- 2. 1 By Wasswa Derrick wasswaderricktimothy7@gmail.com Makerere University 3rd Edition The Bernoulli equation for cylindrical pipes or circular orifices with viscous effects is: 𝑷 + 𝒉𝝆𝒈 + 𝝆 𝑽𝟐 𝟐 + 𝟖𝝁𝒍 𝒓𝟐 𝑽 + 𝟒𝑲𝝁 𝒓 𝑽 + 𝝆𝒍𝑪𝟐 𝒓 𝑽𝟐 = 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 We shall see how to derive it in the text to follow.
- 3. 2 TABLE OF CONTENTS FANNING FRICTION FACTOR/SKIN FRICTION COEFFICIENT(𝑪𝟏)............................... 7 HOW DO WE MEASURE VELOCITY OF EXIT?.......................................................................9 Torricelli flow.............................................................................................................................. 11 To show that using potential energy lost on differential element is a wrong approach ....................................................................................................................................... 12 How do we include viscous effects in Torricelli flow?................................................. 14 How does the velocity manifest itself? ............................................................................. 18 HOW DO WE HANDLE PIPED SYSTEMS?.............................................................................. 23 To show that the Reynolds number is the governing number for flow according to Reynolds Theory................................................................................................................... 23 The nature of 𝑪𝟐.........................................................................................................................29 How do we deal with cases where there is a change of cross-sectional area?....36 THE MODIFIED BERNOULLI EQUATION WITH VISCOUS EFFECTS INCLUDED..... 38 How can we apply the Bernoulli equation above?......................................................... 38 How do we apply the Bernoulli equation to different area pipes? ..........................42 How do we write the Bernoulli equation for a variable cross-sectional area with distance for example for the case of when the pipe is a conical frustrum?........44 HOW DO WE DEAL WITH PRESSURE GRADIENTS? .........................................................46 HEAD LOSS......................................................................................................................................52 THEORY OF MOTION OF PARTICLES IN VISCOUS FLUIDS...........................................58 REFERENCES...................................................................................................................................77
- 4. 3 FUNDAMENTALS OF FLUID FLOW When dealing with describing any type of fluid flow, we have to first solve the Navier Stoke’s equations for that given geometry of pipe and get the velocity profile of the liquid in laminar flow in fully developed state. After getting the velocity profile, we then get the average velocity of that system. Using the average velocity got, we express the head loss ∆ℎ in terms of the average velocity and go ahead and find the fanning friction factor from ∆ℎ = 2𝐶1 𝐿 𝐷 𝑉2 𝑔 Where: ∆ℎ = ℎ𝑒𝑎𝑑 𝑙𝑜𝑠𝑠 𝐶1 = 𝑓𝑎𝑛𝑛𝑖𝑛𝑔 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 The fanning friction factor got will be the basis in using the energy conservation techniques to solve fluid flow problems for both non fully developed and fully developed laminar and transition and turbulent flow and Torricelli flow. Let us demonstrate: Consider flow in a cylindrical pipe due to a pressure gradient: The corresponding Navier Stokes equation in the axial direction is given by: The boundary conditions are 𝑣𝑟 = 0 𝜕𝑣𝑧 𝜕𝑧 = 0 𝑣𝜃 = 0 𝜕𝑣𝑧 𝜕𝑡 = 0 𝑠𝑡𝑒𝑎𝑑𝑦 𝑓𝑙𝑜𝑤 𝑔𝑧 = 0
- 5. 4 𝜕2 𝑣𝑧 𝜕𝜃2 = 0 With all those conditions, the Navier Stoke’s equations reduce to 1 𝑟 𝜕 𝜕𝑟 (𝑟 𝜕𝑣𝑧 𝜕𝑟 ) = 1 𝜇 𝜕𝑃 𝜕𝑧 Since the pressure gradient is a constant, the right-hand side of the equation above is a constant. Multiplying through by r we get: 𝜕 𝜕𝑟 (𝑟 𝜕𝑣𝑧 𝜕𝑟 ) = 1 𝜇 𝜕𝑃 𝜕𝑧 𝑟 Upon integrating once, we get (𝑟 𝜕𝑣𝑧 𝜕𝑟 ) = 1 𝜇 𝜕𝑃 𝜕𝑧 𝑟2 2 + 𝐸 Dividing through by r we get 𝜕𝑣𝑧 𝜕𝑟 = 1 𝜇 𝜕𝑃 𝜕𝑧 𝑟 2 + 𝐸 𝑟 We know that at 𝑟 = 0,the shear stress (𝜇 𝜕𝑣𝑧 𝜕𝑟 ) is finite and so 𝐸 = 0 since if it were not the shear stress would be infinite at 𝑟 = 0. So, we get 𝜕𝑣𝑧 𝜕𝑟 = 1 𝜇 𝜕𝑃 𝜕𝑧 𝑟 2 Integrating once again, we get 𝑣𝑧 = 1 𝜇 𝜕𝑃 𝜕𝑧 𝑟2 4 + 𝐻 Using the no slip condition at 𝑟 = 𝑅 𝑣𝑧 = 0 𝑎𝑡 𝑟 = 𝑅 We get upon substitution 𝐻 = − 1 𝜇 𝜕𝑃 𝜕𝑧 𝑅2 4 So, we get the velocity profile as
- 6. 5 𝑣𝑧 = − 1 4𝜇 𝜕𝑃 𝜕𝑧 (𝑅2 − 𝑟2 ) The above is the velocity profile. We go ahead and find the average velocity as 𝑣𝑎𝑣𝑔 = 1 𝐴 ∬ (𝑣𝑧)𝑟𝑑𝜃𝑑𝑟 2𝜋,𝑅 0,0 Upon integration we get: 𝑣𝑎𝑣𝑔 = −( 𝜕𝑃 𝜕𝑧 ) 𝑅2 8𝜇 𝑣𝑎𝑣𝑔 = −( 𝜕𝑃 𝜕𝑧 ) 𝐷2 32𝜇 We then get the head loss from the average velocity as: − 𝜕𝑃 𝜕𝑧 = 32 𝜇𝑣𝑎𝑣𝑔 𝐷2 Where: 𝐷 = 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑢𝑏𝑒. Upon integrating the formula above we get: − ∫ 𝑑𝑃 𝑃 𝑃0 = 32 𝜇𝑣𝑎𝑣𝑔 𝐷2 ∫ 𝑑𝑧 𝐿 0 We finally get ∆𝑃 = 32 𝜇𝑣𝑎𝑣𝑔 𝐷2 𝐿 Forming an expression of friction head loss, we get ∆ℎ = ∆𝑃 𝜌𝑔 = 32 𝜇𝑣𝑎𝑣𝑔 𝜌𝑔𝐷2 𝐿 Combining the above equation with ∆ℎ = 2𝐶1 𝐿 𝐷 𝑉2 𝑔 We get ∆ℎ = 32 𝜇𝑣𝑎𝑣𝑔 𝜌𝑔𝐷2 𝐿 = 2𝐶1 𝐿 𝐷 𝑉2 𝑔
- 7. 6 We get 𝑪𝟏 = 𝟏𝟔 𝝁 𝝆𝑫𝒗𝒂𝒗𝒈 = 𝟏𝟔 𝑹𝒆 Hence, we have got the friction factor for laminar flow in a cylindrical pipe. We can extend this analysis to other pipe geometries too.
- 8. 7 FANNING FRICTION FACTOR/SKIN FRICTION COEFFICIENT(𝑪𝟏) In the text to follow below, we are going to be using the fanning friction factor(𝑪𝟏) also called the skin friction coefficient in making our calculations. In the figure below, the Darcy friction factor(𝒇) is given. To get the skin friction coefficient from the Darcy friction factor (𝑓), we use the relation below: 𝑓 = 4𝐶1 𝑪𝟏 = 𝟏 𝟒 𝒇 For example, for a cylindrical pipe in the diagram above 𝑓 = 64 𝑅𝑒
- 9. 8 To get the skin friction coefficient/Fanning friction factor, we divide by 4 and get: 𝑪𝟏 = 𝟏𝟔 𝑹𝒆 We can do the same for other geometries in the diagram above.
- 10. 9 HOW DO WE MEASURE VELOCITY OF EXIT? How do we measure velocity in fluid flow? We either measure the flow rate and then divide it by cross sectional area as below 𝑉 = 𝑄 𝐴 Or we can use projectile motion assuming no air resistance and get to know the velocity. Using projectile motion of a fluid out of a hole we can measure its velocity of exit i.e., 𝑅 = 𝑉 × 𝑡 … 𝑎) 𝐻 = 1 2 𝑔𝑡2 … 𝑏) From a) 𝑡 = 𝑅 𝑉 Substituting t into equation b) and making velocity V the subject, we get: 𝑉 = 𝑅√ 𝑔 2𝐻 Where: H is the vertical height of descent and R is the range.
- 11. 10 All the experimental values got in this document were got using the velocity got from projectile motion
- 12. 11 Torricelli flow To derive Torricelli flow equation, we use conservation of energy on a differential element as in the diagram above i.e., we say: 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 𝑏𝑦 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 = 𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 𝑔𝑎𝑖𝑛𝑒𝑑 𝑚𝑔ℎ = 1 2 𝑚𝑉2 𝑉 = √2𝑔ℎ Torricelli flow is observed when there is no pipe on a tank and the velocity of exit is derived to be 𝑉 = √2𝑔ℎ assuming there are no viscous forces.
- 13. 12 To show that using potential energy lost on differential element is a wrong approach To show that it is wrong, let us consider the case of two immiscible liquids allowed to flow through an orifice We see from experiment that the fluid at the bottom i.e., water will flow out first of the orifice in contradiction to saying potential energy lost by the differential element of the floating liquid. In other words, if we were to use potential energy lost, we would say that the differential element of the floating liquid will descend downwards through the denser liquid and flow out which is not observed. So, what is true to say is that the work done by the pressure difference on the differential element at A causes it to gain kinetic energy as shown in the diagram above i.e., 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 = 𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 𝑔𝑎𝑖𝑛𝑒𝑑 For the case of two immiscible liquids as shown in the diagram above: (𝑃1 − 𝑃2)𝑑𝑣 = 1 2 𝑚2𝑉2 Where: 𝑑𝑣 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑎𝑡 𝑡ℎ𝑒 𝑏𝑜𝑡𝑡𝑜𝑚 𝑜𝑓 𝑐𝑜𝑛𝑎𝑡𝑖𝑛𝑒𝑟 = 𝑚2 𝜌2 𝑃1 = 𝐻 + ℎ1𝜌1𝑔 + ℎ2𝜌2𝑔 𝑃2 = 𝐻 Where: 𝐻 = 𝑎𝑡𝑚𝑜𝑠𝑝ℎ𝑒𝑟𝑖𝑐 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒. (𝑃1 − 𝑃2) = ℎ1𝜌1𝑔 + ℎ2𝜌2𝑔
- 14. 13 Upon substituting we get: (ℎ1𝜌1𝑔 + ℎ2𝜌2𝑔) 𝑚2 𝜌2 = 1 2 𝑚2𝑉2 We finally get the velocity as: 𝑉 = √2𝑔 (ℎ1𝜌1 + ℎ2𝜌2) 𝜌2 The same reasoning can be applied to a single liquid scenario flowing out of an orifice by considering the pressure difference at the bottom of container to be giving the differential element at the bottom kinetic energy.
