ABE 325 2021 HWK Channel Stability Restoration and Water Supply - KEY1

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Purdue University *

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325

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Civil Engineering

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Dec 6, 2023

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ABE 32500 Soil and Water Resources Engineering 1 Channel Stability, Restoration and Water Supply Homework (11 pts) 1. (3 pts) Water Supply - Calculate the annual water requirement, storage volume needed, and watershed area required for an impoundment in central Indiana. The supply must serve a poultry grower with 200,000 chickens. Seepage and evaporation losses are 45% of the storage capacity. Estimate annual direct use from Table 11.1: Type of Use Units Annual Water use rate Annual water use Chickens 200,000 birds 0.0011 ha·m/year/100 birds 2.2 ha·m/year Total 2.2 ha·m/year Annual water requirement = 2.2 ha·m/year. Storage volume = Use volume 1-Loss fraction = 4.4 67·9 :;<.=> = 4.00 ha · m Using Figure 11.1 you will find the watershed area required in central Indiana to yield 1 ha·m/year of runoff to be 13 ha, so Watershed area = 4.00 ha · m :?67 67·9 = 52ha 2. (3 pts) Water Supply - Determine the discharge of a 600-mm diameter well that fully penetrates a confined aquifer. The aquifer medium is coarse sand. The top of the aquifer is 28 m below the ground surface. The aquifer is 16 m thick. The water level in an observation well at a distance of 100 m is 7.6 m below the surface. The water level in the well is 24.0 meters below the surface. Equation for steady-state discharge for a well that fully penetrates an extensive, confined aquifer (11.2): ࠵? = 2࠵?࠵?࠵?(ℎ 4 −ℎ : ) log @ (࠵? 4 /࠵? : ) Where d = 16 m, r 1 = 0.300 m, r 2 = 100 m, assume K = 10 -3 m/s (Table 11.2), h 1 = (28 m + 16 m) – (24 m) = 20 m; h 2 = (28 m + 16 m) – (7.6 m) = 36.4 m ࠵? = 2࠵?(10 ;? )(16 ࠵?)(36.4 ࠵? − 20 ࠵?) log @ ( :<< A <.?<< A ) = 0.284 ࠵? ? ࠵? = 284 ࠵?/࠵?
ABE 32500 Soil and Water Resources Engineering 2 3. (2 pts) Channel Restoration - Use Internet resources to determine the hydrologic unit code, drainage area, location and datum of the Wabash River discharge gauge in West Lafayette / Lafayette, IN. Latitude 40°25'30.6", Longitude 86°53'47.4" NAD83 Tippecanoe County, Indiana, Hydrologic Unit 05120108, but renumbered to HUC 03335500 (both numbers are provided by the USGS information page). Drainage area: 7,267 square miles Datum of gage: 503.84 feet above NAVD88.
ABE 32500 Soil and Water Resources Engineering 3 4. (3 pts) Channel Restoration - Compute the required diameter of rock material to form a riffle in a stream with a width to depth ratio of 8:3. The bankfull depth is 1.3 m and the slope of the stream is 0.01 m/m. What would be the diameter required if the channel slope was 0.001 m/m? Width to depth ratio should be based on a stable riffle, so it reflects the conditions of the channel where we have been asked to make our calculations. Start by calculating the width of the channel for the given bank full depth: t = 8 / 3 × 1.3 m = 3.47 m Calculate the hydraulic radius for a parabolic stream channel: ࠵? = 2࠵?࠵? 3 = 2(3.47 ࠵?)(1.3࠵?) 3 = 3.01 ࠵? 4 ࠵? = ࠵? + 8࠵? 4 3࠵? = 3.47 ࠵? + 8(1.3 ࠵?) 4 3(3.47 ࠵?) = 4.77 ࠵? Note: The equation for P for a parabolic cross-section is wrong in the 5 th Edition textbook. ࠵? = ࠵? ࠵? = ࠵?. ࠵?࠵? ࠵? ࠵? ࠵?. ࠵?࠵? ࠵? = ࠵?. ࠵?࠵?࠵? m R can also be approximated using ࠵? = ࠵?࠵? ࠵? = ࠵?(࠵?.࠵?࠵?) ࠵? = ࠵?. ࠵?࠵?࠵? ࠵? Use equation 10.10 to estimate the shear stress from the defined flow: ࠵? ࠵? = ࠵?࠵?࠵? = (࠵?࠵?࠵?࠵?)(࠵?. ࠵?࠵?࠵? ࠵?)(࠵?. ࠵?࠵?) = ࠵?࠵?. ࠵? N/m 2 ( exact ) or = ࠵?࠵?. ࠵? N/m 2 (approx) At incipient motion, the critical shear stress must be equal to the hydraulic shear stress of the stream. Assuming a particle density of 2650 kg/m 3 , solve for the minimum diameter from Equation 10.11 as: ࠵? = ࠵? ࠵? ࠵?࠵?(࠵? ࠵? − ࠵?) = ࠵?࠵?. ࠵? (࠵?. ࠵?࠵?࠵?)(࠵?. ࠵?࠵?)(࠵?࠵?࠵?࠵? − ࠵?࠵?࠵?࠵?) = ࠵?. ࠵?࠵?࠵? m = ࠵?࠵? mm ( exact ) or = ࠵?. ࠵?࠵?࠵? ࠵? = ࠵?࠵?࠵? ࠵?࠵? (approx) If the slope is reduced to 0.001 m/m then the size of the particle can be reduced to: D s=0.001 = ࠵?. ࠵?࠵?࠵?࠵? m = ࠵?. ࠵? mm ( exact ) or = ࠵?. ࠵?࠵?࠵?࠵? ࠵? = ࠵?࠵?. ࠵? ࠵?࠵? (approx)
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