Fluid Mechanics White Seventh Ed Chapter 6 Summary [PDF]

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CHAPTER 6



Always State Assumptions!!! Possible Assumptions:      



Laminar flow (fully developed) Turbulent flow Fully developed Incompressible (Poiseuille) pipe flow Atmospheric pressure (at where ever) Negligible velocity (Very large tank draining)



Subsection Summaries 6.1 Reynolds Number Regimes



𝑅𝑒 =



𝜌𝑉𝑑 𝑉𝑑 = 𝜇 𝑣



𝑣=



𝜇 𝜌



𝑄 = 𝑉𝐴



𝑅𝑒𝑐𝑟𝑖𝑡 ≈ 2300 Viscosity 𝒗 values usually around 1 × 10−4



6.2 Internal vs. External Viscous Flows 𝐿𝑒 𝜌𝑉𝑑 = 𝑔( ) 𝑑 𝜇 Laminar Flow:



𝐿𝑒 ≈ 0.06𝑅𝑒 𝑑



𝐿𝑒 1 1.6(𝑅𝑒) ⁄4 𝑑



For Turbulant Flow:



For



Re ≤ 107



6.3 Head Loss – The Friction Factor



ℎ𝑓 = 𝑓



𝐿 𝑉 2 8𝑓𝐿𝑄 2 = 𝑑 2𝑔 𝜋 2 𝑔𝑑5



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CHAPTER 6 6.4 Laminar Fully Developed Pipe Flow



𝑓𝑙𝑎𝑚 =



64 𝑅𝑒



Flow direction is in direction of falling HGL 𝐻𝐺𝐿1 = 𝑧1 +



𝑝1 𝜌𝑔 𝑝



𝐻𝐺𝐿2 = 𝑧2 + 𝜌𝑔2 Compare 𝐻𝐺𝐿1 and 𝐻𝐺𝐿2  if 𝐻𝐺𝐿1 > 𝐻𝐺𝐿2 the flow is from 1 to 2.



Generalized complete incompressible steady flow energy equation: (



𝑝 𝛼 2 𝑝 𝛼 2 + 𝑉 + 𝑧) = ( + 𝑉 + 𝑧) + ℎ𝑡𝑢𝑟𝑏𝑖𝑛𝑒 − ℎ𝑃𝑢𝑚𝑝 + ℎ𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝜌𝑔 2𝑔 𝜌𝑔 2𝑔 𝑖𝑛 𝑜𝑢𝑡



Single Pipe Flow Problems Known Flow Rate: 1. Use known flow rate to determine Reynolds number 𝑅𝑒 =



𝜌𝑉𝑑 𝑉𝑑 𝑄𝑑 = = 𝜇 𝑣 𝑣𝐴



2. Identify whether flow is laminar of turbulent Laminar < 2300 < Turbulent 3. Use correct expression to determine friction factor 64



Laminar



𝑓 = 𝑅𝑒



Turbulent



𝑓←𝐷



𝜀



(Read off of Moody)



4. Use definition of ℎ𝑓 to determine friction head loss 𝐿 𝑉2 ℎ𝑓 = 𝑓 𝑑 2𝑔 5. Use general energy equation to determine total pressure drop 𝑝 𝛼 2 𝑝 𝛼 2 ( + 𝑉 + 𝑧) = ( + 𝑉 + 𝑧) + ℎ𝑡𝑢𝑟𝑏𝑖𝑛𝑒 − ℎ𝑃𝑢𝑚𝑝 + ℎ𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝜌𝑔 2𝑔 𝜌𝑔 2𝑔 𝑖𝑛 𝑜𝑢𝑡



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CHAPTER 6 Unknown Flow Rate 1. Assume infinite Reynolds Number 2. Obtain friction factor as function of roughness only 𝜀



𝑓←𝐷



(Read off of Moody)



3. Obtain first guess of velocity based on energy conservation 4. Update Reynolds number 5. Update friction factor based on Reynolds number and



𝜀 𝐷



EXAMPLE: Chapter 6 Lecture Notes slide 11 on p.6 After equation for V is derived:



0.05 𝑓



𝑉=√



………… (1)



For Reynolds number 𝑅𝑒 = 50000𝑉 ………… (2)



𝜀



Initial 𝐷 to determine search line:



𝜀 𝐷



=



0.05 10



Assume fully turbulent: 𝑅𝑒 → ∞



= 0.002 𝑓 ← 0.002 ∴ 𝑓 = 0.0235



The assumption of fully turbulent flow is to get initial 𝑓 1. Insert 𝑓 into eqn. (1) to get new 𝑉 ∴ 𝑉 = 1.459 2. Insert new 𝑉 into eqn. (2) to get new 𝑅𝑒 𝜀 𝐷



3. Look up new 𝑅𝑒 on predetermined 0.002 ( ) line to determine new 𝑓 4. Repeat steps 1 to 3 until 𝑓 is stable



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CHAPTER 6



Unknown Diameter: Example 6.10 Find functions that define the diameter of the pipe. In this example:



𝑓=



𝜋2 𝑔ℎ𝑓 𝑑 5 8 𝐿𝑄2



𝑅𝑒 =



4𝑄 𝜋𝑑𝑣



to



𝑑 ≈ 0.655𝑓 1/5



to



𝑅𝑒 =



21800 𝑑



𝜀 6 × 10−5 = 𝑑 𝑑



1. Guess 𝑓 2. calculate 𝑑 𝜀



3. use 𝑑 to calculate 𝑅𝑒 and 𝑑 𝜀



4. use 𝑅𝑒 and 𝑑 to find new 𝑓 5. Repeat steps 1 to 4 until 𝑑 becomes stable 4



CHAPTER 6 Minor or Local Losses in Pipe Systems See Example 6.16 ∆ℎ𝑡𝑜𝑡 = ℎ𝑓 + ∑ ℎ𝑚 =



𝑉 2 𝑓𝐿 ( + ∑ 𝐾) 2𝑔 𝑑



When flow is exiting a pipe to a large reservoir (submerged exit): K =1



ALWAYS!!!!!



Other values for the resistance coefficient k can be read of various graphs in section 6.9 (Minor of Local Losses in Pipe Systems). Fig 6.18a:



Recent measured loss coefficients for 900 elbows (p.401)



Fig 6.18b:



Average loss coefficients for partially open valves (p.402)



Fig 6.20:



Resistance for smooth-walled 450, 900 and 1800 bends At Re = 200 000.



Fig 6.21 a & b: Entrance and exit loss coefficients Fig 6.22:



Sudden expansion and contraction losses



Fig 6.23:



Flow losses with gradual conical expansion



Sudden expansion: 𝐾𝑆𝐸 = (1 −



𝑑2 𝐷2



2



ℎ



𝑚 ) = 𝑉 2 /(2𝑔)



Sudden Contraction: 5



CHAPTER 6 𝐾𝑆𝐶 ≈ 0.42 (1 −



𝑑2 𝐷2



)



Multiple Pipe Systems Ugh… read p.407 to p.413



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