1-7 - Lecture 8 Set Sprinkler Lateral Design I. Dual Pipe...

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Unformatted text preview: Lecture 8 Set Sprinkler Lateral Design I. Dual Pipe Size Laterals • Sometimes it is useful to design a lateral pipe with two different diameters to accomplish either of the following: 1. a reduction in hf 2. an increase in hf • • • • In either case, the basic objective is to reduce pressure variations along the lateral pipe by arranging the friction loss curve so that it more closely parallels the ground slope It is not normally desirable to have more than one pipe size in portable laterals (hand-move, wheel lines), because it usually makes set changes more troublesome For fixed systems with buried laterals, it may be all right to have more than two pipe diameters along the laterals For dual pipe size laterals, approximately 5/8 of the pressure loss due to friction occurs between the lateral inlet and the location of average pressure • Case 1: a lateral on level ground where one pipe size is too small, but the next larger size is too big... • • d1 is the larger diameter, and d2 is the smaller diameter note that (hf)single is much larger than (hf)dual Sprinkle & Trickle Irrigation Lectures Page 85 Merkley & Allen • Case 2: a lateral running downhill where one pipe size is too big, but the next smaller size is too small... • The composite friction loss curve for d1 and d2 more closely parallels the ground slope than the curve with only d1, which means that the pressure variation along the lateral is less with the dual pipe size design II. Location of Average Pressure in Dual Size Laterals • • • Do you believe that 5/8(hf)dual occurs between hl and ha? Consider the analysis shown graphically below (5/8 = 0.625) The plot is for a dual pipe size lateral with D1 = 15 cm, D2 = 12 cm, 100 equally-spaced outlets, 900 m total lateral length, Hazen-Williams C factor of 130, uniform sprinkler flow rate of 0.4 lps, and zero ground slope Merkley & Allen Page 86 Sprinkle & Trickle Irrigation Lectures • • • • • Notice where the ¾ value is on the left-hand ordinate Notice that the head loss from hl to ha is approximately 74% when the pipe size is all D1 (1.0 on the abscissa) and when the pipe size is all D2 (0.0 on the abscissa) The hl values (inlet pressure head) would be different for each point on the curve if it were desired to maintain the same ha for different lateral designs Notice that the distance from the lateral inlet to the location of average pressure head is roughly 40% of the total lateral length, but varies somewhat depending on the ratio of lengths of D1 to D2 (in this example) These calculations can be set up on a spreadsheet to analyze any particular combination of pipe sizes and other hydraulic conditions. Below is an example: Section Flow Distance Diameter (lps) (m) (cm) 1 40.00 9.00 15.00 2 39.60 18.00 15.00 3 39.20 27.00 15.00 4 38.80 36.00 15.00 5 38.40 45.00 15.00 6 38.00 54.00 15.00 Sprinkle & Trickle Irrigation Lectures hf Sum (hf) d(he) head diff from (m) (m) (m) (m) ha (%) 0.31 0.31 0 49.69 12.08 0.31 0.62 0 49.38 11.77 0.30 0.92 0 49.08 11.47 0.29 1.21 0 48.79 11.18 0.29 1.50 0 48.50 10.89 0.28 1.79 0 48.21 10.61 Page 87 hf/(hf)total 0.016 0.032 0.048 0.064 0.079 0.094 Merkley & Allen III. Determining X1 and X2 in Dual Pipe Size Laterals • The friction loss is: ⎛ J FL J F x ⎞ ⎛ J F x ⎞ hf = ⎜ 1 1 − 2 2 2 ⎟ + ⎜ 3 2 2 ⎟ 100 ⎠ ⎝ 100 ⎠ ⎝ 100 (132) where, hf = total lateral friction head loss for dual pipe sizes J1 = friction loss gradient for D1 and Qinlet J2 = friction loss gradient for D1 