Question

A liquid of density 750 kgm^–3 flows smoothly through a horizontal pipe that tapers in
cross-sectional area from A₁ = 1.2 × 10^–2 m² to
\(A_2=\frac{A_1}{2}\)
. The pressure difference between the wide and narrow sections of the pipe is 4500 Pa. The rate of flow of liquid is _____ × 10^–3 m³s^–1.

Answer 1

Correct Answer: 24

To solve the problem of determining the rate of flow of the liquid through the pipe, let's apply the principles of fluid dynamics, specifically the equation of continuity and Bernoulli's equation.

Step 1: Equation of Continuity
The equation of continuity for an incompressible fluid states that the product of cross-sectional area and velocity at any two points along the pipe is constant. Thus, we have:

\(A_1v_1 = A_2v_2\)

Given \(A_1 = 1.2 \times 10^{-2} \, \text{m}^2\) and \(A_2 = \frac{A_1}{2}\), we find \(A_2 = 0.6 \times 10^{-2} \, \text{m}^2\). Therefore, \(v_2 = 2v_1\).

Step 2: Bernoulli’s Equation
Apply Bernoulli’s equation between the wide and narrow sections:

\(\frac{1}{2} \rho v_1^2 + P_1 = \frac{1}{2} \rho v_2^2 + P_2\)

The pressure difference \(P_1 - P_2 = 4500 \, \text{Pa}\). Substituting \(v_2 = 2v_1\), we get:

\(\frac{1}{2} \rho v_1^2 + 4500 = \frac{1}{2} \rho (2v_1)^2\)

Solving for \(v_1\):

\(4500 = \frac{1}{2} \times 750 \times (4v_1^2 - v_1^2)\)

\(4500 = \frac{1}{2} \times 750 \times 3v_1^2\)

\(4500 = 1125v_1^2\)

\(v_1^2 = 4\)

\(v_1 = 2 \, \text{m/s}\)

Then, \(v_2 = 4 \, \text{m/s}\).

Step 3: Calculate Rate of Flow
The rate of flow \(Q\) is given by:

\(Q = A_1v_1 = 1.2 \times 10^{-2} \times 2 = 2.4 \times 10^{-2} \, \text{m}^3/\text{s}\)

Given units in the problem, express as:

\(Q = 24 \times 10^{-3} \, \text{m}^3/\text{s}\)

The computed rate of flow, \(24 \times 10^{-3} \, \text{m}^3/\text{s}\), falls within the given range of 24,24, verifying the solution's correctness.