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?

(P₁ - P₂)f = 4.86*10⁶ Pa

Explanation:

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.

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

Given data :

Volume flow rate = 0.400 m³ / s

Radius of cylindrical pipe = 0.330 m

New radius of pipe ( P₂ ) = 0.110 m

we will apply the equation below

Q =

Q = volumetric flow rate

n = coefficient of viscosity

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