Question

An incompressible liquid is flowing through a tube of non uniform cross section. Derive

the relation between area of cross section of tube and velocity of liquid​

Answer 1

The equation of continuity states that the product of cross-sectional area of the tube and the velocity of the liquid at any point along the tube is constant.

R = vA = constant

• R is the volume flow rate
• A is the area
• v is the flow velocity

Assumption made:

• The tube has a single entry and single exit
• The fluid is non-viscous
• The fluid is incompressible
• The fluid flow is steady

In a small time Δt, fluid covers Δx1 with velocity = v1 in the lower part of the tube.

∴ Δx1 = v1 Δt

Volume of liquid flow = V = A1 Δx1 = A1 v1 Δt

Since, mass(m) = Density(ρ) × Volume(V)

Δm1 = ρ1 A1 v1 Δt         (1)

Mass flux = total mass of the liquid flowing through the given cross sectional area per unit time.

Δm1 / Δt = ρ1 A1 v1         (2)

For the upper end:

Δm2 / Δt = ρ2 A2 v2          (3)

Where v2 = velocity of liquid in upper end

Δt = time

A2 = Cross sectional area of the upper end

Mass flux in the lower end should be equal to the one at the upper end.

ρ1 A1 v1 = ρ2 A2 v2               (4)

∴ ρ A v = constant

Since the fluid in incompressible,

∴ ρ1 = ρ2

⇒ A1 v1 = A2 v2

∴ A v = constant