- 15. 14 How do we include viscous effects in Torricelli flow? To include viscous effects, we conserve energy changes by adding a viscous energy term as below: 𝑃1 = 𝐻 + ℎ𝜌𝑔 𝑃2 = 𝐻 The viscous energy term is: 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒 = 𝐶𝑑 2 𝑚𝑉2 (𝑃1 − 𝑃2)𝑑𝑣 = 1 2 𝑚𝑉2 + 𝐶𝑑 2 𝑚𝑉2 (ℎ𝜌𝑔)( 𝑚 𝜌 ) = 1 2 𝑚𝑉2 + 𝐶𝑑 2 𝑚𝑉2 Where: 𝑪𝒅 = 𝑲𝑪𝟏 Where: 𝐾 = 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑡𝑜 𝑏𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑 𝐶1 = 𝑠𝑘𝑖𝑛 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 For circular orifices/ cylindrical pipes 𝐶1 = 16 𝑅𝑒
- 16. 15 𝑅𝑒 = 𝜌𝑉𝐷 𝜇 = 2𝜌𝑉𝑟 𝜇 𝑪𝟏 = 𝟖𝝁 𝝆𝑽𝒓 Upon substitution in the energy conservation formula, we get: (ℎ𝜌𝑔)( 𝑚 𝜌 ) = 1 2 𝑚𝑉2 + 𝐶𝑑 2 𝑚𝑉2 𝐶𝑑 = 𝐾𝐶1 (ℎ𝜌𝑔)( 𝑚 𝜌 ) = 1 2 𝑚𝑉2 + 𝐾𝐶1 2 𝑚𝑉2 Dividing through by mass m and multiplying through by 2, we get 2𝑔ℎ = 𝑉2 + 𝐾𝐶1𝑉2 Substituting for 𝐶1 = 8𝜇 𝜌𝑉𝑟 We get: 2𝑔ℎ = 𝑉2 + 8𝐾𝜇 𝜌𝑟 𝑉 Rearranging, we get a quadratic formula below: 𝑽𝟐 + 𝟖𝑲𝝁 𝝆𝒓 𝑽 − 𝟐𝒈𝒉 = 𝟎 … … 𝟏) The velocity formula above works for non-piped systems or circular orifices as shown below:
- 17. 16 Back to equation 1) above, we notice that the expression for velocity is a quadratic formula and velocity V is given by: 𝑉 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 We choose the positive velocity i.e. 𝑉 = −𝑏 + √𝑏2 − 4𝑎𝑐 2𝑎 Where: 𝑏 = 8𝐾𝜇 𝜌𝑟 𝑎 = 1 𝑐 = −2𝑔ℎ An expression for V is 𝑽 = − 𝟒𝑲𝝁 𝒓𝝆 + 𝟏 𝟐 √( 𝟖𝑲𝝁 𝝆𝒓 )𝟐 + 𝟖𝒈𝒉 …… . . 𝟐) We can modify the equation above to include a remnant height ℎ0 as observed from experiment., 𝑽 = − 𝟒𝑲𝝁 𝒓𝝆 + 𝟏 𝟐 √( 𝟖𝑲𝝁 𝝆𝒓 )𝟐 + 𝟖𝒈(𝒉 − 𝒉𝟎) …… . . 𝟐)
- 18. 17 When ℎ = ℎ0, the velocity is zero (i.e., the fluid stops flowing). We ask what supports the height ℎ0 in the container? It is the sum of the surface tension pressures at the liquid surfaces that supports ℎ0 as shown below: We say that the liquid pressure ℎ0 is supported by the two menisci i.e., ℎ0𝜌𝑔 = 2𝛾𝑐𝑜𝑠𝜃𝑐 𝑟1 + 2𝛾𝑐𝑜𝑠𝜃𝑑 𝑟 Where: 𝜃𝑐 = 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑖𝑛 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑒𝑟 If 𝜃𝑐 = 𝜃𝑑 , we get 𝒉𝟎 = 𝟐𝜸𝒄𝒐𝒔𝜽𝒄 𝝆𝒈 ( 𝟏 𝒓𝟏 + 𝟏 𝒓 ) If 𝑟1 is very big, then ℎ0 = 2𝛾𝑐𝑜𝑠𝜃𝑐 𝑟𝜌𝑔 Back to the velocity equation, 𝑉 = − 4𝐾𝜇 𝑟𝜌 + 1 2 √( 8𝐾𝜇 𝜌𝑟 )2 + 8𝑔(ℎ − ℎ0) … … . .2) NB: YOU NOTICE THAT TO MEASURE THE CONSTANTS OF FLOW (e.g., K), WE HAVE TO LOOK FOR AN EQUATION FOR WHICH THE FLOW MANIFESTS ITSELF AND THEN WE VARY A FACTOR LIKE RADIUS AND THEN WE SHALL BE ABLE TO CALCULATE THE CONSTANT K
- 19. 18 How does the velocity manifest itself? Factorizing out the term 𝟖𝝁𝑲 𝒓𝝆 from the square root, we get: 𝑉 = − 4𝐾𝜇 𝑟𝜌 + 1 2 √( 8𝐾𝜇 𝜌𝑟 )2 + 8𝑔(ℎ − ℎ0) … … . .2) 𝑉 = − 4𝐾𝜇 𝑟𝜌 + 8𝜇𝐾 2𝑟𝜌 √1 + 8𝑔(ℎ − ℎ0) ( 8𝜇𝐾 𝑟𝜌 )2 𝑉 = − 4𝜇𝐾 𝑟𝜌 + 4𝜇𝐾 𝑟𝜌 √1 + 8𝑔(ℎ − ℎ0) ( 8𝜇𝐾 𝑟𝜌 )2 We get a dimensionless number i.e., 8𝑔(ℎ − ℎ0) ( 8𝜇𝐾 𝑟𝜌 )2 = 𝑟2 𝜌2 𝑔(ℎ − ℎ0) 8𝜇2𝐾2 For small height 𝒉 − 𝒉𝟎 and small radius The term 8𝑔(ℎ − ℎ0) ( 8𝜇𝐾 𝑟𝜌 )2 = 𝑟2 𝜌2 𝑔(ℎ − ℎ0) 8𝜇2𝐾2 ≪ 1 And we can use the approximation (1 + 𝑥)𝑛 ≈ 1 + 𝑛𝑥 for 𝑥 ≪ 1 For which 𝒙 = 8𝑔(ℎ − ℎ0) ( 8𝜇𝐾 𝑟𝜌 )2 = 𝑟2 𝜌2 𝑔(ℎ − ℎ0) 8𝜇2𝐾2 𝒙 𝒂𝒃𝒐𝒗𝒆 𝒊𝒔 𝒕𝒉𝒆 𝒈𝒐𝒗𝒆𝒓𝒏𝒊𝒏𝒈 𝒏𝒖𝒎𝒃𝒆𝒓 And 𝑛 = 1 2
- 20. 19 And we get after the binomial approximation; √1 + 𝑥 ≈ 1 + 1 2 𝑥 𝑓𝑜𝑟 𝑥 ≪ 1 𝑉 = − 4𝜇𝐾 𝑟𝜌 + 4𝜇𝐾 𝑟𝜌 (1 + 4𝑔(ℎ − ℎ0) ( 8𝜇𝐾 𝑟𝜌 )2 ) We finally get the velocity as 𝑽 = 𝒓(𝒉 − 𝒉𝟎)𝝆𝒈 𝟒𝝁𝑲 … . . 𝒂) We can call the equation above equation a) and regime laminar flow When 8𝑔(ℎ − ℎ0) ( 8𝜇𝐾 𝑟𝜌 )2 = 𝑟2 𝜌2 𝑔(ℎ − ℎ0) 8𝜇2𝐾2 𝑖𝑠 𝑐𝑙𝑜𝑠𝑒 𝑡𝑜 1 Velocity V is given by 𝑽 = − 𝟒𝑲𝝁 𝒓𝝆 + 𝟏 𝟐 √( 𝟖𝑲𝝁 𝝆𝒓 )𝟐 + 𝟖𝒈(𝒉 − 𝒉𝟎) Let’s call this equation b) and regime transition flow When 8𝑔(ℎ − ℎ0) ( 8𝜇𝐾 𝑟𝜌 )2 = 𝑟2 𝜌2 𝑔(ℎ − ℎ0) 8𝜇2𝐾2 ≫ 1 We approximate 1 + 8𝑔(ℎ − ℎ0) ( 8𝜇𝐾 𝑟𝜌 )2 ≈ 8𝑔(ℎ − ℎ0) ( 8𝜇𝐾 𝑟𝜌 )2 Velocity 𝑉 = − 4𝜇𝐾 𝑟𝜌 + 4𝜇𝐾 𝑟𝜌 √1 + 8𝑔(ℎ − ℎ0) ( 8𝜇𝐾 𝑟𝜌 )2
- 21. 20 Becomes 𝑉 = − 4𝜇𝐾 𝑟𝜌 + 4𝜇𝐾 𝑟𝜌 √ 8𝑔(ℎ − ℎ0) ( 8𝜇𝐾 𝑟𝜌 )2 𝑽 = − 𝟒𝝁𝑲 𝒓𝝆 + √𝟐𝒈(𝒉 − 𝒉𝟎) When 𝒉 ≫ 𝒉𝟎 We observe 𝑽 = − 𝟒𝝁𝑲 𝒓𝝆 + √𝟐𝒈𝒉 Let’s call this equation c) We can call this regime turbulent flow When the radius is big in turbulent flow, we observe 𝑽 = √𝟐𝒈(𝒉 − 𝒉𝟎) And when ℎ0 is small so that ℎ0 ≈ 0 , the velocity becomes 𝑽 = √𝟐𝒈𝒉 To be able to measure K, we have to find an experiment for which the flow manifests itself as either equation, a), b), or c). Using water which has a low viscosity and varying the radius hole and for height ℎ chosen to be approximately large, it is found that the flow will manifest itself in equation c) (turbulent flow) and plotting a graph of V against √ℎ ,a straight-line graph is got with an intercept, 𝑉 = − 4𝜇𝐾 𝑟𝜌 + √2𝑔ℎ
- 22. 21 The gradient of the above graph is √(𝟐𝒈) the intercept n got is inversely proportional to r and so K can be measured. i.e. 𝑛 = − 4𝜇𝐾 𝑟𝜌 Varying the radius will give a different intercept inversely proportional to r from which K can be got as 𝐾 = − 𝑛𝑟𝜌 4𝜇 Of course, depending on the viscosity of the fluid and height difference (ℎ − ℎ0) and radius r of the orifice, the flow can shift to any equation, a), b), or c). Using water as the fluid and regime c) (turbulent flow) for experiment, it was found that Using viscosity of water as 𝝁 = 𝟖. 𝟗 × 𝟏𝟎−𝟒 𝑷𝒂. 𝒔 𝐊 = 𝟏𝟑𝟑. 𝟔𝟑𝟕𝟓 And so generally for Torricelli flow, there is one friction coefficient 𝑪𝒅 given by: 𝑪𝒅 = 𝑲𝑪𝟏 𝑪𝒅 = 𝟏𝟑𝟑. 𝟔𝟑𝟕𝟓𝑪𝟏 NB: To get the rate of decrease of a fluid in a container, we use the velocity V got i.e., 𝒅𝑽 𝒅𝒕 = −𝑨𝑽 i.e. 𝒅𝒉 𝒅𝒕 = − 𝑨 𝑨𝟎 𝑽 Where: 𝑨𝟎 = 𝒄𝒓𝒐𝒔𝒔 𝒔𝒆𝒄𝒕𝒊𝒐𝒏𝒂𝒍 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒄𝒐𝒏𝒕𝒂𝒊𝒏𝒆𝒓 For example, if the governing number above was such that for all h: 𝑣 = √2𝑔ℎ
- 23. 22 Then 𝑑ℎ 𝑑𝑡 = − 𝐴 𝐴0 𝑉 𝑑ℎ 𝑑𝑡 = − 𝐴 𝐴0 √2𝑔ℎ ∫ 𝑑ℎ √ℎ ℎ ℎ1 = −( 𝐴 𝐴0 )√2𝑔 ∫ 𝑑𝑡 𝑡 0 Where: ℎ = ℎ1 𝑎𝑡 𝑡 = 0 √ℎ1 − √ℎ = ( 𝐴 𝐴0 )𝑡√ 𝑔 2 √𝒉 = √𝒉𝟏 − 𝒕( 𝑨 𝑨𝟎 )√ 𝒈 𝟐 So that will be the equation of height h against time.
- 24. 23 HOW DO WE HANDLE PIPED SYSTEMS? Consider the system below: To show that the Reynolds number is the governing number for flow according to Reynolds Theory For smooth piped systems The governing number of flow equations is the Reynolds number According to Reynold, For laminar flow 𝑅𝑒𝑑 < 2300 I.e. 𝜌𝑉 𝑐𝑑 𝜇 < 2300 2𝜌𝑉 𝑐𝑟 𝜇 < 2300 Where: 𝑉 𝑐 = 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 So 𝑉 𝑐 < 1150 𝜇 𝜌𝑟 In laminar flow
- 25. 24 𝑉 = 𝑟2 𝜌𝑔ℎ 8𝜇𝑙 And 𝑉 𝑐 = 𝑉 So, 𝑟2 𝜌𝑔ℎ 8𝜇𝑙 < 1150 𝜇 𝜌𝑟 𝑟3 𝜌2 𝑔ℎ 9200𝜇2𝑙 < 1 So, the governing condition for laminar flow according to Reynold should be 𝑟3 𝜌2 𝑔ℎ 9200𝜇2𝑙 < 1 As before, let’s conserve energy: Work done by pressure difference on fluid element = Kinetic energy gained + work done against skin friction. Total work done against friction force = 1 2 𝐶𝑑𝑚𝑉2 + ∑ 𝑊 𝑛 2 𝑛=1 OR Total work done against friction force = 1 2 𝐶𝑑𝑚𝑉2 + 𝑊1 + 𝑊2 We consider the axial distance along the length of the pipe to be called z as in the diagram above. 𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑠𝑘𝑖𝑛 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝐹1, 𝑖𝑠 𝑊1 = ∫ 𝐹1𝑑𝑧 𝑙 0 Where: 𝐹1 = 𝑠𝑘𝑖𝑛 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒 1 𝑉𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒, 𝐹1 = 1 2 𝐶1𝐴𝑠𝜌𝑉2 Where: 𝐶1 = 𝑠𝑘𝑖𝑛 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝐶1 = 16 𝑅𝑒 𝐴𝑠 = 2𝜋𝑟∆𝑥
- 26. 25 Since the velocity along the length of the pipe and 𝐴𝑠 are constant independent of position 𝑧 along the pipe, we have: 𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑠𝑘𝑖𝑛 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝐹1 = ∫ 𝐹1𝑑𝑧 𝑙 0 = ∫ ( 1 2 𝐶1𝐴𝑠𝜌𝑉2 ) 𝑑𝑧 𝑙 0 = 1 2 𝐶1𝐴𝑠𝜌𝑉2 ∫ 𝑑𝑧 𝑙 0 𝑊1 = ∫ 𝐹1𝑑𝑧 𝑙 0 = 1 2 𝐶1𝐴𝑠𝜌𝑉2 𝑙 We shall introduce a new friction term to account for Reynolds number as below: 𝑛𝑒𝑤 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒 𝐹2 = 1 2 𝐶2𝐴𝑠𝜌𝑉2 𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑠𝑘𝑖𝑛 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝐹2, 𝑖𝑠 𝑊2 = ∫ 𝐹2𝑑𝑧 𝑙 0 = 1 2 𝐶2𝐴𝑠𝜌𝑉2 𝑙 Since the velocity is constant along the length of the pipe and the terms 𝐶2𝐴𝑠𝜌𝑉2 are constant. Where: 𝐶2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎𝑠 𝑠ℎ𝑎𝑙𝑙 𝑏𝑒 𝑠ℎ𝑜𝑤𝑛 𝑇𝑜𝑡𝑎𝑙 𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑠𝑘𝑖𝑛 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = 1 2 𝐶𝑑𝑚𝑉2 + 1 2 𝐶1𝐴𝑠𝜌𝑉2 𝑙 + 1 2 𝐶2𝐴𝑠𝜌𝑉2 𝑙 𝑚 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 = 𝜋𝑟2 ∆𝑥𝜌 𝐶1 = 16 𝑅𝑒𝑑 = 8𝜇 𝜌𝑉𝑟 𝐶𝑑 = 𝐾𝐶1 𝑅𝑒 = 𝜌𝑉𝑑 𝜇 We conserve energy changes and say: 𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑜𝑛 𝑓𝑙𝑢𝑖𝑑 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑏𝑦 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 𝑔𝑎𝑖𝑛𝑒𝑑 + 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒𝑠 (𝑃1 − 𝑃2)𝑑𝑣 = 1 2 𝑚𝑉2 + ∑ 𝑊 𝑛 2 𝑛=1
