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dimaraw [331]
4 years ago
9

Use the APWA 5600 methodology to compute the 1/25 design discharge for the following catchment. The catchment area consists of 1

0 acres of impervious area, 30 acres of apartments and 35 acres of single family housing. The maximum flow path to the outlet is 1000 feet long with the upper catchment slope having an average slope of 1.5% for approximately 150 feet and the remaining slope of the flow path averaging 1.8%.
Engineering
1 answer:
sweet [91]4 years ago
5 0

Explanation:

For a given flow rate, open channel flow based design requires larger conduit sizes than those dimensioned  based on pressure flow. While it may be more expensive to build designed storm drainage systems

Based on open channel flow, this design procedure provides a margin of safety by providing  headroom in the duct to accommodate an increase in flow above the design discharge. Beneath the majority

Under normal conditions, it is recommended that the size of storm drains be based on a gravity flow to full flow criteria  or almost full. The design of the pressure flow may be justified in certain cases. As the hydraulic calculations are  performed, frequent verification of the existence of the desired flow condition should be performed.

Storm drainage systems can often alternate between pressure and open channel flow conditions from one  section to another (Federal Highway Administration, US Department of Transportation, 1996).

For gravity flow conditions, the Manning formula must be edited as described below.

ܳ ൌ

1,486

݊

ܣܴ ଶ / ଷ ܵ ଵ / ଶ

Where:

Q = Discharge, cubic feet per second

A = flow cross-sectional area, square feet

n = Manning roughness coefficient (see Table 5603-1)

ܴ ൌ

ܣ

ܲ

R = hydraulic radius, feet

S = slope in feet per foot

P = wet perimeter in feet

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The Canadair CL-215T amphibious aircraft is specially designed to fight fires. It is the only production aircraft that can scoop
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3 years ago
A horizontal curve of a two-lane undivided highway (12-foot lanes) has a radius of 678 feet to the center line of the roadway. A
OLEGan [10]

Answer:

maximum speed for safe vehicle operation = 55mph

Explanation:

Given data :

radius ( R ) = 678 ft

old building located ( m )= 30 ft

super elevation = 0.06

<u>Determine the maximum speed for safe vehicle operation </u>

firstly calculate the stopping sight distance

m = R ( 1 - cos \frac{28.655*S}{R} )  ----  ( 1 )

R = 678  

m ( horizontal sightline ) = 30 ft

back to equation 1

30 = 678 ( 1 - cos (28.655 *s / 678 ) )

( 1 - cos (28.655 *s / 678 ) )  = 30 / 678 = 0.044

cos \frac{28.65 *s }{678}  = 1.044

hence ; 28.65 * s = 678 * 0.2956

s = 6.99 ≈ 7 ft

next we will calculate the design speed ( u ) using the formula below

S = 1.47 ut  + \frac{u^2}{30(\frac{a}{3.2} )-G1}  ----  ( 2 )

t = reaction time,  a = vehicle acceleration, G1 = grade percentage

assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0

back to equation 2

6.99 = 1.47 * u * 2.5 + \frac{u^2}{30[(11.2/32.2)-0 ]}

3.675 u  + 0.0958 u^2 - 6.99 = 0

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5 0
3 years ago
If a lever operates at a mechanical disadvantage, it means that the ________.
alekssr [168]

Answer:

The correct answer is option 'B': Load is far from fulcrum and the effort is applied near the fulcrum

Explanation:

A lever works on the principle of balancing of torques. The torque about the fulcrum by the load should be equal to the torque by the applied effort. Since we know that the torque is proportional to both the force and the distance it is applied from the distance from the axis of rotation. A lever is used when we need to lift a heavy load by utilizing this effect of the lever arm.

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4 0
3 years ago
Read 2 more answers
Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa
Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air,  R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes  2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

4 0
4 years ago
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