Fundamentals of Aerodynamics
Fundamentals of Aerodynamics
6th Edition
ISBN: 9781259129919
Author: John D. Anderson Jr.
Publisher: McGraw-Hill Education
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Chapter 7, Problem 7.13P

Note: In the following problems, you will deal with both the International System of Units (SI) (N, kg, m, s, K) and the English Engineering System (lb, slug, ft, s, ° R ). Which system to use will be self-evident in each problem. All problems deal with calorically perfect air as the gas, unless otherwise noted. Also, recall that 1 atm = 2116 lb/ft 2 = 1.01 × 10 5 N/m 2 .

Bernoulli’s equation, Equation ( 3.13 ) , ( 3.14 ) , or ( 3.15 ) , was derived in Chapter 3 from Newton’s second law; it is fundamentally a statement that force = mass × acceleration. However, the terms in Bernoulli’s equation have dimensions of energy per unit volume (check it out), which prompt some argument that Bernoulli’s equation is an energy equation for incompressible flow. If this is so, then it should be derivable from the energy equation for compressible flow discussed in the present chapter. Starting with Equation ( 7.53 ) for inviscid, adiabatic compressible flow, make the appropriate assumptions for an incompressible flow and see what you need to do to obtain Bernoulli’s equation.

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Q1. A curved beam of a circular cross section of diameter "d" is fixed at one end and subjected to a concentrated load P at the free end (Fig. 1). Calculate stresses at points A and C. Given: P = 800 N, d = 30 mm, a 25 mm, and b = 15 mm. Fig.1 P b B (10 Marks)
You are working as an engineer in a bearing systems design company. The flow of lubricant inside a hydrodynamic bearing (p = 0.001 kg m-1 s-1) can be approximated as a parallel, steady, two-dimensional, incompressible flow between two parallel plates. The top plate, representing the moving part of the bearing, travels at a constant speed, U, while the bottom plate remains stationary (Figure Q1). The plates are separated by a distance of 2h = 1 cm and are W = 20 cm wide. Their length is L = 10 cm. By applying the above approximations to the Navier-Stokes equations and assuming that end effects can be neglected, the horizontal velocity profile can be shown to be y = +h I 2h = 1 cm x1 y = -h u(y) 1 dP 2μ dx -y² + Ay + B moving plate stationary plate U 2 I2 L = 10 cm Figure Q1: Flow in a hydrodynamic bearing. The plates extend a width, W = 20 cm, into the page.
Question 1 You are working as an engineer in a bearing systems design company. The flow of lubricant inside a hydrodynamic bearing (µ = 0.001 kg m¯¹ s¯¹) can be approximated as a parallel, steady, two-dimensional, incompressible flow between two parallel plates. The top plate, representing the moving part of the bearing, travels at a constant speed, U, while the bottom plate remains stationary (Figure Q1). The plates are separated by a distance of 2h = 1 cm and are W = 20 cm wide. Their length is L = 10 cm. By applying the above approximations to the Navier-Stokes equations and assuming that end effects can be neglected, the horizontal velocity profile can be shown to be 1 dP u(y) = 2μ dx -y² + Ay + B y= +h Ꮖ 2h=1 cm 1 x1 y = −h moving plate stationary plate 2 X2 L = 10 cm Figure Q1: Flow in a hydrodynamic bearing. The plates extend a width, W = 20 cm, into the page. (a) By considering the appropriate boundary conditions, show that the constants take the following forms: U U 1 dP A =…
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