PIPE FLOWS: LECTURE /04/2017. Yesterday, for the example problem Δp = f(v, ρ, μ, L, D) We came up with the non dimensional relation


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1 /04/07 ECTURE 4 PIPE FOWS: Yesterday, for the example problem Δp = f(v, ρ, μ,, ) We came up with the non dimensional relation f (, ) 3 V or, p f(, ) You can plot π versus π with π 3 as a parameter. Or, You can plot π versus π 3 with π as a parameter. As you need to find Δp versus V, let us first go with π versus π with π 3 as a parameter. From the first row of the data i.e kg / ms, evaluate water p V 4700 From the second row, 700, p.78 0 V
2 ike this, you evaluate for each row of the data. You will get nine data points. For same ratio, you will see that Δp is increasing linearly with. As is present only in the nondimensional parameter3, (where π & π is devoid of it), you can now state f(, 3) g( ) p V i.e., g( ) g or, 3 p V g where, g( ) is a function. (Please note that this g is not acceleration due to gravity) By plotting the available data you will see that 3 p V 0.55 Most of the Newtonian smooth pipe flows correlate in this same manner. This is the principle in many pipe flow problems..75
3 VISCOUS FOWS OR PIPE FOWS: One of the most important application of solving linear momentum & continuity equations are in pipe flows. Recall, you cannot have a fully general solution to the Navier Stokes equation. Based on the nature of the problem & type of fluid flow, we can get particular solutions. As said earlier you may have to neglect or sometimes incorporate the effects of Viscosity Gravity Compressibility Pressure, etc. As a civil engineer, you will be mostly dealing with incompressible fluid (or water) in many practical problems. We will see the application to pipe flows, where as a civil engineer, you may have to design pipe flow network. Pipe flows are generally viscous. Reynold s number plays a prominent part in understanding the pipe flow. Based on the Reynold s number, a pipe flow is classified as aminar Turbulent Example (As adopted from FM White s Fluid Mechanics): From Reynold s experiments on pipe flows of various diameters, it was observed that there is a critical Reynold s number Re critical 300. Below this value, pipe flow is laminar. For a flow through a 5 cm diameter pipe, at what velocity will this critical Reynold s number appear at 0 C. a) For airflow, b) For water flow. The input parameters are: For air, air kg / ms,.6 Re 4 3 air.05 kg / m
4 For water, 0.00 kg / ms, water 998 kg / m water 3 Solution: For pipe flow, the Reynold s number is defined as Re = ρv μ V.05V 0.05 Recritical 300 air or, V m / s air critical V 998V 0.05 Recritical 300 water 0.00 or, V m / s water critical Fully developed pipe flow: Usually in pipe flow design, you take fully developed pipe flow. A fully developed pipe flow is the one in which the effects of viscosity are fully present & the pipe entrance effects are not taken into account. Consider an incompressible pipe flow Entrance Pressure drop Pressure x
5 et us assume that the pipe is connected to an inviscid flow stream. Therefore, there will be no effect of viscosity at the entrance. The velocity profile will look like the first portion. The boundary layer starts growing with respect to pipe direction & merges at length, where the full effects of viscosity is witnessed in velocity profile. From the length onwards the pipe will be having fully developed viscous flow. From dimensional analysis done by other scientists and experts, it is observed that the entrance length is a function of the Reynold s number. In laminar flow: 0.06Re For turbulent flow:.6 Re 4 et us the pipe is connected to an inviscid flow stream. Therefore, there will be no effect of