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Reynold’s Number Reynold’s number is ratio of inersial forces & viscous forces & have no units. N Re = ρDU/µ ρ = Density of fluid (g/cm3) D = Diamerter.

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Presentation on theme: "Reynold’s Number Reynold’s number is ratio of inersial forces & viscous forces & have no units. N Re = ρDU/µ ρ = Density of fluid (g/cm3) D = Diamerter."— Presentation transcript:

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2 Reynold’s Number Reynold’s number is ratio of inersial forces & viscous forces & have no units. N Re = ρDU/µ ρ = Density of fluid (g/cm3) D = Diamerter of pipe (cm) U = Velocity of fluid (cm/sec) µ = Viscosity of fluid (cp, p, g/cm.sec) The velocity of the central arrow is high of all and the arrow which is near the pipe wall have low velocity. The arrows which are near the pipe boundary have low velocity due to increase in roughness or friction.

3 The Law of Conservation of Mass states that mass can be neither created or destroyed. Using the Mass Conservation Law on a steady flow process - where the flow rate do not change over time through a control volume and where the stored mass in the control volume do not change - implements that inflow = outflow This statement is called the Equation of Continuity. Common application where the Equation of Continuity are used are pipes, tubes and ducts with flowing fluids or gases, rivers, overall processes as power plants, diaries, logistics in general, roads, computer networks and semiconductor technology and many more… Equation of Continuity

4 Considering the steady flow of a fluid through a stream tube as shown in figure below 1 2 1000kg/hr 1000kg/hr The mass flow rate a point 1 & at point 2 is constant. So that m 1 = m 2 Vol. Flow rate = Area * velocity (cm 2 *cm/sec) Vol. Flow rate at point 1= U 1 * A 1 Vol. Flow rate at point 2 = U 2 * A 2 Mass flow rate at point 1 = ρ 1 U 1 A 1 Mass flow rate at point 2 = ρ 2 U 2 A 2 Equating mass flow rate at point 1 & 2 ρ 1 U 1 A 1 = ρ 2 U 2 A 2 For incompressible fluids densities are same i.e ρ1 = ρ 2 So, U 1 A 1 = U 2 A 2

5 Mass Velocity An important term is mass velocity designated by G’ or m’. mass velocity m’ =mass flow rate /area = m/A = gm/cm 2 *sec

6 Example 4.1 (McCabe, Smith) Crude oil, specific gravity is 0.887, flows through the piping system as shown in fig. Pipe A is 2 inch, pipe B is 3 inch and each of pipes C is 1.5 inch. An equal quantity of liquid flows through each of pipes C. The flow through pipe A is 30 gal/min. Calculate (a) the mass flow rate in each pipe, (b) the average linear velocity in each pipe, and (c) the mass velocity in each pipe.


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