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 determine the rate of flow, we will use the principle of conservation of mass and Bernoulli's equation. The problem involves a fluid with a known density flowing through a pipe that tapers to half its original cross-sectional area.

First, calculate the cross-sectional area of the narrow section: 
\(A_2 = \frac{A_1}{2} = \frac{1.2 \times 10^{-2}\,\text{m}^2}{2} = 0.6 \times 10^{-2}\,\text{m}^2\)

Using the principle of conservation of mass, the rate of flow through both sections of the pipe is equal. Thus, we have the equation:
\(\rho A_1 v_1 = \rho A_2 v_2\)
where \(\rho\) is the fluid density, and \(v_1\) and \(v_2\) are the fluid velocities in the wide and narrow sections, respectively.

By canceling \(\rho\) and solving for \(v_2\):
\(v_2 = \frac{A_1}{A_2}v_1 = 2v_1\)

Next, apply Bernoulli's equation, which relates pressures, velocity, and height in fluid flow, assuming height difference is zero:
\(\frac{1}{2}\rho v_1^2 + P_1 = \frac{1}{2}\rho v_2^2 + P_2\)

Substitute \(v_2 = 2v_1\) and rearrange to find \(v_1\):\)
\(\frac{1}{2}\rho v_1^2 + 4500\,\text{Pa} = \frac{1}{2}\rho (2v_1)^2\)
\(4500 = 2\rho v_1^2 - \frac{1}{2}\rho v_1^2 = \frac{3}{2}\rho v_1^2\)
\(v_1^2 = \frac{4500 \times 2}{3 \times 750} = \frac{4500}{1125}\)
\(v_1 = \sqrt{4} = 2\,\text{m/s}\)

The rate of flow \((Q)\) is given by \(Q = A_1 v_1\):\)
\(Q = 1.2 \times 10^{-2} \times 2 = 2.4 \times 10^{-2}\,\text{m}^3/\text{s}\)

Converting to the desired unit:
\(Q = 2.4 \times 10^{-3}\,\text{m}^3/\text{s}\)

Therefore, the rate of flow of the liquid is 2.4 × 10⁻³ m³ s⁻¹, which lies within the given range (24,24).

Answer 2

The correct answer is 24

Fig.

Using Bernoulli’s equation
\(P_1 + \frac{1}{2} \rho v^2 = P_2 + \frac{1}{2} \rho 4v^2\)
\(\frac{3}{2}ρv^2=P_1−P_2\)
\(⇒\)\(v = \sqrt{\frac{2(P_1 - P_2)}{3\rho}}\)
\(v = \sqrt\frac{2 \times 4500}{3 \times 750} = 2 \, \text{m/sec}\)
So Q = A₁v = 24 × 10^–3 m³/sec