viscosity at the entrance. The velocity profile will look like the first portion. The boundary layer starts growing with respect to pipe direction & merges at length, where the full effects of viscosity is witnessed in velocity profile. From the length onwards the pipe will be having fully developed viscous flow. From dimensional analysis, it is observed that the entrance length is a function of the Reynold s number. In laminar flow: 0.06Re For turbulent flow:.6 Re 4 Example (Adopted from FM White s Fluid Mechanics) A cm diameter pipe is 0 m long and delivers water at m 3 /s at 0 0 C. What fraction of this pipe is taken as entrance region? Solution: Given Q = m 3 /s
6 V avg = Q A = π =.546 m/s 4 (0.0) For water at 0 0 C, ρ = 998kg m 3 and μ = 0.00kg/ms R e = ρv = = 5080 μ 0.00 Hence the pipe is turbulent. Now.6R e 4 =.6 (5080) = 0.48 m = m You can approximate fully developed pipe flow for this pipe throughout. Head loss Friction Factor In various pipe flow problems, we need to analyse the head loss (i.e. the energy head loss). For pipe flow analyses, let us again use the control volume. The flow is incompressible and steady (assumed). r=r τw Z x z Φ
7 The length of pipe = = x  x The onedimensional continuity equation suggests Q = Q = constant As pipe diameter is same at  and at  (the control volume portion), A = A = A (Area of cross section)and v = v = v The steady flow energy equation (please note that this is NOT INVISCI) is: Where h f is the energy loss. p ρg + α v g + z = p ρg + α v g + z + h f As the flow is steady and also the velocity profile is same throughout, α = α So, h f = z + p ρg Change in height of hydraulic grade line. If we apply momentum equation to the control volume: Consider pressure, gravity and shear forces: F x = p ( π 4 ) + ρg ( π 4 ) sinφ τ w (πr) F x = m (v v ) in steady state condition. As v = v so, F x = 0 i.e. τ w (πr) = pπr + ρgπr sinφ Shear stress, τ w = Rρg [ p ρg + z] τ w = ρgrh f Shear stress is related to the head loss, so, h f = τ w ρgr From literature, it is available about arcyweisbach friction factor f, and how head loss is defined, based on friction factor, i.e. h f = fv g The friction factor in pipe flow analysis is some function of =function(r e, ε, duct shape) Where, ε = wall roughness height (Refer to your lab manual)
8 aminar Fully eveloped Pipe Flow For a fully developed Poiseuille flow in a round pipe of diameter, radius R: And ( dp ) = p+ρg z dx As, v = Q A = u max Since τ w = μ du = 4μV = 8μV dr r=r R As h f = z + p ρg so 8μV = Rρg Or, h f = 3μV ρg [ z + p ρg ] = 8μV ρgr h f = 8μV ρgr = f laminar u = u max ( r R ) where u max = ( dp dx ) R 4μ V R g p + ρg z v = [ ] R 8μ Q = uda = π R v = πr 8μ = R [ p+ρg z ] 8μV = Rρg h f Where, f laminar friction factor = function ( R e ) + ρg z [ p ] Example (Adopted from FM White s Fluid Mechanics) An oil with ρ=900kg/m 3 and μ=0.8kg/ms flows through an inclined pipe. Two sections, section and section are 0 m apart. Assume steady laminar flow. a) Check whether flow is up or down. b) Compute hf between and. c) Compute the discharge Q. d) Velocity, V. e) The Reynolds number. Given following inputs: p=350000pa, z=0.0, p=50000pa, z=0sin40 0 =6.43m, =6cm.
9 Solution: 0m 40 0 The flow of oil will be in the direction of falling hydraulic gradient. As the flow is steady, v = v so v So, (HG) > (HG) = v g g (HG) = p ρg + z = = 39.64m (HG) = p ρg + z = = 34.74m a) The flow occurs from to. b) Head loss between and, h f = = 4.9m c) ischarge, Q for circular pipe, Q = πr 8μ + ρg z [ p ] = πr4 ρg 8μ h f = π (0.03) d) Velocity, V = Q = =.7 m A π (0.03) s e) Reynolds number, R e = ρv = = 80 (aminar flow) μ = m 3 /s
V/ t = 0 p/ t = 0 ρ/ t = 0. V/ s = 0 p/ s = 0 ρ/ s = 0
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