0.01920 mm: 0...! a .0“ 3.1. a.” .O" m' .0..”', 00m. .0” ~ :7 .0“ O. ‘. 303.“ ‘0‘, M 0.0350|0 ..m' .0” :Q 0... o. a. 0”.” 3.6 nun 0.0337“ mm: 0.00 3| 0... 0o 3. 031.73 3.30 2371’ 0.011233. 0.003525 .0” 33 C.“ .0 |. no... Sci, 333:?! ....K’SO mm .0” a C.“ .o 3. 3‘0.“ ‘0” 31“ ...Im‘ ..“CCCJ 0.8 3" C.“ .- |. 3:1.” 6., 3'31} ..OIOOIO «mt-.23 .0” 3‘ O.“ O. I. 35.88 3.” |.'°‘o ...!“ .0003?” C.” u o... a.“ no.» 8.30 tum 0.01m. mm 0.8 .3 .0“ .o'. o o” :0” 13:71? 0.0I7.|. ..m ‘9‘: C. .... .- 8. 3‘3.“ 3.“ Ilml 0.01!!!” ..m .0“ .’ .0“ .0 3. '7‘.“ 3. $3 «a! ...!E‘“ mm .0. .. C.“ .... ‘3'. I.“ 7‘1" 0.01m: 0.0m? ‘0‘ 3‘ 0.“ Col. ”on I." 0m 0.02:!” ..W .0" '- 0.” O.“ “-07 3o“ 3!.“ maul: ..-.” ..33 an. ’3‘ "I: , a 9 m. a U . u C O . — _ _ — _ ‘ _ I I I a... O.” 73.81 3|... “3., 3 3 3 3.8 .0... ”on 383.7 383.. I I 0 “’0. O... V‘LQ 333.7 233.7 ‘ 0 U 3.. ...‘I .3.” 333.7 3‘30, I C O ”.0 0.. 7“.” 332.7 a... O O 7 Q.’ 0.03! .0“ 383.. ”.0. ’ O U .0. Co. $20.80 333.. 3‘... I C . 7.3 Co“. “0‘" 3'... a... . C I. “'0’ .0” “I.“ 330.. ’0. |. i. I! ... 0..“ ‘0”. ”o. ”a, II i. ‘3 .08.. .0. 7"“ ”a. ”a. ‘3 ‘3 33 I... .0” “0‘ ”0' ”o. ‘3 ‘3 ‘. 613.8 .0. ”‘0” mo. ”0. I. I. t. it. .0. ”.703 m3 ”,0. ‘C ‘O 3. ma .0” M“ ”a. ”a. t. .0 '7 IC.‘ .0“ ”0‘. 37.0 “.0 I, I. I. “a. '0’. 770.7” 87.0 “o. i. ‘C i. I'o’ .0”. 13.0. “a. “a. i. i. a m.’ C." M... a... 13.0 a ’3 3| 8.0, Co"| 3‘.” 4 , 2a.! “a. 3! an a 372.. O.” "9.”! no “I I: a a 23.0 .0.” l’ofi.‘ ”0| 8%. a a 20 ”o. .0“ I...“ ml 100.! :. ,. a . 3.8 00". 15.7.! M. ”0‘ a 10 u no.7 Co. -0“, 100.. an. a a 27 :03 Coll. :h.‘ mo. 1.3.! :7 u a ”3.6 .0“ .".' ‘83.. mo, :8 2! fl . .8 9.8“ a... 193.8 sun 8 . ” 0’... C... m‘ m: ”a. ' 3 3| ... Col“ “.8? m. ”-0 3| an a “'0, '0’ -0“ ”a. ”.8 a a a 8.0 .o I“ 8303’ m: ”a. a a a O”.. .0” In.” 393.3 31.0 a 3‘ a ”a. .0 I” 13.3” 238.. I”.. a u u m. C.“ II...” ”Io. ”fie. a a 3" 3... .0 I” 33. 3” ”lo. IWo' ” 30 a ”a, O.“ ta... ”‘0. .‘o’ a SI ” 3.0 9. t“ ‘3.“ 3|.) In.’ 3‘ 3. U 17...: 0.3. 33.9 20:.) ”Io! . n C. :0. .0 "I HuO’. a‘ol ‘.o. .' a .’ ”.7 O.” :MJG 3.3.! m-. .' .1 .: .... on". [3.3” mo. ‘"0. ., .’ .. 330-0 :0 3. 30".”. an. ”a, .. .. ., =0. :.". "c". 3%., ‘no. 3, O. 0. 3|..I 3o$l :00...” m.) In... .. .. .’ .... 30m :30” mo. '”0’ .’ .. ‘0 I’..' ’3” 22.1: 2%.. "0’ .. .. .. .... :13” .3.“ 3a., ’..0. ‘. ‘. ,9 33.. e.” m..., 3 .3 30.3 33 3 I! 00.. ..m v.3” : .3 I”.. 33 :9 : ...! 1.33 3'”-.” :a.’ m.‘ 3 3': 3: O... 3.2,: 1”. :”.O I”.. 3! 2: 90 u.’ :.:3 PL!!! : .o "0.8 W ,. 12 ot.’ 2.33. |2?.,‘, :' .3 3.... fl 9. ’0 L: 3.” 0.3“ 329., :o .8 Case 5: ena TanL 300: w: U06 ‘ 050‘ Q. I. |.4N'— “4 6.0“. U .04 I 5 Is). 5.0.0.... “.... 5708051054 0 5 23272222220 .15) :42: 205.75 92.750 225.24 35.252 227.05 55.552 266.65 55.425 2: .52 65.525 225.24 E .752 202.55 65.252 202.62 25.052 222.46 57.552 222.52 57.262 220.77 56.562 222.44 27.252 200.25 56.652 200.45 56.727 255.22 56.227 255.05 56.224 257.52 55.624 255.25 55.745 257.02 55.245 277.75 55.554 226.55 55.054 226.47 45.002 225.52 34.522 235.26 54.475 224.22 52.575 34.72 54.240 252.56 52.740 252.57 52.525 252.52 52.425 252.52 52.462 252.76 52.56) 252.26 52.275 252.22 52.675 252.55 52.042 250.75 22.542 292.06 52.660 255.50 02.260 250.26 52.224. 205.20 52.524 250.02 52.207 255.56 52.707 255.45 52.565 205.20 52.465 225.04 51.59! 257.65 52.255 225.45 22.542 257.22 62.042 255.22 52.472 257.25 50.572 256.72 60.775 257.52 52.222 256.25 50.622 257.40 52.075 256.25 20.575 257.22 50.554 256.02 50.454 236.55 50.554 255.52 50.254 256.55 50.550 255.72 52.255 235.52 32.522 225.65 $2.222 256.74 52.7%: 255.55 95.272 .26.65 50.776 255.54 50,373 265.55 50.756 255.: 52.245 .3‘5é7 00.700 165 52 55.25. 436.67 g; :3“ 225.52 a;.:g: 144 gun off . design modified NCA05675 52$Cflhuc: FXflIL {trrflltcz ELEMEN? FLO” (1: ;;42| (It: 2 NO (3pc. 207.763 70.76 0.00 0.0 1 325., 700.636 90.26 ’0.05 '0.0 2 .3 2°..J’7 $0.26 '0.60 -GOJ 3 636.0 207.015 0005’ -301, -')o: 6 30’ 207.020 05.56 -0.73 '0.6 5 760.6 235.726 35.02 '0.35 ’0.‘ 5 6.6 225.646 34:57 -o.4z -2.4 f 31).: 206.352 56.63 '0.30 '9.) 5 ' 7.. 