Question

8. What should be the maximum average velocity of water in a tube of diameter 2 cm so that the flow is
laminar? The viscosity of water is 10 Nm's
​

Answer 1

Answer:

0.1 m/s

Explanation:

The formula is ,

Re=pVD/β=VD/v=QD/vA

• V is the mean fluid velocity (SI units: m/s)

• L is a characteristic linear dimension, (traveled length of fluid, or hydraulic radius when dealing with river systems) (m)

• μ is the dynamic viscosity of the fluid (Pa·s or N·s/m² or kg/m·s)

• ν is the kinematic viscosity (ν = μ / ρ) (m²/s)

• β, is the density of the fluid (kg/m³)

• Q is the volumetric flow rate (m³/s)

• A is the pipe cross-sectional area (m²).

• D is the hydraulic diameter of the pipe (m).

Here the formula used is (1)

Now,

When the Reynolds number is less than 2000, flow will be described as laminar

When the Reynolds number is greater than 4000, flow will be described as turbulent

When the Reynolds number is in the range of 2000 to 4000 the flow is considered transitional.

So, we have,

2000 = 103 (v) 2*10-2/0.001

=> v = 0.1 m/s

In this i have converted cm into m

and converted into seconds.

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