1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Dennis_Churaev [7]
3 years ago
15

is released from under a vertical gate into a 2-mwide lined rectangular channel. The gate opening is 50 cm, and the flow rate in

to the channel is 10 m3/s. The channel is lined with reinforced concrete and has aManning roughness coefficient of 0.015, a horizontal slope, and merges with a river where the depth of flow is 3.5 m. Does a hydraulic jump occur in the lined channel
Engineering
1 answer:
SVEN [57.7K]3 years ago
4 0

Answer:

hello your question is incomplete attached below is the complete question

answer: There is a hydraulic jump

Explanation:

First we have to calculate the depth of flow downstream of the gate

y1 = C_{c} y_{g} ----------- ( 1 )

Cc ( concentration coefficient ) = 0.61  ( assumed )

Yg ( depth of gate opening ) = 0.5

hence equation 1 becomes

y1 = 0.61 * 0.5 = 0.305 m

calculate the flow per unit width q

q = Q / b ----------- ( 2 )

Q = 10 m^3 /s

 b = 2 m

hence equation 2 becomes

q = 10 / 2 = 5 m^2/s

next calculate the depth before hydraulic jump y2 by using the hydraulic equation

answer : where  y1 < y2 hence a hydraulic jump occurs in the lined channel

attached below is the remaining part of the solution

You might be interested in
A civil engineer is asked to design a curved section of roadway that meets the following conditions: With ice on the road, when
lianna [129]

Answer:

1. 3.4^{o}

2. 163.3 m

Explanation:

Static friction between road and rubber, μs =0.06

The maximum speed of the car, v = 50 km/h

                                              = (50)(1000/3600) m/s

                                               = 13.89 m/s

The acceleration due to gravity, g = 9.81 m/s^{2}

The frictional force, f = μsN     ...... (1)

The component mg cosθ which balance the normal reaction N

The component mg sinθ acts in an opposite direction to the frictional force f.

        ΣF = mg sinθ-f = 0      ...... (2)

Substitute the equation (1) in equation (2), we get

 ΣF = mgsinθ-μsN = 0

 mgsinθ-μsmgcosθ =0

 μs = sinθ/cosθ

   tanθ = μs

    θ = tan-1( μs) = tan-1(0.06) = 3.4^{o}

(b)The vertical component of the force is

N cosθ = fsinθ+mg

 N cosθ = μsNsinθ+mg

N[cosθ- μs sinθ] = mg     ...... (3)

The horizontal component of the force along the motion of the car is

Nsinθ+fcosθ = ma  (Centripetal acceleration, a = \frac {v^{2}}{r}

  Nsinθ+fcosθ = m(\frac {v^{2}}{r})

   Nsinθ+μsNcosθ = m(\frac {v^{2}}{r})

N[sinθ+μs cosθ] = m(\frac {v^{2}}{r})     ...... (4)    

Dividing the equation (4) with equation (3),

 [sinθ+μscosθ]/[cosθ- μs sinθ] = \frac {v^{2}}{rg}

 cosθ[sinθ/cosθ+μs]/cosθ[1- μs sinθ/cosθ] =\frac {v^{2}}{rg}

[tanθ+μs]/[1-μs tanθ] = \frac {v^{2}}{rg}      

 From part (1), tanθ = μs

 Then the above equation becomes

 \frac {(\mu_s+\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}

\frac {(2\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}

Therefore, the minimum radius of the curvature of the curve is

               r = \frac {v^{2}}{{2 \mu_s/[1-\mu_s^{2}]}g} 

                   = \frac {v^{2}[1-\mu_s^{2}]}{2\mu_s g}

                   = \frac {(13.89 m/s)^{2}[1-(0.06)^{2}]}{(2)(0.06)(9.81)}

                 = 163.3 m

5 0
3 years ago
Help me with this for brainiest:)
kirza4 [7]
Gsvshhenejwnbbwbdbdhebwn
3 0
3 years ago
The water of a 14’ × 48’ metal frame pool can drain from the pool through an opening at the side of the pool. The opening is abo
Tresset [83]

Answer:

Explanation:

Height h = 1.03m

Volume v = 3780 gallons = 3780 * 0.0037851m^3 = 14.3073m^3

Time t = 13.5 mins = 13.5 * 60 = 810 seconds

Length of pool L = 14 inch = 14 * 2.54 = 35.56cm

width of pool b = 48 inch = 48 * 2.54 = 121.92 cm

a.) Consider the bernoulli's equation is given as:

