1. Viscous Flow

2. 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)

3. Section I

4. 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

5. 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)

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

7. (3) (4) (5) Integrating w.r.t. “r” C = Constant
Value obtaned by bounty
Condition
r = R, u =0
(3)
(4)
(5)

8. III Ratio of Maximum Velocity to Average Velocity:

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

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

12. (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 :

13. 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 = N/m2

14. 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) N/m2 , 2) N/m2

15. 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 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) N/m2 per m. , 2) u= 3.50 m/s. 3) Re = 650

16. Laminar Flow Between Two Parallel stationary PlatesD A t B C ∆x

17. (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)

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

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

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

21. III) Ratio of Maximum Velocity to Average Velocity:

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

24. L = x2- x1

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

26. 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) N/m2 per m , 2) 1.33 m/s 3) m3/s

27. 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 = N s/m2
Pressure Gradient:
Shear stress:
Discharge:
Answer: 1) N/m2 per m , 2) N/m2 3) m3/s

28. 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 = m3/s = 5.2 liter/s

29. 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) N/m2 per m 3) N/m2

30. 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 = poise
Ans: 1) Velocity u = m/s 2) Rate of leakage Q = 6.06 x lit/s

31. Section II

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

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

34. 2) A pipe of diameter 20 cm and length 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 = m/s 2) Re = ) f = ) hf = 86.5 m

35. Section III

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

37. 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

38. Hagen Poiseuilli’s Formula
Head loss
Coefficient of Viscosity

39. 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 = m3/s 2) NS/ m2

40. 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 N/Cm2 and the viscosity of liquid is 0.25 poise. Find the rate of flow of liquid through the tube.
Ans: 1) Q = x 10-8 m3/s

41. 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

42. 1 2 3 Stoke’s law, Drag Force Weight of Sphere Force on Sphere
So 1, 2 & 3

43. 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) NS/ m2 or poise

44. 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) NS/ m2 or 36.3 poise

45. Orifice Type Viscometer:
Constant Tem.
Bath
Oil
Measuring Cylinder

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