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![23
SOLUTION
𝑢=𝑢1+𝐶𝑢 𝑥 𝐶𝑢=
𝑢2 − 𝑢1
∆ 𝑥
=
300 −100
2
=100 𝑠
− 1
𝜌=𝜌1+𝐶𝜌 𝑥 𝐶𝜌=
𝜌2 − 𝜌1
∆ 𝑥
=
0.85 −1.2
2
=−0.175 𝑠
− 1
For steady, 2-D, compressible flow:
𝜕(𝜌 𝑢)
𝜕 𝑥
+
𝜕( 𝜌 𝑣)
𝜕 𝑦
=0
𝜕[(𝜌1+𝐶𝜌 𝑥 )(𝑢1+𝐶𝑢 𝑥 )]
𝜕 𝑥
+
𝜕( 𝜌 𝑣)
𝜕 𝑦
=0
𝜌1 𝐶𝑢+𝑢1 𝐶𝜌+2𝐶𝑢 𝐶𝜌 𝑥+
𝜕(𝜌 𝑣)
𝜕 𝑦
=0](https://image.slidesharecdn.com/8-differentialflowanalysis-250905120832-caf0e304/75/8-Differential-Flow-AnalysisNilsson-Riedel-Chapter-3-ppt-pptx-23-2048.jpg)
![26
− ln [𝜌1 𝑢1+(𝜌1 𝐶𝑢+𝑢1 𝐶𝜌 )𝑥+𝐶𝑢 𝐶𝜌 𝑥
2
]=ln 𝑦+𝐶
𝑦=
𝐶
𝜌1𝑢1+( 𝜌1 𝐶𝑢 +𝑢1 𝐶𝜌 ) 𝑥+𝐶𝑢 𝐶𝜌 𝑥
2
To get the height () at the duct exit, find the
streamline of the top wall:
At
At
𝐶 =2 𝜌 1 𝑢1
𝑦=0.9412m](https://image.slidesharecdn.com/8-differentialflowanalysis-250905120832-caf0e304/75/8-Differential-Flow-AnalysisNilsson-Riedel-Chapter-3-ppt-pptx-26-2048.jpg)
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8- Differential Flow AnalysisNilsson Riedel Chapter 3 ppt.pptx
- 1.
- 2. 2 DIFFERENTIAL FLOW ANALYSIS •Lagrangian and Eulerian Flow Descriptions • Acceleration • Streamline • Continuity equation
- 3.
- 4. 4 LAGRANGIAN & EULERIANDESCRIPTIONS • keeps track of the position and velocity of individual particles
- 5. 5 LAGRANGIAN & EULERIANDESCRIPTIONS • The Lagrangian approach is difficult to apply to fluid flow; it requires infinite number of equations for infinite number of particles! • The field variable at a particular location at a particular time is the value of the variable for whichever fluid particle happens to occupy that location at that time. • The flow properties are defined as field variables, functions of space and time. Flow domain
