Viscous Flow

Viscous Flow Section I Section II Section III Flow of Viscous Fluid Circular pipe Two Parallele Pipe Lose of Head due Friction Coefficient of viscosity - Capillary tube method, - orifice type - falling sphere resistance method - Rotating cylinder method Network of pipes (Hardy-cross method)

Section I

r Flow of viscous fluid through circular Y= R-r P A B C D r Flow of viscous fluid through circular (Hagen Poiseuille Equation) for Laminar flow R = Radius of Pipe r = Radius of Fluid element x = Length of fluid element P = pressure on face AB

i) Shear Stress Distribution: Pressure force on fluid elements: ABCD Pressure force on AB = 2) Pressure forec on CD= 3) Shear force on Surface = No Acceleration & Submission of all forces: (1)

II) Velocity Distribution: (2) But from Eq. 1 = 2

(3) (4) (5) Integrating w.r.t. “r” C = Constant Value obtaned by bounty Condition r = R, u =0 (3) (4) (5)

III Ratio of Maximum Velocity to Average Velocity:

X1 L X2 Iv) Drop of pressure for a given length (L) of Pipe: L = x2- x1 X2

Integrating the w.r.t. x - - Loss of Pressure Head:

(Hagen Poiseuille Equation) for Laminar flow Circular Pipe (Hagen Poiseuille Equation) for Laminar flow I) Shear Stress Distribution: II) Velocity Distribution: III Ratio of Maximum Velocity to Average Velocity: Iv) Drop of pressure :

Ex 1) A crude oil of viscosity 0.97 poise and relative density 0.9 is flowing through a horizontal circular pipe of diameter 100 mm and of length 10 m. Calculate the difference of pressure at the two ends of the pipe, if 100 kg of the oil is collected in a tank in 30 s. Assume the laminar flow. Answer: u =0.471 m/s, p1-p2 = 1462.28 N/m2

2) An oil of viscosity 0. 1 Ns/m2 and relative density 0 2) An oil of viscosity 0.1 Ns/m2 and relative density 0.9 is flowing through a circular pipe of diameter 50 mm and the length 300 m. the rate of flow of fluid through the pipe is 3.5 litres/s. Find the pressure drop in a length of 300 m and also the shear stress at the pipe wall. 1) Pressure Drop 2) Shear Stress Answer: 1) 684288 N/m2 , 2) 28.512 N/m2

3) A fluid of viscosity 0.7 Ns/m2 and specific gravity 1.3 is flowing through a circular pipe of diameter 100 mm. The maximum shear stress at the pipe wall is given as 196.2 N/m2, find 1) The pressure gradient 2) the average velocity and 3) Reynold number of the flow. Find The pressure gradient 2) The average velocity and 3) Reynold number of the flow. Answer: 1) - 7848 N/m2 per m. , 2) u= 3.50 m/s. 3) Re = 650

Laminar Flow Between Two Parallel stationary Plates D A t B C ∆x

(1) Pressure force on fluid elements: ABCD Pressure force on AB = 2) Pressure forec on CD= 3) Shear force on Face BC = 3) Shear force on Face AD = No Acceleration & Submission of all forces: (1)

I) Shear Stress Distribution: Max. Shear Stress y=0

II) Velocity Distribution: (2) Integrating w.r.t.. y

Integrating again If y=0, u = 0 C2 = 0 If y=t, u = 0

III) Ratio of Maximum Velocity to Average Velocity:

p1 p2 Y x x1 L x2 IV) Drop of pressure for a given length (L) of Pipe: L = x2- x1 x2

L = x2- x1

Parallel Plates I) Shear Stress Distribution: II) Velocity Distribution: III Ratio of Velocity : Iv) Drop of pressure : v) Rate of Change of flow:

Calculate 1) the pressure gradient along flow 2) the average velocity and 3) the discharge for an oil of viscosity 0.02 Ns/m2 flowing between two stationary parallel plates 1 m wide maintained 10 mm apart. The velocity midway between the plates is 2 m/s. Pressure Gradient: Average Velocity: Discharge: Answer: 1) - 3200 N/m2 per m , 2) 1.33 m/s 3) 0.0133 m3/s

2) Determine 1) the pressure gradient 2) the shear stress at the two horizontal parallel plates and 3) the discharge per meter width for the laminar flow of oil with maximum velocity of 2 m/s between two horizontal parallel fixed plates which are 100 mm apart. Given µ = 2.4525 N s/m2 Pressure Gradient: Shear stress: Discharge: Answer: 1) - 3924 N/m2 per m , 2) 196.2 N/m2 3) 0.0133 m3/s