206.12, 56.35 '0.67 ’9.6 7 306.6 202.09. 67.76 -0.60 -0.2 :3 J, 202.644 '57.67 -o.52 -o.5 ;; .97.: 003.670 67.60 -0.61 '0.) L: 3.1 202.252 07.30 ’00,. .900 L: 4".) 201.650 67.23 '0.57 '0.) 26 5. 202.322 67.2 '2.00 '0.6 23 770.6 2000000 004,0 -00“ .06) 1‘ 610, 299.000 06.52 -l.03 '0.0 27 767.6 259.605 06.36 .0.72 '0.6 26 0.2 60,0),0 0002‘ ‘00:. '0.6 1, 4,,., :”.21: 0‘00, -°.,0 .000 :3 6:05 ;,.o‘.‘ 0000‘ -003‘ -00, :; o“.° 09.0600 0’97: -60“: -000 z: 65.: 297.424 45.42 -1.27 -&.5 :z ’30.: 296.555 050:; -00., ’04, :6 1).: 236.102 36.36 '2.53 ‘0.5 :3 722.2 6,0050, 00.“: .0002 .?03 :0 620. 236.052 36.01 02.2, 00.7 :7 659.7 6,000.3 00.7: '00,, ‘0.6 2, 010, ;,04002 00020 ‘00,, .00, :, ‘70.: 156.696 56.32 '2.03 00.5 25 :2.‘ 251.505 63.75 .0.75 ‘0.6 31 656.5 :".J°. .0007 -049‘ .00‘ 22 :00: :’:.73° 010.0 -00), “0.7 ’1 0.003 190.020 0000‘ .000. -000 )0 210, 252.726 53.36 “2.01 00.3 15 610.5 6,1023: 03000 -l.26 .040 3‘ 210’ 006.00. 03.03 -0005 .000 :1 5,300 157.655 63.10 -5.17 '0.6 30 27.6 6,0000‘ 0200’ -0.63 .000 J, S‘.o‘ 6,20370 0’01. -006, .00‘ 0° 0... 256.360 62.7, '5.33 ‘0.7 62 560.7 156.756 62.56 -5.12 '0.6 42 30.0 10,003. 02.00 .°o,t -900 0) 00,0, .56.:00 63.73 '5.36 .0.7 66 60.0 269.06, 52.60 00.76 00.6 25 660.7 1,040.. 62.57 ‘2.36 -VO, 0‘ :20. 00,0‘01 32.03 ‘00:: -90, 6, 0‘20, 250.657 3 .30 .6.36 .9.7 65 7.. 03.057: 0.0:, ’00,; .ao‘ 0’ ‘37.. 0300;50 014;: '00:, -00, 6’ ’20; 207.552 55.3, °°.63 ’0.) 52 695-6 265.517 62.2 '2.25 '0.7 52 36.0 0.7012, .6000 -000, -04) 53 1200! 200.631 024;: '04:, .00, 50 6‘00 260.156 06.00 '2.07 .90‘ is 265.0 ;"o0°7 020°) '2.;5 -007 26 57.0 256.236 50.36 “0.05 '0.2 57 505-5 237.337 52.56 '2.2, “0.7 55 33., ;.‘o77’ 200.0 -9031 -04, 3, .710‘ .69.205 51.59 ’2.23 00.7 55 10.3 622.37? 0.0;, ‘00:, .04: 51 .030: 257.333 52.57 ‘l.:3 ‘0.: ‘3 6?- 265.662 30.60 '9.37 '0.0 53 263.6 65,4100 :03: ‘1.10 .00: 2‘ 030. 255.766 0.36 .0.65 '0.: e! 666-6 329.:74 52.62 o1.24 -o.7 -4 4).) 