A non-viscous liquid of density $1200kg/m^3$ flows in a streamline through a pipe of varying cross section, as shown with P and Q as centres, $4\times {10}^{-3}{m}^2 and 16\times {10}^{-3}{m}^2$. If the velocity at P is 2 m/sec, find the velocity at Q(in m/s)?

                              

This is a streamline flow problem involving the continuity equation for an incompressible fluid flowing through a pipe of varying cross-section.

Given Information:

• Fluid density (ρ): 1200 kg/m³
• Flow type: Non-viscous, streamline (laminar) flow
• Cross-sectional area at P (A_P): 4 × 10⁻³ m²
• Cross-sectional area at Q (A_Q): 16 × 10⁻³ m²
• Velocity at P (v_P): 2 m/s
• Vertical positions: P is at 3 m height, Q is at 5 m height
• Find: Velocity at Q (v_Q) in m/s

Physical Principle: Continuity Equation

For an incompressible fluid (constant density) flowing through a pipe, the principle of conservation of mass applies. This is expressed by the continuity equation:

A₁v₁ = A₂v₂

Or more specifically for this problem:

A_P × v_P = A_Q × v_Q

This equation states that the volume flow rate (also called discharge rate) remains constant throughout the pipe:

Q = A × v = constant

Where:

• A = cross-sectional area
• v = velocity of fluid
• Q = volume flow rate (m³/s)

Physical Interpretation

Why does velocity change?

• When the pipe widens (larger cross-section), the fluid slows down
• When the pipe narrows (smaller cross-section), the fluid speeds up
• This maintains constant mass flow rate through the pipe

Think of it like squeezing a garden hose: when you partially block the opening (reducing area), water shoots out faster!

Step 1: Apply the continuity equation

A_P × v_P = A_Q × v_Q

Step 2: Substitute the known values

(4 × 10⁻³ m²) × (2 m/s) = (16 × 10⁻³ m²) × v_Q

Step 3: Solve for v_Q

8 × 10⁻³ = 16 × 10⁻³ × v_Q

v_Q = (8 × 10⁻³)/(16 × 10⁻³)

v_Q = 8/16

v_Q = 0.5 m/s

Answer: 0.5 m/s