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A laminar boundary layer occurs over a flat plate at zero incidence tothe flow. The thickness of boundary layer at a section 2 m from theleading edge is 2 mm. The thickness of boundary layer at a section 4 mfrom the leading edge will be:
- a)2 × (2)2 mm
- b)2 × (2)1/2 mm
- c)2 × (2)4/5 mm
- d)2 × (2)1/5 mm
Correct answer is option 'B'. Can you explain this answer?
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A laminar boundary layer occurs over a flat plate at zero incidence to...
Ans. (b) Thickness of boundary layer at 4 mm from leading edge = 2 × (4/2)1/2 = 2 × 21/2
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A laminar boundary layer occurs over a flat plate at zero incidence to...
Solution:
Given:
Thickness of the boundary layer, δ1 = 2 mm at section x1 = 2 m
To find:
Thickness of the boundary layer, δ2 at section x2 = 4 m
Explanation:
- The boundary layer thickness is given by the relation:
δ = 5x/√(Re_x)
where,
x = distance from the leading edge
Re_x = Reynolds number at section x
- For laminar flow over a flat plate, the Reynolds number is given by:
Re_x = (u∞ x)/v
where,
u∞ = free stream velocity
v = kinematic viscosity of the fluid
- As the flow is laminar, the velocity profile is parabolic in nature:
u(y) = (u∞/ν) (δ - y) y
where,
y = distance from the plate surface
- The displacement thickness of the boundary layer is given by:
δ* = ∫0^δ u(y) (1 - u(y)/u∞) dy
- The momentum thickness of the boundary layer is given by:
θ = ∫0^δ u(y) (1 - u(y)/u∞) dy
- Using the above equations, we can find the thickness of the boundary layer at section x2 as follows:
Re_2 = (u∞ x2)/v
δ2 = 5x2/√(Re_2)
- Substituting the values of x2, Re_2 and δ1 in the above equation, we get:
δ2 = 2√2 mm
Therefore, the thickness of the boundary layer at section x2 = 4 m is 2√2 mm, which is approximately equal to 2.83 mm.
Hence, the correct answer is option B.
Given:
Thickness of the boundary layer, δ1 = 2 mm at section x1 = 2 m
To find:
Thickness of the boundary layer, δ2 at section x2 = 4 m
Explanation:
- The boundary layer thickness is given by the relation:
δ = 5x/√(Re_x)
where,
x = distance from the leading edge
Re_x = Reynolds number at section x
- For laminar flow over a flat plate, the Reynolds number is given by:
Re_x = (u∞ x)/v
where,
u∞ = free stream velocity
v = kinematic viscosity of the fluid
- As the flow is laminar, the velocity profile is parabolic in nature:
u(y) = (u∞/ν) (δ - y) y
where,
y = distance from the plate surface
- The displacement thickness of the boundary layer is given by:
δ* = ∫0^δ u(y) (1 - u(y)/u∞) dy
- The momentum thickness of the boundary layer is given by:
θ = ∫0^δ u(y) (1 - u(y)/u∞) dy
- Using the above equations, we can find the thickness of the boundary layer at section x2 as follows:
Re_2 = (u∞ x2)/v
δ2 = 5x2/√(Re_2)
- Substituting the values of x2, Re_2 and δ1 in the above equation, we get:
δ2 = 2√2 mm
Therefore, the thickness of the boundary layer at section x2 = 4 m is 2√2 mm, which is approximately equal to 2.83 mm.
Hence, the correct answer is option B.
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A laminar boundary layer occurs over a flat plate at zero incidence tothe flow. The thickness of boundary layer at a section 2 m from theleading edge is 2 mm. The thickness of boundary layer at a section 4 mfrom the leading edge will be:a)2 × (2)2 mmb)2 × (2)1/2 mmc)2 × (2)4/5 mmd)2 × (2)1/5 mmCorrect answer is option 'B'. Can you explain this answer?
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