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, we use the principle of continuity and Bernoulli's equation. First, the principle of continuity for incompressible flow states that the mass flow rate must be constant throughout the pipe: \(A_1v_1 = A_2v_2\). Given \(A_2 = \frac{A_1}{2}\), we have:

\(1.2 \times 10^{-2}v_1 = \frac{1.2 \times 10^{-2}}{2}v_2\)

\(v_2 = 2v_1\)

Using Bernoulli's equation between the two sections:

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

The pressure difference is \(P_1 - P_2 = 4500\) Pa. Substituting \(v_2 = 2v_1\) and rearranging terms:

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

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

Solving for \(v_1^2\):

\(4500 = 1125v_1^2\)

\(v_1^2 = \frac{4500}{1125}\)

\(v_1^2 = 4\)

So \(v_1 = 2\) m/s. Since \(A_1 = 1.2 \times 10^{-2}\) m\(^2\), the rate of flow \(Q\) is:

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

The question asks for the rate of flow in terms of \(\times 10^{-3}\) m\(^3\)/s, so \(Q\) is 24 \(\times 10^{-3}\) m\(^3\)/s, which falls within the given range (24,24).