and Qinlet - (qa)(x1)/Se J3 = friction loss gradient for D2 and Qinlet - (qa)(x1)/Se F1 = multiple outlet reduction coefficient for L/Se outlets F2 = multiple outlet reduction coefficient for x2/Se outlets x1 = length of D1 pipe (larger size) x2 = length of D2 pipe (smaller size) x1 + x2 = L • • As in previous examples, we assume constant qa As for single pipe size laterals, we will fix hf by ∆h = hf + ∆he = 20% ha and, • (133) hf = 20% ha − ∆he (134) Find d1 and d2 in tables (or by calculation) using Qinlet and... ( J)d1 ≤ Ja ≤ ( J)d2 (135) ⎛ 20% ha − ∆he ⎞ Ja = 100 ⎜ ⎟ FL ⎝ ⎠ (136) for, • • Now there are two adjacent pipe sizes: d1 and d2 Solve for x1 and x2 by trial-and-error, or write a computer program, and make hf = 0.20ha - ∆he (you already have an equation for hf above) Merkley & Allen Page 88 Sprinkle & Trickle Irrigation Lectures IV. Setting up a Computer Program to Determine X1 and X2 • If the Hazen-Williams equation is used, the two F values will be: F1 ≈ 0.351 + 1⎛ 4⎞ 1+ ⎜ ⎟ 2N1 ⎝ 13 N1 ⎠ (137) F2 ≈ 0.351 + 1⎛ 4⎞ ⎜1 + ⎟ 2N2 ⎝ 13 N2 ⎠ (138) where L Se (139) L − x1 Se (140) N1 = N2 = • The three friction loss gradients are: 1.852 ⎛Q ⎞ J1 = K ⎜ 1 ⎟ ⎝C⎠ − D1 4.87 1.852 ⎛Q ⎞ J2 = K ⎜ 2 ⎟ ⎝C⎠ 1.852 ⎛Q ⎞ J3 = K ⎜ 2 ⎟ ⎝C⎠ (141) − D1 4.87 (142) − D24.87 (143) where ⎛L⎞ Q1 = ⎜ ⎟ qa Se ⎠ ⎝ ⎛ L − x1 ⎞ Q2 = ⎜ ⎟ qa Se ⎠ ⎝ • (144) (145) The coefficient K in Eqs. 141-143 is 1,050 for gpm & inches; 16.42(10)6 for lps and cm; or 1.217(10)12 for lps and mm Sprinkle & Trickle Irrigation Lectures Page 89 Merkley & Allen • Combine the above equations and set it equal to zero: f ( x1 ) = α1 ⎡α 2 − α3 (L − x1 ) ⎢ ⎣ 2.852 F2 ⎤ − 0.2ha + ∆he = 0 ⎥ ⎦ (146) where α1 = K 1.852 ⎛q L⎞ α2 = ⎜ a ⎟ ⎝ Se ⎠ α3 = • • • ( − D1 4.87 (147) 100 C1.852 − D1 4.87F1 L − − D24.87 ) (148) 1.852 ⎛ qa ⎞ ⎜ ⎟ ⎝ Se ⎠ (149) The three alpha values are constants Eq. 146 can be solved for the unknown, x1, by the Newton-Raphson method To accomplish this, we need the derivative of Eq. 146 with respect to x1 df ( x1) = dx1 ⎡ S (L − x1)0.852 ⎛ 8 Se ⎞ ⎤ = α1α3 ⎢ 2.852 F2 (L − x1)1.852 − e 1+ ⎜ ⎟⎥ 2 ⎢ ⎝ 13(L − x1) ⎠ ⎥ ⎣ ⎦ • • • • • (150) Note that the solution may fail if the sizes D1 & D2 are inappropriate To make things more interesting, give the computer program a list of inside pipe diameters so that it can find the most appropriate available values of D1 & D2 Note that the Darcy-Weisbach equation could be used instead of HazenWilliams In Eq. 146 you could adjust the value of the 0.2 coefficient on ha to determine its sensitivity to the pipe diameters and lengths The following screenshot is of a small computer program for calculating diameters and lengths of dual pipe size sprinkler laterals Merkley & Allen Page 90 Sprinkle & Trickle Irrigation Lectures V. Inlet Pressure for Dual Pipe Size Laterals 5 1 hl = ha + hf + ∆he + hr 8 2 • • (151) This is the same as the lateral inlet pressure head equation for single pipe size, except that the coefficient on hf is 5/8 instead of 3/4 Remember that for a downhill slope, the respective pressure changes due to friction loss and due to elevation change are opposing VI. Laterals with Flow Control Devices • • • • Pressure regulating valves can be located at the base of each sprinkler: These have approximately 2 to 5 psi (14 to 34 kPa) head loss Also, flow control nozzles (FCNs) can be installed in the sprinkler heads FCNs typically have negligible head loss For a lateral on level ground, the minimum pressure is at the end: Sprinkle & Trickle Irrigation Lectures Page 91 Merkley & Allen • The lateral inlet pressure head, hl, is determined such that