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dedylja [7]
3 years ago
10

Water at 25°C and 1 atm is flowing over a long flat plate with a velocity of 14 m/s. Determine the distance from the leading edg

e of the plate where the flow becomes turbulent, and the thickness of the boundary layer at that location. The density and dynamic viscosity of water at 1 atm and 25°C are rho = 997 kg/m3 and μ = 0.891 × 10–3 kg/m·s.
Physics
1 answer:
Gre4nikov [31]3 years ago
7 0

Answer:

L=31.9 mm

δ = 0.22 mm

Explanation:

Given that

v= 14 m/s

ρ=997 kg/m³

μ= 0.891 × 10⁻3 kg/m·s

As we know that when Reynolds number grater than 5 x 10⁵ then flow will become turbulent.

Re=\dfrac{\rho vL}{\mu}

L=\dfrac{Re\mu}{\rho v}

L=\dfrac{5\times 10^5\times 0.891\times 10^{-3}}{ 14 \times 997}\ m

L=0.0319 m

L=31.9 mm

The  thickness of the boundary layer at that location L given as

\delta =\dfrac{5L}{\sqrt{Re}}

\delta =\dfrac{5\times0.0319}{\sqrt{5\times 10^5}}

δ = 0.00022 m

δ = 0.22 mm

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c) Check Explanation.

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The ball not bouncing back to a height of 8 m shows energy loss at some point in the total motion of the ball (most likely at the collision). If kinetic energy was conserved, the ball would bounce back up to the height at which it fell from (8 m) after the collision with the floor.

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If the ball had not bounced at all, this means it lost all of its kinetic energy to the floor, and this points to a perfectly inelastic collision between the ball and the floor as they stick together after collision.

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