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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 heat transfer rate due to free convection from a vertical surface, 1 m high and 0.6 m wide, to quiescent air that is 20 K co
andreyandreev [35.5K]

Answer:

The ratio of heat transfer rate is 0.88

Explanation:

Given;

Case1 :

height of vertical surface, L = 1 m

width of vertical surface, w = 0.6 m

Case 2:

height of vertical surface, L = 0.6 m

width of vertical surface, w = 1 m

At an assumed film temperature of air = 300 K

then, read off from heat transfer table, temperature inverse β, surface area flow rate v, and Pr, to determine Rayleigh number for the two cases.

β = 1/300 = 0.00333 K⁻¹

v = 15.89 x 10⁻⁶ m²/s

Pr = 0.69

Case 1, L = 1 m

R_a = \frac{g\beta TL^3P_r}{v^2}

R_a = \frac{9.8*0.00333* 20*1^3*0.69}{(15.89x10^{-6})2} \\\\R_a = 1.784 *10^9

Case 2, L = 0.6 m

R_a = \frac{g\beta TL^3P_r}{v^2} \\\\R_a = \frac{9.8*0.00333* 20*0.6^3*0.69}{(15.89*10^{-6})^2}\\\\ R_a = 3.853 *10^8

From the values of Rayleigh numbers above, case 1 is Turbulent flow while case 2 is laminar flow

Thus: C₁ = 0.1, n₁ = ¹/₃

          C₂ = 0.59, n₂ = 1/4

Ratio of heat transfer rate is given as:

\frac{q_1}{q_2} = \frac{h_1 \delta T}{h_2 \delta T} \\\\\frac{q_1}{q_2} = \frac{h_1}{h_2} \\\\But, \frac{hL}{k} = CR_a^n L, \ \ h=\frac{k}{L}(CR_a^n L)\\\\\frac{q_1}{q_2} = \frac{C_1R_a_1^n L_2}{C_2R_a_2^n L_1} = \frac{0.1(1.784*10^9)^{\frac{1}{3}} *0.6}{0.59(3.853*10^8)^{\frac{1}{4}} *1} \\\\\frac{q_1}{q_2} = \frac{72.76}{82.66} = 0.88

Therefore, the ratio of heat transfer rate is 0.88

4 0
3 years ago
How to clean a snowblower carburetor without removing it.
Anarel [89]

Answer:

Spray carburetor cleaner on the inside of the bowl and wipe the liquid, dirt, and concentrated fuel off of it. Now take the main jet, spray the cleaner through it and wipe off the dirt. Then take a copper wire, scrub it through the tiny holes in the jet to complete the cleaning process.

Explanation:

8 0
3 years ago
Using the background information provided and your answers to questions 1 & 2, look at the hex packet below and identify the
yanalaym [24]

Answer:

Goal address: Based on the above Ethernet outline design, we get that the goal address begins from the first hex worth and is of size 12 hex values (got from question 1). In light of the bundle given, we get that the goal address is 6c40 0889 c448.  

Source address: Similarly, the source address begins after the goal address and is of size 12 hex values (got from question 1). Thus, the appropriate response is f832 e4a7 fb38.  

Type/Length: The sort/length begins after the source address and is of size 4 hex values (got from question 1). Thus, the appropriate response is 0806.  

Data (Payload): The data(payload) begins after the sort/length and finishes not long before FCS(Checksum) which is of 4 bytes for example 4 * 2 = 8 hex qualities. So the information comprises of everything between the 'type/length' and 'CRC'. We persuade the CRC to be c0a8 01f2. Henceforth, the appropriate response is 0001 0800 0604 0002 f832 e4a7 bf38 c0a8 0101 6c40 0889 c448.  

Question 4:  

The Ethernet parcel type characterizes the convention utilized for sending the bundle information. We realize that type 0x0800 demonstrates the IPv4 convention, 0x0806 shows an ARP convention, and 0x86DD demonstrates an IPv6 convention. In light of the appropriate response we got in Question 3, we realize that the sort/length esteem for the given parcel is 0806, which implies that convention utilized for sending the information bundle was ARP convention.

8 0
3 years ago
Info security:
il63 [147K]

Answer:

True

Explanation:

Dual home host - it is referred to as the firewall that is incorporated with two or more networks. out of these two networks, one is assigned to the internal network and the other is for the network. The main purpose of the dual-homed host is to ensure that no Internet protocol traffic is induced between both the network.

The most simple example of a dual-homed host is a computing motherboard that is provided with two network interfaces.

7 0
4 years ago
Substance Specific Heat (cal/g°C) Specific Heat (J/g°C) water 1.00 4.18 steam 0.96 4.02 alcohol 0.59 2.47 ice 0.50 2.09 wood 0.4
Jobisdone [24]

Answer:

C) 43,2°C

Explanation:

<em>Sensible heat</em> is the amount of thermal energy that is required to change the temperature of an object, the equation for calculating the heat change is  given by:

Q=msΔT

where:

  • Q, heat that has been absorbed or realeased by the substance [J]
  • m, mass of the substance [g]
  • s, specific heat capacity [J/g°C]
  • ΔT, changes in the substance temperature [°C]

To solve the problem, we clear ΔT of the equation and then replace our data:

Q=890 [J],

m=16,6 [g],

s=2,74 [J/g°C]

Q=msΔT.......................ΔT=Q/ms

ΔT=\frac{890 J}{16,6g*2,47\frac{J}{gC}}=21,7°C

As:

ΔT=Tfinal-Tinitial

Tfinal=ΔT+Tinitial

Tfinal=21,7+21,5=43,2°C

The final temperature of the ethanol is 43,2°C.

4 0
4 years ago
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