ME 305 Fluid Mechanics I. Part 8 Viscous Flow in Pipes and Ducts


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1 ME 305 Fluid Mechanics I Part 8 Viscous Flow in Pipes and Ducts These presentations are prepared by Dr. Cüneyt Sert Mechanical Engineering Department Middle East Technical University Ankara, Turkey Please ask for permission before using them. You are NOT allowed to modify them. 81
2 Flow in Pipes and Ducts Flow in closed conduits (circular pipes and noncircular ducts) are very common. 82
3 Flow in Pipes and Ducts (cont d) We assume that pipes/ducts are completely filled with fluid. Other case is known as open channel flow. Typical systems involve pipes/ducts of various sizes connected to each other by various fittings, elbows, tees, etc. Valves are used to control flow rate. Fluid is usually forced to flow by a fan or pump. We need to perform frictional head loss (pressure drop) calculations. They mostly depend on experimental results and empirical relations. First we need to be able to differentiate between laminar and turbulent flows. 83
4 Laminar vs. Turbulent Flows Laminar flow is characterized by smooth streamlines and highly ordered motion. Fluid flows as if there are immiscible layers of fluid. Turbulent flow is highly disordered. Usually there are unsteady, random fluctuations on top of a mean flow. Most flows encountered in practice are turbulent. Laminar flows occur in small pipes (tubes) and with viscous fluids. Laminar Movies Reynolds Experiment Pipe D Transitional Q = VA ν Re D = VD ν = ρvd μ Turbulent Laminar vs. Turbulent Velocity Profiles 84
5 Laminar vs. Turbulent Flows (cont d) For pipe flow transition from laminar to turbulent flow occurs at a critical Reynolds number of Re = Re D < 2300 : Laminar Re D > 2300 : Turbulent Inertial forces Viscous forces = ρv2 L 2 μvl = ρvl μ = VL ν Characteristic velocity Characteristic length At large Reynolds numbers inertial forces, which are proportional to fluid s density and square of its velocity, are much higher than the viscous forces. Viscous forces can not regularize random fluctuations. At small Reynolds numbers viscous forces are high enough to suppress random fluctuations and keep the flow in order. Exercise : Estimate the typical range of Reynolds numbers of the flow inside the pipe that supplies water to your shower. 85
6 Pressure Loss (Pressure Drop, Head Loss) in Pipe Flow As a fluid flows in a straight, constant diameter pipe its pressure drops due to viscous effects, known as major pressure loss. Additional pressure drops occur due to other components such as valves, bends, tees, sudden expansions, sudden contractions, etc. These are known as minor pressure losses. In Chapter 6 we studied analytical solution of HagenPoiseuille flow, which is the steady, fullydeveloped, laminar flow in a circular pipe. Rate of pressure drop for this flow was shown to be constant. Pressure drop over a pipe section of length L is given by Δp = f L D ρv 2 2 Darcy friction factor Dynamic pressure where f = 64/Re D for laminar flow. 86
7 Major Pressure Loss in Laminar Pipe Flow 1 2 L flow D Δp = f L D ρv 2 2 Consider the EBE between sections 1 and 2. p 1 ρg + V 2 1 2g + z 1 = p 2 ρg + V 2 2 2g + z 2 + h f Friction head is related to the pressure drop as h f = Δp ρg = f L D V 2 2g The above important equation is known as the DarcyWeisbach equation. 87
8 Major Pressure Loss in Turbulent Pipe Flow Analytical solution of HagenPoiseuille flow is valid for laminar flows. It is not possible to solve NavierStokes equations analytically for turbulent flows. Instead we need to determine pressure drop values experimentally. For a turbulent pipe flow pressure drop depends on Δp = Δp L, D, V, ε, μ, ρ A BuckinghamPi analysis yields the following relation New parameter : Surface roughness Δp ρv 2 /2 = F L D, Re D, ε D which is usually expressed as Δp = f Re D, ε D L D ρv
9 Major Pressure Loss in Turbulent Pipe Flow (cont d) For turbulent pipe flows friction factor is a function of Reynolds number and relative surface roughness. f Re D, ε D Relative surface roughness In 1933 Nikuradse performed very detailed experiments to determine friction factor for turbulent flows. Later Colebrook presented this experimental data as the following equation Colebrook Formula 1f ε/d = 2 log Re D f An easy to use graphical representation of this equation is known as the Moody diagram, given in the next slide). Be careful in reading the logarithmic horizontal and vertical axes of the Moody diagram. 89
