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In-s [12.5K]
3 years ago
8

Water flows around a 6-ft diameter bridge pier with a velocity of 12 ft/s. Estimate the force (per unit length) that the water e

xerts on the pier. Assume that the flow can be approximated as an potential fluid flow around the front half of the cylinder, but due to flow seperation, the average pressure on the rea half is constant and approximately equal to 1/2 the pressure at point A.
Engineering
1 answer:
jolli1 [7]3 years ago
3 0

Answer: hello the diagram related to your question is missing please the third image is the missing part of the question

Fx = 977.76 Ib/ft

Explanation:

<u>Estimate the force that water exerts on the pier </u>

V = 12 ft/s

D( diameter ) = 6 ft

first express the force  on the first half of the cylinder  as

Fx1 =  - -2\int\limits^\pi _\frac{\pi }{2}   {Ps*cos\beta *a} \, d\beta   ---------------- ( 1 )

where ; Fy = 0

Ps = Po + 1/2 Pv^2 ( 1 - 4 sin^2β )  ------------- ( 2 )

Input equation (2)  into equation ( 1 )         (note :  assuming Po = 0 )

attached below is the remaining part of the solution

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Find the mathematical equation for SF distribution and BM diagram for the beam shown in figure 1.​
Novosadov [1.4K]

Answer:

i) SF: v(x) = \frac{(w_0* x )^2}{2L}

ii) BM : = \frac{(w_0*x)^3}{6L}

Explanation:

Let's take,

\frac{y}{w_0} = \frac{x}{L}

Making y the subject of formula, we have :

y = \frac{x}{L} * w_0

For shear force (SF), we have:

This is the area of the diagram.

v(x) = \frac{1}{2} * y = \frac{1}{2} * \frac{x}{L} * w_0

= \frac{(w_0* x )^2}{2L}

The shear force equation =

v(x) = \frac{(w_0* x )^2}{2L}

For bending moment (BM):

BM = v(x) * \frac{x}{3}

= \frac{(w_0* x )^2}{2L}  * \frac{x}{3}

= \frac{(w_0*x)^3}{6L}

The bending moment equation =

= \frac{(w_0*x)^3}{6L}

5 0
4 years ago
Recommend the types of engineers needed to collaborate on a city project to build a skateboard park near protected wetlands.
4vir4ik [10]

Answer:

A civil engineer.

Explanation:

Civil engineering is the science that deals with the design, creation and maintenance of constructions for civil use on the earth's surface. Thus, this specialty seeks to adapt soils to the needs of life in society, creating buildings, bridges, and all other constructions adapted to civil life, while taking care of the correct use of soils and the correct distribution of spaces and resources. to be used for such constructions.

4 0
3 years ago
Why is the back-work ratio much higher in the brayton cycle than in the rankine cycle?
zloy xaker [14]

The back-work ratio much higher in the Brayton cycle than in the Rankine cycle because a gas cycle is the Brayton cycle, while a steam cycle is the Rankine cycle. Particularly, the creation of water droplets will be a constraint on the steam turbine's efficiency. Since gas has a bigger specific volume than steam, the compressor will have to work harder while using gas.

<h3>What are modern Brayton engines?</h3>

Even originally Brayton exclusively produced piston engines, modern Brayton engines are virtually invariably of the turbine variety. Brayton engines are also gas turbines.

<h3>What is the ranking cycle?</h3>

A gas cycle is the Brayton cycle, while the Ranking cycle is a steam cycle. The production of water droplets will especially decrease the steam turbine's performance. Gas-powered compressors will have to do more work since gas's specific volume is greater than steam's.

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To know more about Rankine cycle, visit: brainly.com/question/13040242

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4 0
1 year ago
Air at 80 kPa and 10°C enters an adiabatic diffuser steadily with a velocity of 150 m/s and leaves with a low velocity at a pre
il63 [147K]

Answer:

The exit temperature is 293.74 K.

Explanation:

Given that

At inlet condition(1)

P =80 KPa

V=150 m/s

T=10 C

Exit area is 5 times the inlet area

Now

A_2=5A_1

If consider that density of air is not changing from inlet to exit then by using continuity equation

A_1V_1=A_2V_2

So   A_1\times 150=5A_1V_2

V_2=30m/s

Now from first law for open system

h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w

Here Q=0 and w=0

h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}

When air is treating as ideal gas  

h=C_pT

Noe by putting the values

h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}

1.005\times 283+\dfrac{150^2}{2000}=1.005\times T_2+\dfrac{30^2}{2000}

T_2=293.74K

So the exit temperature is 293.74 K.

7 0
3 years ago
List, in ascending order, the cutoff frequencies for the first ten modes of a rectangular waveguide, normalized to the cutoff fr
Alisiya [41]

Answer:

Fcte10 < Fcte01 = Fcte0 < Fctm11 < Fcte21 = Fctm21 < Fcte12 = Fctm12 < Fcte22

Explanation:

Assuming a = 2b

Attached below is the required steps to the solution

The cutoff frequencies for the first ten modes of a rectangular waveguide listed in ascending order  is :

Fcte10 < Fcte01 = Fcte0 < Fctm11 < Fcte21 = Fctm21 < Fcte12 = Fctm12 < Fcte22

5 0
3 years ago
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