255.067 66.60 0.36 0.2 67 522.; 227.251 42.22 -L.27 .9,: 94 34-0 154.377 .c.sz -o.44 -e.z .9 94.7 237.365 52.75 '2.27 -§.7 70 65.5 149.04; 54.2; 9.:5 0.: ’2 23.4 205.966 52.76 ‘5.05 '9.6 2 63.. :44.>44 3;.24 -2.ss o9.s 12 2-6 APPENDIX C SOURCE CODE OF THE SIHULATION MODEL ”SYSTEM” RE" IIOIOIIIIIIOIIIIIIIDIIllllliiliilfiiiilll.IIOIIIICIIIIIIIIIIIIIDIIIOI RBI ' PROCEAN m I RBI | , - REH 0 THIS PROCRAH CALCULATES THE DISCHARGE-HEAD (Q-H) RELATIONSHIP AT I REH | ‘l'i-E PUHP POR THE CENTER PM"?! STSTEH BASED ON THE mow I RD! | RELATIOUSHIP DE 041 AT TIE-E PIYCI' POINT (CALCULATED BY PROGRAM - REH 0' -mORx-) ,SUCTION HEAD,mIVERT HEAD, AND THE LENGTH I RD! ' OF THE KAI}! PIPE LINE :- Rm "DE 041 RELATIONSHIP DATA OBTAIE-ZD PROII THIS PROCRAN IS USED To - REH 0 PLOT THE SYSTEM CURVE SO THAT THE OPERATING POINT OP I! RSI! ' THE PUHP CAN BE DETERMINED 1. RE)! ' _ I RBI ' HRITTEN IN TURBO-BASIC - RE“ IIOICIIOIIOIIII.OIUOIIIDSIIIODIICIIDIIIIIIIIDIIOIIIIOOIIIIIDCICOIII. CLS:IHPUT"DELIVERY HEAD (ET)=-,DH INPUI'SCCTIOII HEAD (PT)=-,SH FIXED=SH+DH ' IHPUT'ROUGFJIZSS OP THE PIPE (INGES)=',E IUPUT'DIMGI'ZR OP THE PIPE (INCIES)=',D IWDIM-E’I‘ER OP THE SUCTION PIPE (IECIIES)=',DS HAIHDI=DI 12:§JCTDI =DS/12 . AREA 8 (3. “IMWHAINDI‘fl :AREAS=(3. “INNSUCTDI 2) EBYD=E/D:EBYSD:EIDS INPUI'LEmTH OP THE DELIVERY PIPE .(PEET)='.L INPUT-COEPPICIENT OP THE STSTEH AT THE PIVO‘I‘:',K INPIrr-EIPONEUT OP THE STSTIII AT THE mars-J. VHC:385.9'(D‘II) OPEN '0',M,"SYSTD!.DA‘1" IWWA‘ER MICE OF DISCHARGE I” GPH:',GPH1 INPUT'UPPER MICE OP DISCHARGE IN GP!=',GPHZ QSUH=GPH1-((GPH2-GPH1)I10) CLS ' PRIIHH,CHR$(18):IPRINTM, CRR$(27) '2'; P8111791, CHE: ( 27) "6" :PRIHTH , PRINTNJABUZ) USING"ELEVATION HEAD (FT)=OH.I";DH PRI!-1'N1,TAB( 12) USING"LIFT PROH HATER LEVEL TO GROUND LEVEL (FT):IH.I";SH PRINTHJABHZ) USING"ROUGHNESS OP T'f-E PIPE (INCHES):I.MNH";E PRINTHJAN 12) USING'DIM‘EZTER OF TIE PIPE (INCHES):H.HH";D 14S 146 PRIN'I'M,TAB( 12) USING"DIAKETER OF THE SUCTION PIPE=H.HH";DS PRINT“ ,TAB(12) USING'LEJ.GTH OF THE DELIVERY PIPE (FEET): HM. I';L PRINT“ ,TAB( 12) USIIIG"COE FFICIENT OP THE SYSTEM AT THE PIVUI'dU. 