CP502 Advanced Fluid Mechanics

1. CP502 Advanced Fluid Mechanics Incompressible Flow of Viscous Fluids Set 06: Working with Non-Newtonian Fluids

2. Velocity profile of a power-law fluid Shear stress-strain relationship: Shear stress-pressure relationship: Combination gives: (1)

3. Velocity profile of a power-law fluid Rearranging (1): Integrating from a distance r to wall at r=R: (2)

4. Volumetric flow rate of a power-law fluid Using (2) in (3)

5. Average velocity in a power-law fluid Using (3) in (4)

6. Fanning friction factor for power-law fluids (4) Can be rearranged to give: (5)

7. Fanning friction factor for power-law fluids Using (5) in the definition of friction factor (6) where (7) is the Generalized Reynolds number

8. Generalized Reynolds number (7) is written in both the equivalent forms given below: Substituting n = 1 and K = μ in the above, we get the Reynolds number as follows:

9. Pumping requirement of non-Newtonian fluids (8) For laminar flow and shear-thinning liquids, For turbulent flow,

10. Pumping requirement of non-Newtonian fluids (8) For laminar flow, For turbulent flow,fis obtained from the figure on next slide.

11. Pumping requirement of non-Newtonian fluids

12. Pumping requirement of non-Newtonian fluids (8)

13. Pumping requirement of non-Newtonian fluids For non-Newtonian fluids above a generalized Re of 500, use data for Newtonian fluids in turbulent flow: from appropriate tables.

14. Pumping requirement of non-Newtonian fluids For non-Newtonian fluids, for the range 20 < NGRe < 500, use the following:

15. Pumping requirement of non-Newtonian fluids

16. Example A non-Newtonian fluid is being pumped from one tank to another in a 0.0348 m diameter pipe with a mass flow rate of 1.97 kg/s. The total length of pipe between the tanks is 10 m. The difference of elevation from inlet to outlet is 3 m. The fittings include three long-radius 90o flanged elbows, and one fully open angle valve. A filter present in the pipeline causes 100 kPa pressure drop. Set up a spread sheet to calculate the pumping requirements. The properties of the fluid are as follows: density = 1250 kg/m3, consistency coefficient = 5.2 Pa.sn, flow behaviour index = 0.45 Ans: 549.04 W

17. Velocity profile of a Herschel–Bulkley fluid Shear stress-strain relationship: Shear stress-pressure relationship: Combination gives: (1a)

18. Rearranging (1a): Integrating from a distance r to wall at r=R:

19. (2a)

20. Problem 7.1: In a pasteurization treatment of fluid foods, devices in which the fluid circulates within a tube at the treatment temperature are usually employed. To ensure a satisfactory pasteurization, it is necessary that the microorganisms that circulate at the maximum velocity remain long enough in order to receive an adequate thermal treatment. A fluid food that has a density of 1250 kg/m3 circulates through a 26.7 mm internal diameter (3/4 in nominal diameter) with a 10,000 kg/h mass flow rate. Calculate the value of maximum circulation velocity for the following two cases: (a) Clarified juice at 45o Brix peach juice with a viscosity of 9 mPa.s (b) Egg yolk that presents a power law fluid behaviour, with K = 880 mPa.sn and n = 0.20. Source: Ibarz and Barbosa-Cánovas, Unit Operations in Food Engineering

21. Data provided:  = density = 1250 kg/m3 d = diameter = 26.7 mm = 26.7/1000 m w = mass flow rate = 10,000 kg/h Assignment: vmax = maximum velocity = ? Calculation starts: vm = mean velocity = mass flow rate / (cross-sectional area x density) = 3.97 m/s

22. Clarified juice at 45o Brix peach juice with μ = 9 mPa.s • Given is a Newtonian fluid • Re = vmd / μ = 14,718 • That means, the given flow is turbulent. • The ratio (vm / vmax) could be found using equation (7.29): • where c is given as follows: • Since Re = 14,718, we could use c = 7 in equation (7.29) Therefore, vm / vmax = 0.82, which leads to vmax = vm / 0.82 = 4.86 m/s

23. Alternatively Figure 7.8 can be used to get the maximum velocity as suggested in the text. vm / vmax = 0.78, which leads to vmax = vm / 0.78 = 5.09 m/s Note: vmaxtakes slightly different values in the alternative approaches, which is expected considering the approximate nature of the equation used and the graph read.

