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The friction factor in the relation h = AfLV2/2gd for laminar flow in a circular pipe was found to be 0.01. This corresponds to a flow Reynolds number equal to
  • a)
    400    
  • b)
    800
  • c)
    1600    
  • d)
    2000
Correct answer is option 'C'. Can you explain this answer?
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Friction factor and laminar flow in a circular pipe

In fluid dynamics, the friction factor is a dimensionless quantity that characterizes the frictional resistance to fluid flow in a pipe. It is denoted by the symbol "f". The friction factor depends on various factors including the pipe geometry, fluid properties, and the flow regime.

Flow Reynolds number and its relation with the friction factor

The flow Reynolds number (Re) is a dimensionless quantity that represents the ratio of inertial forces to viscous forces in a flowing fluid. It is defined as the product of the fluid velocity (V), the characteristic length (L), and the fluid density (ρ), divided by the fluid viscosity (μ):

Re = ρVL/μ

For laminar flow in a circular pipe, the friction factor can be calculated using the relation:

h = (AfLV^2)/(2gd)

Where:
h is the head loss
A is the cross-sectional area of the pipe
f is the friction factor
L is the length of the pipe
V is the average fluid velocity
g is the acceleration due to gravity
d is the diameter of the pipe

Calculation of friction factor and flow Reynolds number

Given that the friction factor (f) is 0.01, we can use the above equation to calculate the flow Reynolds number. Rearranging the equation, we have:

f = 2gdh / (AfLV^2)

Since we have the value of the friction factor (f), we plug it into the equation:

0.01 = 2gdh / (ALV^2)

We can rewrite the equation in terms of the flow Reynolds number (Re):

0.01 = 2gdh / (ALV^2) = 2gdh / (AL(VL/ρ)^2) = 2gρhL / (A(ρVL)^2)

Simplifying the equation further, we get:

0.01 = 2gρhL / (A(ρVL)^2) = 2gρhL / (ρ^2AV^2L^2) = 2gh / (AV^2)

Now, we can relate the flow Reynolds number (Re) to the friction factor (f) and substitute the values into the equation:

Re = ρVL / μ = ρVL / (μ/ρ) = ρ^2AVL / (μ/ρ) = ρAVL^2 / μ = 2gh / (AV^2)

Therefore, we can conclude that for laminar flow in a circular pipe, a friction factor of 0.01 corresponds to a flow Reynolds number of 1600 (option C).
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The friction factor in the relation h = AfLV2/2gd for laminar flow in a circular pipe was found to be 0.01. This corresponds to a flow Reynolds number equal toa)400 b)800c)1600 d)2000Correct answer is option 'C'. Can you explain this answer?
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