Chapter05

Mechanics of Fluids

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
89 CHAPTER 5 The Differential Forms of the Fundamental Laws 5.1 0 = - + ∂ρ ρ t d V V ndA c v c s . . . . $ . v Using Gauss’ theorem: 0 = - + ∇⋅ - = + ∇⋅ - ∂ρ ρ ∂ρ ρ t d V V d V t V dV c v c v c v . . . . . . ( ) ( ) . v v v v Since this is true for all arbitrary control volumes (i.e., for all limits of integration), the integrand must be zero: ∂ρ ρ t V = v v ( ) . 0 This can be written in rectangular coordinates as - = + + ∂ρ ρ ρ ρ t x u y v z w ( ) ( ) ( ). This is Eq. 5.2.2. The other forms of the continuity equation follow. 5.2 & & . m m m t in out element - = ( 29 ρ θ ρ ρ θ v rd dz v r v dr r dr d dz r r r ( ) ( ) - + + + - + ρ ρ ∂θ ρ θ θ θ θ v drdz v v d drdz ( ) + + - + + = + ρ θ ρ ρ θ ρ θ v r dr d dr v z v dz r dr d dr t r dr d drdz z z z 2 2 2 ( ) . Subtract terms and divide by rd drdz q : - - + - - + = + ρ ρ ∂θ ρ ρ ρ θ v r r v r dr r v r z v r dr r t r dr r r r z ( ) ( ) ( ) / / . 1 2 2 Since dr is an infinitesimal, ( )/ ( / . r dr r r dr r + = + = 1 2 1 and Hence, ∂ρ ρ ∂θ ρ ρ ρ θ t r v r v z v r v r z r + + + + = ( ) ( ) ( ) . 1 1 This can be put in various forms.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
90 5.3 & & . m m m t in out element - = ρ θ θ φ ρ ρ θ θ φ v rd r d v r v dr r dr d r dr d r r r ( ) sin ( ) ( ) ( )sin - + + + + + - + + ρ θ φ ρ ∂θ ρ θ θ φ θ θ θ v dr r dr d v v d dr r dr d 2 2 sin ( ) sin + + - + + ρ θ ρ ∂φ ρ φ θ φ φ φ v dr r dr d v v d dr r dr d 2 2 ( ) = + ρ θ θ φ t r dr drd d 2 2 sin Because some areas are not rectangular, we used an average length ( / ). r dr + 2 Now, subtract some terms and divide by rd d dr q f : - - - + - + ρ θ ρ θ ρ θ ∂θ ρ θ θ v v r v r dr r v r dr r r r r sin ( )sin ( ) ( ) 2 2 - + = + ∂φ ρ ∂ρ θ φ ( ) v r dr r t r dr r 2 2 2 Since dr is infinitesimal ( ) / ( / )/ . r dr r r r dr r + = + = 2 2 1 and Divide by r q and there results ∂ρ ρ ∂θ ρ θ ∂φ ρ ρ θ φ t r v r v r v r v r r + + + + = ( ) ( ) ( ) 1 1 2 0 5.4 For a steady flow t = 0. Then, with v w = = 0 Eq. 5.2.2 yields ρ ρ ρ x u du dx u d dx ( ) . = + = 0 0 or Partial derivatives are not used since there is only one independent variable. 5.5 Since the flow is incompressible D Dt r = 0. This gives 3 23 1 20 0 1 200 ˆ ˆ ˆˆ cos 2 sin2 rr pp p i i ii r qq ¶q  ∴∇ = + = --   v or u x w z ∂ρ + = 0. Also, v v ∇⋅ = V 0, or u x w z + = 0.
Background image of page 2
91 5.6 Given: ∂ρ t z = 0 0 , . Since water can be considered to be incompressible, we demand that D Dt r = 0.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This document was uploaded on 03/14/2008.

Page1 / 14

Chapter05 - CHAPTER 5 The Differential Forms of the...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online