Solution 1 field area the field is given to be

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Solution: 1. Field Area: The field is given to be rectangular. Note that there are 43,560 ft 2 /acre. The irrigated area is the length of the mainline (1,320 ft) multiplied by twice the length of one lateral (2 x 600 ft): (1,320)(1,200) 36.4 acres 43,560 = 2. System Capacity: Use Eq. 5.4 and the given data: s Ad (36.4)(1.9) Q 453 453 373 gpm fT (7)(12) = = = 3. Velocity Checks: Table 8.4: 5-inch aluminum pipe has an inside diameter of 4.900 inches (0.408 ft). Note that the maximum recommended velocity, in general, for sprinkler systems is 5 to 7 fps. 3(a). Full system capacity in the mainline: s s Q 3 2 Q 4(373 gpm) V 6 A (60 s/min)(7.481gal/ft )(0.408 ft) = = = π .36 fps 3(b). Half system capacity in the mainline: s s Q /2 3 2 Q 2(373 gpm) V 3 2A (60 s/min)(7.481gal/ft )(0.408 ft) = = = π .18 fps 3(c). Half system capacity through a hydrant valve: s hydrant 3 2 Q 2(373 gpm) V 4 2A (60 s/min)(7.481gal/ft )(0.333 ft) = = = π .77 fps All of the above velocities are below 7 fps, so they are found to be acceptable.

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