3) An oil of viscosity 10 poise flows between two parallel fixed plates which are kept at a distance of 50 mm apart. Find the rate of oil between the plates if the drop of pressure in a length of 1.2 m be 0.3 N/cm3. The width of the plates is 200 mm. Answer: 1) u = 0.52 m/s, 2) Q = 0.0052 m3/s = 5.2 liter/s

4) Water at 15 oC flows between two large parallel plates at distance of 1.6 mm apart. Determine a) The maximum velocity b) the pressure drop per unit length and c) The shear stress at the walls of the average velocity is 0.2 m/s . The viscosity of water at is given as 0.01 poise. Maximum Velocity: The pressure drop: Shear Stress: Answer: 1) 0.3 m/s, 2) 937.44 N/m2 per m 3) 0.749 N/m2

4) The radial clearance between a hydraulic plunger and cylinder walls is 0.1 mm the length of the plunger is 300 mm and diameter 100 mm. find the velocity of leakage and rate of leakage past the plunger at an instant when the difference of the pressure between the two ends of the plunger is 9 m of water . Take µ = 0.0127 poise Ans: 1) Velocity u = 0.193 m/s 2) Rate of leakage Q = 6.06 x 10 -3 lit/s

Section II

Lose of Head due Friction: Loss of Head hf Loss of Head due to Fricition hf (Hagen Poiseuille Equation)

1) Water is flowing through a 200mm diameter pipe with coefficient of friction f = 0.04. The shear stress at a point 40 mm from the pipe axis is 0.00981 N/cm2. Calculate the shear stress at the pipe wall. Ans: 1) Re = 400 2) 0.0245 N/ cm2

2) A pipe of diameter 20 cm and length 10000 m is laid at a slope of 1 in 200. An oil of Sp.gr. 0.9 and Viscosity 1.5 poise is pumped up at the rate of 20 liters per S. Find the head loss due to friction. Ans: 1) u = 0.6366 m/s 2) Re = 763.89 3) f = 0.02094 4) hf = 86.5 m

Section III

Coefficient of Viscousity - Capillary tube method, - Falling sphere resistance method Orifice type Viscometer - Rotating cylinder method

Capillary Tube Method: Constant Head Tank h = Difference of pressure head for length L D = Diameter of Capillary tube L = Length of tube ρ = Density of fluid µ = Coefficient of viscosity Measuring Tank

Hagen Poiseuilli’s Formula Head loss Coefficient of Viscosity

Ex 1) The viscosity of an oil of Sp. gr Ex 1) The viscosity of an oil of Sp.gr. 0.9 is measured by a capillary tube of diameter 50 mm. the difference of pressure head between two points 2 m apart is 0.5 m of water. The mass of oil collected in a measuring tank is 60 kg in 100 s. Find the viscosity of oil. Ans: 1) Q = 0.000667 m3/s 2) 0.5075 NS/ m2

2) A capillary tube of diameter 2 mm and length 100 mm is used for measuring viscosity of a liquid. The difference of pressure between the two ends of the tube is 0.6867 N/Cm2 and the viscosity of liquid is 0.25 poise. Find the rate of flow of liquid through the tube. Ans: 1) Q = 107.86 x 10-8 m3/s

Falling Sphere Resistance Method: U Constant Tem. Bath Fixed Mark d Sphere L = Distance travelled by sphere in Viscous fluid t = time taken by Sphere ρs = Density of Sphere ρf = Density of fluid W = Weight of Sphere Fb= force acting on Sphere

1 2 3 Stoke’s law, Drag Force Weight of Sphere Force on Sphere So 1, 2 & 3

3) A sphere of diameter 2 mm falls 150 mm in 20 s in a viscous liquid 3) A sphere of diameter 2 mm falls 150 mm in 20 s in a viscous liquid. The density of the sphere is 7500 kg/m3 and of liquid is 900 kg/m3. Find the co-efficient of viscosity of the liquid. Ans: 1) 1.917 NS/ m2 or 19.17 poise

4) Find the viscosity of a liquid of sp. gr 4) Find the viscosity of a liquid of sp.gr. 0.8 when a gas bubble of diameter 10mm rises steadily through the liquid at a velocity 1.2 cm/s. Neglect the weight of the bubble. Fall rises Ans: 1) 3.63 NS/ m2 or 36.3 poise

Orifice Type Viscometer: Constant Tem. Bath Oil Measuring Cylinder

Prepared by, Dr Dhruvesh Patel www.drdhruveshpatel.com Source: www.google.com