P_1+\rho gh_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gh_2 + \frac{1}{2}\rho v_2^2 ...(1)

consider the equation of bernoulli at the top of the pool

P_0+\rho gh_1 + \frac{1}{2}\rho v_1^2 =constant ...(2)

where P_1=P_0 atm pressure

At the top of the pool v_1=0m/s, substitute in V_1 in equation (2)

P_0+\rho gh_1 =constant ...(3)

Hence equation (3) serves as the bernoullis equation at the top.

b.) Consider the equation of bernoulli's at the opening of the pool

P_2+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(4)\\P_0+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(5)

where P_2=P_0 atm pressure and h_2=0m

P_0+\rho v_1^2 =constant ...(6)

Hence equation (6) serves as the bernoullis equation of water at the opening of the pool.

c.) Consider the equation (3) and (4)

        P_0+\rho gh_1 =P_0+\rho v_1^2\\\\\frac{1}{2}\rho v_2^2=\rho gh_1\\v_2^2=2gh_1\\v_2=(\sqrt{2gh_1})m/s...(7)    

Hence velocity is v_2=(\sqrt{2gh_1})m/s

d.) consider (7)

v_2=(\sqrt{2(9.81)(1.03)})=4.4954m/s(approx)

This is the norminal value of velocity  

e.) consider the equation of flow rate interval of v and t

flow(t)=\frac{dv}{dt}(m^3/s) hence this is the flow rate

f.) Consider the equation cross sectional area in terms of V,v2 and t

AV_2=\frac{v}{t}\\\\A=\frac{v}{v_2t}(m^2)...8

hence this serves as the cross sectional area.

g.) Consider the equation of area from equation (8)

A=\frac{v}{v_2t}\\=\frac{14.3073}{4.4954\times 810}=0.003929=0.00393m^2=39.3cm^2

6 0
3 years ago
A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2
Svetllana [295]

79 f t^{3} is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

V_{1}=100\ \mathrm{ft}^{3}

First has a natural water content of 25% = \frac{25}{100} = 0.25

Shrinkage limit, w_{1}=12 \%=\frac{12}{100}=0.12

G_{s}=2.70

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

V \propto[1+e]

\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}  ------> eq 1

e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}

The above equation is at S_{r}=1,

e_{1}=w_{1} \times G_{s}

Applying the given values, we get

e_{1}=0.25 \times 2.70=0.675

Shrinkage limit is lowest water content

e_{2}=w_{2} \times G_{s}

Applying the given values, we get

e_{2}=0.12 \times 2.70=0.324

Applying the found values in eq 1, we get

\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904

V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}

7 0
3 years ago
How to update android 4.4.2 to 5.1​
faust18 [17]

Answer:

try settings and go to updates?

Explanation:

8 0
3 years ago
Other questions:
  • Remy noticed that after oiling his skateboard wheels, it was easier to reach the speeds he needed to perform tricks. How did the
    6·1 answer
  • 5.5 A scraper with a 275 hp diesel engine will be used to excavate and haul earth for a highway project. An evaluation of the jo
    10·1 answer
  • An engineer is considering time of convergence in a new Layer 3 environment design. Which two attributes must be considered? (Ch
    15·1 answer
  • Air in a 10 ft3 cylinder is initially at a pressure of 10 atm and a temperature of 330 K. The cylinder is to be emptied by openi
    10·2 answers
  • Why do engineers (and others) use the design process?
    13·1 answer
  • Write a script (Program 2) to perform t he following matrix operations. Use output commands to clearly output each problem with
    15·1 answer
  • Steam enters an adiabatic turbine at 6 MPa, 600°C, and 80 m/s and leaves at 50 kPa, 100°C, and 140 m/s. If the power output of t
    14·1 answer
  • A(94,0,14) B(52,56,94) C(10,6,48) D(128,64,10)
    6·1 answer
  • A 1.9-mm-diameter tube is inserted into an unknown liquid whose density is 960 kg/m3, and it is observed that the liquid rises 5
    7·1 answer
  • Whose responsibility is it to provide direction on correct ladder usage?<br> select the best option.
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!