Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 10, Problem 62P

Chapter 10, Problem 62P,

Expert Solution & Answer
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To determine

The laplace equation of velocity potential function in spherical polar coordinate.

Answer to Problem 62P

The laplace equation of velocity potential function in spherical polar coordinate is (1 cos2θ sin2θ)[( 2ϕ r 2)+1r2( 2ϕ r 2)]=0.

Explanation of Solution

Given information:

The distance from origin is r and angle of inclination between the radial vector and axis of rotation is θ.

The radial vector about the coordinate axis is shown in Figure (1).

  Fluid Mechanics: Fundamentals and Applications, Chapter 10, Problem 62P

  Figure-(1)

Write the expression for component of vector r along the x-axis.

  x=rcosθ   ...... (I)

Write the expression for component of vector r along the y-axis.

  y=rsinθ   ...... (II)

Write the expression for Laplace equation for the velocity potential function Cartesian coordinate.

  2ϕ=02ϕx2+2ϕy2=0   ...... (III)

Write expression for Laplace equation as function of r and θ.

  [( 2 ϕ r 2 )( x 2 r 2 )+( 2 ϕ θ 2 )( x 2 θ 2 )]+[( 2 ϕ r 2 )( y 2 r 2 )+( 2 ϕ θ 2 )( y 2 θ 2 )]=0[( 2 ϕ r 2 ) ( x r ) 2+( 2 ϕ θ 2 ) ( x θ ) 2]+[( 2 ϕ r 2 ) ( y r ) 2+( 2 ϕ θ 2 ) ( y θ ) 2]=0   ...... (IV)

Differentiate Equation (I) with respect to r.

  xr=r(rcosθ)=cosθ   ...... (V)

Differentiate Equation (I) with respect to θ.

  xθ=θ(rcosθ)=rsinθ   ...... (VI)

Differentiate Equation (II) with respect to r.

  yr=r(rsinθ)=sinθ   ...... (VII)

Differentiate Equation (II) with respect to θ.

  yθ=θ(rsinθ)=rcosθ   ...... (VIII)

Substitute cosθ for xr, rsinθ for xθ, sinθ for yr and rcosθ for yθ in Equation (IV).

  ( 2 ϕ r 2 ) ( cosθ )2+( 2 ϕ θ 2 ) ( rsinθ )2+( 2 ϕ r 2 ) ( sinθ )2+( 2 ϕ θ 2 ) ( rcosθ )2=0( 2 ϕ r 2 )(1 cos 2 θ+1 sin 2 θ)+( 2 ϕ r 2 )1r2(1 cos 2 θ+1 sin 2 θ)=0( 2 ϕ r 2 )(1 cos 2 θ sin 2 θ)+( 2 ϕ r 2 )1r2(1 cos 2 θ sin 2 θ)=0(1 cos 2 θ sin 2 θ)[( 2 ϕ r 2 )+1 r 2( 2 ϕ r 2 )]=0

Conclusion:

The laplace equation of velocity potential function in spherical polar coordinate is (1 cos2θ sin2θ)[( 2ϕ r 2)+1r2( 2ϕ r 2)]=0.

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Chapter 10 Solutions

Fluid Mechanics: Fundamentals and Applications

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