Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 4, Problem 4.3P
To determine

(a)

The acceleration ax.

The acceleration ay.

Expert Solution
Check Mark

Answer to Problem 4.3P

The acceleration ax is 18units.

The acceleration ay is 26units.

Explanation of Solution

Given information:

The two dimensional velocity field is,

V=(x2y2+x)i(2xy+y)j

Here, the variables are x, y and z.

The given point is (x,y)=(1,2).

Write the general expression for velocity.

V=ui+vj+wk

Here, the velocity function for the i coordinate is u, the velocity function for the j coordinate is v, the velocity function for k coordinate is w.

Write the expression for the acceleration in x-direction.

ax=uux+vuy+wuz ..... (I)

Here, the velocity gradient for u in x-coordinate is ux, the velocity gradient for u in y-coordinate is uy and the velocity gradient for u in z-coordinate is uz.

Write the expression for the acceleration in y-direction.

ay=uvx+vvy+wvz ..... (II)

Here, the velocity gradient for v in x-coordinate is vx, the velocity gradient for v in y-coordinate is vy and the velocity gradient for v in z-coordinate is vz.

Substitute (x2y2+x) for u, (2xy+y) for v and 0 for w in Equation (I).

ax=[( x 2 y 2 +x) ( x 2 y 2 +x ) x+[ ( 2xy+y )] [ x 2 y 2 +x] y+( 0) ( x 2 y 2 +x ) z]=(x2y2+x)(2x+1)+[(2xy+y)](2y)+0=(x2y2+x)(2x+1)+[(2xy+y)](2y) ..... (III)

Substitute (x2y2+x) for u, (2xy+y) for v and 0 for w in Equation (II).

ay=[( x 2 y 2 +x) [ ( 2xy+y )] x+[ ( 2xy+y )] [ ( 2xy+y )] y+( 0) [ ( 2xy+y )] z]=(x2y2+x)(2y)+[(2xy+y)](2x1)=(x2y2+x)(2y)+[(2xy+y)(2x+1)] ..... (IV)

Calculation:

Substitute 1 for x and 2 for y in Equation (III).

ax=[( 1)222+1][2(1)+1]+[(2( 1)( 2)+2)](2×2)=2×3+24=246=18units

Substitute 1 for x and 2 for y in Equation (IV).

ay=(1222+1)(2(2))+[(2( 1)( 2)+2)(2( 1)+1)]=(2)(4)+18=26units

Conclusion:

Thus, the acceleration ax is 18units.

Thus, the acceleration ay is 26units.

To determine

(b)

The velocity component in the direction θ=40°C.

Expert Solution
Check Mark

Answer to Problem 4.3P

The velocity component in the direction θ=40° is 5.3882units.

Explanation of Solution

Given information:

The angle for the velocity component is θ=40°.

Write the expression for two-dimensional velocity field.

V=(x2y2+x)i(2xy+y)j ..... (V)

Write the expression for the normal vector at any angle.

n=cosθi+sinθj ..... (VI)

Here, the angle with the x-axis is θ.

Write the expression for the magnitude of velocity in direction of n.

V=Vn ..... (VII)

Calculation:

Substitute 1 for x and 2 for y in Equation (V).

V=(1222+1)i(2(1)(2)+2)j=2i6j

Substitute 40° for θ in Equation (VI).

n=cos(40°)i+sin(40°)j=0.766i+0.6427j

Substitute 2i6j for V and 0.766i+0.6427j for n in Equation (VII).

V=(2i6j)(0.766i+0.6427j)=(2)(0.766)+(6)(0.6427)=1.5323.8562=5.3882units

Conclusion:

Thus, the velocity component in the direction θ=40° is 5.3882units.

To determine

(c)

The direction of maximum velocity.

Expert Solution
Check Mark

Answer to Problem 4.3P

The direction of maximum velocity is 55.3°.

Explanation of Solution

Given data The direction of maximum velocity will be equal to the direction of maximum acceleration.

Write the expression for direction of maximum acceleration.

tanθ=( a y a x )θ=tan1( a y a x ) ..... (VIII)

Calculation:

Substitute 26 for ay and 18 for ax in Equation (VIII).

θ=tan1( 26 18)=tan1(1.444)=55.3°

Conclusion:

Thus, the direction of maximum velocity is 55.3°.

To determine

(d)

The direction of maximum acceleration.

Expert Solution
Check Mark

Answer to Problem 4.3P

The direction of maximum acceleration is 55.3°.

Explanation of Solution

Write the expression for direction of maximum acceleration.

θ=tan1(ayax) ..... (IX)

Calculation:

Substitute 26 for ay and 18 for ax in Equation (IX).

θ=tan1( 26 18)=tan1(1.444)=55.3°

Conclusion:

Thus, the direction of maximum velocity is 55.3°.

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Chapter 4 Solutions

Fluid Mechanics

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