Chapter5 - 57:020 Mechanics of Fluids and Transport...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 1 Chapter 5 Finite Control Volume Analysis 5.1 Continuity Equation RTT can be used to obtain an integral relationship expressing conservation of mass by defining the extensive property B = M such that β = 1. B = M = mass β = dB/dM = 1 General Form of Continuity Equation ∫∫ ρ + ρ = = CV CS dA V V d dt d 0 dt dM or ρ = ρ CV CS V d dt d dA V net rate of outflow rate of decrease of of mass across CS mass within CV Simplifications: 1. Steady flow: 0 V d dt d CV = ρ 2. V = constant over discrete dA (flow sections): ρ = ρ CS CS A V dA V
Background image of page 1

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

View Full DocumentRight Arrow Icon
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 2 3. Incompressible fluid ( ρ = constant) CS CV d VdA dV dt ⋅= ∫∫ conservation of volume 4. Steady One-Dimensional Flow in a Conduit: = ρ CS 0 A V −ρ 1 V 1 A 1 + ρ 2 V 2 A 2 = 0 for ρ = constant Q 1 = Q 2 Some useful definitions: Mass flux ρ = A dA V m & Volume flux = A dA V Q Average Velocity A / Q V = Average Density ρ = ρ dA A 1 Note: mQ ≠ ρ & unless ρ = constant
Background image of page 2
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 3 Example *Steady flow *V 1,2,3 = 50 fps *@ © V varies linearly from zero at wall to V max at pipe center *find 4 m & , Q 4 , V max 0 *water, ρ w = 1.94 slug/ft 3 ρ = = ρ CV CS V d dt d 0 dA V i.e., - ρ 1 V 1 A 1 - ρ 2 V 2 A 2 + ρ 3 V 3 A 3 + ρ 4 A 4 4 dA V = 0 ρ = const. = 1.94 lb-s 2 /ft 4 = 1.94 slug/ft 3 ρ = 4 4 4 dA V m & = ρ V(A 1 + A 2 – A 3 ) V 1 =V 2 =V 3 =V=50f/s = () 2 2 2 5 . 1 2 1 4 50 144 94 . 1 + π × × = 1.45 slugs/s 4 m &
Background image of page 3

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

View Full DocumentRight Arrow Icon
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 4 Q 4 = 75 . m 4 = ρ & ft 3 /s = 4 A 4 4 dA V velocity profile Q 4 = ∫∫ θ π o r 0 2 0 o max dr rd r r 1 V max 2 o 2 o max r 0 o 3 r 0 2 max r 0 o 2 max r 0 o max V r 3 1 3 1 2 1 r V 2 r 3 r 2 r V 2 dr r r r V 2 rdr r r 1 V 2 o 0 o o π = π = π = π = π = V max = 86 . 2 r 3 1 Q 2 o 4 = π fps V 4 V 4 ( θ ) dA 4 2 o max 2 o 4 r V r 3 1 A Q V π π = = = max V 3 1
Background image of page 4
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 5 5.2 Momentum Equation Derivation of the Momentum Equation Newton’s second law of motion for a system is time rate of change of the momentum of the system = sum of external forces acting on the system Since momentum is mass times velocity, the momentum of a small particle of mass ߩܸ݀ is ܸ ߩܸ݀ and the momentum of the entire system is ׬ ܸ ߩܸ݀ ௦௬௦ . Thus, ܦ ܦݐ ܸ ߩܸ݀ ൌ෍ܨ ௦௬௦ ௦௬௦ Recall RTT: ܦܤ ௦௬௦ ܦݐ ߲ ߲ݐ න ߚߩܸ݀ ஼௏ ൅ න ߚߩܸ ڄ݀ܣ ஼ௌ With ܤ ௦௬௦ ൌܯܸ and ߚൌ ௗ஻ ೞ೤ೞ ௗெ ൌܸ , ܦ ܦݐ ܸ ߩܸ݀ ߲ ߲ݐ න ܸ ߩܸ݀ ஼௏ ൅ න ܸ ߩܸ ஼ௌ ௦௬௦
Background image of page 5

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

View Full DocumentRight Arrow Icon
57:020 Mechanics of Fluids and Transport Processes Chapter 5
Background image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 42

Chapter5 - 57:020 Mechanics of Fluids and Transport...

This preview shows document pages 1 - 7. Sign up to view the full document.

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