Reynold number (Fluid Mechanics)(FM)
- 1. SHREE SA’D VIDYA MANDAL INSTITUTE OF TECHNOLOGY DEPARTMENT OF CIVIL ENGINEERING
- 3. -:PRESENTED BY:- Name Arvindsai Dhaval Fahim Patel Navazhushen Patel M.Asfak Patel Enrollment no. 130454106002 130454106001 140453106005 140453106008 140453106007
- 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. 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.
- 6. 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.
- 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 1. 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-
- 10. 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. ν VxD R μ ρxVxD R ForceViscous ForceInertia R N N N
- 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. 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.
- 14. EXAMPLE
- 15. 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
- 17. • Head loss due to Friction:- m5.128 0.30x9.81x2 (4.52)x60x10x6.156x4 2.g.d 4.f.l.V hf 23- 2
- 18. 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
- 20. Head loss due to Friction:- • Power required:- kW5.02P 1000 9.48x0.06x9.81x900 μ hfxQxxP P waterofm9.48 02x9.81x0.2 (1.91x30x0.0051x4 2.g.d 4.f.l.V hf g )2 2
- 21. 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/s D =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
- 22. Reynolds number(RN):- .TransitionisFlow4000)Rn(20003021 0.1 0.20x1.59x950 μ DxVxρ RN ..
- 23. 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