- 27. 26 (𝑃1 − 𝑃2)𝑑𝑣 = 1 2 𝑚𝑉2 + 1 2 𝐶𝑑𝑚𝑉2 + 1 2 𝐶1𝐴𝑠𝜌𝑉2 𝑙 + 1 2 𝐶2𝐴𝑠𝜌𝑉2 × 𝑙 Where: 𝑑𝑣 = 𝑚 𝜌 We have ignored the surface tension effects for now. Substitute for 𝐶1 and for 𝐶𝑑 as before and get: (ℎ𝜌𝑔) 𝑚 𝜌 = 1 2 𝑚𝑉2 + 1 2 ( 8𝐾𝜇 𝜌𝑉𝑟 )𝑚𝑉2 + 1 2 ( 8𝜇 𝜌𝑉𝑟 )2𝜋𝑟∆𝑥𝜌𝑉2 𝑙 + 1 2 𝐶22𝜋𝑟∆𝑥𝜌𝑉2 × 𝑙 𝑚 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 = 𝜋𝑟2 ∆𝑥𝜌 Dividing through by mass m and simplifying we get: 2𝑔ℎ = 𝑉2 + 16𝜇𝑙 𝑟2𝑉𝜌 𝑉2 + 8𝐾𝜇 𝜌𝑟𝑉 𝑉2 + 2𝑙𝐶2 𝑟 𝑉2 𝑉2 (1 + 8𝐾𝜇 𝜌𝑟𝑉 + 2𝑙 𝑟 𝐶2) + 16𝜇𝑙 𝑟2𝜌 𝑉 − 2𝑔ℎ = 0 𝑽𝟐 (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) + 𝟐𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲)𝑽 − 𝟐𝒈𝒉 = 𝟎 We get velocity as a quadratic formula where: 𝑉 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 We choose the positive velocity as below: 𝑉 = −𝑏 + √𝑏2 − 4𝑎𝑐 2𝑎 Where: 𝑏 = 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾) 𝑎 = (1 + 2𝑙 𝑟 𝐶2) 𝑐 = −2𝑔ℎ 𝑉 = −𝑏 + √𝑏2 − 4𝑎𝑐 2𝑎
- 28. 27 𝑽 = − 𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲) (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) + √( 𝟐𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲))𝟐 + (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐)(𝟖𝒈𝒉) 𝟐 (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) The above is the velocity V. Pouiselle /Laminar flow can be demonstrated: First of all, we factorize the term 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾) out of the square root 𝑉 = − 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾) 2 (1 + 2𝑙 𝑟 𝐶2) + 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾) 2 (1 + 2𝑙 𝑟 𝐶2) √1 + (1 + 2𝑙 𝑟 𝐶2) 8𝑔ℎ ( 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾))2 For long pipes and small radius The term (1 + 2𝑙 𝑟 𝐶2)8𝑔ℎ ( 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾))2 ≪ 1 And we can use the approximation (1 + 𝑥)𝑛 ≈ 1 + 𝑛𝑥 for 𝑥 ≪ 1 Or √1 + 𝑥 ≈ 1 + 1 2 𝑥 for 𝑥 ≪ 1 𝑥 = (1 + 2𝑙 𝑟 𝐶2) 8𝑔ℎ ( 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾))2 In laminar flow also 2𝑙 𝑟 𝐶2 ≫ 1 and 8𝑙 𝑟 ≫ 4𝐾
- 29. 28 so that 1 + 2𝑙 𝑟 𝐶2 ≈ 2𝑙 𝑟 𝐶2 And 8𝑙 𝑟 + 4𝐾 ≈ 8𝑙 𝑟 So (1 + 2𝑙 𝑟 𝐶2)8𝑔ℎ ( 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾))2 ≈ 𝑟4 𝜌2 ( 2𝑙 𝑟 𝐶2) 256𝜇2𝑙2 × 8𝑔ℎ 𝑟4 𝜌2 ( 2𝑙 𝑟 𝐶2) 256𝜇2𝑙2 × 8𝑔ℎ = 𝑟3 𝜌2 𝑔ℎ𝐶2 16𝜇2𝑙 ≪ 1 For laminar flow, recalling the condition 𝑟3 𝜌2 𝑔ℎ 9200𝜇2𝑙 < 1 And comparing with 𝑟3 𝜌2 𝑔ℎ𝐶2 16𝜇2𝑙 ≪ 1 We get 𝐶2 16 = 1 9200 𝐶2 = 1.739 × 10−3 this proves that 𝐶2 is a constant since the critical Reynolds number for laminar flow is also a constant. Continuing from above to demonstrate the Pouiselle flow, Using the binomial expansion and after making the above substitutions, We use the binomial approximation √1 + 𝑥 ≈ 1 + 1 2 𝑥 𝑓𝑜𝑟 𝑥 ≪ 1 And get:
- 30. 29 √1 + (1 + 2𝑙 𝑟 𝐶2) 8𝑔ℎ ( 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾))2 ≈ 1 + (1 + 2𝑙 𝑟 𝐶2) 4𝑔ℎ ( 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾))2 1 + (1 + 2𝑙 𝑟 𝐶2) 4𝑔ℎ ( 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾))2 ≈ 1 + ( 2𝑙 𝑟 𝐶2) 4𝑔ℎ ( 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 ))2 = 1 + 𝑟4 𝜌2 256𝜇2𝑙2 × ( 2𝑙 𝑟 𝐶2)4𝑔ℎ 2 (1 + 2𝑙 𝑟 𝐶2) 𝑉 = − 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾) + 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾)√1 + (1 + 2𝑙 𝑟 𝐶2) 8𝑔ℎ ( 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾))2 2( 2𝑙 𝑟 𝐶2)𝑉 = − 16𝜇𝑙 𝑟2𝜌 + 16𝜇𝑙 𝑟2𝜌 (1 + 𝑟4 𝜌2 256𝜇2𝑙2 × ( 2𝑙 𝑟 𝐶2)4𝑔ℎ) Simplifying, we get velocity V as: 𝑽 = 𝒓𝟐 𝝆𝒈𝒉 𝟖𝝁𝒍 And the flow rate Q as: 𝑸 = 𝝅 𝟖 𝒓𝟒 𝝁 𝝆𝒈𝒉 𝒍 The term 𝑟3𝜌2𝑔ℎ 9200𝜇2𝑙 is a dimensionless number and it should demarcate when Pouiselle flow begins according to Reynold’s theory. The nature of 𝑪𝟐 For laminar flow 𝑅𝑒 < 𝑅𝑒𝑐𝑟 Where: 𝑅𝑒𝑐𝑟 = 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑅𝑒𝑦𝑛𝑜𝑙𝑑𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 𝑓𝑜𝑟 𝐿𝑎𝑚𝑖𝑛𝑎𝑟 𝑓𝑙𝑜𝑤 𝜌𝑉 𝑐𝑑 𝜇 < 𝑅𝑒𝑐𝑟 2𝜌𝑉 𝑐𝑟 𝜇 < 𝑅𝑒𝑐𝑟 Where: 𝑉 𝑐 = 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 In laminar flow
- 31. 30 𝑉 = 𝑟2 𝜌𝑔ℎ 8𝜇𝑙 And 𝑉 𝑐 = 𝑉 So, 2𝜌𝑟 𝜇 𝑉 < 𝑅𝑒𝑐𝑟 2𝜌𝑟 𝜇 × 𝑟2 𝜌𝑔ℎ 8𝜇𝑙 < 𝑅𝑒𝑐𝑟 𝑟3 𝜌2 𝑔ℎ 4𝜇2𝑙(𝑅𝑒𝑐𝑟) < 1 Comparing with what we got earlier 𝑟3 𝜌2 𝑔ℎ𝐶2 16𝜇2𝑙 ≪ 1 We get 𝑟3 𝜌2 𝑔ℎ 4𝜇2𝑙(𝑅𝑒𝑐𝑟) = 𝑟3 𝜌2 𝑔ℎ𝐶2 16𝜇2𝑙 𝐶2 4 = 1 (𝑅𝑒𝑐𝑟) 𝑪𝟐 = 𝟒 (𝑹𝒆𝒄𝒓) 𝑪𝟐 𝒊𝒔 𝒂 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒂𝒔 𝒔𝒉𝒂𝒍𝒍 𝒃𝒆 𝒔𝒉𝒐𝒘𝒏 𝒆𝒙𝒑𝒆𝒓𝒊𝒎𝒆𝒏𝒕𝒂𝒍𝒍𝒚 We can use the expression of 𝐶2 above to draw a similar expression for entrance length. It is known that the entrance length is given by: 𝑳𝒆 𝑫 = 𝟎. 𝟎𝟓𝟕𝟓𝑹𝒆 Using one of the conditions for laminar flow shown below: 2𝑙 𝑟 𝐶2 ≫ 1 Substituting for 𝐶2 we get
- 32. 31 2𝑙 𝑟 × 4 (𝑅𝑒𝑐𝑟) ≫ 1 And get 4𝑙 𝐷 × 4 (𝑅𝑒𝑐𝑟) ≫ 1 𝒍 𝑫 ≫ 𝟏 𝟏𝟔 (𝑹𝒆𝒄𝒓) The critical point is 𝑙 𝐷 = 1 16 (𝑅𝑒𝑐𝑟) 𝒍 𝑫 = 𝟎. 𝟎𝟔𝟐𝟓(𝑹𝒆𝒄𝒓) Comparing with the expression for entrance length, they look similar 𝑳𝒆 𝑫 = 𝟎. 𝟎𝟓𝟕𝟓𝑹𝒆 Though for the entrance length the Reynold number is allowed to vary but, in the expression derived above the critical Reynolds number is used which is a fixed value. We can use a similar argument to describe the entrance length for rough pipes in laminar flow knowing the expression of the friction factor for rough pipes.
- 33. 32 NB. We shall see that experiment doesn’t obey Reynold’s theory exactly and we have to make some modifications. We shall see that 𝐶2 takes on a different value from the one got using Reynold number as from experiment and so the critical Reynolds number will also change. To show the experimental deviation from Reynold’s theory we have to look at the equation below as derived above: 𝑽𝟐 (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) + 𝟐𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲)𝑽 − 𝟐𝒈𝒉 = 𝟎 We can rearrange the equation above and get: [2𝑔ℎ − 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾)𝑉] 𝑉2 = (1 + 2𝑙 𝑟 𝐶2) Let 𝑃 = [2𝑔ℎ − 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾)𝑉] 𝑉2 𝑃 = (1 + 2𝑙 𝑟 𝐶2) From experiment when the graph of P (for fixed h) against length L was plotted, a straight-line graph was observed as predicted by the equation but the value of 𝐶2 got was different from that derived by Reynold’s theory. Notice that the intercept got is one (1) as from theory Experimental results for a pipe of radius 0.0035 mm are shown below for a graph of P (for fixed h) against length L. 𝐶2 was found to be 𝐶2 = 0.0058541 from the graph below
- 34. 33 y = 3.3452x + 1 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 0.2 0.4 0.6 0.8 1 1.2 P length L(m) A graph of P against length L for pipe of diameter 7mm
- 35. 34 For laminar flow. We have shown that 𝐶2 is a constant from the graph above. Using 𝐶2 we can get the critical Reynolds number for laminar flow as below: So, the Critical Reynolds number for laminar flow becomes 683.2819 since For laminar flow 𝑅𝑒 < 𝑅𝑒𝑐𝑟 Where: 𝑅𝑒𝑐𝑟 = 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑅𝑒𝑦𝑛𝑜𝑙𝑑𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 𝑓𝑜𝑟 𝐿𝑎𝑚𝑖𝑛𝑎𝑟 𝑓𝑙𝑜𝑤 𝜌𝑉 𝑐𝑑 𝜇 < 𝑅𝑒𝑐𝑟 2𝜌𝑉 𝑐𝑟 𝜇 < 𝑅𝑒𝑐𝑟 Where: 𝑉 𝑐 = 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 In laminar flow 𝑉 = 𝑟2 𝜌𝑔ℎ 8𝜇𝑙 And 𝑉 𝑐 = 𝑉 So, 2𝜌𝑟 𝜇 𝑉 < 𝑅𝑒𝑐𝑟 2𝜌𝑟 𝜇 × 𝑟2 𝜌𝑔ℎ 8𝜇𝑙 < 𝑅𝑒𝑐𝑟 𝑟3 𝜌2 𝑔ℎ 4𝜇2𝑙(𝑅𝑒𝑐𝑟) < 1 Comparing with 𝑟3 𝜌2 𝑔ℎ𝐶2 16𝜇2𝑙 ≪ 1 We get 𝑟3 𝜌2 𝑔ℎ 4𝜇2𝑙(𝑅𝑒𝑐𝑟) = 𝑟3 𝜌2 𝑔ℎ𝐶2 16𝜇2𝑙
- 36. 35 𝐶2 4 = 1 (𝑅𝑒𝑐𝑟) 𝑅𝑒𝑐𝑟 = 4 𝐶2 = 4 0.0058541 = 𝟔𝟖𝟑. 𝟐𝟖𝟏𝟖 𝑹𝒆𝒄𝒓 = 𝟔𝟖𝟑. 𝟐𝟖𝟏𝟖 Since 𝐶2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 so the critical Reynolds number is a constant. Generally, the governing number for flow rate regime is 𝑮𝒐𝒗𝒆𝒓𝒏𝒊𝒏𝒈 𝑵𝒖𝒎𝒃𝒆𝒓 = (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐)𝟖𝒈𝒉 ( 𝟐𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲))𝟐 And the velocity equation is: 𝑽 = − 𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲) (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) + √( 𝟐𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲))𝟐 + (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐)(𝟖𝒈𝒉) 𝟐 (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐)
- 37. 36 How do we deal with cases where there is a change of cross-sectional area? We say, 𝐴1𝑉1 = 𝐴2𝑉2 And get: 𝑉2 = 𝐴1𝑉1 𝐴2 We know the general expression of the velocity 𝑉1 as developed before: 𝑽𝟏 = − 𝝁 𝒓𝟏𝝆 ( 𝟖𝒍 𝒓𝟏 + 𝟒𝑲) (𝟏 + 𝟐𝒍 𝒓𝟏 𝑪𝟐) + √( 𝟐𝝁 𝒓𝟏𝝆 ( 𝟖𝒍 𝒓𝟏 + 𝟒𝑲))𝟐 + (𝟏 + 𝟐𝒍 𝒓𝟏 𝑪𝟐)𝟖𝒈(𝒉 − 𝒉𝟎)) 𝟐(𝟏 + 𝟐𝒍 𝒓𝟏 𝑪𝟐) We can then get 𝑉2 as 𝑽𝟐 = 𝑨𝟏𝑽𝟏 𝑨𝟐 𝑽𝟐 = −( 𝑨𝟏 𝑨𝟐 ) 𝝁 𝒓𝟏𝝆 ( 𝟖𝒍 𝒓𝟏 + 𝟒𝑲) (𝟏 + 𝟐𝒍 𝒓𝟏 𝑪𝟐) + ( 𝑨𝟏 𝑨𝟐 ) √( 𝟐𝝁 𝒓𝟏𝝆( 𝟖𝒍 𝒓𝟏 + 𝟒𝑲))𝟐 + (𝟏 + 𝟐𝒍 𝒓𝟏 𝑪𝟐)(𝟖𝒈(𝒉 − 𝒉𝟎)) 𝟐(𝟏 + 𝟐𝒍 𝒓𝟏 𝑪𝟐) We shall use the derivation above in the analysis to follow.