- 6. 6 EXAMPLE A steady, incompressible,two-dimensional velocity field is given by Determine if there are any stagnation points in this flow field and, if so, where? At any stagnation point: 𝑢 𝑣 ⃗ 𝑉 =(𝑢 ,𝑣)=(0.5+0.8 𝑥) ⃗ 𝑖+(1.5−0.8 𝑦)⃗ 𝑗 𝑢=0.5+0.8 𝑥=0→𝑥=−0.625m 𝑣=1.5−0.8 𝑦=0→ 𝑦=1.875m Solution
- 7. 7 ACCELERATION FIELD 1 𝑢𝑣 𝑤 ⃗ 𝑎= 𝜕⃗ 𝑉 𝜕𝑡 +𝑢 𝜕⃗ 𝑉 𝜕𝑥 +𝑣 𝜕⃗ 𝑉 𝜕 𝑦 +𝑤 𝜕⃗ 𝑉 𝜕 𝑧 𝑢 𝑣 𝑤 𝑥 𝑧 𝑦
- 8. 8 LOCAL & CONVECTIVEACCELERATION ⃗ 𝑎= 𝑑⃗ 𝑉 𝑑𝑡 = 𝜕 ⃗ 𝑉 𝜕𝑡 +𝑢 𝜕 ⃗ 𝑉 𝜕 𝑥 +𝑣 𝜕⃗ 𝑉 𝜕 𝑦 +𝑤 𝜕⃗ 𝑉 𝜕𝑧 Local acceleration Convective (advective) acceleration A B 𝜕⃗ 𝑉 𝐴 𝜕𝑡 𝑢 𝜕 ⃗ 𝑉 𝜕 𝑥 𝜕⃗ 𝑉 𝐵 𝜕𝑡 Total (material or substantial derivative)
- 9. 9 ACCELERATION COMPONENTS ⃗ 𝑎= 𝜕⃗ 𝑉 𝜕𝑡 +𝑢 𝜕⃗ 𝑉 𝜕𝑥 +𝑣 𝜕⃗ 𝑉 𝜕 𝑦 +𝑤 𝜕⃗ 𝑉 𝜕𝑧 (𝑎𝑥 ,𝑎𝑦 ,𝑎𝑧 )= 𝜕(𝑢,𝑣 ,𝑤) 𝜕𝑡 +𝑢 𝜕(𝑢,𝑣,𝑤) 𝜕 𝑥 +𝑣 𝜕(𝑢,𝑣 ,𝑤) 𝜕 𝑦 +𝑤 𝜕(𝑢,𝑣 ,𝑤) 𝜕 𝑧 𝒂𝒙= 𝝏𝒖 𝝏𝒕 +𝒖 𝝏𝒖 𝝏𝒙 +𝒗 𝝏𝒖 𝝏𝒚 +𝒘 𝝏𝒖 𝝏 𝒛 𝒂𝒚= 𝝏𝒗 𝝏𝒕 +𝒖 𝝏𝒗 𝝏 𝒙 +𝒗 𝝏𝒗 𝝏 𝒚 +𝒘 𝝏𝒗 𝝏 𝒛 𝒂𝒛= 𝝏𝒘 𝝏𝒕 +𝒖 𝝏𝒘 𝝏𝒙 +𝒗 𝝏𝒘 𝝏 𝒚 +𝒘 𝝏𝒘 𝝏 𝒛
- 10. 10 EXAMPLE Calculate the accelerationat ⃗ 𝑎=(𝑎𝑥 ,𝑎𝑦 ) 𝒂𝒙= 𝝏𝒖 𝝏𝒕 +𝒖 𝝏𝒖 𝝏𝒙 +𝒗 𝝏𝒖 𝝏𝒚 𝒂𝒚= 𝝏 𝒗 𝝏𝒕 +𝒖 𝝏𝒗 𝝏 𝒙 +𝒗 𝝏 𝒗 𝝏 𝒚
- 11. 11 SOLUTION 𝑢=0.5+0.8 𝑥 , 𝜕𝑢 𝜕𝑥 =0.8 , 𝜕𝑢 𝜕 𝑦 =0 𝑣 =1.5 −0.8 𝑦 , 𝜕𝑣 𝜕𝑥 =0 , 𝜕𝑣 𝜕 𝑦 =−0.8 𝑎𝑥 =0+(0.5+0.8𝑥) (0.8)+(1.5− 0.8 𝑦) (0)=0.4+0.64 𝑥 𝑎𝑦=0+(0.5+0.8 𝑥)(0)+(1.5−0.8 𝑦)(−0.8)=−1.2+0.64 𝑦 𝑎𝑥 (2,3)=1.68 𝑚/ 𝑠2 𝑎𝑦 (2,3)=0.72𝑚/𝑠2
- 12. 12 EXAMPLE 4-3 𝑎𝑥 (2,3)=1.68𝑚/ 𝑠2 𝑎𝑦 (2,3)=0.72𝑚/𝑠2 𝑎=√𝑎𝑥 2 +𝑎𝑦 2 =√1.682 +0.722 𝑚/𝑠2 𝑎=1.828𝑚/𝑠2 𝜃=tan − 1 (𝑎𝑦 𝑎𝑥 )=tan − 1 (0.72 1.68 ) 𝜃=23.2𝑜
- 13.
