For a forced vortex (rigid-body rotation) of a liquid in a horizontal plane with angular velocity ω, what is the radial pressure gradient relation?

Introduction / Context:
Vortex motions are common in rotating machinery and mixing vessels. In a forced vortex, the fluid rotates as a solid body with tangential velocity v = ω r. The radial pressure variation must balance the required centripetal acceleration of the fluid elements.

Given Data / Assumptions:

• Steady rotation about a vertical axis; analysis in a horizontal plane (z constant).
• Incompressible fluid with density ρ.
• Negligible radial flow (pure rotation).

Concept / Approach:
Centripetal acceleration of a rotating element is v^2 / r = ω^2 r. The radial pressure gradient supplies this inward acceleration: (1/ρ) * dp/dr = v^2 / r = ω^2 r. Hence dp/dr = ρ * ω^2 * r, showing that pressure increases quadratically with radius.

Step-by-Step Solution:
Use equilibrium in the radial direction: dp/dr = ρ * v^2 / r.With v = ω r, substitute to obtain dp/dr = ρ * ω^2 * r.Integrate if needed: p(r) = p(0) + (ρ * ω^2 / 2) * r^2.

Verification / Alternative check:
Laboratory observations show parabolic free surfaces in open vessels, consistent with p ∝ r^2 in the liquid below.

Why Other Options Are Wrong:
ρ ω r: missing one power of ω.ρ ω^2 / r: incorrect dependence on r.ρ v^2: dimensionally inconsistent (missing 1/r).Zero gradient: contradicts required centripetal force.

Common Pitfalls:

• Confusing forced vortex (v = ω r) with free vortex (v ∝ 1/r), where dp/dr = ρ k^2 / r^3.
• Ignoring gravity; here analysis is strictly in a horizontal plane where z is constant.

Final Answer:
dp/dr = ρ * ω^2 * r