- 38. 37 Where: 𝒉𝟎 = 𝟐𝜸𝒄𝒐𝒔𝜽𝒄 𝝆𝒈 ( 𝟏 𝒓𝟎 + 𝟏 𝒓𝟐 ) 𝜃𝑐 = 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 𝑎𝑛𝑔𝑙𝑒
- 39. 38 THE MODIFIED BERNOULLI EQUATION WITH VISCOUS EFFECTS INCLUDED. We are going to look at cylindrical pipes. Recalling the conservation of energy technique used before to get the velocity as below: 2𝑔ℎ = 𝑉2 + 16𝜇𝑙 𝑟2𝑉𝜌 𝑉2 + 8𝐾𝜇 𝜌𝑟𝑉 𝑉2 + 2𝑙𝐶2 𝑟 𝑉2 2𝑔ℎ = 𝑉2 + 16𝜇𝑙 𝑟2𝜌 𝑉 + 8𝐾𝜇 𝜌𝑟 𝑉 + 2𝑙𝐶2 𝑟 𝑉2 Multiplying through by 𝜌 and dividing through by 2, we get: 𝜌𝑔ℎ = 𝜌 𝑉2 2 + 8𝜇𝑙 𝑟2 𝑉 + 4𝐾𝜇 𝑟 𝑉 + 𝜌𝑙𝐶2 𝑟 𝑉2 Finally, we get for cylindrical pipe or circular orifice: 𝑷 + 𝒉𝝆𝒈 + 𝝆 𝑽𝟐 𝟐 + 𝟖𝝁𝒍 𝒓𝟐 𝑽 + 𝟒𝑲𝝁 𝒓 𝑽 + 𝝆𝒍𝑪𝟐 𝒓 𝑽𝟐 = 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 OR 𝑷 + 𝒉𝝆𝒈 + 𝝆 𝑽𝟐 𝟐 + 𝝆𝑪𝑫 𝑽𝟐 𝟐 + 𝝆𝒍𝑪𝟏 𝒓 𝑽𝟐 + 𝝆𝒍𝑪𝟐 𝒓 𝑽𝟐 = 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 Or 𝑷 + 𝒉𝝆𝒈 + 𝝆 𝑽𝟐 𝟐 + 𝟖𝝁𝒍 𝒓𝟐 𝑽 + 𝟒𝑲𝝁 𝒓 𝑽 + 𝝆𝒍𝑪𝟐 𝒓 𝑽𝟐 = 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 How can we apply the Bernoulli equation above? Considering the Torricelli flow first: Let us first consider a circular orifice on a tank:
- 40. 39 Using the Bernoulli equation, we get 𝑷𝒙 + 𝒉𝒙𝝆𝒈 + 𝝆 𝑽𝒙 𝟐 𝟐 + 𝟖𝝁𝒍𝒙 𝒓𝒙 𝟐 𝑽𝒙 + 𝟒𝑲𝝁 𝒓𝒙 𝑽𝒙 + 𝝆𝒍𝒙𝑪𝟐 𝒓𝒙 𝑽𝒙 𝟐 = 𝑷𝒚 + 𝒉𝒚𝝆𝒈 + 𝝆 𝑽𝒚 𝟐 𝟐 + 𝟖𝝁𝒍𝒚 𝒓𝒚 𝒚 𝑽𝒚 + 𝟒𝑲𝝁 𝒓𝒚 𝑽𝒚 + 𝝆𝒍𝒚𝑪𝟐 𝒓𝒚 But 𝑙𝑥 = 0 𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑞𝑢𝑖𝑑 𝑖𝑛 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑒𝑟 𝑖𝑠 𝑎𝑡 𝑟𝑒𝑠𝑡 𝑙𝑥 represents the wetted length the fluid moves. ℎ𝑥 = ℎ ℎ𝑦 = 0 𝑙𝑦 = 0 𝑃𝑥 = 𝐻 − 2𝛾𝑐𝑜𝑠𝜃𝑥 𝑟𝑥 𝑃𝑦 = 𝐻 + 2𝛾𝑐𝑜𝑠𝜃𝑦 𝑟𝑦 𝐻 = 𝑎𝑡𝑚𝑜𝑠𝑝ℎ𝑒𝑟𝑖𝑐 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑉 𝑦 = 𝑉 When the cross-sectional area of the container is large so that the rate of change of height of the surface level is negligible, then: 𝑉 𝑥 = 0 Upon substitution of all the above we get:
- 41. 40 ℎ𝑥𝜌𝑔 − 2𝛾𝑐𝑜𝑠𝜃𝑥 𝑟𝑥 − 2𝛾𝑐𝑜𝑠𝜃𝑦 𝑟𝑦 = 𝜌 𝑉 𝑦 2 2 + 4𝐾𝜇 𝑟 𝑉 𝑦 Or (ℎ − ℎ0)𝜌𝑔 = 𝜌 𝑉 𝑦 2 2 + 4𝐾𝜇 𝑟 𝑉 𝑦 Where: ℎ0 = 2𝛾𝑐𝑜𝑠𝜃𝑥 𝑟𝑥𝜌𝑔 + 2𝛾𝑐𝑜𝑠𝜃𝑦 𝑟𝑦𝜌𝑔 If 𝜃𝑥 = 𝜃𝑦 ℎ0 = 2𝛾𝑐𝑜𝑠𝜃𝑥 𝜌𝑔 ( 1 𝑟𝑥 + 1 𝑟𝑦 ) Where we can go ahead and get the velocity of exit from the quadratic formula which is what we got before for Torricelli flow. i.e., 𝑽𝟐 + 𝟖𝑲𝝁 𝒓𝝆 𝑽 − 𝟐𝒈(𝒉 − 𝒉𝟎) = 𝟎 How can we apply the Bernoulli equation for cylindrical pipes?
- 42. 41 𝑷𝒙 + 𝒉𝒙𝝆𝒈 + 𝝆 𝑽𝒙 𝟐 𝟐 + 𝟖𝝁𝒍𝒙 𝒓𝒙 𝟐 𝑽𝒙 + 𝟒𝑲𝝁 𝒓𝒙 𝑽𝒙 + 𝝆𝒍𝒙𝑪𝟐 𝒓𝒙 𝑽𝒙 𝟐 = 𝑷𝒚 + 𝒉𝒚𝝆𝒈 + 𝝆 𝑽𝒚 𝟐 𝟐 + 𝟖𝝁𝒍𝒚 𝒓𝒚 𝒚 𝑽𝒚 + 𝟒𝑲𝝁 𝒓𝒚 𝑽𝒚 + 𝝆𝒍𝒚𝑪𝟐 𝒓𝒚 𝑽𝒚 𝟐 But ℎ𝑥 = ℎ ℎ𝑦 = 0 𝑃𝑥 = 𝐻 − 2𝛾𝑐𝑜𝑠𝜃𝑥 𝑟𝑥 𝑃𝑦 = 𝐻 + 2𝛾𝑐𝑜𝑠𝜃𝑦 𝑟𝑦 𝐻 = 𝑎𝑡𝑚𝑜𝑠𝑝ℎ𝑒𝑟𝑖𝑐 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑙𝑥 = 0 𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑖𝑠 𝑎𝑡 𝑟𝑒𝑠𝑡 𝑙𝑦 = 𝑙 𝑉 𝑦 = 𝑉 𝑟𝑦 = 𝑟 When the cross-sectional area of the container is large, 𝑉 𝑥 = 0 Upon substitution of all the above we get: ℎ𝑥𝜌𝑔 − 2𝛾𝑐𝑜𝑠𝜃𝑦 𝑟𝑦 − 2𝛾𝑐𝑜𝑠𝜃𝑥 𝑟𝑥 = 𝜌 𝑉 𝑦 2 2 + 8𝜇𝑙 𝑟2 𝑉 𝑦 + 4𝐾𝜇 𝑟 𝑉 𝑦 + 𝜌𝑙𝐶2 𝑟 𝑉 𝑦 2 (𝒉 − 𝒉𝟎)𝝆𝒈 = 𝝆 𝑽𝟐 𝟐 + 𝟖𝝁𝒍 𝒓𝟐 𝑽 + 𝟒𝑲𝝁 𝒓 𝑽 + 𝝆𝒍𝑪𝟐 𝒓 𝑽𝟐 Where: ℎ0 = 2𝛾𝑐𝑜𝑠𝜃𝑥 𝜌𝑔 ( 1 𝑟𝑥 + 1 𝑟𝑦 ) From the equation above, we can go ahead and find the velocity of exit 𝑉 = 𝑉 𝑦 which is what we derived before.
- 43. 42 How do we apply the Bernoulli equation to different area pipes? Again, we use the modified Bernoulli equation as below: 𝑷𝒙 + 𝒉𝒙𝝆𝒈 + 𝝆 𝑽𝒙 𝟐 𝟐 + 𝟖𝝁𝒍𝒙 𝒓𝒙 𝟐 𝑽𝒙 + 𝟒𝑲𝝁 𝒓𝒙 𝑽𝒙 + 𝝆𝒍𝒙𝑪𝟐 𝒓𝒙 𝑽𝒙 𝟐 = 𝑷𝒚 + 𝒉𝒚𝝆𝒈 + 𝝆 𝑽𝒚 𝟐 𝟐 + 𝟖𝝁( 𝒍𝟏 𝒓𝟏 𝟐 + 𝒍𝟐 𝒓𝟐 𝟐 ) 𝑽𝒚 + 𝟒𝑲𝝁 𝒓𝟐 𝑽𝒚 + 𝝆𝑪𝟐 ( 𝒍𝟏 𝒓𝟏 + 𝒍𝟐 𝒓𝟐 ) 𝑽𝒚 𝟐 But 𝑙𝑥 = 0 ℎ𝑥 = ℎ ℎ𝑦 = 0 𝑃𝑥 = 𝐻 − 2𝛾𝑐𝑜𝑠𝜃𝑥 𝑟𝑥 𝑃𝑦 = 𝐻 + 2𝛾𝑐𝑜𝑠𝜃𝑦 𝑟𝑦 𝐻 = 𝑎𝑡𝑚𝑜𝑠𝑝ℎ𝑒𝑟𝑖𝑐 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 When the cross-sectional area of the container is large so that the rate of fall of the surface level is negligible, 𝑉 𝑥 = 0 And we finally get ℎ𝑥𝜌𝑔 = 𝜌 𝑉 𝑦 2 2 + 8𝜇( 𝑙1 𝑟1 2 + 𝑙2 𝑟2 2)𝑉 𝑦 + 4𝐾𝜇 𝑟2 𝑉 𝑦 + 𝜌𝐶2( 𝑙1 𝑟1 + 𝑙2 𝑟2 )𝑉 𝑦 2 Before we can get 𝑉 𝑦 we have to ask what will 𝑉 𝑦 be when 𝑙2 𝑖𝑠 𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑡𝑜 0 ?
- 44. 43 𝑉 𝑦 will be given by: ℎ𝑥𝜌𝑔 = 𝜌 𝑉 𝑦 2 2 + 8𝜇 ( 𝑙1 𝑟1 2) 𝑉 𝑦 + 4𝐾𝜇 𝑟2 𝑉 𝑦 + 𝜌𝐶2 ( 𝑙1 𝑟1 ) 𝑉 𝑦 2 𝒏 But remember that when 𝑙2 𝑖𝑠 𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑡𝑜 0 , the area at the exit will be 𝐴2 and so the velocity will be given by 𝑉 = 𝐴1 𝐴2 𝑉 𝑦 To get the velocity above, we have to make a substitution in equation n above as: 𝑉 𝑦 = 𝐴2 𝐴1 𝑉 Upon substitution in the equation n above, we get: ℎ𝑥𝜌𝑔 = 𝜌 𝑉2 2 ( 𝐴2 𝐴1 )2 + 8𝜇 ( 𝑙1 𝑟1 2) ( 𝐴2 𝐴1 )𝑉 + 4𝐾𝜇 𝑟2 ( 𝐴2 𝐴1 )𝑉 + 𝜌𝐶2 ( 𝑙1 𝑟1 ) ( 𝐴2 𝐴1 )2 𝑉2 We finally get the velocity as 𝑉 = ( 𝑨𝟏 𝑨𝟐 ) [ − 𝜇 𝑟1𝜌 ( 8𝑙1 𝑟1 + 4𝐾) (1 + 2𝑙1 𝑟1 𝐶2) + √( 2𝜇 𝑟1𝜌 ( 8𝑙1 𝑟1 + 4𝐾))2 + (1 + 2𝑙1 𝑟1 𝐶2)(8𝑔(ℎ − ℎ0)) 2 (1 + 2𝑙1 𝑟1 𝐶2) ] 𝑉 = [ − 𝜇 𝑟1𝜌 ( 8𝑙1 𝑟1 + 4𝐾) (1 + 2𝑙1 𝑟1 𝐶2) ( 𝑨𝟏 𝑨𝟐 ) + ( 𝑨𝟏 𝑨𝟐 ) √( 2𝜇 𝑟1𝜌 ( 8𝑙1 𝑟1 + 4𝐾))2 + (1 + 2𝑙1 𝑟1 𝐶2)(8𝑔(ℎ − ℎ0)) 2 (1 + 2𝑙1 𝑟1 𝐶2) ] As required. In fact, we already showed this velocity before. The factor we were interested in to show was: ( 𝑨𝟏 𝑨𝟐 ) So going back to the velocity equation, we have to incorporate the above factor so that when we reduce 𝑙2 𝑡𝑜 0 , we arrive at the required velocity above as shown below: ℎ𝑥𝜌𝑔 = 𝜌 𝑉 𝑦 2 2 + 8𝜇( 𝑙1 𝑟1 2 + 𝑙2 𝑟2 2)𝑉 𝑦 + 4𝐾𝜇 𝑟2 𝑉 𝑦 + 𝜌𝐶2( 𝑙1 𝑟1 + 𝑙2 𝑟2 )𝑉 𝑦 2 We substitute:
- 45. 44 𝑉 𝑦 = 𝐴2 𝐴1 𝑉 And get: ℎ𝑥𝜌𝑔 = 𝜌 𝑉2 2 ( 𝐴2 𝐴1 )2 + 8𝜇( 𝑙1 𝑟1 2 + 𝑙2 𝑟2 2)( 𝐴2 𝐴1 )𝑉 + 4𝐾𝜇 𝑟2 ( 𝐴2 𝐴1 )𝑉 + 𝜌𝐶2( 𝑙1 𝑟1 + 𝑙2 𝑟2 )( 𝐴2 𝐴1 )2 𝑉2 We have to include the surface tension effects and the equation becomes, (𝒉 − 𝒉𝟎)𝝆𝒈 = 𝝆 𝑽𝟐 𝟐 ( 𝑨𝟐 𝑨𝟏 )𝟐 + 𝟖𝝁( 𝒍𝟏 𝒓𝟏 𝟐 + 𝒍𝟐 𝒓𝟐 𝟐)( 𝑨𝟐 𝑨𝟏 )𝑽 + 𝟒𝑲𝝁 𝒓𝟐 ( 𝑨𝟐 𝑨𝟏 )𝑽 + 𝝆𝑪𝟐( 𝒍𝟏 𝒓𝟏 + 𝒍𝟐 𝒓𝟐 )( 𝑨𝟐 𝑨𝟏 )𝟐 𝑽𝟐 Where: ℎ0 = 2𝛾𝑐𝑜𝑠𝜃𝑥 𝜌𝑔 ( 1 𝑟𝑥 + 1 𝑟𝑦 ) We can go ahead and find the velocity V from the above. How do we write the Bernoulli equation for a variable cross-sectional area with distance for example for the case of when the pipe is a conical frustrum? We can write the Bernoulli equation as an integral as below: 𝑷 + 𝒉𝝆𝒈 + 𝝆 𝑽𝟐 𝟐 + 𝟖𝝁𝒍 𝒓𝟐 𝑽 + 𝟒𝑲𝝁 𝒓 𝑽 + 𝝆𝒍𝑪𝟐 𝒓 𝑽𝟐 = 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝑃 + ℎ𝜌𝑔 + 𝜌 𝑉2 2 + 8𝜇𝜋𝑉 ∫ ( 1 𝐴 )𝑑𝑥 𝑙 0 + 4𝐾𝜇 𝑟 𝑉 + 𝜌𝐶2𝑉2 ∫ ( 1 𝑟 )𝑑𝑥 𝑙 0 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Where: 𝐴𝑠 = 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎
- 46. 45 𝐴𝑇 = 𝑡𝑜𝑡𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 Or 𝑃 + ℎ𝜌𝑔 + 𝜌 𝑉2 2 + 8𝜇𝜋𝑉 ∫ ( 1 𝐴 )𝑑𝑥 𝑙 0 + 4𝐾𝜇 𝑟 𝑉 + 𝜌𝐶2𝑉2 ∫ ( 1 𝑟 )𝑑𝑥 𝑙 0 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Where: 𝐴𝑠 = 𝑤𝑒𝑡𝑡𝑒𝑑 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑓𝑟𝑢𝑠𝑡𝑟𝑢𝑚 𝐴𝑇 = 𝑤𝑒𝑡𝑡𝑒𝑑 𝑡𝑜𝑡𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑓𝑟𝑢𝑠𝑡𝑟𝑢𝑚 In applying the formula above recall that the area and radius r of the conical frustrum vary with distance x. i.e. 𝐴 = 𝐴2 𝑥 𝑙 + [1 − 𝑥 𝑙 ]𝐴1 And 𝑟 = 𝑟2 𝑥 𝑙 + [1 − 𝑥 𝑙 ]𝑟1 When the area is not varying, then we arrive back to the original expression. Using the friction factors for other geometries like the rectangular ducts, we can use energy conservation techniques used above to develop the general equation of velocity and even develop the Bernoulli equation for rectangular ducts.
