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Chapter 6: Problem 2

Show that when laminar flow occurs with mean velocity \(u_{\mathrm{m}}\) between extensive stationary flat plates the mean kinetic energy divided by mass of the fluid is \(1.543 u_{\mathrm{m}}^{2} / 2\).

Short Answer

Expert verified
The mean kinetic energy per unit mass is \(1.543 \frac{u_m^2}{2} \).

Step by step solution

01

- Understand the problem

Recognize that we need to calculate the mean kinetic energy per unit mass for laminar flow between two parallel plates and show it equals to \(1.543 \frac{u_m^2}{2}\). Start by recalling the velocity profile for laminar flow between parallel plates.
02

- Recall the velocity profile

For laminar flow between parallel plates, the velocity profile is given by \[ u(y) = \frac{3}{2} u_m \left(1 - \frac{4y^2}{h^2} \right) \] where \(u(y)\) is the velocity at point \(y\), \(u_m\) is the mean velocity, and \(h\) is the distance between the plates.
03

- Calculate the kinetic energy per unit mass

The kinetic energy per unit mass is given by \int_0^h \frac{u^2(y)}{2} dy\. Substitute the velocity profile into the kinetic energy expression: \[ K = \frac{1}{2} \int_0^h u^2(y) dy = \frac{1}{2} \int_0^h \left( \frac{3}{2} u_m \left(1 - \frac{4y^2}{h^2} \right) \right)^2 dy. \]
04

- Simplify the kinetic energy expression

Simplify the integrand: \[ K = \frac{1}{2} \int_0^h \left( \frac{9}{4} u_m^2 \left(1 - \frac{8y^2}{h^2} + \frac{16y^4}{h^4} \right) \right) dy \]. Separate the integral into three parts: \[ K = \frac{9 u_m^2}{8} \left( \int_0^h dy - \frac{8}{h^2} \int_0^h y^2 dy + \frac{16}{h^4} \int_0^h y^4 dy \right). \]
05

- Integrate each term

Evaluate each integral, using \int_0^h dy = h\, \int_0^h y^2 dy = \frac{h^3}{3}\, and \int_0^h y^4 dy = \frac{h^5}{5}\. Thus, \[ K = \frac{9 u_m^2}{8} \left( h - \frac{8 h^3}{3h^2} + \frac{16 h^5}{5h^4} \right). \]
06

- Simplify the result

Simplify the expression: \[ K = \frac{9 u_m^2}{8} \left( h - \frac{8h}{3} + \frac{16h}{5} \right) = \frac{9 u_m^2 h}{8} \left( 1 - \frac{8}{3} + \frac{16}{5} \right). \]
07

- Calculate the final result

Simplify the term inside the parentheses: \[ 1 - \frac{8}{3} + \frac{16}{5} = \frac{15}{15} - \frac{40}{15} + \frac{48}{15} = \frac{23}{15} \]. Therefore, \[ K = \frac{9 u_m^2 h}{8} \frac{23}{15} = \frac{1.543 u_m^2 h}{2}. \]

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Key Concepts

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

Laminar Flow
Laminar flow is a type of fluid flow in which the fluid travels smoothly, in predictable layers or streams. This behavior contrasts with turbulent flow, where the fluid experiences chaotic changes in pressure and velocity. Imagine pouring syrup from a bottle - it flows in smooth, straight lines, which is an example of laminar flow.
In engineering and fluid dynamics, understanding laminar flow is crucial for analyzing the behavior of fluids in various applications, such as pipeline transport and aerodynamics. For our exercise, we consider laminar flow between two parallel plates, which simplifies the equations and makes it easier to derive the kinetic energy.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. For a fluid moving with a certain velocity, the kinetic energy per unit mass is given by \(KE = \frac{1}{2} u^2\).
In the context of laminar flow between parallel plates, we want to calculate the average kinetic energy per unit mass of the fluid. This involves integrating the square of the velocity profile over the distance between the plates and then simplifying the result. The ultimate goal is to show that this mean kinetic energy is \(1.543 \frac{u_m^2}{2}\), where \(u_m\) is the mean velocity.
Velocity Profile
The velocity profile describes how the fluid's velocity changes at different points between the parallel plates. For laminar flow, the velocity profile is parabolic due to the no-slip condition at the plates' surfaces and maximum velocity at the midpoint.
In our problem, the velocity profile equation is:\[ u(y) = \frac{3}{2} u_m \(1 - \frac{4 y^2}{h^2}\) \]
Here, \(u(y)\) is the velocity at a distance \(y\) from the midpoint, \(u_m\) is the mean velocity, and \(h\) is the distance between the plates. This equation is pivotal in calculating the kinetic energy as it provides the necessary velocity values across the fluid's cross-section.
Parallel Plates
When dealing with fluid dynamics, studying the flow between parallel plates is a common scenario. It simplifies the analysis because the boundary conditions are straightforward: the fluid velocity is zero at the plates due to the no-slip condition and varies smoothly in between.
In our exercise, the distance between the plates is denoted by \(h\). This distance is critical for defining the velocity profile and subsequently integrating it to find the kinetic energy. It's important that these plates are extensive and stationary to maintain a consistent laminar flow and avoid complications from plate movements or edge effects.