135"“ PRINTH,TAB(12) USING”EXPONENT OP THE SYSTDI AT THE PIVUT=L ("l'fl PRIN'I'H,:PRINTH, . PRINT“ T,AB(12)_ "QSUN FIXED H -PRIC HEAD ERIC HEAD VEL HEAD OPR HEAD TOTAL HEAD“ PRIII'I'AI,TAB(12)_ "GPH FT DEL( IT) SUC'N’PT) PT PT PT PRINTII,TAB(12)_ . FOR Ial T0 II QSUH=QSUH+((GPH2-GPH1)I10) VELO=QSUI1/(M8. BPAREA) VELOS=OSUH/(m18. 8'AREA3) =(VEL09ILTINDI)/(o. 00001059) READS: (VELOSPSUCTDIHM. 00001059) FRIC’!’ =1. 325 ' (Loom. 27*EBYD + s. 72’((1/REI: 0) 0.9)) (-2)) PRIcrs=1. 325 ' (Loom. 279mm 5. 72'((1/RE1205) 0.9)) (-2)) mEEPRICTEL / (32. 2-(uu8. 8‘2 2)-(3. 1A159 2)'(HAIHDI 5)) RszaaPalc-IS'SH / (32. 2mm. 8‘2)'(3. 111159‘ 2)'(SUCTDI 5)) EHEA.D=1. 1-R-(OSUH 2) . PHEADS: 1. "RSNOSUH 2) VEUIEAD=(QSUI-1‘2)IVHC SPRNEAD=(OSUH/E)‘( 11A) HEAM I )= FIXEmVEUfEADISPRIEADIFHEADIE-‘HEADZ PRINTM ,‘I'AB(12) USINC_ "NM III. II IIII. III'I' IIII. IIII III.IIII IIII.III IIII.IIII';_ OSUH, FIXED, FI'IEAD, PHEADS, VELHEAD ,SPRHEAD,IIEAD(I) PRIWRUII COMPLETE. OUTPUT FILE IS 'SYSTD‘I.DAT" STOP END APPENDIX D SOURCE CODE OF THE AUXILLARY SPRINKLER SIHULATION HODEL "AUXISPR' RE" .III.IIIIIIIIIIIIIIDIICIIIIIIIINEIIIIICIIIIIIIIIIIIIIIIEIDOIIIIICIOC REM P PMRAN AUXISPR P REM P I REM P TRIS PROGRAM DE’I'EIIHINES TIE POSITION AND REQUIRED CAPACITY OF P REP! P TIE AUXILIARY SPRINKLERS THAT ARE TO BE ADDED IN TIE P REM P CENTER PIVOT IRRIGATION SYSTEM FOR THE P REM P NODIFICATIOEI OF THE DESIGN 80 P REM P THAT TIE PUHP ALNAYS OPERATES AT A FIXED OPERATING POINT P REM P I RD! P NRI‘I'I’EN IN TURBO-BASIC P REM IIINIIOIICIIOIIIIIIIIIIIIII.IIIIIIIIIICIIOIIDIIGIIIIIDIIIDIIINIIOUII REM STATEHEI-IT SOD CONTAINS TIE DISTANCE OF THE PLUGS FROM TIE PIVOT POINT REH IN TIE SYSTEN DIN L(100),TOTL(100).OOUT(50).Q(100) OPEN '0',I1,'OUTLET.DAT' INPUT'END GUN DISCHARGE = “:08 INPUTPLENGTII OF TIE LATERAL LINE = ”;TL INPUT'NININUN DISCHARGE TO PASS TIIRU TIE AUXILLARY SPRINKLERS = P;OIIIN REM INPUT TIE PQSITION OF TIE PLUGS IN TIE SXSEN AT DATA STATEMENT 500 RESTORE 500 TOIL(O)=O FOR I = I TO 100 READ TOTLU) IP TOTL(I) = 