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faust18 [17]
3 years ago
14

Identify the fluid property responsible for the development of the velocity boundary layer and the kinds of fluids. For the idea

lized inviscid fluids, there is no velocity boundary layer. For what kinds of fluids will there be no velocity boundary layer in a pipe?
Engineering
1 answer:
jarptica [38.1K]3 years ago
4 0

Answer:

The fluid property responsible for the development of velocity boundary layer is majorly the fluid's viscosity.

For non-viscous fluids (in theory, because no fluid is entirely non-viscous), there will be no velocity boundary layer.

Explanation:

The velocity boundary layer is the thin layer of viscous fluid that is in direct contact with the pipe surface. The velocity of fluid in this layer is 0 as fluid doesn't move in this layer.

This phenomenon is due to the viscosity of the fluid. Viscosity of the fluid refers to the internal friction that exists between fluid layers, so, the layer of fluid in contact with non-moving, static surface of the pipe experiences friction that causes this layer to not move, causing the fluid velocity to vary from 0 at this surface to the maximum value at the centre of the pipe, before the velocity begins to drop again until it reaches 0 at the other end of the circular pipe.

Since viscosity is the primary cause of this, non-viscous or inviscid fluids are saved from this phenomenon as their flows do not have the velocity boundary layer.

Although, a completely non-viscous or inciscid fluid is an idealized concept because all fluids will experience some sort of viscosity (no matter how small) between their fluid layers. Hence, a velocity boundary layer, no matter how thin (or of minute thickness), will exist in the flow of real fluids.

But, an idealized non-viscous or inviscid fluid will not have a velocity boundary layer.

Hope this Helps!!!

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3 years ago
H. Blasius correlated data on turbulent friction factor in smooth pipes. His equation f s m o o t h ≈ 0.3164 Re − 1 / 4 fsmooth≈
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Answer:

Therefore the angle  the pipe needed to make the static pressure constant along the pipe is θ = 4° 16'

Explanation:

The first step to take is to calculate the the velocity of flow through a pipe

Q =Av

Where Q = is the discharge through pipe

A = Area of the pipe

v = the flow of velocity

We substitute 0.001 m^3/s for Q and 0.03 m for D

Q= Av

0.001=Av

Substitute π/4 D² for A

0.001 = π/4 D² (v)

v = 0.004/πD²

D = he diameter of the pipe

substitute 3 cm  for D

v=  0.004/π * [3 cm * 1 m/100 cm]²

v =1.414 m/s

Obtain fluid properties from the table Kinematic viscosity and Dynamic of water

p =1000 kg /m³

μ= 1.002 * 10^ ⁻³ N.s/m³

Thus,

we write the expression to determine  the Reynolds number of flow

Re = pvD/μ

Re = is the Reynolds number

p =density

μ = dynamic viscosity at 20⁰C

We then substitute 1000 kg /m³ in place of p, 1.002 * 10^ ⁻³ N.s/m³ for μ,

1.414 m/s for v and 0.03 m for D

Thus,

Re = 1000 * 1.414 * 0.03/ 1.002 * 10^ ⁻³ = 42335

The next step is to calculate the friction factor form the Blasius equation

f = 0.3164 (Re)^1/4

f = friction factor

We substitute 42335 for Re

f = 0.3164 (42335)1/4

=0.022

The next step is to write the expression to determine the friction head loss

hl = flv²/2gD

hl = head loss

l = length of pipe

g=  acceleration due to gravity

We then again substitute 0.022 for f, 1.414 m/s for v, 0.03 m for D, and 9.8 m/s² for g.

so,

hl = flv²/2gD

hl/L = 0.022 * 1.414²/2 * 9.81 * 0.03

sinθ = 0.07473

θ = 4° 16'

Therefore the angle  the pipe needed to make the static pressure constant along the pipe is θ = 4° 16'

3 0
3 years ago
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