- 47. 46 HOW DO WE DEAL WITH PRESSURE GRADIENTS? Assume constant cross-sectional area and equal spacing as shown of length 𝑙. Using the velocity equation derived before: 𝑉2 (1 + 2𝑙 𝑟 𝐶2) + 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾)𝑉 − 2𝑔ℎ = 0 𝑉2 (1 + 2𝑙 𝑟 𝐶2) + 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾)𝑉 = 2𝑔ℎ Assume 𝑉1 = 𝑉2 = 𝑉3 = 𝑉4 = 𝑉 The equations of head loss become: 𝑉2 2𝑔 (1 + 2𝑙 𝑟 𝐶2) + 𝜇 𝑟𝑔𝜌 ( 8𝑙 𝑟 + 4𝐾)𝑉 = ℎ1 − ℎ2 𝑉2 2𝑔 (1 + 2𝑙 𝑟 𝐶2) + 𝜇 𝑟𝑔𝜌 ( 8𝑙 𝑟 + 4𝐾)𝑉 = ℎ2 − ℎ3 𝑉2 2𝑔 (1 + 2𝑙 𝑟 𝐶2) + 𝜇 𝑟𝑔𝜌 ( 8𝑙 𝑟 + 4𝐾)𝑉 = ℎ3 − ℎ4 𝑉2 2𝑔 (1 + 2𝑙 𝑟 𝐶2) + 𝜇 𝑟𝑔𝜌 ( 8𝑙 𝑟 + 4𝐾)𝑉 = ℎ4
- 48. 47 ℎ1 − ℎ2 𝑙 = ℎ2 − ℎ3 𝑙 = ℎ3 − ℎ4 𝑙 = ℎ4 𝑙 = 𝑉2 2𝑔𝑙 (1 + 2𝑙 𝑟 𝐶2) + 𝜇 𝑟𝑔𝑙𝜌 ( 8𝑙 𝑟 + 4𝐾)𝑉 = 𝑚 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Since 𝑙 is the same throughout. Adding all the equations of head loss above we get Equation b) below. 𝑽𝟐 𝟐𝒈 (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) + 𝝁 𝒓𝒈𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲) 𝑽 = 𝒉𝟏 𝟒 … . . 𝒃) Where 𝑚 = 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑚𝑙 = ℎ1 4 We see that the uniform pressure gradient is only achieved because of the fixed equal length intervals. 𝑉2 2𝑔𝑙 (1 + 2𝑙 𝑟 𝐶2) + 𝜇 𝑟𝑔𝑙𝜌 ( 8𝑙 𝑟 + 4𝐾)𝑉 = 𝑚 𝑉2 (1 + 2𝑙 𝑟 𝐶2) + 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾)𝑉 − 2𝑔𝑚𝑙 = 0 We can get the velocity below: 𝟐 (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) 𝑽 = − 𝟐𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲) + √( 𝟐𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲))𝟐 + 𝟖𝒈𝒎𝒍(𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) or 𝑽 = − 𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲) (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) + √( 𝟐𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲))𝟐 + 𝟖𝒈𝒎𝒍(𝟏 + 𝟐𝒍 𝒓 𝑪𝟐))) 𝟐 (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) Again, it can be shown after making the assumptions as above that when 8𝑔𝑚𝑙(1 + 2𝑙 𝑟 𝐶2) ( 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾))2 ≪ 1 Or since
- 49. 48 𝑚𝑙 = ℎ1 4 8𝑔 ℎ1 4 (1 + 2𝑙 𝑟 𝐶2) ( 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾))2 ≪ 1 We use the binomial approximation √1 + 𝑥 ≈ 1 + 1 2 𝑥 𝑓𝑜𝑟 𝑥 ≪ 1 And get: 𝑽 = 𝟐𝒈𝒎𝒍 𝟏𝟔𝝁𝒍 𝒓𝟐𝝆 𝑉 = 𝑟2 𝜌𝑔𝑚 8𝜇 𝑚 = 𝑑ℎ 𝑑𝑥 𝑸 = 𝝅𝒓𝟒 𝟖𝝁 𝒅𝑷 𝒅𝒙 We notice that Pouiselle flow arrives due to equal spacing of the tubes but we notice that nonlinear pressure gradients can also be created provided non equal spacing We notice ℎ = −𝑚𝑥 + ℎ1 𝑉2 2𝑔 (1 + 2𝑙 𝑟 𝐶2) + 𝜇 𝑟𝑔𝜌 ( 8𝑙 𝑟 + 4𝐾)𝑉 = ℎ1 − ℎ2 𝑉2 2𝑔 (1 + 2𝑙 𝑟 𝐶2) + 𝜇 𝑟𝑔𝜌 ( 8𝑙 𝑟 + 4𝐾)𝑉 = ℎ2 − ℎ3 𝑉2 2𝑔 (1 + 2𝑙 𝑟 𝐶2) + 𝜇 𝑟𝑔𝜌 ( 8𝑙 𝑟 + 4𝐾)𝑉 = ℎ3 − ℎ4 𝑉2 2𝑔 (1 + 2𝑙 𝑟 𝐶2) + 𝜇 𝑟𝑔𝜌 ( 8𝑙 𝑟 + 4𝐾)𝑉 = ℎ4 Adding all
- 50. 49 𝟒𝑽𝟐 𝟐𝒈 (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) + 𝟒𝝁 𝒓𝒈𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲) 𝑽 = 𝒉𝟏 We can get V. For turbulent flow 2(1 + 2𝑙 𝑟 𝐶2)𝑉 = − 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾) + √[ 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾)]2 + 8𝑔𝑚𝑙(1 + 2𝑙 𝑟 𝐶2) Factorizing ( 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾))2 out of the square root, we get: 2(1 + 2𝑙 𝑟 𝐶2)𝑉 = − 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾) + 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾) √1 + (8𝑔𝑚𝑙)(1 + 2𝑙 𝑟 𝐶2) ( 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾))2 Also, in turbulent flow 8𝑔𝑚𝑙(1 + 2𝑙 𝑟 𝐶2) ( 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾))2 ≫ 1 Where: 𝑚𝑙 = ℎ1 4 OR 8𝑔 ℎ1 4 (1 + 2𝑙 𝑟 𝐶2) ( 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾))2 ≫ 1 𝑽 = − 𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲) (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) + √ (𝟐𝒈𝒎𝒍) (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) Or 𝑽 = − 𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲) (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) + √ (𝟐𝒈 𝒉𝟏 𝟒 ) (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐)
- 51. 50 Got by adding up the equations of head loss above Using the equation below for turbulent flow: 𝑽 = − 𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲) (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) + √ (𝟐𝒈𝒎𝒍) (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) In terms of the flow rate and pressure gradient, we get 𝑸 = 𝑨 − 𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲) (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) + 𝑨√ 𝟐 𝝆 𝒅𝑷 𝒅𝒙 (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) The equation says that the flow rate Q is directly proportional to the square root of the pressure gradient with an intercept. when: √ (2𝑔𝑚𝑙) (1 + 2𝑙 𝑟 𝐶2) ≫ − 𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾) (1 + 2𝑙 𝑟 𝐶2) Where: 𝑚𝑙 = ℎ1 4 Then − 𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾) (1 + 2𝑙 𝑟 𝐶2) + √ (2𝑔𝑚𝑙) (1 + 2𝑙 𝑟 𝐶2) ≈ √ (2𝑔𝑚𝑙) (1 + 2𝑙 𝑟 𝐶2) So, we get the velocity as 𝑽 = √ (𝟐𝒈𝒎𝒍) (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) After rearranging, we get 𝑸𝟐 = 𝑨𝟐 𝟐𝒍 𝝆(𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) 𝒅𝑷 𝒅𝒙
- 52. 51 If 𝟐𝒍 𝒓 𝑪𝟐 ≫ 𝟏 Then 1 + 2𝑙 𝑟 𝐶2 ≈ 2𝑙 𝑟 𝐶2 upon substitution, we get 𝑄2 = 𝐴2 𝑟 𝜌(𝐶2) 𝑑𝑃 𝑑𝑥 𝑸𝟐 = 𝑨𝟐 𝑫 𝟐𝝆(𝑪𝟐) 𝒅𝑷 𝒅𝒙 Which is also an equation for fully turbulent flow.
- 53. 52 HEAD LOSS Back to systems below: The head loss is given by: 𝒉 = 𝟒𝒇 𝒍 𝑫 × 𝑽𝟐 𝟐𝒈 … … … … . .1) OR 𝒉 = 𝟐𝒇 𝒍 𝑫 × 𝑽𝟐 𝒈 where we substitute for the correct friction factor and get the flow rate. But in our derivations, we get the head loss as below: generally, 𝑉2 (1 + 2𝑙 𝑟 𝐶2) + 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾)𝑉 = 2𝑔(ℎ1 − ℎ2) (ℎ1 − ℎ2) = ℎ𝑒𝑎𝑑𝑙𝑜𝑠𝑠 rearranging ℎ1 − ℎ2 = [ 𝑉2 2𝑔 (1 + 2𝑙 𝑟 𝐶2) + 𝜇 𝑟𝑔𝜌 ( 8𝑙 𝑟 + 4𝐾)𝑉] ℎ1 − ℎ2 = 𝑉2 2𝑔 [(1 + 2𝑙 𝑟 𝐶2) + 2𝜇 𝑟𝑉𝜌 ( 8𝑙 𝑟 + 4𝐾)] from equation 1) above
- 54. 53 ℎ1 − ℎ2 = 4𝑓 𝑙 𝐷 × 𝑉2 2𝑔 = 𝑉2 2𝑔 [(1 + 2𝑙 𝑟 𝐶2) + 2𝜇 𝑟𝑉𝜌 ( 8𝑙 𝑟 + 4𝐾)] 4𝑓 𝑙 𝐷 = [(1 + 2𝑙 𝑟 𝐶2) + 2𝜇 𝑟𝑉𝜌 ( 8𝑙 𝑟 + 4𝐾)] 4𝑓 = 𝐷 𝑙 + 4𝐶2 + 4𝜇 𝑙𝑉𝜌 ( 8𝑙 𝑟 + 4𝐾) 𝒇 = 𝑫 𝟒𝒍 + 𝑪𝟐 + 𝝁 𝒍𝑽𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲) For laminar flow 𝐷 4𝑙 ≈ 0 and 8𝑙 𝑟 + 4𝑘 ≈ 8𝑙 𝑟 and 𝐶2 ≈ 0 𝑓 = 8𝜇 𝑉𝑟𝜌 𝒇 = 𝟏𝟔 𝑹𝒆𝒅 For turbulent flow, the governing equation was 𝑽(𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) = − 𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲) + √𝟐𝒈(𝟏 + 𝟐𝒍 𝒓 𝑪𝟐(𝒉𝟏 − 𝒉𝟐) [𝑉(1 + 2𝑙 𝑟 𝐶2) + 𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾)]2 = 2𝑔(1 + 2𝑙 𝑟 𝐶2)(ℎ1 − ℎ2) 𝑉2 [(1 + 2𝑙 𝑟 𝐶2) + 2𝜇 𝑟𝑉𝜌 ( 8𝑙 𝑟 + 4𝐾)]2 = 2𝑔(1 + 2𝑙 𝑟 𝐶2)(ℎ1 − ℎ2) Therefore, head loss ∆𝒉 = (𝒉𝟏 − 𝒉𝟐) is ∆𝒉 = 𝑽𝟐 𝟐𝒈 [𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) + 𝟐𝝁 𝒓𝑽𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲)]𝟐 (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) Compare with ∆ℎ = 4𝑓 𝑙 𝐷 × 𝑉2 2𝑔
- 55. 54 4𝑓 𝑙 𝐷 = [1 + 2𝑙 𝑟 𝐶2) + 2𝜇 𝑟𝑉𝜌 ( 8𝑙 𝑟 + 4𝐾)]2 (1 + 2𝑙 𝑟 𝐶2 + 𝛽𝑙 (𝑟 + 𝑙) ) 𝑓 = 𝐷 4𝑙 [(1 + 2𝑙 𝑟 𝐶2) + 2𝜇 𝑟𝑉𝜌 ( 8𝑙 𝑟 + 4𝐾)]2 (1 + 2𝑙 𝑟 𝐶2) We get this expression for the friction coefficient 𝒇 = 𝑫 𝟒𝒍 × [(𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) + 𝟒 𝑹𝒆𝒅 ( 𝟖𝒍 𝒓 + 𝟒𝑲)]𝟐 (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) Comparing the equation below for smooth pipes in turbulent flow with the Blasius equation, they should give the same value i.e., ∆𝒉 = 𝑽𝟐 𝟐𝒈 [𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) + 𝟐𝝁 𝒓𝑽𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲)]𝟐 (𝟏 + 𝟐𝒍 𝒓 𝑪𝟐) Compare with: The Blasius Friction factor is: 𝑓 = 0.079 𝑅𝑒0.25 For turbulent flow: 𝑅𝑒 < 100,000 And the Blasius equation is: Blasius predicts that turbulent flow equation is [2] ∆𝒉 = 𝟎. 𝟐𝟒𝟏𝝆𝟎.𝟕𝟓 𝜇𝟎.𝟐𝟓 𝝆𝒈𝑫𝟒.𝟕𝟓 × 𝑸𝟏.𝟕𝟓 𝒍 𝑾𝒉𝒆𝒓𝒆 𝑫 = 𝒅𝒊𝒂𝒎𝒆𝒕𝒆𝒓 𝒐𝒇 𝒑𝒊𝒑𝒆 The two equations should predict the same flow rate or head loss. A. For rough pipes For rough pipes, the friction coefficient is given by: 𝟏 √𝒇 = 𝟒. 𝟎𝒍𝒐𝒈𝟏𝟎 𝑫 𝒆 + 𝟐. 𝟐𝟖
- 56. 55 We notice that the friction factor is independent of the Reynolds number and a constant for a given diameter for high Reynolds numbers. From the equation of head loss, ℎ = 4𝑓 𝑙 𝐷 × 𝑉2 2𝑔 Rearranging, we get: 𝑸𝟐 = 𝑫 𝟐𝝆𝒇 × 𝑨𝟐 × 𝒅𝑷 𝒅𝒙 This is the formula for flow rate for which we substitute the friction factor Recalling from the formulas derived before replacing 𝐶2 with 𝐶4 and using the formula below: 2 (1 + 2𝑙 𝑟 𝐶4) 𝑉 = − 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾) + 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾) √1 + (1 + 2𝑙 𝑟 𝐶4)8𝑔ℎ [ 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾)]2 𝑤ℎ𝑒𝑟𝑒 𝐶4 = 𝑓 = 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑓𝑜𝑟 𝑟𝑜𝑢𝑔ℎ 𝑝𝑖𝑝𝑒𝑠 2 (1 + 2𝑙 𝑟 𝐶4) 𝑉 = − 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾) + 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾)√1 + (1 + 2𝑙 𝑟 𝐶4) 8𝑔ℎ ( 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾))2 For turbulent flow (1 + 2𝑙 𝑟 𝐶4)8𝑔ℎ ( 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾))2 ≫ 1 And 1 + (1 + 2𝑙 𝑟 𝐶4)8𝑔ℎ ( 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾))2 ≈ (1 + 2𝑙 𝑟 𝐶4)8𝑔ℎ ( 2𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾))2 Substituting 𝐶4 = 𝑓
- 57. 56 𝑉 = − 𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾) (1 + 2𝑙 𝑟 𝑓) + √ 2𝑔ℎ (1 + 2𝑙 𝑟 𝑓) For turbulent flow, when, √ 𝟐𝒈𝒉 (𝟏 + 𝟐𝒍 𝒓 𝒇) ≫ 𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲) (𝟏 + 𝟐𝒍 𝒓 𝒇) The condition above is sufficient in getting the fully developed turbulent flow. Then − 𝜇 𝑟𝜌 ( 8𝑙 𝑟 + 4𝐾) (1 + 2𝑙 𝑟 𝑓) + √ 2𝑔ℎ (1 + 2𝑙 𝑟 𝑓) ≈ √ 2𝑔ℎ (1 + 2𝑙 𝑟 𝑓) Velocity becomes 𝑉 = √ 2𝑔ℎ (1 + 2𝑙 𝑟 𝑓) And if 2𝑙 𝑟 𝑓 ≫ 1 1 + 2𝑙 𝑟 𝑓 ≈ 2𝑙 𝑟 𝑓 𝑉 = √ 2𝑔ℎ ( 2𝑙 𝑟 𝑓) Rearranging We get
- 58. 57 𝑸𝟐 = 𝑫 𝟐𝝆𝒇 × 𝑨𝟐 × 𝒅𝑷 𝒅𝒙 Which is the same as that we got by rearranging the head loss. The condition 2𝑙 𝑟 𝑓 ≫ 1 Is equivalent to finding the entrance length in laminar flow for rough pipes where: 𝟏 √𝒇 = 𝟒. 𝟎𝒍𝒐𝒈𝟏𝟎 𝑫 𝒆 + 𝟐. 𝟐𝟖 So generally, for rough pipes the velocity is given by: 𝟐(𝟏 + 𝟐𝒍 𝒓 𝒇)𝑽 = − 𝟐𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲) + √([ 𝟐𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲)]𝟐 + (𝟏 + 𝟐𝒍 𝒓 𝒇)(𝟖𝒈𝒉)) or 𝑽 = − 𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲) (𝟏 + 𝟐𝒍 𝒓 𝒇) + √[ 𝟐𝝁 𝒓𝝆 ( 𝟖𝒍 𝒓 + 𝟒𝑲)]𝟐 + (𝟏 + 𝟐𝒍 𝒓 𝒇)(𝟖𝒈𝒉) 𝟐 (𝟏 + 𝟐𝒍 𝒓 𝒇) Where 𝑓 is given by: 𝟏 √𝒇 = 𝟒. 𝟎𝒍𝒐𝒈𝟏𝟎 𝑫 𝒆 + 𝟐. 𝟐𝟖 The derivation of the above formula of velocity can be got from our analysis we did before concerning derivation of the Reynolds number. We can extend the above energy conservation techniques for flow in a siphon and even derive the Darcy flow equation for porous media.