999 THEN OUTNO=I-I:I:IOO: COM 10 10 NEXT I FOR I = I TO OUTNO QIII=Q€P(1-((TOTL(II/TL)‘2I) NEXT I J=1:QUP=QE:FLAG=O ' PRINTJ‘I,TAB( IZI'DFSICN DATA :":PRINTH, PRINTPI,TA.B( 12) USING'SYSTDI LENGTH = PIPJ FT”;TL PRINTII,TAB( 12) USING"DISCHARGE THRU THE END GUN : PIPJ GPH";QE PRINTI1,TAE(12) USINC_ "HINIHUH DISCHARGE THRU THE SPRINKLER =II GPH”;QHIN I47 148 PRINT“.:PRINTH,TAB(12)_ n ” NO PLUG DISTANCE PROH DISCHARGE SPRINKLER NOZZLE EXPECTED PRINTII,TAB(12) '- no PIVOT (FT) (GPN) MODEL SIZE RRESIESI)‘ PRINTH,TAB(12)_ ' ‘ " FOR 1‘; 1 TO_OUTNO-l Q:QUP-Q(I) IF Q0.5 THE-3) Uzu-J DS : FOR K31 To 15 IF [:1 m1: WATS 12, 28: PRINT”"'PLEASE “AI”... PRINT I 8&2ka VAR=0: VARI=O: VARZ: 0: CO!13=0: RESULT=O: DSUH=0 H=U U.‘I°(O O1'TIHI'3(K)) PRINT”, FRI ‘J'NZ :TABUZ) USII'G'TIHER SETH!!!) d" 1";9'100/31 PRIIJ'I'32,TAB(12)" S DISTANCE DEPTH" PRIHTRE ,TAE(12)" RD (E?) (INCH) PRIHT:2,TAE(12)'-.. ------.. - FOR J: 1 TO I P(J)=J'25:SU}21=0:VAR=O:VABI=0:VARZ=O:CON$=O:RESW.T=O FOR I = 1 To SP8 1P L=2 AIID P(J)<781 Tm 0(SPR,2)80 EISEIP L=2 AHD P(.I)=>787 m 0(SPR,2)=OLAST EHD IP uRATIo(I)=(P(J)-L(I,L))/R(I.L) IP nRATIO(I)=<-1 THEN GOTO 30 IP DIRATIO(I):) 1 THEH cam 10 COSH=(L(I,L)‘2 + P(J)‘2 - R(I,L)‘2)/(2'L(I,L)'P(J)) sum: (1- (COSH 2)) o. 5 HT=Am(smI/cosn) own ,L) =(61. 2789'Q(I, '1)”. .375)/(wr'R(I, L)'L(I, 1.)) nRATIom : L(I ,,L)/R(I L VAR (u-(nRATImI)!(nRATIO(I)+nRATIO(I))))/( I (nRATIOm 2_)_ . CALL ImCRATIORwr, VAR ,RESULT) SUHH a OHM” ,L)'((1-DIRATIO(I) 2)‘O.5)'RESULT + sum NEXT I n DEPTHU) : ( /H)'SUHH PRIHT52,TAB(12) usmmm ma." IM.MI*;J,J*25,DEPTH(J) DSUH : DSUH + DEPTH(J) NEXT J AVGD=DSUHIN DEPmS(x,L)=AVCD IP AVOD>2 THEH ms NEXT x “O 151 IF L:2 THEN VALU:SPR-I ELSE VALU:SPR END IF COVER: -L(VALU, L)+R(VALU, L)'0. 