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Why is flow separation in flow over cylinders delayed in turbulent flow?

Short Answer

Expert verified
Answer: The primary reason for the delay in flow separation in turbulent flow is the increased momentum transfer in the boundary layer due to turbulence. This enhanced mixing, caused by rapid fluctuations and eddies, enables the fluid to maintain its momentum for a longer distance around the cylinder, pushing the separation point further downstream and resulting in reduced drag.

Step by step solution

01

Introduction to Laminar and Turbulent Flow

In fluid dynamics, there are two different types of flow regimes: laminar and turbulent flow. Laminar flow occurs when a fluid flows smoothly, with uniform or no fluctuations in velocity. In contrast, turbulent flow has chaotic and rapid variations in velocity and pressure. Turbulent flow has many scales of motion, ranging from large-scale eddies to small-scale turbulence.
02

Flow Separation in Flow over Cylinders

Flow separation is a phenomenon that occurs when fluid flowing over a surface, like a cylinder, experiences an abrupt change in pressure, causing the fluid to separate from the surface and create a recirculation region behind the body. This separation can result in a significant increase in drag and loss of lift forces, affecting the overall performance of the system.
03

Flow Separation in Laminar Flow

In laminar flow, the fluid particles flow smoothly in parallel layers, with the boundary layer (a thin layer of fluid close to the body's surface) remaining attached to the surface, as seen at low Reynolds numbers. However, as the flow goes around the cylinder, an adverse pressure gradient develops because of the change in pressure distribution, causing the fluid momentum to decrease. This eventually leads to flow separation, with the formation of a low-pressure recirculation region behind the cylinder.
04

Flow Separation in Turbulent Flow

Turbulent flow, on the other hand, is characterized by fluctuations and eddies. These fluctuations transport momentum throughout the flow, allowing for the boundary layer to have more momentum or kinetic energy. This increased momentum helps the fluid overcome the adverse pressure gradient, delaying the flow separation as the flow moves around the cylinder. Consequently, turbulent flow separates at a later point on the cylinder's surface compared to laminar flow.
05

Delay in Flow Separation in Turbulent Flow

The primary reason for the delay in flow separation in the turbulent flow is the increased momentum transfer in the boundary layer due to turbulence. The rapid fluctuations and eddies in the turbulent flow mix the fluid more effectively compared to the laminar flow, allowing the fluid to maintain its momentum for a longer distance around the cylinder. This enhanced mixing delays flow separation by pushing the separation point further downstream, resulting in a narrower wake and reduced drag.

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Most popular questions from this chapter

A 0.4-m-diameter spherical tank of negligible thickness contains iced water at \(0^{\circ} \mathrm{C}\). Air at \(25^{\circ} \mathrm{C}\) flows over the tank with a velocity of \(3 \mathrm{~m} / \mathrm{s}\). Determine the rate of heat transfer to the tank and the rate at which ice melts. The heat of fusion of water at \(0^{\circ} \mathrm{C}\) is \(333.7 \mathrm{~kJ} / \mathrm{kg}\).

Air \(\left(1 \mathrm{~atm}, 5^{\circ} \mathrm{C}\right)\) with a free-stream velocity of \(2 \mathrm{~m} / \mathrm{s}\) flows in parallel with a stationary thin \(1-\mathrm{m} \times 1-\mathrm{m}\) flat plate over the top and bottom surfaces. The flat plate has a uniform surface temperature of $35^{\circ} \mathrm{C}\(. Determine \)(a)\( the average friction coefficient, \)(b)$ the average convection heat transfer coefficient, and (c) the average convection heat transfer coefficient using the modified Reynolds analogy, and compare with the result obtained in \((b)\).

A thin, square, flat plate has \(1.2 \mathrm{~m}\) on each side. Air at \(10^{\circ} \mathrm{C}\) flows over the top and bottom surfaces of a very rough plate in a direction parallel to one edge, with a velocity of $48 \mathrm{~m} / \mathrm{s}$. The surface of the plate is maintained at a constant temperature of \(54^{\circ} \mathrm{C}\). The plate is mounted on a scale that measures a drag force of \(1.5 \mathrm{~N}\). Determine the total heat transfer rate from the plate to the air.

The top surface of the passenger car of a train moving at a velocity of $95 \mathrm{~km} / \mathrm{h}\( is \)2.8-\mathrm{m}\( wide and \)8-\mathrm{m}$ long. The top surface is absorbing solar radiation at a rate of $380 \mathrm{~W} / \mathrm{m}^{2}\(, and the temperature of the ambient air is \)30^{\circ} \mathrm{C}$. Assuming the roof of the car to be perfectly insulated and the radiation heat exchange with the surroundings to be small relative to convection, determine the equilibrium temperature of the top surface of the car. Answer: \(37.5^{\circ} \mathrm{C}\)

Air at \(25^{\circ} \mathrm{C}\) flows over a 5 -cm-diameter, \(1.7-\mathrm{m}\)-long smooth pipe with a velocity of $4 \mathrm{~m} / \mathrm{s}\(. A refrigerant at \)-15^{\circ} \mathrm{C}$ flows inside the pipe, and the surface temperature of the pipe is essentially the same as the refrigerant temperature inside. The drag force exerted on the pipe by the air is (a) \(0.4 \mathrm{~N}\) (b) \(1.1 \mathrm{~N}\) (c) \(8.5 \mathrm{~N}\) (d) \(13 \mathrm{~N}\) (e) \(18 \mathrm{~N}\) (For air, use $\nu=1.382 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \rho=1.269 \mathrm{~kg} / \mathrm{m}^{3}$ )

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