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const2013 [10]
3 years ago
7

The following statements are about the laminar boundary layer over a flat plate. For each statement, answer whether the statemen

t is true or false.
1. At a given x-location, if the Reynolds number were to increase, the boundary layer thickness would also increase.
A. True B. False
2. As outer flow velocity increases, so does the boundary layer thickness.
A. True B. False
3. As the fluid viscosity increases, so does the boundary layer thickness.
A. True B. False
4. As the fluid density increases, so does the boundary layer thickness.
A. True B. False
5. The boundary layer equations are approximations of the Navier-Stokes equation.
A. True B. False
6. The curve representing boundary layer thickness as function of x is a streamline.
A. True B. False
7. The boundary layer approximation bridges the gap between the Euler equation and the Navier-Stokes equation.
A. True B. False
Engineering
1 answer:
Nikolay [14]3 years ago
6 0

Answer:

1. B. False

2.  B. False

3. A. True

4. B. False

5. A. True

6. A. True

7. A. True

Explanation:

1. B. False

The relation of Reynolds' number, Reₓ to boundary layer thickness δ at a point x is given by the relation

\delta = \dfrac{x \times C}{\sqrt{Re_x} }

That is the boundary layer thickness is inversely proportional to the square root of the Reynolds' number so that if the Reynolds' number were to increase, the boundary layer thickness would decrease

Therefore, the correct option is B. False

2.  B. False

From the relation

Re_x = \dfrac{U_o \times x}{v}

As the outer flow velocity increases, the boundary layer thickness diminishes

3. A. True

As the viscous force is increased the boundary layer thickness increases

4. B. False

Boundary layer thickness is inversely proportional to velocity

5. A. True

The boundary layer model developed by Ludwig Prandtl is a special case of the Navier-Stokes equation

6. A. True

Given a definite boundary layer thickness, the curve representing the boundary layer thickness is a streamline

7. A. True

The boundary layer approximation by Prandtl Euler bridges the gap between the Euler (slip boundary conditions) and Navier-Stokes (no slip boundary conditions) equations.

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If a construction company is considering a new type of material to use in their construction, which factors would they
Amanda [17]

I would honestly select every one of the given options. Gor a company evaluating this new material it would be very valuable to hit each of these factors.

4 0
3 years ago
A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of wid
dmitriy555 [2]

Complete Question:

A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.

Answer:

T_{min} = 26 mins 40 secs

Explanation:

Reduction in depth, Δd = 20 mm

Depth of cut, d_c = 0.5 mm

Number of passes necessary for this reduction, n = \frac{\triangle d}{d_c}

n = 20/0.5

n = 40 passes

Tool width, w = 5 mm

Width of metal plate, W = 200 mm

For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times

Speed of tool, v = 100 mm/s

Time/pass = \frac{40*400}{400} \\Time/pass = 40 sec

minimum time required to reduce the depth of the plate by 20 mm:

T_{min} = number of passes * Time/pass

T_{min} = n * Time/pass

T_{min} = 40 * 40

T_{min} =  1600 = 26 mins 40 secs

3 0
4 years ago
Read 2 more answers
Name the seven physical qualities for which standards have been developed.
SIZIF [17.4K]

Answer:

HUMAN DEVELOPMENT

MOTOR BEHAVIOR

EXERCISE SCIENCE

MEASUREMENT AND EVALUATION

HISTORY AND PHILOSOPHY

UNIQUE ATTRIBUTES OF LEARNERS

CURRICULUM THEORY AND DEVELOPMENT

Explanation:

6 0
4 years ago
Calculate the load, PP, that would cause AA to be displaced 0.01 inches to the right. The wires ABAB and ACAC are A36 steel and
Nataly [62]

Answer:

P = 4.745 kips

Explanation:

Given

ΔL = 0.01 in

E = 29000 KSI

D = 1/2 in  

LAB = LAC = L = 12 in

We get the area as follows

A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²

Then we use the formula

ΔL = P*L/(A*E)

For AB:

ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB

For AC:

ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC

Now, we use the condition

ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in

⇒  ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in

Knowing that   PAB*Cos 30°+PAC*Cos 30° = P

we have

(2.107*10⁻⁶ in/lbf)*P = 0.01 in

⇒  P = 4745.11 lb = 4.745 kips

The pic shown can help to understand the question.

5 0
4 years ago
Atmospheric air at 25 °C and 8 m/s flows over both surfaces of an isothermal (179C) flat plate that is 2.75m long. Determine the
vekshin1

Answer:

Re=100,000⇒Q=275.25 \frac{W}{m^2}

Re=500,000⇒Q=1,757.77\frac{W}{m^2}

Re=1,000,000⇒Q=3060.36 \frac{W}{m^2}

Explanation:

Given:

For air      T_∞=25°C  ,V=8 m/s

  For surface T_s=179°C

     L=2.75 m    ,b=3 m

We know that for flat plate

Re⇒Laminar flow

Re>30\times10^5⇒Turbulent flow

<u> Take Re=100,000:</u>

 So this is case of laminar flow

  Nu=0.664Re^{\frac{1}{2}}Pr^{\frac{1}{3}}

From standard air property table at 25°C

  Pr= is 0.71  ,K=26.24\times 10^{-3}

So    Nu=0.664\times 100,000^{\frac{1}{2}}\times 0.71^{\frac{1}{3}}

Nu=187.32   (\dfrac{hL}{K_{air}})

187.32=\dfrac{h\times2.75}{26.24\times 10^{-3}}

     ⇒h=1.78\frac{W}{m^2-K}

heat transfer rate =h(T_∞-T_s)

                           =275.25 \frac{W}{m^2}

<u> Take Re=500,000:</u>

So this is case of turbulent flow

  Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}

Nu=0.037\times 500,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}

Nu=1196.18  ⇒h=11.14 \frac{W}{m^2-K}

heat transfer rate =h(T_∞-T_s)

                             =11.14(179-25)

                           = 1,757.77\frac{W}{m^2}

<u> Take Re=1,000,000:</u>

So this is case of turbulent flow

  Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}

Nu=0.037\times 1,000,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}

Nu=2082.6  ⇒h=19.87 \frac{W}{m^2-K}

heat transfer rate =h(T_∞-T_s)

                             =19.87(179-25)

                           = 3060.36 \frac{W}{m^2}

7 0
3 years ago
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