75 H:IHT(COVER/25) IF H-(ODVER/25)>o.5 THEN H:R-I CALL APPunifOTalty(fl,K,Dl-ZPTH().P().WIF°RH) UHIP(L):DHIPORH PRIR'Isz, *1 TIHER AVERAOE- PRIImz,'SEITIHC DEPTHUR)‘ FENDER,"- FOH K : 1 TO 15 PRIH'MZ, USING" I" HJH";TIHES(K),DEP‘mS(K,L) IE DEPmS(x,L)>2 man [:15 NEXT x DEPT(O)=0.I FOR 1:1 T0 '9 DEPT(I)=DEPT(I-I)+0.I NEXT I ADDS:O:I:1 FOR 1:1 TO ALL+I IF FLGI=1 THEN ADDS=ADDS+0.10 DEPT(I)=ADD3 END IF IE FLCI:1 AND I31 AND L=2 THEN DEPT(1):DEPTHS(I,1) ADDS=0.1'INT(DEPT(1)'100/10) END IF IF I: I AND L:I AND DEPTHS(1,1)>DEPT(I) THEN DEPT(1)= DEPTH3(I, L) ADDS: 0. I'INT‘DEPT(1)'10) FLC:1:FLGI:1 IF Lzl THEN ALL=ALL-((ADDS-0.2)'IOI END IF IF DEPTHS(K,L)2 THEN 1:15 NEXT I POR 1:1 TO ALL TIHEEC(P,I,2)=EOL(P)/VEL(L2) 153 TIHEEG(P,I,1):EOL(P)/VEL(I,1) TIMETOT(I,P)=TIHEEG(P,I,1)+TIKEEG(P,I,2) NEXT I NEXT P PRINToa :PRINTI2 PRINT32:TAD(12) USING“ NO OP SPRHNHLERS INCLUDING END GUN: II';SPR PRINT92,TAB(12) USING_ " SPEED OP THE LAST TONER AT 1007 NOVEHENT (PT/HR): III.IJ';U PRINTaz TAD(12) USING_ ~ DISTANCE PROH PIVOT TO-LAST TONER (FT) =III.I ';TL PRINTaz TAB(12) USINO_ .. APPLICATION UNIPORHITV (END GUN ON) =HJ';UNIF(2) PRINTIE TAB(12) USING_ 9 APPLICATION UNIPORHITT (END GUN OPP)=II.I';UNIP(I) PRINTIZ,:PRIHTOZ, POR P : 1 TO 3 TOTUNIP : EG(P)'UNIF(2)+(1-EG(P))’UHIF(1) , PRINTIZ, PRIHT$2,TAB(12) USING_ . PERCENTAGE OP DISTANCE COVERED NITH GUN ON : III 1';EG(P)'1OO PRINT92,TAD(12)_ "DEPTH END GUN ON END GUN OPP f PRIHT82 TAn(12) . IN ' TIHE'R TIHE TIHER TIHE UNIPORHITX TOTAL TINE' PRINT-:2 TAE(12)_ “INCH, z NR 1 NR 7 HR“ PRINT92,TAB(12)_ . . _. .--... ...... -... ----.. --.. — FOR I : I To ALL PRINTIE,TAE(12) USING “03.0! III.II II.03' 019.0: o:.:o 10.59 lll.ll';_ DEPT(I),TIHEX(I,2),TIHE£G(P,I,2),TIHEX(I,1),?1HEEG(P,I,1)L_ TOTUNIP,TIHETOT(I,P) NEXT I print02,:pr1nt02, NEXT P PRINT~DONE~=STOP SUB APPunITDruILy(N,x,D(2).P(1).”NIPORN) SUHI:O : SUHz : 0 : SUH3 : O mRJ:ImN SUHI : P(J)'D(J,K) + SUHI SUHZ : NJ) +30"? NEXT J CONS : SUHI/SUHz POR J : I TO N SUH3 : P(J)'(ABS(D(J,K)-COHS)) + SUH3 154 NEXT J ' UNIPORH : 100'(1.0 - (SUH3/SUNI)) END SUB SUB IWEGRATION(VI,VAR,RESULT) H : HT/SO SUH:O SUH : (1-VAR'SIH(O)'SIH(O))‘O.5 50 FOR I = 1 TO 50 I? I = 50 THEN ~ A:(1-VAR'SIN((I'H)!2)'SIH((I’H)/2)) IF A<0.0 THEN 1:0 SUN : SUN 9 A‘DJ ELSE A=(1-VAR9SIH((I'H)/2)'SIH((I'H)/2)) IF A