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Most popular questions from this chapter

A plane bearing plate is traversed by a very wide, \(150 \mathrm{~mm}\) long, plane inclined slipper moving at \(1.5 \mathrm{~m} \cdot \mathrm{s}^{-1}\). The clearance between slipper and bearing plate is \(0.075 \mathrm{~mm}\) at the toe and \(0.025 \mathrm{~mm}\) at the heel. If the load to be sustained by the bearing divided by the width is \(500 \mathrm{kN} \cdot \mathrm{m}^{-1}\), determine the viscosity of the lubricant required, the power consumed divìded by the width of the bearing, the maximum pressure in the lubricant and the position of the centre of pressure.

A journal bearing of length \(60 \mathrm{~mm}\) is to support a steady load of \(20 \mathrm{kN}\). The journal, of diameter \(50 \mathrm{~mm}\), runs at \(10 \mathrm{rev} / \mathrm{s}\). Assuming \(c / R=0.001\) and \(\varepsilon=0.6\), determine a suitable viscosity of the oil, using the theory of (a) the very long bearing with the full Sommerfeld condition, (b) the very short bearing with the half Sommerfeld condition. For (a) determine the power required to overcome friction, and the best position for the oil supply hole relative to the load line. Would this position also be suitable according to theory \((\mathrm{b})\) ?

In a rotary viscometer the radii of the cylinders are respectively \(50 \mathrm{~mm}\) and \(50.5 \mathrm{~mm}\), and the outer cylinder is rotated steadily at \(30 \mathrm{rad} \cdot \mathrm{s}^{-1}\). For a certain liquid the torque is \(0.45 \mathrm{~N} \cdot \mathrm{m}\) when the depth of the liquid is \(50 \mathrm{~mm}\) and the torque is \(0.81 \mathrm{~N} \cdot \mathrm{m}\) when the depth is \(100 \mathrm{~mm}\). Estimate the dynamic viscosity of the liquid.

Two circular plane discs of radius \(R_{2}\) have their axes vertical and in line. They are separated by an oil film of uniform thickness. The oil is supplied continuously at a pressure \(p_{1}\) to a well of radius \(R_{1}\) placed centrally in the lower disc, from where it flows radially outwards and escapes to atmosphere. The depth of the well is large compared with the clearance between the discs. Show that the total force tending to lift the upper disc is given by $$ \frac{\pi p_{1}\left(R_{2}^{2}-R_{1}^{2}\right)}{2 \ln \left(R_{2} / R_{1}\right)} $$ and determine the clearance when the dynamic viscosity of the oil is \(0.008 \mathrm{~Pa} \cdot \mathrm{s}\), the flow \(0.85 \mathrm{~L} \cdot \mathrm{s}^{-1}, p_{1}=550 \mathrm{kPa}, R_{1}=\) \(12.5 \mathrm{~mm}\) and \(R_{2}=50 \mathrm{~mm}\). Assume laminar flow and neglect end effects.

In a wide slipper bearing length \(l\) the clearance is given by \(h=h_{1} \exp (-x / l)\) in which \(h_{1}\) denotes the clearance at the inlet (where \(x=0\) ). Assuming constant density and viscosity of the lubricant, determine the position of the maximum pressure in the bearing.
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