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 = 
----------- ( 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
Answer:
Q=0.95 W/m
Explanation:
Given that
Outer diameter = 0.3 m
Thermal conductivity of material
So the mean conductivity
So heat conduction through cylinder
Q=0.95 W/m
Answer:
σ =5.39Mpa
Explanation:
step one:
The flexure strength is defined as the tendency with which unreinforced concrete yield to bending forces
Flexural strength test Flexural strength is calculated using the equation:
σ = FL/ (bd^2 )----------1
Where
σ = Flexural strength of concrete in Mpa
F= Failure load (in N).
L= Effective span of the beam
b= Breadth of the beam
step two:
Given data
F=40.45 kN= 40450N
b=0.15m
d=0.15m
L=0.45m
step three:
substituting into the expression we have
σ = 40450*0.45/ (0.15*0.15^2 )
σ =18202.5/ (0.15*0.15^2 )
σ =18202.5/ (0.15*0.0225 )
σ =18202.5/0.003375
σ =5393333.3
σ =5393333.3/1000000
σ =5.39Mpa
Therefore the flexure strength of the concrete is 5.39Mpa
Answer:
Using the formula
V =20y/(x^2+y^2)^1/2 - 20x/(x^2+y^2)^1/2
Hence fluid speed at x axis =20x/(x^2+y^2)^1/2
While the fluid speed at y axis =20y/(x^2+y^2)^1/2
Now the angle at 1, 5
We substitute into the formula above
V= 20×5/(1+25)^1/2= 19.61
For x we have
V = 20× 1/(1+25)^1/2= 3.92
Angle = 19.61/3.92= 5.0degrees
Angel at 5, and 2
We substitute still
V = 20×5/(2+25)^1/2=19.24
At 2 we get
V= 20×2/(2+25)^1/2=7.69
Dividing we get 19.24/7.69=2.5degrees
At 1 and 0
V = 20/(1)^1/2=20
At 0, v =0
Angel at 2 and 0 = 20degrees
At 5 and 2
V = 100/(25+ 4)^1/2=18.56
At x = 2
40/(√29)=7.43
Angle =18.56/7.43 = 2.49degrees.
Answer:
minimum radius of bend.
Explanation:
The sharpest bend that can be placed in a piece of metal without critically weakening the part is called the minimum radius of bend.