- 14. 14 STREAMLINE A curve thatis everywhere tangent to the instantaneous local velocity vector. Streamlines cannot be directly observed experimentally except in steady flow fields.
- 15.
- 16. 16 EXAMPLE 4-4 For thefollowing steady, incompressible 2-D velocity field, plot the streamlines in the right half of the flow (): 𝑢=0.5+0.8 𝑥 ,𝑣=1.5 − 0.8 𝑦 solution ln (0.5+0.8 𝑥) 0.8 = ln (1.5 − 0.8 𝑦) − 0.8 +𝐶 ln 1 (0.5+0.8 𝑥) =ln (1.5 −0.8 𝑦)+ln 𝐶 ⃗ 𝑉 =(𝑢 ,𝑣)=(0.5+0.8 𝑥) ⃗ 𝑖+(1.5−0.8 𝑦)⃗ 𝑗
- 17. 17 𝐶 0.5+0.8𝑥 =1.5 − 0.8𝑦 The last equation draws a family of streamlines, each of them passes through a different point in the flow field and hence corresponds to a different value of the constant . 𝑦 =1.875 − 𝐶 0.8 ( 0.5+0.8 𝑥 )
- 18.
- 19. 19 CONSERVATION OF MASS(CONTINUITY EQUATION) • Assume one dimensional, unsteady, compressible flow. • The conservation of mass principle implies: 𝑦 𝑥 𝑑𝑥 𝜌 𝜌𝑢𝑑𝑦𝑑𝑧+ 𝜕(𝜌𝑢𝑑𝑦𝑑𝑧) 𝜕𝑥 𝑑𝑥 𝑢 𝑧 𝑑𝑦 𝑑𝑧 𝜌𝑢𝑑𝑦 𝑑𝑧 ˙ 𝑚=𝜌 ∙𝑉 ∙ 𝐴 𝑚=𝜌𝑑𝑥 𝑑𝑦 𝑑𝑧 𝜌 𝑢𝑑𝑦 𝑑𝑧 −(𝜌 𝑢 𝑑𝑦 𝑑𝑧 + 𝜕(𝜌 𝑢 𝑑𝑦 𝑑𝑧 ) 𝜕 𝑥 𝑑𝑥)= 𝜕(𝜌 𝑑𝑥 𝑑𝑦 𝑑𝑧 ) 𝜕𝑡
- 20. 20 CONSERVATION OF MASS(CONTINUITY EQUATION) 𝜌 𝑢𝑑𝑦 𝑑𝑧 −(𝜌 𝑢 𝑑𝑦 𝑑𝑧 + 𝜕(𝜌 𝑢 𝑑𝑦 𝑑𝑧 ) 𝜕 𝑥 𝑑𝑥)= 𝜕(𝜌 𝑑𝑥 𝑑𝑦 𝑑𝑧 ) 𝜕𝑡 − 𝜕 ( 𝜌 𝑢 𝑑𝑦 𝑑𝑧 ) 𝜕 𝑥 𝑑𝑥 = 𝜕 ( 𝜌 𝑑𝑥 𝑑𝑦 𝑑𝑧 ) 𝜕𝑡 𝜕 𝜌 𝜕 𝑡 + 𝜕 ( 𝜌 𝑢) 𝜕 𝑥 =0
