Fox and McDonald's Introduction to Fluid Mechanics
Fox and McDonald's Introduction to Fluid Mechanics
9th Edition
ISBN: 9781118912652
Author: Philip J. Pritchard, John W. Mitchell
Publisher: WILEY
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Chapter 6, Problem 1P

An incompressible frictionless flow field is given by V = ( A x + B y ) i ^ + ( B x A y ) j ^ , where A = 2 s−1 and B = 2 s−1, and the coordinates are measured in meters. Find the magnitude and direction of the acceleration of a fluid particle at point (x, y) = (2, 2). Find the pressure gradient at the same point, if g = g j ^ and the fluid is water.

Expert Solution & Answer
Check Mark
To determine

The magnitude of the acceleration of a fluid particle at point (2,2).

The direction of the acceleration of a fluid particle at point (2,2).

The pressure gradient of fluid at point (2,2).

Explanation of Solution

Given:

The flow field (V) is (Ax+By)i^+(BxAy)j^.

The constants A and B are 2s1 and 2s1.

Fluid particle point (x,y) is (2,2)m.

Consider the density of water (ρ) is 999kg/m3.

Calculations:

From the flow field (V), the components are:

  u=Ax+Byv=BxAy

Calculate the acceleration of the particle along x direction (ax).

  ax=uxu+vyu=(Ax+By)x(Ax+By)+(BxAy)y(Ax+By)=A(Ax+By)+(BxAy)B=(A2+B2)x

  ax=[(2s1)2+(2s1)2](2m)=16m/s2

Calculate the acceleration of the particle along y direction (ay).

  ay=uxv+vyv=(Ax+By)x(BxAy)+(BxAy)y(BxAy)=B(Ax+By)(BxAy)A

  ay=(A2+B2)y=[(2s1)2+(2s1)2](2m)=16m/s2

Calculate the resultant acceleration of a fluid particle at point.

  a=(ax)2+(ay)2=(16m/s2)2+(16m/s2)2=22.627m/s2

Thus, the magnitude of the acceleration of a fluid particle at point (2,2) is 22.627m/s2_.

Calculate the direction of the acceleration of fluid particle (θ).

  θ=tan1(ayax)=tan1(16m/s216m/s2)=45°

Thus, the direction of the acceleration of a fluid particle at point (2,2) is 45°_.

Calculate the pressure gradient in x direction.

  xp=ρax=(999kg/m3)×16m/s2×Ns2kgm×1kPa1000N/m2=15.984kPa/m

Thus, the pressure gradient of fluid at point (2,2) in x direction is 15.984kPa/m_.

Calculate the pressure gradient in y direction.

  yp=ρgyρay=(999kg/m3)×(9.81m/s2+16m/s2)×Ns2kgm×1kPa1000N/m2=25.784kPa/m

Thus, the pressure gradient of fluid at point (2,2) in y direction is 25.784kPa/m_.

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Chapter 6 Solutions

Fox and McDonald's Introduction to Fluid Mechanics

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