Connect 1 Semester Access Card For Fluid Mechanics Fundamentals And Applications
Connect 1 Semester Access Card For Fluid Mechanics Fundamentals And Applications
3rd Edition
ISBN: 9780077670245
Author: CENGEL
Publisher: McGraw-Hill Education
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Chapter 4, Problem 24P

A steady, incompressible, two-dimensional (in the xy-plane) velocity field is given by V = ( 0.523 1.88 x + 3.94 y ) i + ( 2.44 + 1.26 x + 1.88 y ) j Calculate the acceleration at the point (x.y) = (-1.55, 2.07).

Expert Solution & Answer
Check Mark
To determine

The acceleration at the point (1.55,2.07).

Answer to Problem 24P

The acceleration of the flow is 23.7699i+13.664j with a magnitude of 27.417.

Explanation of Solution

Given information:

Write the expression for the velocity field.

  V=(0.5231.88x+3.94y)i+(2.44+1.26x+1.88y)j   ...... (I)

Here, the variables are x, y, the velocity field is V, and the unit vectors are i and j.

Write the expression for the velocity component along x direction.

  u=0.523+1.88x+3.94y   ...... (II)

Here, the variables are x,y and the velocity component along x direction is u.

Write the expression for the velocity component in y direction.

  v=2.44+1.26x+1.88y  ...... (III)

Here, the variables are x,y and the velocity component along y direction is v.

Write the expression for the acceleration of a flow along horizontal axis at point (1.55,2.07).

  ax=ut+uux+vuy   ...... (IV)

Here, the velocity component in x direction is u, the velocity component in x direction is v, and the acceleration is ax.

Write the expression for the acceleration of a flow along vertical axis at point (1.55,2.07).

  ay=vt+uvx+vvy   ...... (V)

Here, the velocity component in x direction is u, the velocity component in y direction is v, and the acceleration is ay.

Write the expression for the magnitude of the acceleration.

  a=ax2+ay2   ...... (VI)

Here, the magnitude of the acceleration is a.

Write the expression for the vector notation of the acceleration.

  a=axi+ayj   ...... (VII)

Calculation:

Differentiate Equation (II) with respect to x.

  ux=(0.5231.88x+3.94y)xux=1.88

Differentiate Equation (II) with respect to y.

  uy=(0.5231.88x+3.94y)xuy=3.94

Differentiate Equation (II) with respect to t.

  ut=(0.5231.88x+3.94y)xut=0

Differentiate Equation (III) with respect to x.

  vx=(2.44+1.26x+1.88y)xvx=1.26

Differentiate Equation (III) with respect to y.

  vy=(2.44+1.26x+1.88y)xuy=1.88

Differentiate Equation (III) with respect to t.

  vt=(2.44+1.26x+1.88y)xut=0

Substitute 1.55 for x and 2.07 for y in Equation (II).

  u=0.523(1.88(1.55))+(3.94(2.07))=0.523+2.914+8.1558=11.5928

Substitute 1.55 for x and 2.07 for y in Equation (III).

  v=2.44+(1.26(1.55))+(1.88×2.07)=2.441.953+3.8916=0.5014

Substitute 11.5928 for y, 0.5014 for v, 0 for ut, 1.88 for ux and 3.94 for uy in Equation (IV).

  ax=0+((11.5928)(1.88))+((0.5014)(3.94))=21.79441.9755=23.7699

Substitute 11.5928 for y, 0.5014 for v, 0 for vt, 1.26 for vx and 1.88 for vy in Equation (V).

  ay=0+((11.5298)(1.26))+(0.5014)(1.88)=14.5275480.942632=13.58

Substitute 23.7699 for ax and 13.664 for ay in Equation (VI).

  a=( 23.7699)2+13.6642=565.008+186.704=27.417

Substitute 23.7699 for ax and 13.664 for ay in Equation (VII).

  a=23.7699i+13.664j

Conclusion:

The acceleration of the flow is 23.7699i+13.664j with a magnitude of 27.417.

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Chapter 4 Solutions

Connect 1 Semester Access Card For Fluid Mechanics Fundamentals And Applications

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