- 21. 21 3-D CONTINUITY EQUATION 𝜕𝜌 𝜕𝑡 + 𝜕(𝜌 𝑢) 𝜕 𝑥 + 𝜕( 𝜌 𝑣 ) 𝜕 𝑦 + 𝜕 (𝜌 𝑤 ) 𝜕 𝑧 =0 𝜕(𝜌 𝑢) 𝜕 𝑥 + 𝜕( 𝜌 𝑣) 𝜕 𝑦 + 𝜕( 𝜌 𝑤) 𝜕 𝑧 =0 For steady flow: For incompressible flow: 𝜕𝑢 𝜕𝑥 + 𝜕𝑣 𝜕 𝑦 + 𝜕𝑤 𝜕 𝑧 =0
- 22. 22 EXAMPLE: DESIGN OFA COMPRESSIBLE CONVERGING DUCT
- 23. 23 SOLUTION 𝑢=𝑢1+𝐶𝑢 𝑥 𝐶𝑢= 𝑢2− 𝑢1 ∆ 𝑥 = 300 −100 2 =100 𝑠 − 1 𝜌=𝜌1+𝐶𝜌 𝑥 𝐶𝜌= 𝜌2 − 𝜌1 ∆ 𝑥 = 0.85 −1.2 2 =−0.175 𝑠 − 1 For steady, 2-D, compressible flow: 𝜕(𝜌 𝑢) 𝜕 𝑥 + 𝜕( 𝜌 𝑣) 𝜕 𝑦 =0 𝜕[(𝜌1+𝐶𝜌 𝑥 )(𝑢1+𝐶𝑢 𝑥 )] 𝜕 𝑥 + 𝜕( 𝜌 𝑣) 𝜕 𝑦 =0 𝜌1 𝐶𝑢+𝑢1 𝐶𝜌+2𝐶𝑢 𝐶𝜌 𝑥+ 𝜕(𝜌 𝑣) 𝜕 𝑦 =0
- 24. 24 𝜕(𝜌 𝑣) 𝜕 𝑦 =−𝜌1𝐶𝑢 −𝑢1 𝐶𝜌 −2𝐶𝑢 𝐶𝜌 𝑥 Integrate w.r.t. 𝜌 𝑣=−(𝜌1 𝐶𝑢+𝑢1 𝐶𝜌)𝑦 −2𝐶𝑢 𝐶𝜌 𝑥 𝑦+ 𝑓 (𝑥) Boundary conditions: at for all values of 𝑓 (𝑥)=0 𝑣= −(𝜌1 𝐶𝑢+𝑢1 𝐶𝜌 ) 𝑦 −2𝐶𝑢 𝐶𝜌 𝑥𝑦 𝜌1+𝐶𝜌 𝑥
- 25. 25 The 2-D streamlineequation: 𝑑𝑥 𝑢 = 𝑑𝑦 𝑣 𝑑𝑥 𝑢1+𝐶𝑢 𝑥 = 𝑑𝑦 −(𝜌1 𝐶𝑢+𝑢1𝐶𝜌 ) 𝑦 −2𝐶𝑢 𝐶𝜌 𝑥𝑦 𝜌1+𝐶𝜌 𝑥 −(𝜌1 𝐶𝑢+𝑢1𝐶𝜌 )−2𝐶𝑢 𝐶𝜌 𝑥 (𝜌1+𝐶𝜌 𝑥)(𝑢1+𝐶𝑢 𝑥) 𝑑𝑥= 𝑑𝑦 𝑦 − (𝜌1 𝐶𝑢 +𝑢1 𝐶𝜌 )+2𝐶𝑢 𝐶𝜌 𝑥 𝜌1 𝑢1+(𝜌1 𝐶𝑢+𝑢1 𝐶𝜌 )𝑥+𝐶𝑢 𝐶𝜌 𝑥 2 𝑑𝑥= 𝑑𝑦 𝑦 ∫❑ ∫❑
- 26. 26 − ln [𝜌1𝑢1+(𝜌1 𝐶𝑢+𝑢1 𝐶𝜌 )𝑥+𝐶𝑢 𝐶𝜌 𝑥 2 ]=ln 𝑦+𝐶 𝑦= 𝐶 𝜌1𝑢1+( 𝜌1 𝐶𝑢 +𝑢1 𝐶𝜌 ) 𝑥+𝐶𝑢 𝐶𝜌 𝑥 2 To get the height () at the duct exit, find the streamline of the top wall: At At 𝐶 =2 𝜌 1 𝑢1 𝑦=0.9412m
- 27. 27 SUMMARY • Differential FlowAnalysis –Lagrangian and Eulerian Flow Descriptions –Acceleration –Streamline –Continuity equation