Terms Associated with Boundary Layer MCQ Quiz in मल्याळम - Objective Question with Answer for Terms Associated with Boundary Layer - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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Latest Terms Associated with Boundary Layer MCQ Objective Questions

Top Terms Associated with Boundary Layer MCQ Objective Questions

Terms Associated with Boundary Layer Question 1:

Energy thickness (δ**) in hydrodynamic boundary layer is calculated by:

  1. 0δuU(1uU)dy
  2. 0δ(uU)2[1uU]dy
  3. 0δuU[1(uU)2]dy
  4. 0δuU[1(uU)3]dy

Answer (Detailed Solution Below)

Option 3 : 0δuU[1(uU)2]dy

Terms Associated with Boundary Layer Question 1 Detailed Solution

Displacement thickness (δ*) =0δ(1uU)dy

Momentum thickness (θ) =0δuU[1uU]dy

Energy thickness (δ**) =0δuU[1(uU)2]dy

Shape factor (H) =δθ

Terms Associated with Boundary Layer Question 2:

The velocity distribution in the boundary layer is given by uU=(yδ)1/7

[u=point velocity at distance yU=free stream velocityδ=nominal thickness]

What would be the displacement thickness(δ*)?

  1. δ5
  2. δ6
  3. δ7
  4. δ8

Answer (Detailed Solution Below)

Option 4 : δ8

Terms Associated with Boundary Layer Question 2 Detailed Solution

Concept:

For a particular velocity profile (u/U):

1) Displacement thickness (δ*)

δ=oδ(1uU)dy

2) Momentum thickness (θ)

θ=oδuU(1uU)dy

3) Energy thickness (δ**)

δ=oδuU[1(uU)2]dy

 

Calculation:

For the given velocity profile:

uU=(yδ)1/7

Displacement thickness (δ*)

δ=oδ[1uU]dy

δ=oδ[1(yδ)1/7]dy

δ=oδ(y78×y87δ17)dy

Putting the value of upper and lower limits, we get

δ* = δ - 7δ / 8

δ* = δ / 8

Terms Associated with Boundary Layer Question 3:

Consider the laminar flow of water over a flat plate of length 1 m. If the boundary layer thickness at a distance of 0.25 m from the leading edge of the plate is 8 mm, the boundary layer thickness (in mm), at a distance of 0.75 m, is _______

Answer (Detailed Solution Below) 13.6 - 14.1

Terms Associated with Boundary Layer Question 3 Detailed Solution

Explanation:

For laminar flow over a flat plate

Blausius equation is:

δx=5(Rex)12δx=5(ρVxμ)12

δx=constantx {∴ δ, v & μ are constant}

δx=constantδ1x1=δ2x2δ2=x2x1δ1δ2=0.750.25×8

⇒ δ2 = 13.86 mm

Terms Associated with Boundary Layer Question 4:

For the laminar flow of air (Prandtl number = 0.71) over a flat plate at zero incidence, the thicknesses of the velocity boundary layer δ and the thermal boundary layer δt are such that

  1. δt > δ
  2. δt = δ
  3. δt < δ
  4. Insufficient info

Answer (Detailed Solution Below)

Option 1 : δt > δ

Terms Associated with Boundary Layer Question 4 Detailed Solution

Concept:

δ = the thickness of the hydrodynamic boundary layer; the region of flow where the velocity is less than 99% of the far-field velocity.

δT = the thickness of the thermal boundary layer; the region of flow where the local temperature nearly reaches the value (99%) of the bulk flow temperature

δδT=(Pr)13

Calculation:

Given: Pr = 0.71

Pr=(δδT)3

δδT=(0.71)13=0.89

Terms Associated with Boundary Layer Question 5:

The velocity distribution in a turbulent boundary layer is given by uv=(yδ)17. What is the displacement thickness δ*

  1. δ
  2. δ/7
  3. 7δ/8
  4. δ/8

Answer (Detailed Solution Below)

Option 4 : δ/8

Terms Associated with Boundary Layer Question 5 Detailed Solution

Concept:

δ=0δ(1uv)dy=0δ[1(yδ)17]dyt=yδas y=0 t =0 and as y=δ t =1 orδdt=dyδ=01[1t17]δdt=δ01[1t17]dtδ=δ[tt8787]01=δ[178]

= δ/8 

Terms Associated with Boundary Layer Question 6:

A fluid is flowing over a flat plate. At a distance of 8 cm from the leading edge, the Reynolds number is found to be 25600. The thickness of the boundary layer at this point is

  1. 1.5 mm
  2. 2.5 mm
  3. 4.0 mm
  4. 5.0 mm

Answer (Detailed Solution Below)

Option 2 : 2.5 mm

Terms Associated with Boundary Layer Question 6 Detailed Solution

Concept:

The thickness of the boundary layer is given by δ=5xRex

For turbulent flow delta=0.379xRe15

Where δ = Boundary layer thickness, x = Distance of boundary layer from the leading edge, Rex = Reynold’s number at the distance x from the leading edge

