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Margaret [11]
3 years ago
11

Water drains at a constant rate through a saturated soil column with a diameter of 1.5 feet and a height of 3 feet. The hydrauli

c head is maintained at 5 feet at the top of the column and 0.81 feet at the bottom. After a period of one hour, 199 cubic inches of water has drained through the column. What is the hydraulic conductivity of the soil in the column in cubic feet per hour
Engineering
1 answer:
Debora [2.8K]3 years ago
6 0

Storm Water Design Criteria Manual — January 2016

Chapter 2 | Page | 2.3

2.6 TECHNICAL CRITERIA

The storm drainage planning and design should follow the criteria within this manual.

The following two items should be considered during the design process.

1)

Diversion of storm water away from the natural watercourse will not be allowed, except

within the property boundaries controlled by the developer under the following conditions: a)

The storm water is returned to its natural flowing watercourse prior to leaving the developer's

property, b) For watersheds greater than twenty (20) acres, a timing analysis of the existing

and diverted hydrograph must be performed to confirm that the peak flow rate has not been

increased at the point that it reenters the watercourse, as a result of the diversion.

2)

All developments shall provide adequate drainage outfall at the lower end of the site into an

existing street, alley, drainage, easements or right-of-way, or to the centerline of an existing

natural drain. Where a proposed street, storm drain, or open channel does not discharge into a

natural low or into an existing adequate drainage easement, then facilities and drainage

easements of adequate width — to contain the design discharge — shall be constructed and

dedicated to the centerline of an existing natural low within the same watershed. However,

when the natural low lies within the developer's property, the developer will only be required

to plat an easement to the centerline of the natural low; provided that the easement is able to

accommodate the facilities that will be built in conjunction with future development of that

property.

2.7 FLOODPLAIN MANAGEMENT

Floodplain management has changed over the years due to the National Flood Insurance Act

of 1968. Changes from the program included USGS maps being used with the delineation of

floodplains, HUD issuing flood insurance maps in the late 1970’s, and the founding of FEMA

in 1979. The National Flood Insurance Program was ultimately placed under FEMA. As the

City began participating in this action in the late 1970’s, flood insurance can be purchased

through insurance carriers for buildings in Bexar County.

Floodplain management is used to minimize flooding of buildings, reduce flood losses, and

improve the quality of life — and safety of the citizens of Bexar County.

2.8 IMPLEMENTATION

The provisions of this manual shall apply to any application for subdivision plat, master

development plan, capital improvement project, or building permit approval except as

otherwise provided by Chapter 35 of the UDC. A storm water management plan shall be

provided as set forth in Appendix “A” of this manual.

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3 years ago
Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,
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Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0

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v_0 = Initial velocity of ball A

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(v_A)_n' = Final velocity of ball A

v_B' = Final velocity of ball B

e = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0

Coefficient of restitution is given by

e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0

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Adding the above two equations we get

2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0

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(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0

From the conservation of momentum along the plane of contact we have

(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}

v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}

Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0.

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