- 59. 58 THEORY OF MOTION OF PARTICLES IN VISCOUS FLUIDS Before we look at modelling a falling sphere, let us first look at a graph of drag coefficient against Reynolds number [1] for a sphere: Consider a falling sphere: The drag force is given by: 𝐹 = 1 2 𝐶𝐷𝐴𝜌𝑉2 Where: 𝐴 = 𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑒𝑑 𝑎𝑟𝑒𝑎 The forces acting on it are shown below:
- 60. 59 𝑊 = 𝑤𝑒𝑖𝑔ℎ𝑡 = 𝑚𝑔 = 𝜌𝑠𝑉0𝑔 𝑈 = 𝑢𝑝𝑡ℎ𝑟𝑢𝑠𝑡 = 𝜌𝑉0𝑔 𝐹𝑑 = 𝑑𝑟𝑎𝑔 𝑓𝑜𝑟𝑐𝑒 = 1 2 𝐶2𝐴𝜌𝑉2 𝐹𝑣 = 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒 = 1 2 𝐶𝑑𝐴𝜌𝑉2 Where: 𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 𝜌𝑠 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑝ℎ𝑒𝑟𝑒 𝑉0 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑝ℎ𝑒𝑟𝑒 𝑉 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑠𝑝ℎ𝑒𝑟𝑒 𝐶2 = 𝑑𝑟𝑎𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑖𝑛 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 𝑓𝑙𝑜𝑤 We say: 𝒎 𝒅𝑽 𝒅𝒕 = 𝑾 − 𝑼 − 𝟏 𝟐 𝑪𝒅𝑨𝝆𝑽𝟐 − 𝟏 𝟐 𝑪𝟐𝑨𝝆𝑽𝟐 As before: 𝐶2 = 𝑑𝑟𝑎𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑖𝑛 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 𝑓𝑙𝑜𝑤 𝐶𝑑 = 𝑑𝑟𝑎𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑖𝑛 𝑙𝑎𝑚𝑖𝑛𝑎𝑟 𝑓𝑙𝑜𝑤 𝐶𝑑 = 24 𝑅𝑒 = 24𝜂 𝜌𝑉𝑑 = 12𝜂 𝜌𝑉𝑟 𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑
- 61. 60 𝐴 = 𝜋𝑟2 For a sphere we shall use 𝐶2 = 0.4 which is the value of 𝐶2 𝑓𝑜𝑟 𝑅𝑒𝑦𝑛𝑜𝑙𝑑 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 500 < 𝑅𝑒𝑑 < 105 As in the diagram above of drag against Reynolds number. 𝑅𝑒 = 𝜌𝑉𝑑 𝜂 𝑚 = 4 3 𝜋𝑟3 𝜌𝑠 𝜌𝑠 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑠𝑝ℎ𝑒𝑟𝑒 Substituting, we get: 𝒎 𝒅𝑽 𝒅𝒕 = 𝒎𝒈 − 𝝆𝑽𝟎𝒈 − 𝟏 𝟐 𝑪𝒅𝑨𝝆𝑽𝟐 − 𝟏 𝟐 𝑪𝟐𝑨𝝆𝑽𝟐 Dividing through by m and multiplying through by 2, we get 𝟐 𝒅𝑽 𝒅𝒕 = 𝟐( 𝝆𝒔 − 𝝆 𝝆𝒔 )𝒈 − 𝟗𝜼 𝒓𝟐𝝆𝒔 𝑽 − 𝟑𝑪𝟐𝝆 𝟒𝒓𝝆𝒔 𝑽𝟐 NB The above differential equation can be solved to get the velocity as a function of time. What happens when the body stops accelerating (i.e., at terminal velocity)? 𝑑𝑉 𝑑𝑡 = 0 We get in steady state (i.e., when the acceleration is zero), we reach terminal velocity 0 = 2( 𝜌𝑠 − 𝜌 𝜌𝑠 )𝑔 − 9𝜂 𝑟2𝜌𝑠 𝑉 − 3𝐶2𝜌 4𝑟𝜌𝑠 𝑉2 3𝐶2𝜌 4𝑟𝜌𝑠 𝑉2 + 9𝜂 𝑟2𝜌𝑠 𝑉 − 2( 𝜌𝑠 − 𝜌 𝜌𝑠 )𝑔 = 0 This is a quadratic formula and the terminal velocity can be got as: 3𝐶2𝜌 2𝑟𝜌𝑠 𝑉 = − 9𝜂 𝑟2𝜌𝑠 + √(( 9𝜂 𝑟2𝜌𝑠 )2 + 6𝐶2𝜌𝑔 𝑟𝜌𝑠 ( 𝜌𝑠 − 𝜌 𝜌𝑠 ))
- 62. 61 𝑉 = − 6𝜂 𝑟𝐶2𝜌 + √( 36𝜂2 𝑟2𝐶2 2 𝜌2 + 8𝑟𝜌𝑠𝑔 3𝐶2𝜌 ( 𝜌𝑠 − 𝜌 𝜌𝑠 )) 𝑽 = − 𝟔𝜼 𝒓𝑪𝟐𝝆 + √( 𝟑𝟔𝜼𝟐 𝒓𝟐𝑪𝟐 𝟐 𝝆𝟐 + 𝟖𝒓(𝝆𝒔 − 𝝆)𝒈 𝟑𝑪𝟐𝝆 ) The above is the terminal velocity. We are going to show that provided some condition is met, the terminal velocity can be either Stoke’s flow or turbulent flow. Coming back to the equation above below: 3𝐶2𝜌 2𝑟𝜌𝑠 𝑉 = − 9𝜂 𝑟2𝜌𝑠 + √(( 9𝜂 𝑟2𝜌𝑠 )2 + 6𝐶2𝜌𝑔 𝑟𝜌𝑠 ( 𝜌𝑠 − 𝜌 𝜌𝑠 )) In the velocity equation above, let us factorize 9𝜂 𝑟2𝜌𝑠 out of the square root and get 3𝐶2𝜌 2𝑟𝜌𝑠 𝑉 = − 9𝜂 𝑟2𝜌𝑠 + 9𝜂 𝑟2𝜌𝑠 √(1 + 2𝐶2𝑔𝜌𝜌𝑠𝑟3 27𝜂2 ( 𝜌𝑠 − 𝜌 𝜌𝑠 )) The governing term 2𝐶2𝑔𝜌𝜌𝑠𝑟3 27𝜂2 ( 𝜌𝑠 − 𝜌 𝜌𝑠 ) If the term below under the square root 2𝐶2𝑔𝜌𝜌𝑠𝑟3 27𝜂2 ( 𝜌𝑠 − 𝜌 𝜌𝑠 ) ≪ 1 We shall arrive at Stoke’s flow. We can use the binomial approximation and get (1 + 𝑥)𝑛 ≈ 1 + 𝑛𝑥 for 𝑥 ≪ 1 Where: 𝑛 = 1 2 𝑥 = 2𝐶2𝑔𝜌𝜌𝑠𝑟3 27𝜂2 ( 𝜌𝑠 − 𝜌 𝜌𝑠 )
- 63. 62 We use the binomial approximation √1 + 𝑥 ≈ 1 + 1 2 𝑥 𝑓𝑜𝑟 𝑥 ≪ 1 √(1 + 2𝐶2𝑔𝜌𝜌𝑠𝑟3 27𝜂2 )( 𝜌𝑠 − 𝜌 𝜌𝑠 ) ≈ (1 + 𝐶2𝑔𝜌𝜌𝑠𝑟3 27𝜂2 ( 𝜌𝑠 − 𝜌 𝜌𝑠 )) Upon substitution in the velocity equation, we get 3𝐶2𝜌 2𝑟𝜌𝑠 𝑉 = − 9𝜂 𝑟2𝜌𝑠 + 9𝜂 𝑟2𝜌𝑠 √(1 + 𝐶2𝑔𝜌𝜌𝑠𝑟3 27𝜂2 ( 𝜌𝑠 − 𝜌 𝜌𝑠 )) 3𝐶2𝜌 2𝑟𝜌𝑠 𝑉 = − 9𝜂 𝑟2𝜌𝑠 + 9𝜂 𝑟2𝜌𝑠 (1 + 𝐶2𝑔𝜌𝜌𝑠𝑟3 27𝜂2 ( 𝜌𝑠 − 𝜌 𝜌𝑠 )) Therefore, upon simplification, the terminal velocity will be 𝑉 = 2 9 𝑟2 𝜌𝑠𝑔 𝜂 ( 𝜌𝑠 − 𝜌 𝜌𝑠 ) 𝑽 = 𝟐 𝟗 𝒓𝟐 𝒈 𝜼 (𝝆𝒔 − 𝝆) Which is Stoke’s flow. Also, if 2𝐶2𝑔𝜌𝜌𝑠𝑟3 27𝜂2 ( 𝜌𝑠 − 𝜌 𝜌𝑠 ) ≫ 1 We can say (1 + 2𝐶2𝑔𝜌𝜌𝑠𝑟3 27𝜂2 ( 𝜌𝑠 − 𝜌 𝜌𝑠 )) ≈ 2𝐶2𝑔𝜌𝜌𝑠𝑟3 27𝜂2 ( 𝜌𝑠 − 𝜌 𝜌𝑠 ) The velocity becomes: 3𝐶2𝜌 2𝑟𝜌𝑠 𝑉 = − 9𝜂 𝑟2𝜌𝑠 + 9𝜂 𝑟2𝜌𝑠 √(1 + 2𝐶2𝑔𝜌𝜌𝑠𝑟3 27𝜂2 ( 𝜌𝑠 − 𝜌 𝜌𝑠 )) 3𝐶2𝜌 2𝑟𝜌𝑠 𝑉 = − 9𝜂 𝑟2𝜌𝑠 + 9𝜂 𝑟2𝜌𝑠 √ 2𝐶2𝑔𝜌𝜌𝑠𝑟3 27𝜂2 ( 𝜌𝑠 − 𝜌 𝜌𝑠 ) Upon simplification, the velocity becomes:
- 64. 63 𝑉 = − 6𝜂 𝑟𝐶2𝜌 + √ 8𝑔𝑟𝜌𝑠 3𝐶2𝜌 ( 𝜌𝑠 − 𝜌 𝜌𝑠 ) 𝑽 = − 𝟔𝜼 𝒓𝑪𝟐𝝆 + √ 𝟖𝒈𝒓 𝟑𝑪𝟐𝝆 (𝝆𝒔 − 𝝆) The above is the terminal velocity in turbulent flow If − 6𝜂 𝑟𝐶2𝜌 ≈ 0 Then the terminal velocity becomes: 𝑽 = √ 𝟖𝒈𝒓(𝝆𝒔 − 𝝆) 𝟑𝑪𝟐𝝆 The above is the terminal velocity in turbulent flow: It can be got by saying: 𝑚𝑔 − 𝑈 = 1 2 𝐶0𝐴𝜌𝑉2 𝑈 = 𝑢𝑝𝑡ℎ𝑟𝑢𝑠𝑡 = 4 3 𝜋𝑟3 𝜌𝑔 4 3 𝜋𝑟3(𝜌𝑠 − 𝜌)𝑔 = 1 2 𝐶0𝜋𝑟2 𝜌𝑉2 Since 𝐶0 = 0.4 In turbulent flow We get 𝑉 = √ 8 3𝐶0 𝑟𝑔 (𝜌𝑠 − 𝜌) 𝜌 LET US SOLVE THE DIFFERENTIAL EQUATION BELOW AS GOT ABOVE; 𝟐 𝒅𝑽 𝒅𝒕 = 𝟐( 𝝆𝒔 − 𝝆 𝝆𝒔 )𝒈 − 𝟗𝜼 𝒓𝟐𝝆𝒔 𝑽 − 𝟑𝑪𝟐𝝆 𝟒𝒓𝝆𝒔 𝑽𝟐 𝑑𝑉 𝑑𝑡 = 𝐴𝑔 − 𝐵𝑉 − 𝐶𝑉2
- 65. 64 Where: 𝐴 = 2( 𝜌𝑠 − 𝜌 𝜌𝑠 ) 𝐵 = 9𝜂 𝑟2𝜌𝑠 𝐶 = 3𝐶2𝜌 4𝑟𝜌𝑠 ∫ 𝑑𝑉 𝐴𝑔 − 𝐵𝑉 − 𝐶𝑉2 = 1 2 ∫ 𝑑𝑡 ∫ 𝑑𝑉 −𝐶(𝑉2 + 𝐵 𝐶 𝑉 − 𝐴 𝐶 𝑔) = 1 2 ∫ 𝑑𝑡 ∫ 𝑑𝑉 𝑉2 + 𝐵 𝐶 𝑉 − 𝐴 𝐶 𝑔 = − 𝐶 2 ∫ 𝑑𝑡 Let 𝐵 𝐶 = 𝑚 𝐴 𝐶 𝑔 = 𝑛 ∫ 𝑑𝑉 𝑉2 + 𝑚𝑉 − 𝑛 = − 𝐶 2 ∫ 𝑑𝑡 𝑉2 + 𝑚𝑉 − 𝑛 = (𝑉 + 𝑚 2 )2 − 𝑚2 4 − 𝑛 = (𝑉 + 𝑚 2 )2 − ( 𝑚2 4 + 𝑛) ∫ 𝑑𝑉 (𝑉 + 𝑚 2 )2 − ( 𝑚2 4 + 𝑛) = − 𝐶 2 ∫ 𝑑𝑡 Let 𝑃 = ( 𝑚2 4 + 𝑛) ∫ 𝑑𝑉 (𝑉 + 𝑚 2 )2 − (√𝑃)2 = − 𝐶 2 ∫ 𝑑𝑡 ∫ 𝑑𝑉 (𝑉 + 𝑚 2 − √𝑃)(𝑉 + 𝑚 2 + √𝑃) = − 𝐶 2 ∫ 𝑑𝑡
- 66. 65 1 (𝑉 + 𝑚 2 − √𝑃)(𝑉 + 𝑚 2 + √𝑃) = 𝐿 (𝑉 + 𝑚 2 − √𝑃) + 𝐾 (𝑉 + 𝑚 2 + √𝑃) 𝐿 = 1 2√𝑃 𝐾 = −1 2√𝑃 1 2√𝑃 ∫ 𝑑𝑉 (𝑉 + 𝑚 2 − √𝑃) − 1 2√𝑃 ∫ 𝑑𝑉 (𝑉 + 𝑚 2 + √𝑃) = − 𝐶 2 ∫ 𝑑𝑡 ln(𝑉 + 𝑚 2 − √𝑃) − 𝑙𝑛 (𝑉 + 𝑚 2 + √𝑃) = −𝐶√𝑃𝑡 + 𝐷 𝐷 is an integration constant 𝑎𝑡 𝑡 = 0 , 𝑉 = 0 Upon substitution, we get ln ( 𝑚 2 − √𝑃 𝑚 2 + √𝑃 ) = 𝐷 The velocity equation becomes: ln[( 𝑚 2 + √𝑃 𝑚 2 − √𝑃 )( 𝑉 + 𝑚 2 − √𝑃 𝑉 + 𝑚 2 + √𝑃 )] = −𝐶√𝑃𝑡 𝑚 = 𝐵 𝐶 = 6𝜂 𝑟𝐶2𝜌 𝐴 = 2( 𝜌𝑠 − 𝜌 𝜌𝑠 ) 𝐶 = 3𝐶2𝜌 4𝑟𝜌𝑠 𝑃 = ( 𝑚2 4 + 𝑛) 𝑛 = 𝐴 𝐶 𝑔 = 8𝑟𝑔 3𝐶2𝜌 (𝜌𝑠 − 𝜌) Therefore
