A pressure difference of 6.00 x 104 Pa is required to maintain a volume flow rate of 0.400 m3 /s for a viscous fluid flowing through a section of cylindrical pipe that has a radius 0.330 m. What pressure difference is required to maintain the same volume flow rate if the radius of the pipe is decreased to 0.110 m?

To determine how variables affect the flow rate of an incompressible fluid undergoing laminate flow in a cylindrical tube, we use Poiseuille's equation.

Q = Πr⁴ / 8η * [(p₁ - p₂) / L]

Q = 0.40 m³/s

P₁ - P₂ = 6*10⁴ pa

r₁ = 0.33m

r₂ = 0.11m

According to Poiseuille's law, the pressure difference is inversely proportional to the radius of the pipe raised to power of 4.

P₁ - P₂ = 1 / r⁴

(P₁ - P₂)i / (P₁ - P₂)f = R⁴f / Ri⁴

Saving for (P₁ - P₂)f

(P₁ - P₂)f = (Rf⁴ / Ri⁴) * (P₁ - P₂)i

(P₁ - P₂)f = [(0.33)⁴ / (0.11)⁴] = 6*10⁴

(P₁ - P₂)f = 4.86*10⁶pa.

batolisis batolisis · Answer 2 · 2022-01-14T13:47:06+01:00

The pressure difference required to maintain the same volume flow rate :

4.86 * 10⁶ Pa

Given data :

Volume flow rate = 0.400 m³ / s

Radius of cylindrical pipe = 0.330 m

New radius of pipe ( P₂ ) = 0.110 m

Determine the pressure difference required

we will apply the equation below

Q = [tex]\frac{\pi R^{4} }{8n} * \frac{(P_{1}-P_{2}) }{L}[/tex]

Q = volumetric flow rate

n = coefficient of viscosity

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