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Artist 52 [7]
3 years ago
12

Consider steady, incompressible, laminar flow of a Newtonian fluid in an infinitely long round pipe annulus of inner radius Ri a

nd outer radius Ro Ignore the effects of gravity. A constant negative pressure gradient _P/_x is applied in the x-direction, (∂P/∂x)=(P1-P2)/(x2-x1) where x1 and x2 are two arbitrary locations along the x-axis, and P1 and P2 are the pressures at those two locations. The pressure gradient may be caused by a pump and/or gravity. Note that we adopt a modified cylindricaloordinate system here with x instead of z for the axial component, namely, ( r, θ ,x) and (Ur, Uθ ,Ux) Derive an expression for the velocity field in the annular space in the pipe.
Engineering
1 answer:
svetlana [45]3 years ago
8 0

Answer:

is it multiple choice?

Explanation:

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2. What is the original length of the rectangular bar if the deformation is 0.005 in with a force of 1000 lbs and an area of 0.7
Ugo [173]

Answer:

18.75in

Explanation:

Modulus of elasticity = Stress/Strain

Since stress = Force/Area

Given

Force = 1000lb

Area = 0.75sqin

Stress = 1000/0.75

Stress = 1333.33lbsqin

Strain

Strain = Stress/Modulus of elasticity

Strain = 1333.33/5,000,000

Strain = 0.0002667

Also

Strain = extension/original length

extension = 0.005in

Original length = extension/strain

Original length = 0.005/0.0002667

Original length = 18.75in

Hence the original length of the rectangular bar is 18.75in

6 0
3 years ago
Please help me I don’t understand
Lapatulllka [165]

Answer:

A battery changes chemical energy to <u>electrical</u> energy

A drum changes mechanical energy to <u>sound</u> energy

Explanation:

A battery is a device that stores chemical energy and converts it to electrical energy.

Beating a drum represents a mechanical to sound energy  conversion.

4 0
3 years ago
Read 2 more answers
A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is
-BARSIC- [3]

Answer:

\mathbf{C_{10} = 137.611 \ kN}

Explanation:

From the information given:

Life requirement = 40 kh = 40 40 \times 10^{3} \ h

Speed (N) = 520 rev/min

Reliability goal (R_D) = 0.9

Radial load (F_D) = 2600 lbf

To find C10 value by using the formula:

C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}

where;

x_D = \text{bearing life in million revolution} \\  \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}

\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}

The Weibull parameters include:

x_o = 0.02

(\theta - x_o) = 4.439

b= 1.483

∴

Using the above formula:

C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}

C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}

C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}

C_{10} = 30962.449 \ lbf

Recall that:

1 kN = 225 lbf

∴

C_{10} = \dfrac{30962.449}{225}

\mathbf{C_{10} = 137.611 \ kN}

7 0
3 years ago
All the fnaf UNC charecters
astra-53 [7]

Answer:

T.Freddy.

T.Bonnie.

T.Chica.

G.Freddy.

N. Freddy.

F.Foxy.

Mr.Hippo.

R.Freddy.

Explanation:

8 0
3 years ago
Read 2 more answers
A cone penetration test was carried out in normally consolidated sand, for which the results are summarized below: Depth (m) Con
Cerrena [4.2K]

Answer:

hello your question is incomplete attached below is the missing equation related to the question  

answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°

Explanation:

<u>Determine the friction angle at each depth</u>

attached below is the detailed solution

To calculate the vertical stress = depth * unit weight of sand

also inverse of Tan = Tan^-1

also qc is in Mpa while σ0 is in kPa

Friction angle at each depth

2 meters = 40.389°

3.5 meters  = 38.987°

5 meters = 38.022°

6.5 meters = 39.869°

8 meters = 40.265°

6 0
3 years ago
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