CP 502 Advanced Fluid Mechanics Incompressible Flow of

CP 502 Advanced Fluid Mechanics Incompressible Flow of Viscous Fluids Set 06: Working with Non-Newtonian Fluids Prof. R. Shanthini 03 July 2019

Velocity profile of a power-law fluid Shear stress-strain relationship: Shear stress-pressure relationship: Combination gives: (1) Prof. R. Shanthini 03 July 2019

Velocity profile of a power-law fluid Rearranging (1): Integrating from a distance r to wall at r=R: (2) Prof. R. Shanthini 03 July 2019

Volumetric flow rate of a power-law fluid Using (2) in (3) Prof. R. Shanthini 03 July 2019

Average velocity in a power-law fluid Using (3) in (4) Prof. R. Shanthini 03 July 2019

Fanning friction factor for power-law fluids (4) Can be rearranged to give: (5) Prof. R. Shanthini 03 July 2019

Fanning friction factor for power-law fluids Using (5) in the definition of friction factor (6) where (7) Prof. R. Shanthini 03 July 2019 is the Generalized Reynolds number

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: Prof. R. Shanthini 03 July 2019

Pumping requirement of non-Newtonian fluids (8) For laminar flow and shear-thinning liquids, For turbulent flow, Prof. R. Shanthini 03 July 2019

Pumping requirement of non-Newtonian fluids (8) For laminar flow, For turbulent flow, f is obtained from the figure on next slide. Prof. R. Shanthini 03 July 2019

Pumping requirement of non-Newtonian fluids Prof. R. Shanthini 03 July 2019

Pumping requirement of non-Newtonian fluids (8) Prof. R. Shanthini 03 July 2019

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. Prof. R. Shanthini 03 July 2019

Pumping requirement of non-Newtonian fluids For non-Newtonian fluids, for the range 20 < NGRe < 500, use the following: Prof. R. Shanthini 03 July 2019

Pumping requirement of non-Newtonian fluids Prof. R. Shanthini 03 July 2019

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 90 o flanged elbows, and one fully open angle valve. A filter present in the pipeline causes 100 k. Pa pressure drop. Set up a spread sheet to calculate the pumping requirements. The properties of the fluid are as follows: density = 1250 kg/m 3, consistency coefficient = 5. 2 Pa. sn, flow behaviour index = 0. 45 Prof. R. Shanthini 03 July 2019 Ans: 549. 04 W

Velocity profile of a Herschel–Bulkley fluid Shear stress-strain relationship: Shear stress-pressure relationship: Combination gives: (1 a) Prof. R. Shanthini 03 July 2019

Rearranging (1 a): Integrating from a distance r to wall at r=R: Prof. R. Shanthini 03 July 2019

(2 a) Prof. R. Shanthini 03 July 2019

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/m 3 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 45 o Brix peach juice with a viscosity of 9 m. Pa. s (b) Egg yolk that presents a power law fluid behaviour, with K = 880 m. Pa. sn and n = 0. 20. Prof. R. Shanthini 03 July 2019 Source: Ibarz and Barbosa-Cánovas, Unit Operations in Food Engineering

Data provided: = density = 1250 kg/m 3 d = diameter = 26. 7 mm w = mass flow rate = 26. 7/1000 m = 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 Prof. R. Shanthini 03 July 2019

(a) Clarified juice at 45 o Brix peach juice with μ = 9 m. Pa. 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 Prof. R. Shanthini 03 July 2019 vmax = vm / 0. 82 = 4. 86 m/s

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 Prof. R. Shanthini 03 July 2019 Note: vmax takes slightly different values in the alternative approaches, which is expected considering the approximate nature of the equation used and the graph read.

(b) Egg yolk that presents a power law fluid behaviour, with K = 880 m. Pa. sn and n = 0. 20. Given is a power law fluid Re. G = = 37, 828 That means, the given flow is turbulent. The ratio (vm / vmax) could be found using Figure 7. 9. Prof. R. Shanthini 03 July 2019

Reading at Re. G = 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 Prof. R. Shanthini 03 July 2019

Problem 7. 2: A fluid food whose density is 1200 kg/m 3 circulates at 25 o. C 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 69 o Brix with a viscosity of 324 m. Pa. s (b) Nonclarified rasberry juice (without pectin elimination) of 41 o Brix that has a pseudoplastic behaviour, with n = 0. 73 and K = 1. 6 Pa. sn (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. (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 Prof. R. Shanthini 03 July 2019 Source: Ibarz and Barbosa-Cánovas, Unit Operations in Food Engineering

Data provided: = density = 1200 kg/m 3 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 Prof. R. Shanthini 03 July 2019 ?

(a) Concentrated peach juice at 69 o Brix with μ = 324 m. Pa. 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, Prof. R. Shanthini 03 July 2019

(b) Nonclarified rasberry juice of 41 o Brix that has a pseudoplastic behaviour, with n = 0. 73 and K = 1. 6 Pa. sn Given is a power law fluid Re. G = (equation 7. 7) = 70. 8 Re. G, critical = (depends on n; equation 7. 8) =2262 Re. G < Re. G, critical (laminar flow) alpha = (equation 7. 54) = (2 n+1)(5 n+3)/[3(3 n+1)2] = 0. 536 for power law fluids in laminar flow KE = 1623. 6 J/h Prof. R. Shanthini 03 July 2019

(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 Re. G = (equation 7. 7) = 62. 3 Re. G, critical (Figure 7. 4) depends on the generalized Hedstrom number (He. G) and the flow index He. G = (equation 7. 12 b) = 1093. 6 Re. G, critical = 2200 (approximately) Prof. R. Shanthini 03 July 2019

Re. G < Re. G, 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 Prof. R. Shanthini 03 July 2019

(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 Re. B = (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 Prof. R. Shanthini 03 July 2019

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/m 3. Prof. R. Shanthini 03 July 2019 Source: Ibarz and Barbosa-Cánovas, Unit Operations in Food Engineering