Question

Glycerine of density 1.25 × 10³ kg m^-3 is flowing through the conical section of pipe. The area of cross-section of the pipe at its ends is 10 cm2 and 5 cm2 and pressure drop across its length is 3 Nm^-2. The rate of flow of glycerine through the pipe is x x10^-5 m3 s^-1. The value of x is____.

Answer 1

Correct Answer: 4

To determine the rate of flow and the value of x, we start by analyzing the given situation using the principle of conservation of mass and Bernoulli's equation.

The continuity equation for fluid flow through a pipe is given by:

A₁v₁ = A₂v₂

where A₁ and A₂ are the cross-sectional areas, and v₁ and v₂ are the velocities of the fluid at these sections.

For this problem:

A₁ = 10 cm² = 10 × 10^-4 m²
A₂ = 5 cm² = 5 × 10^-4 m²

The Bernoulli's equation is:

P₁ + 0.5ρv₁² = P₂ + 0.5ρv₂²

The pressure drop is P₁ - P₂ = 3 Nm^-2. With the density of glycerine, ρ = 1.25 × 10³ kg/m³, substitute this into Bernoulli's equation:

3 = 0.5 × 1.25 × 10³ × (v₂² - v₁²)

v₂² - v₁² = 2 × 3 / (0.5 × 1.25 × 10³) = 4.8 × 10^-3

Using continuity equation, v₁ = 0.5 × v₂.

Substituting, (v₂)² - (0.5v₂)² = 4.8 × 10^-3

v₂² - 0.25v₂² = 4.8 × 10^-3

0.75v₂² = 4.8 × 10^-3

v₂² = 6.4 × 10^-3

v₂ = √(6.4 × 10^-3) = 0.08 m/s

Using continuity again: v₁ = 0.5 × 0.08 = 0.04 m/s

Flow rate Q is given by Q = A₂v₂, thus:

Q = 5 × 10^-4 × 0.08 = 4 × 10^-5 m³/s

From Q = x × 10^-5 m³/s, equating gives x = 4.

The value of x, 4, falls within the expected range of 4,4, confirming the correctness of our calculations.