--- title: "Sprinkler layout wetting pattern" subject: "Irrigation and Drainage Engineering" chapter: 1 key: "semester_6/irrigation_and_drainage_engineering:sprinkler-wetting-pattern-plot" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/sprinkler-wetting-pattern-plot site: BCT Engineering --- # Sprinkler layout wetting pattern Irrigation and Drainage Engineering · Chapter 1: Introduction - Asked in: 2072 Magh, 5 marks - Video: [Auto CAD learning. Lect 4. Sprinkler irrigation system Design, Sprinklers quantity. By Engr. Waheed](https://www.youtube.com/watch?v=KJxc3xpTRmE) — Landscape maintenance and development - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/sprinkler-wetting-pattern-plot - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-1 ## Question A farmer with his 40 m × 40 m plot plans to irrigate his field using 4 sprinklers having a throw distance as 10 m and each placed 20 m apart. Prepare a sketch of wetting pattern of these sprinklers. Write your comments on the moisture pattern and suggest measure to improve it if required. ## Short answer Placing 4 sprinklers 20 m apart with 10 m throw wets only 78.5% (1257 m²) of the 1600 m² plot, leaving 343 m² dry at centre and corners because circles barely touch. To improve, use a4x4 layout at10 m spacing or increase throw to ~14.2 m for full overlap, or use triangular spacing. ## Answer #### Layout and wetting pattern The four sprinklers are placed 20 m apart, i.e. at the centres of the four 20 m x 20 m quarters of the plot, at (10,10), (30,10), (10,30) and (30,30) m. Each wets a circle of radius 10 m. ```text 40 m +--------------------+ | .--. .--. | | ( S1 ) ( S2 ) | | '--' '--' | | .--. .--. | | ( S3 ) ( S4 ) | | '--' '--' | 0 +--------------------+ 0 20 40 m neighbouring circles touch at only one point; the 4 corners and the centre are dry ``` #### Calculation $$ \begin{aligned} A_{plot} &= 40\times 40 = 1600\ \text{m}^2\\ A_{wetted} &= 4\times\pi (10)^2 = 1256.6\ \text{m}^2\\ \text{Coverage} &= 1256.6/1600 = 78.5\%\\ \text{Dry area} &\approx 343\ \text{m}^2\ (21.5\%) \end{aligned} $$ #### Comments - Neighbouring circles only touch; there is no overlap, so the centre of the plot, the four corners and the spaces between the circles stay dry. - The depth of water is highest near each sprinkler and falls to zero at the 10 m edge, so the distribution is non-uniform (poor uniformity coefficient). - About 21% of the plot gets no water. #### Improvement 1. **Reduce the spacing** to about 50-60% of the wetted diameter (20 m), i.e. 10-12 m square spacing, so that neighbouring circles overlap. A 3 x 3 layout (13.3 m spacing, 9 sprinklers) just covers the whole plot; a 4 x 4 layout (10 m spacing, 16 sprinklers) gives good overlap and uniformity. 2. **Or increase the throw radius** to at least $10\sqrt{2} = 14.14$ m (about 14.2 m) with the same four sprinklers, so that the circles meet at the centre and corners of each quarter. 3. Use a triangular layout for better uniformity, and use part-circle sprinklers along the edges so that water is not thrown outside the plot. 4. Avoid operating in high wind. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Where are the four sprinklers placed? A. At the four corners B. All at the centre C. At (10,10), (30,10), (10,30), (30,30) m D. Along one edge Answer: C. They are 20 m apart at centres of the quarters. (Key point: Layout) 2. What fraction of the plot is wetted? A. 100% B. 50% C. 95% D. About 78.5% Answer: D. 4 x pi x 10^2 = 1256.6 m^2 of 1600 m^2. (Key point: Coverage) 3. Why is the pattern poor? A. Circles only touch with no overlap, leaving dry centre, corners and gaps B. Circles overlap too much C. Sprinklers are too close D. Water falls evenly Answer: A. Non-uniform depth and about 21% dry area. (Key point: Comment) 4. Minimum throw radius for four sprinklers to meet at the plot centre? A. 20 m B. 10 x sqrt(2) = 14.14 m C. 10 m D. 7.07 m Answer: B. The centre of the plot is 14.14 m from each sprinkler. 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