- 67. 66 𝑃 = 36𝜂2 𝑟2𝐶2 2 𝜌2 + 8𝑟𝑔(𝜌𝑠 − 𝜌) 3𝐶2𝜌 Upon substitution, the velocity becomes ln[( 𝑚 2 + √𝑃 𝑚 2 − √𝑃 )( 𝑉 + 𝑚 2 − √𝑃 𝑉 + 𝑚 2 + √𝑃 )] = −𝐶√𝑃𝑡 ln[ ( 6𝜂 𝑟𝐶2𝜌 + √ 36𝜂2 𝑟2𝐶2 2 𝜌2 + 8𝑟𝑔(𝜌𝑠 − 𝜌) 3𝐶2𝜌 6𝜂 𝑟𝐶2𝜌 − √ 36𝜂2 𝑟2𝐶2 2 𝜌2 + 8𝑟𝑔(𝜌𝑠 − 𝜌) 3𝐶2𝜌 )( 𝑉 + 6𝜂 𝑟𝐶2𝜌 − √ 36𝜂2 𝑟2𝐶2 2 𝜌2 + 8𝑟𝑔(𝜌𝑠 − 𝜌) 3𝐶2𝜌 𝑉 + 6𝜂 𝑟𝐶2𝜌 + √ 36𝜂2 𝑟2𝐶2 2 𝜌2 + 8𝑟𝑔(𝜌𝑠 − 𝜌) 3𝐶2𝜌 ) ] = − 3𝐶2𝜌 4𝑟𝜌𝑠 √ 36𝜂2 𝑟2𝐶2 2 𝜌2 + 8𝑟𝑔(𝜌𝑠 − 𝜌) 3𝐶2𝜌 𝑡 ( 𝟔𝜼 𝒓𝑪𝟐𝝆 + √ 𝟑𝟔𝜼𝟐 𝒓𝟐𝑪𝟐 𝟐 𝝆𝟐 + 𝟖𝒓𝒈(𝝆𝒔 − 𝝆) 𝟑𝑪𝟐𝝆 𝟔𝜼 𝒓𝑪𝟐𝝆 − √ 𝟑𝟔𝜼𝟐 𝒓𝟐𝑪𝟐 𝟐 𝝆𝟐 + 𝟖𝒓𝒈(𝝆𝒔 − 𝝆) 𝟑𝑪𝟐𝝆 )( 𝑽 + 𝟔𝜼 𝒓𝑪𝟐𝝆 − √ 𝟑𝟔𝜼𝟐 𝒓𝟐𝑪𝟐 𝟐 𝝆𝟐 + 𝟖𝒓𝒈(𝝆𝒔 − 𝝆) 𝟑𝑪𝟐𝝆 𝑽 + 𝟔𝜼 𝒓𝑪𝟐𝝆 + √ 𝟑𝟔𝜼𝟐 𝒓𝟐𝑪𝟐 𝟐 𝝆𝟐 + 𝟖𝒓𝒈(𝝆𝒔 − 𝝆) 𝟑𝑪𝟐𝝆 ) = 𝒆 − 𝟑𝑪𝟐𝝆 𝟒𝒓𝝆𝒔 √ 𝟑𝟔𝜼𝟐 𝒓𝟐𝑪𝟐 𝟐𝝆𝟐 + 𝟖𝒓𝒈(𝝆𝒔−𝝆) 𝟑𝑪𝟐𝝆 𝒕 The velocity can be got by making V the subject of the formula above. 𝐴𝑡 𝑡 = ∞ 𝑜𝑟 𝑎𝑡 𝑠𝑡𝑒𝑎𝑑𝑦 𝑠𝑡𝑎𝑡𝑒 When the exponential term below 3𝐶2𝜌 4𝑟𝜌𝑠 √ 36𝜂2 𝑟2𝐶2 2 𝜌2 + 8𝑟𝑔(𝜌𝑠 − 𝜌) 3𝐶2𝜌 𝑡 ≈ ∞ The exponential becomes zero Since 𝑒−∞ = 0 and we get 𝑉 + 6𝜂 𝑟𝐶2𝜌 − √ 36𝜂2 𝑟2𝐶2 2 𝜌2 + 8𝑟𝑔(𝜌𝑠 − 𝜌) 3𝐶2𝜌 = 0 𝑉 = − 6𝜂 𝑟𝐶2𝜌 + √ 36𝜂2 𝑟2𝐶2 2 𝜌2 + 8𝑟𝑔(𝜌𝑠 − 𝜌) 3𝐶2𝜌 Which is what we got before as the terminal velocity. Knowing the velocity at a particular depth h, we can get the time taken to fall to depth h. Or
- 68. 67 We can make velocity the subject of the formula in the expression above of velocity as a function of time and then integrate knowing that 𝑑𝑥 𝑑𝑡 = 𝑉 To get 𝑥 as a function of time t. Similarly, we can use energy conservation techniques to get the velocity as a function of height h and then using the expression above, we can tell the time taken to achieve a particular velocity or height h. This is what we are going to do below: Consider a falling sphere: If there were viscous effects in an unbounded medium, we conserve energy changes and say: 𝐏𝐨𝐭𝐞𝐧𝐭𝐢𝐚𝐥 𝐞𝐧𝐞𝐫𝐠𝐲 𝐜𝐡𝐚𝐧𝐠𝐞 = 𝐊𝐢𝐧𝐞𝐭𝐢𝐜 𝐞𝐧𝐞𝐫𝐠𝐲 𝐠𝐚𝐢𝐧𝐞𝐝 + 𝐰𝐨𝐫𝐤 𝐝𝐨𝐧𝐞 𝐚𝐠𝐚𝐢𝐧𝐬𝐭 𝐯𝐢𝐬𝐜𝐨𝐮𝐬 𝐟𝐨𝐫𝐜𝐞𝐬 𝐚𝐧𝐝 𝐮𝐩𝐭𝐡𝐫𝐮𝐬𝐭 𝑚𝑔ℎ = 1 2 𝑚𝑉2 + 𝜌𝑉0𝑔𝑙 + 1 2 𝐶𝑑𝐴𝜌𝑉2 × 𝑙 + 1 2 𝐶2𝐴𝜌𝑉2 × 𝑙 𝑉0 = 4 3 𝜋𝑟3 Where: 𝐶2 = 𝑑𝑟𝑎𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑖𝑛 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 𝑓𝑙𝑜𝑤 = 0.4 𝐶𝑑 = 𝑑𝑟𝑎𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑖𝑛 𝑙𝑎𝑚𝑖𝑛𝑎𝑟 𝑓𝑙𝑜𝑤 𝐶𝑑 = 24 𝑅𝑒 = 24𝜂 𝜌𝑉𝑑 = 12𝜂 𝜌𝑉𝑟
- 69. 68 𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝑙 = ℎ 𝒉 is the vertical depth below the point of release 𝐴 = 𝜋𝑟2 𝐶2 = 𝑑𝑟𝑎𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑖𝑛 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 𝑓𝑙𝑜𝑤 For a sphere 𝐶2 = 0.4 Where: 𝑅𝑒 = 𝜌𝑉𝑑 𝜂 𝑚𝑔ℎ = 1 2 𝑚𝑉2 + 𝜌𝑉0𝑔ℎ + 1 2 𝐶𝑑𝐴𝜌𝑉2 × ℎ + 1 2 𝐶2𝐴𝜌𝑉2 × ℎ 𝑚 = 4 3 𝜋𝑟3 𝜌𝑠 𝜌𝑠 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑠𝑝ℎ𝑒𝑟𝑒 𝐴 = 𝜋𝑟2 Substituting, we get: 4 3 𝜋𝑟3(𝜌𝑠 − 𝜌)𝑔ℎ = 1 2 × 4 3 𝜋𝑟3 𝜌𝑠𝑉2 + 1 2 × 12𝜇 𝜌𝑉𝑟 × 𝜋𝑟2 𝜌𝑉2 × ℎ + 1 2 𝐶2𝜋𝑟2 𝜌𝑉2 × ℎ Simplifying we get 𝑉2 (1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ) + 9𝜂ℎ 𝑟2𝜌𝑠 𝑉 − 2𝑔( 𝜌𝑠 − 𝜌 𝜌𝑠 )ℎ = 0 In the expression above, if h is large such that 1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ ≈ 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ We get 𝑉2 ( 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ) + 9𝜂ℎ 𝑟2𝜌𝑠 𝑉 − 2𝑔( 𝜌𝑠 − 𝜌 𝜌𝑠 )ℎ = 0 And get 𝑉2 ( 3𝐶2𝜌 4𝑟𝜌𝑠 ) + 9𝜂 𝑟2𝜌𝑠 𝑉 − 2𝑔( 𝜌𝑠 − 𝜌 𝜌𝑠 ) = 0
- 70. 69 The above is a quadratic equation and the velocity V got will be independent of height h hence it will be the terminal velocity as got before. 𝑽 = − 𝟔𝜼 𝒓𝑪𝟐𝝆 + √( 𝟑𝟔𝜼𝟐 𝒓𝟐𝑪𝟐 𝟐 𝝆𝟐 + 𝟖𝒓(𝝆𝒔 − 𝝆)𝒈 𝟑𝑪𝟐𝝆 ) Okay now coming back to 𝑉2 (1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ) + 9𝜂ℎ 𝑟2𝜌𝑠 𝑉 − 2𝑔( 𝜌𝑠 − 𝜌 𝜌𝑠 )ℎ = 0 The above is a quadratic formula and the solution is: 2 (1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ) 𝑉 = − 9𝜂ℎ 𝑟2𝜌𝑠 + √(( 9𝜂ℎ 𝑟2𝜌𝑠 )2 + 8𝑔ℎ( 𝜌𝑠 − 𝜌 𝜌𝑠 )(1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ))… … . . 𝑴 𝑽 = − 𝟗𝜼𝒉 𝟐𝒓𝟐𝝆𝒔 (𝟏 + 𝟑𝑪𝟐𝝆 𝟒𝒓𝝆𝒔 𝒉) + √(( 𝟗𝜼𝒉 𝒓𝟐𝝆𝒔 )𝟐 + 𝟖𝒈( 𝝆𝒔 − 𝝆 𝝆𝒔 )𝒉(𝟏 + 𝟑𝑪𝟐𝝆 𝟒𝒓𝝆𝒔 𝒉)) 𝟐 (𝟏 + 𝟑𝑪𝟐𝝆 𝟒𝒓𝝆𝒔 𝒉) The above is the velocity of a sphere in a viscous fluid at depth h from the initial point h is the vertical depth from the point of release. Laminar flow occurs when 𝑅𝑒 < 𝑅𝑒𝑐𝑟 Where: 𝑅𝑒𝑐𝑟 is the critical Reynolds number below which laminar flow acts We shall calculate the value of 𝑅𝑒𝑐𝑟 in the text to follow. For laminar or Stokes’s flow 𝑉 = 2 9 𝑟2 (𝜌𝑠 − 𝜌)𝑔 𝜂 𝑅𝑒 = 𝜌𝑉𝑑 𝜂 = 𝜌 × 2 9 𝑟2 (𝜌𝑠 − 𝜌)𝑔 𝜂 𝜂 × 2𝑟 < 𝑅𝑒𝑐𝑟 Therefore
- 71. 70 𝟒𝒓𝟑 𝝆(𝝆𝒔 − 𝝆)𝒈 𝟗𝜼𝟐𝑹𝒆𝒄𝒓 < 𝟏 That is the condition for laminar flow or Stoke’s flow Going back to equation M and factorizing out (( 9𝜂ℎ 𝑟2𝜌𝑠 )2 we get 2 (1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ)𝑉 = − 9𝜂ℎ 𝑟2𝜌𝑠 + 9𝜂ℎ 𝑟2𝜌𝑠 √(1 + 8𝑔( 𝜌𝑠 − 𝜌 𝜌𝑠 )ℎ(1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ)( 𝑟2𝜌𝑠 9𝜂ℎ )2) If the term 8𝑔( 𝜌𝑠 − 𝜌 𝜌𝑠 )ℎ( 𝑟2 𝜌𝑠 9𝜂ℎ )2 (1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ) ≪ 1 Is very small, we can use the approximation (1 + 𝑥)𝑛 ≈ 1 + 𝑛𝑥 for 𝑥 ≪ 1 I.e., if 8𝑔 ℎ ( 𝜌𝑠 − 𝜌 𝜌𝑠 )( 𝑟2 𝜌𝑠 9𝜂 )2 (1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ) ≪ 1 And if 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ > 1 So that (1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ) ≈ 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ We get 8𝑔 ℎ ( 𝜌𝑠 − 𝜌 𝜌𝑠 )( 𝑟2 𝜌𝑠 9𝜂 )2 ( 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ) ≪ 1 And get 𝟐𝑪𝟐𝒓𝟑 𝝆(𝝆𝒔 − 𝝆)𝒈 𝟐𝟕𝜼𝟐 < 𝟏 Comparing with the condition for laminar flow derived before 𝟒𝒓𝟑 𝝆(𝝆𝒔 − 𝝆)𝒈 𝟗𝜼𝟐𝑹𝒆𝒄𝒓 < 𝟏 We get
- 72. 71 4 9𝑅𝑒𝑐𝑟 = 2𝐶2 27 Substituting 𝐶2 = 0.4 We get 𝑹𝒆𝒄𝒓 = 𝟏𝟓 The implication is that the critical Reynolds number for laminar flow is 15 The governing number of falling for a sphere is: 𝑵𝒖𝒎𝒃𝒆𝒓 = 𝟖𝒈 𝒉 ( 𝝆𝒔 − 𝝆 𝝆𝒔 )(𝟏 + 𝟑𝑪𝟐𝝆 𝟒𝒓𝝆𝒔 𝒉)( 𝒓𝟐 𝝆𝒔 𝟗𝜼 )𝟐 Or 𝑵𝒖𝒎𝒃𝒆𝒓 = 𝟒𝒈(𝝆𝒔 − 𝝆)𝒓𝟑 𝜼𝟐 [ 𝟐𝒓𝝆𝒔 𝟖𝟏𝒉 + 𝑪𝟐𝝆 𝟓𝟒 ] If 4𝑔(𝜌𝑠 − 𝜌)𝑟3 𝜂2 [ 2𝑟𝜌𝑠 81ℎ + 𝐶2𝜌 54 ] ≪ 1 2 (1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ)𝑉 = − 9𝜂ℎ 𝑟2𝜌𝑠 + 9𝜂ℎ 𝑟2𝜌𝑠 √(1 + 8𝑔( 𝜌𝑠 − 𝜌 𝜌𝑠 )ℎ(1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ)( 𝑟2𝜌𝑠 9𝜂ℎ )2) We use the binomial approximation √1 + 𝑥 ≈ 1 + 1 2 𝑥 𝑓𝑜𝑟 𝑥 ≪ 1 √(1 + 8𝑔( 𝜌𝑠 − 𝜌 𝜌𝑠 )ℎ(1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ)( 𝑟2𝜌𝑠 9𝜂ℎ )2) ≈ (1 + 4𝑔( 𝜌𝑠 − 𝜌 𝜌𝑠 )ℎ(1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ)( 𝑟2 𝜌𝑠 9𝜂ℎ )2 ) And get 2(1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ)𝑉 = − 9𝜂ℎ 𝑟2𝜌𝑠 + 9𝜂ℎ 𝑟2𝜌𝑠 (1 + 4𝑔( 𝜌𝑠 − 𝜌 𝜌𝑠 )ℎ(1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ)( 𝑟2 𝜌𝑠 9𝜂ℎ )2 ) Making the substitution 1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ ≈ 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ we get 2 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ𝑉 = − 9𝜂ℎ 𝑟2𝜌𝑠 + 9𝜂ℎ 𝑟2𝜌𝑠 (1 + 4𝑔( 𝜌𝑠 − 𝜌 𝜌𝑠 )ℎ 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ( 𝑟2 𝜌𝑠 9𝜂ℎ )2 )
