Change the depth of gravel on the ground surface to 0m. How does this affect the design? Explain why?
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a.) Determine the dimensions of a reinforced concrete box culvert that is to be placed under a roadway, and the gradual transition needed for a 15-ft wide rectangular flood control channel at a slope of 0.0001 ft. Transition structures are used to transition flow from an open channel to a culvert and then transition flow from the downstream of the culvert back to the channel. Gradual transitions are used when the flow is subcritical throughout the culvert. The culvert has a length of 150 ft and is to convey a design discharge of 135 cfs. Use a Manning’s n of 0.02. Straight line transitions are to be used so the expansion and contraction coefficients, Ce and Cc, are respectively, 0.5 and 0.3. Assume that section 1 defines the flood-control channel just before the contraction upstream of the culvert and section 4 defines the flood-control channel after the expansion downstream of the culvert. Sections 2 and 3 are located at the upstream and downstream ends of the culvert. The headloss in the contraction is Cc(V2²/2g - V1²/2g) and the headloss in the expansion is Ce(V3²/2g - V4²/2g). The transition lengths (LT) are specified using a 4.5:1 flare so that LT = 4.5(W4 - W3)/2 and LT = 4.5(W1 - W2)/2, where W is the widths at the respective sections. Choose y3 ~ 1.1yc to have a subcritical depth. One important aspect of the culvert and transition design is to make sure that the backwater effect (y1 - y4) does not exceed a preferred or required amount (freeboard). For the purposes of this problem, that amount is 1 ft. You may design and analyze this culvert either by hand, spreadsheet, or HEC-RAS. b.) Analyze the culvert that you designed. The streambed elevation at the upstream end of the culvert is 1000 ft above mean sea level. Will this culvert have inlet or outlet control for the design discharge? What does this mean with regards to your design? c.) Perform a water surface profile analysis of the culvert and transition.
Sri K.
Krishna G.
Solⁱ Given that the depth of flow ranges 0.6 m < H < 0.9 m. ∴ Considering => H = (0.6 + 0.9) / 2 = 0.75 m Now verify the equation & above consideration Q = S⁰ᐧ⁵ (BH)⁵/3 / r (B+2H)%/3 => 0.0002⁰ᐧ⁵ × (20 × 0.75)⁵/3 / 0.03 × (20 + 2 × 0.75)%/3 => 5.67 m3/sec Now, Considering a slightly less depth. => (0.6 + 0.75) / 2 = 0.675 m Q = 0.0002⁰ᐧ⁵ (20 × 0.675)⁵/3 / 0.03 × (20 + 2 × 0.675)%/3 = 4.74 m3/sec Now, A new depth b/w 0.675 & 0.75 m Avg of depth will be = (0.675 + 0.75) / 2 = 0.7125 m Q = 0.0002⁰ᐧ⁵ × (20 × 0.7125)⁵/3 / 0.3 × (20 + 2 × 0.7125)%/3 => 0.014 × 84.5 / 0.03 × 7.55 => 5.22 m3/sec Now, Taking Avg depth b/w 0.675 & 0.7125 => (0.675 + 0.7125) / 2 = 0.69375 Q = 0.14 × (20 × 0.69375)⁵/3 / 0.03 × (20 + 2 × 0.69375)%/3 => 4.90 m3/sec Now Taking bisectⁱ of 0.69375 & 0.7125 m Xₘ = (0.69375 + 0.7125) / 2 = 0.703 Q = 0.014 × (20 × 0.703)⁵/3 / 0.03 × (20 × 0.703)%/3 = 4.95 m3/sec ∴ The depth will be greater than 0.703 m & less than 0.71 m Ans.
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