10 Moody Diagram to Read Friction Factor for Pipe Flows Right of this line is known as wholly turbulent flow, where f is no longer a function of Re D 0.05 f Laminar flow Laminar to turbulent transition Smooth pipe (ε = 0) Re D ε/d 810
11 Moody Diagram (cont d) To use the Moody diagram first we read the roughness of the pipe of interest from a table (see the next slide). Next we calculate the relative roughness (ε/d) and Reynolds number, and read the friction factor value f. f = 64/Re D relation for laminar flows appears in the Moody diagram as a straight line. Even for smooth pipes (ε = 0) friction factor is not zero, as expected. For high Re D values f becomes independent of Re D. This region is known as wholly turbulent flow. Between the laminar and turbulent regions, there is a transition region (2100 < Re D < 4000), where the data is not very reliable. We do not expect to read f with more than 10 % accuracy from the Moody diagram. 811
12 Moody Diagram (cont d) For pipes which are in service for a long time, roughness values given in tables for new pipes should be used with caution. As an alternative to the Moody diagram, following explicit formula can be used to calculate f. 1f = 1.8 log ε/d Re D 812
13 Minor Head Losses Pressure drops due to the flow through valves, bends, tees, sudden area changes, etc. are known as minor losses. Minor head losses can be calculated as h f = k V2 2g Head loss coefficient Head loss coefficient is usually obtained from a figure or table, similar to the ones given in the textbooks. For certain piping system elements minor losses are given in terms of an equivalent length h f = f L e V 2 D 2g where L e is the length of a straight pipe section that would create a major loss equal to the minor loss created by the element. f is obtained from the Moody diagram. 813
14 Examples A loss calculation in a pipe problem can be of three types Type 1: L and D of the pipe are known. Find p (or h f ) for a specified Q. Type 2: L and D of the pipe are known. Find Q for a specified p. Type 3: L of the pipe is known. Find D for a specified Q and p. Type 1 problems are easy. But Type 2 and Type 3 problems require an iterative solution because Re D and f cannot be computed directly. Solution starts with an initial guess for f and/or D. Usually couple of iterations are enough. Exercise : (Type 3 problem) Heated air at 1 atm and 35 is to be transported in a 150 m long plastic pipe at a rate of 0.35 m 3 /s. If the head loss in the pipe is not to exceed 20 m, determine the minimum diameter of the duct. Exercise : (Type 2 problem) Reconsider the previous example. Now the duct length is doubled while its diameter is maintained constant. If the total head loss is to remain constant, determine the new flow rate through the duct. 814
15 Examples (cont d) Exercise : The pump shown below is used to deliver water to a distribution system. There is no flow to or from reservoir A, but it is used to damp the pressure oscillations and protect the pump. The length of the suction and dischage pipes are 100 m and 150 m, respectively. The diameter and roughness of all pipes are 0.1 m and 0.5 mm, respectively. Head loss coefficient for the inlet to the pipe is 0.5. Gate valves are fully open and 90 o standard elbows are used. Volumetric flow rate of water to reservoir B is 0.02 m 3 /s. Determine a) head of the pump, b) height of the water level h in reservoir B. 20 m Elbow Pump Reservoir A Reservoir B h 3 m Gate valves Elbow Water 815
16 Examples (cont d) Exercise : Sketch of the warm water heating system of a studio apartment is shown in the figure. Volumetric flow rate of water through the piping system with a diameter of 1.25 cm is 0.2 L/s. Friction factor is Head loss coefficients for the elbow piece, heater valve, radiator, gate valve and boiler are 2.0, 5.0, 3.0, 1.0 and 3.0, respectively. Determine the power required to drive the circulation pump, which has an efficiency of 75%. 10 m Elbow 1 m Radiator Heater valve Gate valves Elbow Boiler Pump 8 m Elbow 816
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