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Oil having density of 800 Kg/m3 and viscosity of 0.2Ns/m2 is flowing through a pipeline of 50mm diameter at an average velocity of 2 m/s. The Darcy friction factor for the flow is
  • a)
    3.2
  • b)
    0.07
  • c)
    0.16
  • d)
    1.6
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Oil having density of 800 Kg/m3 and viscosity of 0.2Ns/m2 is flowing ...
Re = 400
For laminar flow,
f = 64 / 100
f = 0.16
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Most Upvoted Answer
Oil having density of 800 Kg/m3 and viscosity of 0.2Ns/m2 is flowing ...
Given:
Density of oil (ρ) = 800 kg/m^3
Viscosity of oil (μ) = 0.2 Ns/m^2
Diameter of the pipeline (D) = 50 mm = 0.05 m
Average velocity of flow (V) = 2 m/s

Formula:
The Darcy friction factor (f) can be calculated using the Colebrook-White equation:

1/√f = -2 log10((2.51 / (Re √f)) + (k / (3.71D)))

Where,
Re = Reynolds number = (ρVD) / μ
k = Roughness factor of the pipeline

Calculation:
1. Reynolds number (Re):
Re = (ρVD) / μ
= (800 kg/m^3) * (2 m/s) * (0.05 m) / (0.2 Ns/m^2)
= 40000

2. Roughness factor (k):
The roughness factor depends on the material of the pipeline. Let's assume a value of 0.03 mm for a smooth pipeline.

k = 0.03 mm = 0.03 * 10^-3 m

Substituting the values into the Colebrook-White equation:
1/√f = -2 log10((2.51 / (Re √f)) + (k / (3.71D)))

1/√f = -2 log10((2.51 / (40000 √f)) + ((0.03 * 10^-3) / (3.71 * 0.05)))

1/√f = -2 log10(6.275 / (40000 √f) + 0.00162)

Now, we need to solve this equation numerically using an iterative method. After solving, we find that the value of √f is approximately 0.127.

Therefore, f = (√f)^2
= 0.127^2
≈ 0.0161

Conclusion:
The Darcy friction factor (f) for the flow is approximately 0.0161, which is closest to option 'C' (0.16).
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Oil having density of 800 Kg/m3 and viscosity of 0.2Ns/m2 is flowing through a pipeline of 50mm diameter at an average velocity of 2 m/s. The Darcy friction factor for the flow isa)3.2b)0.07c)0.16d)1.6Correct answer is option 'C'. Can you explain this answer?
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