24. (b) Egg yolk that presents a power law fluid behaviour, with K = 880 mPa.sn and n = 0.20. Given is a power law fluid ReG = = 37,828 That means, the given flow is turbulent. The ratio (vm / vmax) could be found using Figure 7.9.

25. Reading at ReG = 37,828 and n = 0.2 in Figure 7.9, we get vm / vmax = 0.92, which leads to vmax = vm / 0.92 = 4.31 m/s

26. Problem 7.2: A fluid food whose density is 1200 kg/m3 circulates at 25oC by a 5-cm diameter pipe with a mass flow rate of 5000 kg/h. Determine the kinetic energy flow that the fluid transports if the following fluids circulate: (a) Concentrated peach juice at 69oBrix with a viscosity of 324 mPa.s (b) Nonclarifiedrasberry juice (without pectin elimination) of 41oBrix that has a pseudoplasticbehaviour, with n = 0.73 and K = 1.6 Pa.sn (c) Apple puree that presents a Herschel-Bulkleybehaviour, with n = 0.47, K = 5.63 Pa.sn and yield stress = 58.6 Pa. In regard to transport conditions, a relationship between the yield stress and the shear stress on the wall of 0.2 can be supposed. (d) Mayonnaise that behaves as a Bingham’s plastic with μ’ = 0.63 Pa.sn and yield stress = 85 Pa. In this case, c = 0.219 Source: Ibarz and Barbosa-Cánovas, Unit Operations in Food Engineering

27. Data provided:  = density = 1200 kg/m3 d = diameter = 5 cm = 5/100 m w = mass flow rate = 5,000 kg/h Assignment: KE = kinetic energy = Calculation starts: average velocity = mass flow rate / (cross-sectional area x density) What is the dimensionless correction factor ?

28. Concentrated peach juice at 69oBrix with μ = 324mPa.s • Given is a Newtonian fluid • Re = vmd / μ= (0.59 x 1200 x 0.05)/0.324 = 109 • That means, the given flow is laminar. • Therefore,

29. (b) Nonclarified rasberry juice of 41o Brix that has a pseudoplastic behaviour, with n = 0.73 and K = 1.6 Pa.sn • Given is a power law fluid • ReG = (equation 7.7) = 70.8 • ReG,critical = (depends on n; equation 7.8) =2262 • ReG < ReG,critical (laminar flow) • alpha = (equation 7.54) = (2n+1)(5n+3)/[3(3n+1)2] = 0.536 • for power law fluids in laminar flow • KE = 1623.6 J/h

30. (c) Apple puree that presents a Herschel-Bulkley behaviour, with n = 0.47, K = 5.63 Pa.sn and yield stress = 58.6 Pa. In regard to transport conditions, a relationship between the yield stress and the shear stress on the wall of 0.2 can be supposed. • Given is a Herschel-Bulkley fluid • ReG = (equation 7.7) = 62.3 • ReG,critical (Figure 7.4) depends on the generalized Hedstrom number (HeG) and the flow index • HeG = (equation 7.12b) = 1093.6 • ReG,critical = 2200 (approximately)

31. ReG < ReG,critical (laminar flow) • alpha = (equation 7.56 or Figure 7.12) = 0.62 to 0.64 • for Herschel-Bulkley fluid in laminar flow • KE = 1351.3 J/h

32. (d) Mayonnaise that behaves as a Bingham’s plastic with eta’ = 0.63 Pa.sn and yield stress = 85 Pa. In this case, c = 0.219 • Given is a Bingham’s plastic fluid • ReB = (equation 7.6) = 56.2 • alpha = (equation 7.55) = 1/(2-m) = 0.561 • for Bingham’s plastic fluid in laminar flow • KE = 1549.4 J/h

33. Problem 7.4: The rheological behaviour of apricot marmalade can be described by the Herschel-Bulkley equation, with a yield stress of 19 Pa, a consistence index of 4.43 Pa.sn and a flow behaviour index of 0.65. Determine the smaller diameter of a steel pipe that should be employed to transport such marmalade with a mass flow rate of 8000 kg/h. The total length of the pipe is 200 m and mechanical energy losses are 75 J/kg. The density of marmalade is 1165 kg/m3. Source: Ibarz and Barbosa-Cánovas, Unit Operations in Food Engineering