Calculation:

Given:

Since in this given case, Reynold’s number is 25600 which is much lesser than the limit for laminar flow (2 × 105), so the flow is laminar.

 x = 8 cm = 0.08 m

δ=5xRex=5×0.0825600=0.4160=1400=0.0025m=2.5mm

Terms Associated with Boundary Layer Question 7:

A fluid of constant density ρ flows a stationary, smooth flat plate with an incipient free stream velocity U as shown in the figure. The thickness of the boundary layer at the section r – s is δ. The velocity distribution within the boundary layer is approximated by uU=yδ. The plate width perpendicular to the plane of figures is w. The momentum flux into the control volume across the section q-r is 

603f9b5ebd2e7f1a42bed82e 16335953231651

  1. ρU2δ2w4
  2. ρU2δw2
  3. ρU2δw3
  4. ρU2δw6

Answer (Detailed Solution Below)

Option 1 : ρU2δ2w4

Terms Associated with Boundary Layer Question 7 Detailed Solution

Concept:

The momentum flux across a section is given by

Momentum flux = mass × flow velocity

Calculation:

At section p-q,

Flow velocity = U

Mass flow rate = ρ × (δ × w) × U

⇒ mp-q = ρUδw 

At section r-s,

Mass flow rate will be:

mrs=0δρ×w×Uδ×ydy

mrs=ρUδw2

At section p-s,

The surface ps is flat plate and no fluid can pass through it.

⇒ mp-s = 0;

Now

At section q-r,

Flow velocity = U

By law of conservation of mass,

mp-q = mq-r + mr-s + mp-s

⇒ mq-r­ = mp-q – (mp-s + mr-s­)

mqr=ρUδw(0+ρUδw2)=ρUδw2

Momentum flux will be:

Momentum flux = Mass flow rate × velocity

Since the velocity at section q-r is not constant we need to integrate it.

Momentum =  0δUyδ×(massflowrate)dy

Momentum = 0δUyδ×ρUδw2dy

Momentum = ρU2w20δydy

Momentum = ρU2w2×δ22

Momentum = ρU2δ2w4

 

Terms Associated with Boundary Layer Question 8:

Consider a steady, two-dimensional, incompressible flow over a flat at zero angle of incidence with respect to the uniform free stream. The boundary layer thickness is 1 mm at a location where the local Reynolds number is 1000. If the free stream velocity of the flow alone is increased by a factor of 4, then the boundary layer thickness at the same location, in mm will be

  1. 0.25
  2. 0.5
  3. 2
  4. 4

Answer (Detailed Solution Below)

Option 2 : 0.5

Terms Associated with Boundary Layer Question 8 Detailed Solution

Concept:

For laminar boundary on a flat plate,

The boundary layer thickness at a distance x from leading-edge is given as

δx=5Rex

Calculation:

Given

When Rex = 1000, δ = 1 mm;

1x=51000x=210mm

Now free stream velocity is increased by four,

We have Re = ρVD/μ

Now V1 = 4V Re1 = 4Re

Re1 = 4000;

Now the boundary layer thickness will be

δ=5×2104000=0.5mm

 The new boundary layer thickness is 0.5 mm;

Terms Associated with Boundary Layer Question 9:

The turbulent boundary layer thickness over a flat plate varies as

  1. x17

  2. x15

  3. x12

  4. x45

  5. x43

Answer (Detailed Solution Below)

Option 4 :

x45

Terms Associated with Boundary Layer Question 9 Detailed Solution

Explanation:

Boundary-Layer:  

  • When a fluid of ambient velocity flows over a flat stationary plate, the bottom layer of fluid directly contacts with the solid surface and its velocity reaches to zero.
  • Due to the cohesive forces between two layers, the bottom layer offers resistance to the adjacent layer and due to this reason, the velocity gradient develops in a fluid.
  • A thin region over a surface velocity gradient is significant, known as the boundary layer.

The thickness of the boundary layer is given by

The Laminar boundary layer for flat plate given by Blasius equation is :

δ=5xRe  

Re=ρUxμRex

δ=5xReδx

Where, x = distance where the boundary layer is to be found, Re = Reynolds no, ρ = density of the fluid, V = velocity of the fluid

µ = dynamic viscosity fluid

for laminar flow, the Reynolds number should be less than 5 × 105.
The Turbulent boundary layer for the flat plate is given as

For turbulent flow  Re > 5 x 105

δ=0.379x(Re)15δx115δx45

Terms Associated with Boundary Layer Question 10:

The ratio of boundary layer thickness to displacement thickness for the given velocity distribution is ________ UU=(yδ)1/7

Answer (Detailed Solution Below) 8

Terms Associated with Boundary Layer Question 10 Detailed Solution

Explanation:

Displacement thickness,

δ=0δ(1UU)dy=0δ1(yδ)17dy

δ=[y]0δ1δ1/7[y87(87)]0δ

δ=δ1δ17(78)δ87=δ(178)

δ=δ8δδ=8
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