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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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2 years ago
Mercury has a radial acceleration of 3.96 × 10−2 m/s2 and its orbital period is T = 88 days. What is the radius of Mercury’s orb
Maslowich

Answer: 58,045,522,878.8 meters

Explanation:

Ok, the data we have is

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Radial acceleration = ar = 3.96x10^-2 m/s^2

And we know that the equation for the radial acceleration is:

ar = v^2/r = r*w^2

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ar = r*(2*pi/T)^2

Now we solve this for r.

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