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 108P

Calculate the nine components of the viscous stress tensor in cylindrical coordinates (see Chap. 9) for the velocity field of Prob 10-107. Discuss your results.

Expert Solution & Answer
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To determine

The nine component of the viscous stress tensor in cylindrical coordinates.

Answer to Problem 108P

The nine component of the viscous stress tensor in cylindrical coordinates is τij=[000000000].

Explanation of Solution

Given information:

The radial velocity component is 0, angular velocity component is ωr and z-direction velocity component is 0.

Write the expression for all the nine component of viscous stress tensor in cylindrical coordinates.

  τij=[ τ rr τ rθ τ rz τ θr τ θθ τ θz τ zr τ zθ τ zz]   ...... (I)

Here, viscous stress tensor in i-j plane is τij.

Write the expression for viscous stress tensor in rr plane.

  τrr=2μurr   ...... (II)

Here, dynamic viscosity is μ.velocity in radial direction is ur.

Write the expression for viscous stress tensor in rθ plane.

  τrθ=μ(rr( u θ r)+1rurθ)   ...... (III)

Here, velocity in angular direction is uθ.

Write the expression for viscous stress tensor in rz plane.

  τrz=μ(urz+uzr)   ...... (IV)

Here, velocity in z-direction is uz

Write the expression for viscous stress tensor in θr plane.

  τθr=μ(rr( u θ r)+1rurθ)   ...... (V)

Write the expression for viscous stress tensor in θθ plane.

  τθθ=2μ(1rurθ+urr)   ...... (VI)

Write the expression for viscous stress tensor in θz plane.

  τθz=μ(uθz+1ruzθ)   ...... (VII)

Write the expression for viscous stress tensor in zr plane.

  τzr=μ(urz+uzr)   ...... (VIII)

Write the expression for viscous stress tensor in zθ plane.

  τzθ=μ(uθz+1ruzθ)   ...... (IX)

Write the expression for viscous stress tensor in zz plane.

  τzz=2μuzz   ...... (X)

Substitute 2μurr for τrr, μ(urz+uzr) for τrz, μ(rr( u θ r)+1rurθ) for τrθ, μ(rr( u θ r)+1rurθ) for τθr, 2μ(1rurθ+urr) for τθθ, μ(urz+uzr) for τzr, μ(uθz+1ruzθ) for τθz, μ(uθz+1ruzθ) for τzθ, 2μuzz for τzz in Equation (I).

  τij=[2μ u r rμ( r r ( u θ r )+ 1 r u r θ )μ( u r z + u z r )μ( r r ( u θ r )+ 1 r u r θ )2μ( 1 r u r θ + u r r )μ( u θ z + 1 r u z θ )μ( u r z + u z r )μ( u θ z + 1 r u z θ )2μ u z z]   ...... (XI)

Substitute 0 for ur, ωr for uθ and 0 for uz in Equation (XI).

   τ ij =[ 2μ ( 0 ) r μ( r r ( ωr r )+ 1 r ( 0 ) θ ) μ( ( 0 ) z + ( 0 ) r ) μ( r r ( ωr r )+ 1 r ( 0 ) θ ) 2μ( 1 r ( 0 ) θ + ( 0 ) r ) μ( ( ωr ) z + 1 r ( 0 ) θ ) μ( ( 0 ) z + ( 0 ) r ) μ( ( ωr ) z + 1 r ( 0 ) θ ) 2μ ( 0 ) z ]

   =[ 2μ( 0 ) μ( r( 0 )+ 1 r ( 0 ) ) μ( 0+0 ) μ( r( 0 )+ 1 r ( 0 ) ) 2μ( 1 r ( 0 )+0 ) μ( 0+ 1 r ( 0 ) ) μ( 0+0 ) μ( 0+ 1 r ( 0 ) ) 2μ( 0 ) ]

   =[ 0 0 0 0 0 0 0 0 0 ]

Here, all component of viscous stress tensor is zero, so flow is inviscid and no viscous stress is present in flow field.

Conclusion:

The nine component of the viscous stress tensor in cylindrical coordinates is τij=[000000000].

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