SHREE SA’D VIDYA MANDAL INSTITUTE OF TECHNOLOGY

DEPARTMENT OF CIVIL ENGINEERING

Subject:-Fluid Mechanics

Topic:-Reynolds experiment

-:PRESENTED BY:-

Name Arvindsai

Dhaval

Fahim Patel

Navazhushen Patel

M.Asfak Patel

Enrollment no.130454106002

130454106001

140453106005

140453106008

140453106007

INTRODUCTIONProf. Osborne Reynolds conducted

the experiment in the year 1883.This was conducted to demonstrate

the existence of two types of flow :-1.Laminar Flow2.Turbulent Flow

1.Laminar Flow:-

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.

2.Turbulent Flow:- 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.

What is Reynolds Number ?

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

APPARATUS

Observation by Reynolds1. At low velocity,

the dye will move in a line parallel to the tube and also it does not get dispersed.

2. At velocity little more than before the dye moves in a wave form.

3. At more velocity the dye will no longer move in a straight-line.

FORMULAS

Where,ρ =density of liquid (Kg/m3 ) V=mean velocity of liquid m/SD=diameter of pipe(m)µ=dynamic velocity(N.S/m2 ) =kinematic viscosity(m2 /S)

Where

Reynold number is a dimensionless quantity.

νVxDR

μρxVxDR

Force ViscousForce InertiaR

N

N

N

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.

If Reynold number i.e. 2000 < 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.

EXAMPLE

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 0.302

=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

Reynolds number(RN):-

/sm 10 x 6.156

RN1/40.079

f

Turbulent. is Flow4000) ( 27,120

4)-10 x (0.5

0.30) x (4.52

νVxD

R

μρxVxD

R

24-

N

N

• Head loss due to Friction:-

m 5.128

0.30 x 9.81 x 2

(4.52) x 60 x 10 x 6.156 x 4

2.g.d4.f.l.V

hf

23-

2

Example 2:- An oil of viscosity 0.9 and viscosity 0.06 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 0.202 = 0.0314 m2

= 0.06 m3/s d = 200 cm ρ = 0.9 x1000 =

900kg/m3 = 0.20 m L = 500 m µ = 0.6 poise = 0.006 Ns/m3

Reynolds number(RN):-

0.0051

1/4(0.079)/RN f

m/s 1.91

Q/AV

Turbulent is Flow4000) ( 57,300

(0.006

0.20 x 1.91 x 900

Head loss due to Friction:-

• Power required:-

kW 5.02 P1000

9.48 x 0.06 x 9.81 x 900

μhf x Qx x P

P

waterof m 9.48

02x9.81x0.2(1.91 x 30 x 0.0051 x 4

2.g.d4.f.l.V

hf

g

)2

2

Example 3:- oil of Sp. Gr 0.095 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 0.202

=0.314 m2

= 0.05m3/sD =20 cm = 0.20 m ρ=0.95 x 1000

=950 kg/m3 µ = 1.0poise = 0.1 Ns/m3 V= Q/A=0.05/0.0314

=1.59m/s

Reynolds number(RN):-

.Transition is Flow 4000)Rn(2000 3021

0.1

0.20 x 1.59 x 950

μD x V x ρ

R N

..

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/secd= 20 cm=0.20 m x

0.202=0.314m2

µ = 0.38 Ns/m3

ρ =950 kg/m3

Reynolds number(RN):-

Laminar. is Flow(2000) 960

0.38

0.20 x 2 x 912

μD x V x ρ

R N