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garri49 [273]
3 years ago
12

Water discharging into a 10-m-wide rectangular horizontal channel from a sluice gate is observed to have undergone a hydraulic j

ump. The flow depth and velocity before the jump are 0.8m and 7m/s, respectively. Determine (a) the flow depth and the Froude number after the jump (b) the head loss (c) the dissipation ratio.

Engineering
1 answer:
elixir [45]3 years ago
3 0

Answer:

A) Flow depth = 2.46 m, Froude number after jump = 0.464

B) head loss = 0.572 m

C) dissipation ratio = 0.173

Explanation:

Given data :

Velocity before jump ( v1 ) = 7 m/s

flow depth before jump ( y1 ) = 0.8 m

g = 9.81 m/s

Esi = 3.3 m ( calculated )

attached below is a detailed solution of the problem

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8 0
3 years ago
Three groups of students are given study outlines for 6 weeks. One group studies 2 hours a night, a second group studies 1 hour
katrin2010 [14]

Answer:

The constant here is the study outline

Explanation:

In scientific research, the constant variable is that part/variable of the experiment that does not change or is set not to change. Examples include temperature, environment or height.

Assuming the scenery described in this question is an experiment. All the groups presented are bound by a constant during the experiment. The constant here is the study outline. The study outline provided to the students is not going to change.

NOTE: There could be confusion as regards the answer being the final exam grade but that will be the dependent variable as that will be the outcome of the experiment while the time spent to study will be the independent variable.

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3 years ago
Which of the following openings does not require guardrails?
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Answer:

Potholes

Explanation:

7 0
3 years ago
An equimolar liquid mixture of benzene an toluene is separated into two product streams by distillation. Inside the column a liq
mamaluj [8]

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7 0
3 years ago
Fluid originally flows through a tube at a rate of 100 cm^3/s. To illustrate the sensitivity of the Poiseuille flow rate to vari
ankoles [38]

Answer:

a) 0.01

b) 150 cm^3/s

c) 300 cm^3/s

d) 25 cm^3/s

Explanation:

a) We know that :  

     Q=ΔP/R

     R=8ηl/π*r^4

Givens:

     r^2 = 0.1 r_1  

Plugging known information to get :  

      Q=ΔP/R

         =ΔP*π*r^4/8*η*l

Q_2/r_2^4 =Q_1/r_1^4

           Q_2=Q_1/r_1^4*r_2^4

                  =Q_1/r_1^4*r*0.0001*r_1^4

          Q_2 = 0.01

b) From the rate flow of the fluid we know that :  

            Q=ΔP/R                                   (1)

              F=η*Av/l                                 (2)

             R=8*ηl/π*r^4                           (3)

<em>Where: </em>

ΔP is the change in the pressure .  

r is the raduis of the tube .

l is the length of the tube .

η is the coefficient of the vescosity of the fluid .

R is the resistance of the fluid .

Givens: Q1 = 100 cm^3/s , ΔP= 1.5

Plugging known information into EQ.1 :  

                 Q=ΔP/R

     Q_2/ΔP2=Q_1/ΔP

              Q_2=150 cm^3/s

c) we know that :

      F = η*Av/l  

can be written as :  

     ΔP = F/A = η*v/l  

Givens: η_2 = 3η_1  

           Q=ΔP/R  

           Q=η*v/l*R

Q_2/η_2=Q_1/η_1

       Q_2=300 cm^3/s

d) We know that :  

           Q=ΔP/R

           R=8*ηl/π*r^4  

Givens: l_2 = 4*l_1

Plugging known information to get :  

              Q=ΔP/R  

              Q=ΔP*π*r^4/8*ηl

   Q_2/l_2=Q_1/l_1

          Q_2 = 25 cm^3/s

5 0
4 years ago
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