1. SHREE SA’D VIDYA MANDAL INSTITUTE OF TECHNOLOGY
DEPARTMENT OF CIVIL ENGINEERING

2. Subject:- Fluid Mechanics
Topic:-Reynolds experiment

3. -:PRESENTED BY:- Name Arvindsai Dhaval Fahim Patel Navazhushen Patel
M.Asfak Patel
Enrollment no

4. INTRODUCTION
Prof. Osborne Reynolds conducted the experiment in the year 1883.
This was conducted to demonstrate the existence of two types of flow :-
1.Laminar Flow
2.Turbulent Flow

5. Factors responsible for laminar flow are:-
Laminar flow is defined as that type of flow in which the fluid particles move along well-defined paths or stream lines and all the stream lines are straight and parallel.
Factors responsible for laminar flow are:-
High viscosity of fluid.
Low velocity of flow.
Less flow area.
For example,
Flow through pipe of uniform cross-section.

6. 2.Turbulent Flow:- For example,
Turbulent flow is defined as that type of flow in which the fluid particles move is a zigzag way. The Fluid particles crosses the paths of each other.
For example,
Flow in river at the time of flood.
Flow through pipe of different cross- section.

7. What is Reynolds Number ?
The ratio of inertia force to viscous force is said to be the Reynolds number (RN).

8. APPARATUS

9. Observation by Reynolds
At low velocity, the dye will move in a line parallel to the tube and also it does not get dispersed.
At velocity little more than before the dye moves in a wave form.
At more velocity the dye will no longer move in a straight-line.

10. Reynold number is a dimensionless quantity.
FORMULAS
Where,
ρ =density of liquid (Kg/m3 )
V=mean velocity of liquid m/S
D=diameter of pipe(m)
µ=dynamic velocity(N.S/m2 )
=kinematic viscosity(m2 /S)
Where
Reynold number is a dimensionless quantity.

11. Types Of Flows Based On Reynold Number:-
If Reynold number, RN < 2000 the flow is laminar flow.
If Reynold number, RN > 4000 the flow is turbulent flow.

12. If Reynold number i.e < RN < 4000,we observe a flow in which we can see both laminar and turbulent flow to gather. This flow is called Transition flow.
RN = 2300 is usually accepted as the value at transition , RN that exists anywhere in the transition region is called the critical Reynolds number.

14. EXAMPLE

15. Solution:- Q=320 liters/second
Example 1 :- An oil of viscosity 0.5 stoke is flowing through a pipe of 30 cm in diameter at a rate of 320 liters per second. Find the head loss due to friction for the pipe length of 60 cm.
Solution:-
Q=320 liters/second
=0.32 m3/s
d=30 cm=0.30 m x =0.070m2
L=60 m
=0.5 stoke
= 0.5 x 10-4 m2/s V= Q/A=0.32/0.0707
=4.52 m/s

16. Reynolds number(RN):-

17. Head loss due to Friction:-

18. Example 2:- An oil of viscosity 0. 9 and viscosity 0
Example 2:- An oil of viscosity 0.9 and viscosity poise is flowing through a pipe of diameter 200 mm at the rate of 60 liters per second. Find the head loss due to friction for a 500 m length of pipe. find the power required to maintain this flow
Solution:-
Q = 60 liters/second x = m2
= 0.06 m3/s
d = 200 cm ρ = 0.9 x1000 = 900kg/m3
= 0.20 m
L = 500 m
µ = 0.6 poise = Ns/m3

19. Reynolds number(RN):-

20. Head loss due to Friction:-
Power required:-

21. Example 3:- oil of Sp. Gr is flowing through a pipe of 20 cm in diameter. if a rate of flow 50 liters/second and viscosity of oil is 1 poise , decide the type of flow.
Solution:-
Q = 50 liters/second x =0.314 m2
= 0.05m3/s
D =20 cm = 0.20 m ρ=0.95 x 1000 =950 kg/m3
µ = 1.0poise
= 0.1 Ns/m V= Q/A=0.05/0.0314
=1.59m/s

22. Reynolds number(RN):-

23. Example 4:- Liquid is flowing through a pipe of 200 mm in diameter
Example 4:- Liquid is flowing through a pipe of 200 mm in diameter. Tube with mean viscosity of oil 2 m/sec. If density of liquid is 912 kg/ m3 and viscosity is 0.38 N.S/m2 the type of flow.
Solution:-
V= 2 m/sec
d= 20 cm=0.20 m x 0.202=0.314m2
µ = 0.38 Ns/m3
ρ =950 kg/m3

24. Reynolds number(RN):-