Question

Calculate the frictional pressure drop in pascals for olive oil flowing through a commercial pipe having an inside diameter of 0.0525m and a length of 76.2m. The velocity of the fluid is 1.22 m/s. Use the friction factor method. Is the flow laminar or turbulent?

Answer 1

Answer: 90.5 kN/m²

Explanation:

Given

Diameter, d = 0.0525 m

Length, L = 76.2 m

Velocity, v = 1.22 m/s

N = DVρ/μ

Where

Density of olive oil, ρ = 919 kg/m³

Viscosity of olive oil, μ = 0.084 pa.s

So that,

N = [0.0525 * 1.22 * 919] / 0.084

N = 58.862 / 0.084

N = 700.74 due to this number being lower than 2100, we can say that the flow is Laminar

Frictional pressure drop = 4fρLv²/2D

Where f = 16/N = 0.0228

F = [4 * 0.0228 * 919 * 76.2 * 1.22²] / 2 * 0.0525

F = 9505.72 / 0.105

F = 90530.67 N/m²

F = 90.5 kN/m²

Answer 2

Answer:

F = 90.65KN/m²

Explanation:

Given  that,

d = 0.0525 m

L = 76.2 m

v = 1.22 m/s

R = DVρ/μ

$R = \frac{0.0525 \times 1.22 \times 919}{0.084}$

$R = \frac{58.862}{0.084}$

$R = 700.7$

Laminar Flow

$f =\frac{16}{R} \\\\f = \frac{16}{700.7} = 0.02283$

Frictional pressure drop =

4fρLv²/2D

$F = \frac{4fpLv^2}{2D}$

$F = \frac{4\times0.22283\times919\times76,2\times1.22\62}{2\times0.0525} \\\\F = 90650 N/m^2$

F = 90.65KN/m²