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A fluid flows through a horizental pipe having two different cross-sections of area \(A\) and \(2 A\). If the pressure at the thin cross-section is \(p\) and fluid velocity is \(v\), the velocity and pressure at the thicker cross-section is (take the density of fluid as \(\rho\) ) (a) \(\frac{v}{2}, p+\frac{1}{2} \rho v^{2}\) (b) \(\frac{v}{4}, p+\frac{3}{8} \rho v^{2}\) (c) \(\frac{v}{2}, p+\frac{3}{8} \rho v^{2}\) (d) \(v_{2} p+\frac{3}{4} \rho v^{2}\)

Short Answer

Expert verified
The velocity is \(\frac{v}{2}\) and the pressure is \(p + \frac{3}{8} \rho v^2\).

Step by step solution

01

Apply the Continuity Equation

The Continuity Equation for fluid flow through a pipe states that the product of the cross-sectional area \(A\) and velocity \(v\) at any two points is constant for incompressible flow. If \(v_1\) is the velocity at the thin cross-section with area \(A\), and \(v_2\) is the velocity at the thicker cross-section with area \(2A\), then:\[ A \cdot v_1 = 2A \cdot v_2 \]Given \(v_1 = v\), solving for \(v_2\) gives:\[ v_2 = \frac{v}{2} \].
02

Apply Bernoulli's Equation

Bernoulli's Equation relates the pressure \(p\), fluid velocity \(v\), and height \(h\) at any two points along a streamline of an incompressible, non-viscous flow. Since the pipe is horizontal, the height difference is zero, simplifying the equation to:\[ p_1 + \frac{1}{2} \rho v_1^2 = p_2 + \frac{1}{2} \rho v_2^2 \]Substitute \(v_1 = v\), \(v_2 = \frac{v}{2}\), and solve for \(p_2\), the pressure at the thicker cross-section:\[ p + \frac{1}{2} \rho v^2 = p_2 + \frac{1}{2} \rho \left( \frac{v}{2} \right)^2 \]\[ p_2 = p + \frac{1}{2} \rho v^2 - \frac{1}{2} \rho \left( \frac{v^2}{4} \right) \]Simplifying:\[ p_2 = p + \frac{1}{2} \rho v^2 - \frac{1}{8} \rho v^2 = p + \frac{3}{8} \rho v^2 \].
03

Derive the Final Answers for Velocity and Pressure

From Step 1, we found that the velocity at the thicker cross-section \(v_2\) is \(\frac{v}{2}\). From Step 2, we determined that the pressure at the thicker cross-section \(p_2\) is \(p + \frac{3}{8} \rho v^2\). Thus, the velocity and pressure at the thicker cross-section are \(\frac{v}{2}\) and \(p + \frac{3}{8} \rho v^2\), respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuity Equation
The Continuity Equation is a fundamental aspect of fluid dynamics, focusing on the principle of conservation of mass within a fluid flow. It states that for an incompressible and steady-flowing fluid, the mass flow rate is constant from one cross-section of a pipe to another.

When a fluid flows through pipes with varying diameters, like a thin to a thick section, its velocity and cross-sectional area change accordingly to maintain a constant flow rate. Mathematically, this is expressed as:
  • For two sections, 1 and 2: \(A_1 \cdot v_1 = A_2 \cdot v_2\)
Here, \(A\) represents the cross-sectional area and \(v\) the flow velocity. When the area increases, velocity decreases, and vice versa.

In our exercise, the velocity at the thin section of the pipe is reduced by half at the thick section, as derived: \(v_2 = \frac{v}{2}\). This concept highlights how adjusting pipe dimensions affects fluid speed.
Bernoulli's Equation
Bernoulli's Equation is vital in understanding energy conservation in fluid dynamics for incompressible, non-viscous flows. It relates pressure, kinetic energy per unit volume, and potential energy per unit volume at different points along a streamline.

The principle is given by:
  • \(p + \frac{1}{2} \rho v^2 + \rho gh = \text{constant}\)
For horizontal pipes, height changes are negligible (\(\rho gh\)), so the equation simplifies.

Using Bernoulli's Equation in our horizontal pipe problem, we found how velocity and pressure interact:
  • At the thin section: \(p + \frac{1}{2} \rho v^2\)
  • At the thick section: \(p_2 + \frac{1}{2} \rho \left(\frac{v}{2}\right)^2\)
This resulted in pressure increasing by \(\frac{3}{8} \rho v^2\) at the thicker cross-section.
Incompressible Flow
Incompressible flow is a key concept where the fluid density \(\rho\) remains constant throughout its motion. This simplifies many equations in fluid mechanics, making calculations more manageable.

For many liquids, such as water, incompressibility is a valid assumption because their density doesn't change significantly under pressure.
  • Continuity Equation assumes incompressibility for mass conservation.
  • Bernoulli's Equation uses constant density for simplifying energy conservation along a streamline.
In the given exercise, assuming the fluid is incompressible allowed us to utilize these relationships to determine velocity and pressure changes as the fluid moves through sections of varying diameter.

Understanding incompressible flow is foundational, allowing us to apply these principles broadly across various fluid dynamics problems. It underpins the idea of constant density and its effect on flow behaviors.

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Most popular questions from this chapter

The work done in increasing the size of a rectangular soap film with dimensions \(8 \mathrm{~cm} \times 3.75 \mathrm{~cm}\) to \(10 \mathrm{~cm} \times 6 \mathrm{~cm}\) is \(2 \times 10^{-4} \mathrm{~J}\). The surface tension of the film in \(\mathrm{Nm}^{-1}\) is (a) \(1.65 \times 10^{-2}\) (b) \(3.3 \times 10^{-2}\) (c) \(6.6 \times 10^{-2}\) (d) \(8.25 \times 10^{-2}\)

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