- 73. 72 𝑽 = 𝟐 𝟗 𝒓𝟐 (𝝆𝒔 − 𝝆)𝒈 𝜼 𝒅𝒉 𝒅𝒕 = 𝟐 𝟗 𝒓𝟐 (𝝆𝒔 − 𝝆)𝒈 𝜼 Using the number below: 𝑵𝒖𝒎𝒃𝒆𝒓 = 𝟒𝒈(𝝆𝒔 − 𝝆)𝒓𝟑 𝜼𝟐 [ 𝟐𝒓𝝆𝒔 𝟖𝟏𝒉 + 𝑪𝟐𝝆 𝟓𝟒 ] We can tell when Stoke’s flow or laminar flow begins by substituting the changing increasing value of h in the number above until h is such that the number is far less than one and then there, we can say the sphere is in laminar flow. Also given a fixed height h for example a fluid in a container, we can determine the radius and density of the sphere for which Stoke’s flow will be observed. To get the time taken to reach Stoke’s flow, we can integrate the velocity equation below: 𝑽𝟏 = − 𝟗𝜼𝒉 𝟐𝒓𝟐𝝆𝒔 (𝟏 + 𝟑𝑪𝟐𝝆 𝟒𝒓𝝆𝒔 𝒉) + √(( 𝟗𝜼𝒉 𝒓𝟐𝝆𝒔 )𝟐 + 𝟖𝒈( 𝝆𝒔 − 𝝆 𝝆𝒔 )𝒉(𝟏 + 𝟑𝑪𝟐𝝆 𝟒𝒓𝝆𝒔 𝒉)) 𝟐 (𝟏 + 𝟑𝑪𝟐𝝆 𝟒𝒓𝝆𝒔 𝒉) As 𝑑ℎ 𝑑𝑡 = 𝑉1 From an initial height to a height when Stoke’s flow begins or we can use another simpler method as will be shown later. After that time on to afterwards, the sphere will undergo terminal velocity as: 𝑽 = 𝟐 𝟗 𝒓𝟐 (𝝆𝒔 − 𝝆)𝒈 𝜼 𝒅𝒉 𝒅𝒕 = 𝟐 𝟗 𝒓𝟐 (𝝆𝒔 − 𝝆)𝒈 𝜼 This is the formula for terminal velocity of a sphere i.e., Stoke’s law for laminar flow/fall. The integration of the above velocity equation is difficult, so we shall see an alternative method later in the text later.
- 74. 73 Also, when 8𝑔 ℎ ( 𝜌𝑠 − 𝜌 𝜌𝑠 )(1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ)( 𝑟2 𝜌𝑠 9𝜂 )2 ≫ 1 Or when 𝟒𝒈(𝝆𝒔 − 𝝆)𝒓𝟑 𝜼𝟐 [ 𝟐𝒓𝝆𝒔 𝟖𝟏𝒉 + 𝑪𝟐𝝆 𝟓𝟒 ] ≫ 𝟏 Then from 2 (1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ)𝑉 = − 9𝜂ℎ 𝑟2𝜌𝑠 + 9𝜂ℎ 𝑟2𝜌𝑠 √(1 + 8𝑔( 𝜌𝑠 − 𝜌 𝜌𝑠 )ℎ(1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ)( 𝑟2𝜌𝑠 9𝜂ℎ )2) 1 + 8𝑔( 𝜌𝑠 − 𝜌 𝜌𝑠 )ℎ(1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ)( 𝑟2 𝜌𝑠 9𝜂ℎ )2 ≈ 8𝑔 ℎ ( 𝜌𝑠 − 𝜌 𝜌𝑠 )(1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ)( 𝑟2 𝜌𝑠 9𝜂 )2 Upon substitution, we get; 2 (1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ) 𝑉 = − 9𝜂ℎ 𝑟2𝜌𝑠 + 9𝜂ℎ 𝑟2𝜌𝑠 √(8𝑔( 𝜌𝑠 − 𝜌 𝜌𝑠 )ℎ(1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ)( 𝑟2𝜌𝑠 9𝜂ℎ )2) 2𝑉 = − 9𝜂ℎ 𝑟2𝜌𝑠 (1 + 3𝐶2𝜌 2𝑟𝜌𝑠 ℎ) + √ (8𝑔ℎ) (1 + 3𝐶2𝜌 2𝑟𝜌𝑠 ℎ) ( 𝜌𝑠 − 𝜌 𝜌𝑠 ) 𝑽 = −𝟗𝜼𝒉 𝟐𝒓𝟐𝝆𝒔 (𝟏 + 𝟑𝑪𝟐𝝆 𝟒𝒓𝝆𝒔 𝒉) + √ 𝟐𝒈𝒉 (𝟏 + 𝟑𝑪𝟐𝝆 𝟒𝒓𝝆𝒔 𝒉) ( 𝝆𝒔 − 𝝆 𝝆𝒔 ) When h is large such that 1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ ≈ 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ The velocity becomes: 𝑉 = −9𝜂ℎ 2𝑟2𝜌𝑠 ( 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ) + √ 2𝑔ℎ ( 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ) ( 𝜌𝑠 − 𝜌 𝜌𝑠 )
- 75. 74 𝑉 = −9𝜂 2𝑟2𝜌𝑠 ( 3𝐶2𝜌 4𝑟𝜌𝑠 ) + √ 2𝑔 ( 3𝐶2𝜌 4𝑟𝜌𝑠 ) ( 𝜌𝑠 − 𝜌 𝜌𝑠 ) … … 𝑳 𝑉 = −6𝜂 𝐶2𝑟𝜌 + √ 8 3𝐶2 𝑟𝑔 𝜌𝑠 𝜌 ( 𝜌𝑠 − 𝜌 𝜌𝑠 ) 𝑽 = −𝟔𝜼 𝑪𝟐𝒓𝝆 + √ 𝟖 𝟑𝑪𝟐 𝒓𝒈 (𝝆𝒔 − 𝝆) 𝝆 The above velocity is the terminal velocity reached which is what we got before for turbulent flow. If −6𝜂 𝐶2𝑟𝜌 ≪ 1 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙 Then −6𝜂 𝐶2𝑟𝜌 ≈ 0 Then we get 𝑽 = √ 𝟖 𝟑𝑪𝟐 𝒓𝒈 (𝝆𝒔 − 𝝆) 𝝆 Comparing with the governing equation for turbulent flow drag, So Comparing with 𝑉 = √ 8 3𝐶2 𝑟𝑔 (𝜌𝑠 − 𝜌) 𝜌 𝐶0 = 𝐶2 So, we have proved that 𝐶2 is the drag coefficient in turbulent flow. Again, we can use the number:
- 76. 75 𝑵𝒖𝒎𝒃𝒆𝒓 = 𝟒𝒈(𝝆𝒔 − 𝝆)𝒓𝟑 𝜼𝟐 [ 𝟐𝒓𝝆𝒔 𝟖𝟏𝒉 + 𝑪𝟐𝝆 𝟓𝟒 ] And substitute in the increasing value of h and then determine the point h when the number will be far greater than 1 and also when 1 + 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ ≈ 3𝐶2𝜌 4𝑟𝜌𝑠 ℎ . At this point, terminal velocity will be reached and from that point afterwards, the sphere will obey 𝑑ℎ 𝑑𝑡 = 𝑉𝑇 Where 𝑉𝑇 = 𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 This is the equation for turbulent flow for high Reynolds number Generally, the equation of velocity is: 𝑽 = − 𝟗𝜼𝒉 𝟐𝒓𝟐𝝆𝒔 (𝟏 + 𝟑𝑪𝟐𝝆 𝟒𝒓𝝆𝒔 𝒉) + √(( 𝟗𝜼𝒉 𝒓𝟐𝝆𝒔 )𝟐 + 𝟖𝒈( 𝝆𝒔 − 𝝆 𝝆𝒔 )𝒉(𝟏 + 𝟑𝑪𝟐𝝆 𝟒𝒓𝝆𝒔 𝒉)) 𝟐 (𝟏 + 𝟑𝑪𝟐𝝆 𝟒𝒓𝝆𝒔 𝒉) The equation above also works for transition flow also which is in-between laminar and turbulent flow. The equation above can be integrated from an initial height ℎ0 to a given height h and the time taken for the sphere to fall can be found as 𝑑ℎ 𝑑𝑡 = 𝑉3 The integration would be difficult but we can use the method below: Recall we got the velocity as a function of time as: 𝐥𝐧[ ( 𝟔𝜼 𝒓𝑪𝟐𝝆 + √ 𝟑𝟔𝜼𝟐 𝒓𝟐𝑪𝟐 𝟐 𝝆𝟐 + 𝟖𝒓𝒈(𝝆𝒔 − 𝝆) 𝟑𝑪𝟐𝝆 𝟔𝜼 𝒓𝑪𝟐𝝆 − √ 𝟑𝟔𝜼𝟐 𝒓𝟐𝑪𝟐 𝟐 𝝆𝟐 + 𝟖𝒓𝒈(𝝆𝒔 − 𝝆) 𝟑𝑪𝟐𝝆 ) ( 𝑽 + 𝟔𝜼 𝒓𝑪𝟐𝝆 − √ 𝟑𝟔𝜼𝟐 𝒓𝟐𝑪𝟐 𝟐 𝝆𝟐 + 𝟖𝒓𝒈(𝝆𝒔 − 𝝆) 𝟑𝑪𝟐𝝆 𝑽 + 𝟔𝜼 𝒓𝑪𝟐𝝆 + √ 𝟑𝟔𝜼𝟐 𝒓𝟐𝑪𝟐 𝟐 𝝆𝟐 + 𝟖𝒓𝒈(𝝆𝒔 − 𝝆) 𝟑𝑪𝟐𝝆 ) ] = − 𝟑𝑪𝟐𝝆 𝟒𝒓𝝆𝒔 √ 𝟑𝟔𝜼𝟐 𝒓𝟐𝑪𝟐 𝟐 𝝆𝟐 + 𝟖𝒓𝒈(𝝆𝒔 − 𝝆) 𝟑𝑪𝟐𝝆 𝒕 Knowing the velocity as a function of h as above, we can substitute the known velocity at height h and then tell the time taken to reach that velocity (or height) from the equation above of velocity against time. If we were working in a vacuum so that 𝝆 = 𝟎 and 𝜼 = 𝟎 , we get
- 77. 76 𝑚 𝑑𝑉 𝑑𝑡 = 𝑚𝑔 𝒅𝑽 𝒅𝒕 = 𝒈 Which is independent of the body dimensions. So, in a vacuum, bodies will fall at the same rate. We can also calculate the velocity when the gravity is varying using: 𝒈 = √( 𝑮𝑴 𝒓𝟐 )
- 78. 77 REFERENCES [1] C. E. R. E. G. L. James R.Welty, "DRAG," in Fundamentals of Momentum, Heat and Mass Transfer, Oregon, John Wiley & Sons, Inc., 2008, p. 141. [2] Chegg, "Chegg," Chegg, 2022. [Online]. Available: https://www.chegg.com/homework- help/questions-and-answers/class-showed-hagen-poiseuille-equation-analytical-expression- pressure-drop-laminar-flow-re-q18503972. [Accessed 11 4 2022].