the minimum pressure in the lateral is enough to provide ha at each sprinkler... hl = ha + hf + ∆he + hr + hcv (152) where hcv is the pressure head loss through the flow control device • For a lateral with flow control devices, the average pressure is not equal to the nominal sprinkler pressure havg ≠ ha • If the pressure in the lateral is enough everywhere, then ⎛q ⎞ ha = ⎜ a ⎟ ⎝ Kd ⎠ • (153) 2 (154) where ha is the pressure head at the sprinklers Below is a sketch of the hydraulics for a downhill lateral with flow control devices Merkley & Allen Page 92 Sprinkle & Trickle Irrigation Lectures VII. Anti-Drain Valves • • • • Valves are available for preventing flow through sprinklers until a certain minimum pressure is reached These valves are installed at the base of each sprinkler and are useful where sprinkler irrigation is used to germinate seeds on medium or high value crops The valves help prevent seed bed damage due to low pressure streams of water during startup and shutdown But, for periodic-move, the lines still must be drained before moving Sprinkle & Trickle Irrigation Lectures Page 93 Merkley & Allen Gravity-Fed Lateral Hydraulic Analysis I. Description of the Problem • A gravity-fed sprinkler lateral with evenly spaced outlets (sprinklers), beginning at a distance Se from the inlet: • The question is, for known inlet head, H0, pipe diameter, D, sprinkler spacing, Se, ground slope, So, sprinkler discharge coefficient, Kd, riser height, hr, and pipe material (C factor), what is the flow rate through each sprinkler? Knowing the answer will lead to predictions of application uniformity In this case, we won’t assume a constant qa at each sprinkler • • II. Friction Loss in the Lateral Hazen-Williams equation: hf = JL 100 1.852 6⎛Q⎞ J = 16.42 (10 ) ⎜ ⎟ ⎝C⎠ (155) D−4.87 (156) for Q in lps; D in cm; J in m/100 m; L in m; and hf in m. Merkley & Allen Page 94 Sprinkle & Trickle Irrigation Lectures • Between two sprinklers, 1.852 JSe ⎛Q⎞ 4 hf = = 16.42 (10 ) Se ⎜ ⎟ 100 ⎝C⎠ D−4.87 (157) or, hf = hw Q1.852 (158) where Q is the flow rate in the lateral pipe between two sprinklers, and hw = 16.42 (10 ) SeC−1.852D−4.87 4 (159) III. Sprinkler Discharge typically, q = Kd h (160) where q is the sprinkler flow rate in lps; h is the pressure head at the sprinkler in m; and Kd is an empirical coefficient: Kd = KoA, where A is the cross sectional area of the inside of the pipe IV. Develop the System of Equations • • Suppose there are only four sprinklers, evenly spaced (see the above figure) Suppose that we know H0, Kd, C, D, hr, So, and Se ⎛q ⎞ (Q − Q ) H1 = hr + ⎜ 1 ⎟ = hr + 1 2 2 Kd ⎝ Kd ⎠ 2 q1 = K d H1 − hr → 2 (161) q2 = K d H2 − hr → ⎛q ⎞ (Q − Q ) H2 = hr + ⎜ 2 ⎟ = hr + 2 2 3 Kd ⎝ Kd ⎠ q3 = K d H3 − hr → ⎛q ⎞ (Q − Q ) H3 = hr + ⎜ 3 ⎟ = hr + 3 2 4 Kd ⎝ Kd ⎠ q4 = K d H4 − hr → ⎛ q4 ⎞ Q42 H4 = hr + ⎜ ⎟ = hr + 2 Kd ⎝ Kd ⎠ 2 2 2 (162) 2 (163) 2 Sprinkle & Trickle Irrigation Lectures Page 95 (164) Merkley & Allen • Pressure heads can also be defined independently in terms of friction loss along the lateral pipe H1 = H0 − hw Q11.852 − ∆he (165) H2 = H1 − hw Q21.852 − ∆he (166) H3 = H2 − hw Q31.852 − ∆he (167) H4 = H3 − hw Q41.852 − ∆he (168) where, ∆he = Se − So2 (169) +1 and So is the ground slope (m/m) • • • • The above presumes a uniform, constant ground slope Note that in the above equation, ∆he is always positive. So it is necessary to multiply the result by –1 (change the sign) whenever So < 0. Note also that So < 0 means the lateral runs in the downhill direction Combining respective H equations: ( Q1 − Q2 ) 2 = H − hw Q11.852 − ∆he − hr ( Q 2 − Q3 ) 2 = H − ∆he − hr (171) ( Q3 − Q 4 ) 2 = H − hw Q31.852 − ∆he − hr (172) K d2 K d2 K d2 Q4 2 K d2 Merkley & Allen 0 1.852 1 − h w Q2 2 = H3 − hw Q41.852 − ∆he − hr Page 96 (170) (173) Sprinkle & Trickle Irrigation Lectures • Setting the equations equal to zero: f1 = H0 ( Q1 − Q2 )2 − h − K2 d 1.852 w Q1 f2 ( Q1 − Q2 )2 − ( Q2 − Q3 )2 − h = f3 ( Q2 − Q3 )2 − ( Q3 − Q4 )2 − h = K2 d K2 d K2 d f4 K2 d ( Q3 − Q4 )2 − Q42 − h = K2 d K2 d − ∆he − hr = 0 1.852 w Q2 1.852 w Q3 1.852 w Q4 (174) − ∆he − hr = 0 (175) − ∆he − hr = 0 (176) − ∆he − hr = 0 (177) The system of equations can be put into matrix form as follows: ⎡ ∂f1 ⎢ ∂Q ⎢1 ⎢ ∂f2 ⎢ ∂Q ⎢1 ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ∂f1 ∂Q2 ∂f2 ∂Q2 ∂f2 ∂Q3 ∂f3 ∂Q2 ∂f3 ∂Q3 ∂f4 ∂Q3 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ∂f3 ⎥ ⎥ ∂Q4 ⎥ ∂f4 ⎥ ⎥ ∂Q4 ⎦ ⎡ ⎤ ⎡⎤ ⎢ δQ1 ⎥ ⎢ f1 ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ δQ2 ⎥ ⎢ f2 ⎥ ⎢ ⎥ =⎢ ⎥ ⎢ ⎥ ⎢⎥ δQ3 ⎥ ⎢ ⎢ f3 ⎥ ⎢ ⎥ ⎢⎥ ⎢ δQ ⎥ ⎢ f4 ⎥ 4 ⎢ ⎥ ⎢⎥ ⎣⎦ ⎣ ⎦ (178) where the two values in the first row of the square matrix are: ∂f1 −2 ( Q1 − Q2 ) 0.852 = − 1.852hw Q1 2 ∂Q1 Kd (179) 2 ( Q1 − Q2 ) ∂f1 = ∂Q2 K2 d (180) Sprinkle & Trickle Irrigation Lectures Page 97 Merkley & Allen The two values in the last row of the square matrix, for n sprinklers, are: 2 ( Qn−1 − Qn ) ∂fn = ∂Qn−1 K2 d (181) ∂fn −2Qn−1 0 = − 1.852hw Qn.852 2 ∂Qn Kd (182) and the three values in each intermediate row of the matrix are: 2 ( Qi−1 − Qi ) ∂fi = ∂Qi−1 K2 d (183) 2 ( Qi+1 − Qi−1 ) ∂fi = − 1.852hw Qi0.852 2 ∂Qi Kd (184) 2 ( Qi − Qi+1 ) ∂fi = ∂Qi+1 K2 d (185) where i is the row number • • • • • • • This is a system of nonlinear algebraic equations The square matrix is a Jacobian matrix; all blank values are zero Solve for Q1, Q2, Q3, and Q4 (or up to Qn, in general) using the NewtonRaphson method, Gauss elimination, and backward substitution (or other solution method for a linear set of equations) Knowing the flow rates, you can go back and directly calculate the pressure heads one by one The problem could be further generalized by allowing for different pipe sizes in the lateral, by including minor losses, by allowing variable elevation changes between sprinkler positions, etc. However, it is still a problem of solving for x unknowns and x equations For pumped systems (not gravity, as above), we could include a mathematical representation of the pump characteristic curve to determine the lateral hydraulic performance; that is, don’t assume a constant H0, but replace it by some function Merkley & Allen Page 98 Sprinkle & Trickle Irrigation Lectures V. Brute-Force Approach • • There is a computer program that will do the above calculations for a gravityfed lateral with multiple sprinklers But, if you want to write your own program in a simpler way, you can do the calculations by “brute-force” as follows: 1. 2. 3. 4. 5. Guess the pressure at the end of the lateral Calculate q for the last sprinkler Calculate hf over the distance Se to the next sprinkler upstream Calculate ∆he over the same Se Get the pressure at that next sprinkler and calculate the sprinkler flow rate 6. Keep moving upstream to the lateral inlet 7. If the head is more than the available head, reduce the end pressure and start over, else increase the pressure and start over • Below is a screenshot of a computer program that will do the above calculations for a gravity-fed lateral with multiple sprinklers Sprinkle & Trickle Irrigation Lectures Page 99 Merkley & Allen Merkley & Allen Page 100 Sprinkle & Trickle Irrigation Lectures ...
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