Steady Incompressible Flow in Pressure Conduits Laminar and

Steady Incompressible Flow in Pressure Conduits

Laminar and Turbulent Flow in Pipes Laminar Paths of Particles don’t obstruct each other Viscous forces are dominant Velocity of fluid particles only changes in magnitude Lateral component of velocity is zero Turbulent Paths do intersect each other Inertial forces are dominant Velocity of fluid particles change in magnitude and direction Lateral components do exist.

Laminar and Turbulent Flow in Pipes If we measure the head loss in a given length of uniform pipe at different velocities , we will find that, as long as the velocity is low enough to secure laminar flow, the head loss, due to friction, is directly proportional to the velocity, as shown in Fig. 8. 1. But with increasing velocity, at some point B, where visual observation of dye injected in a transparent tube would show that the flow changes from laminar to turbulent, there will be an abrupt increase in the rate at which the head loss varies. If we plot the logarithms of these two variables on linear scale or in other words, if we plot the values directly on log-log paper, we will find that, after passing a certain transition region (BCA in Fig. 8. 1), the lines will have slopes ranging from about 1. 75 to 2.

Laminar and Turbulent Flow in Pipes Thus we see that for laminar flow the drop in energy due to friction varies as V, while for turbulent flow the friction varies as Vn, where n ranges from about 1. 75 to 2. The lower value of 1. 75 for turbulent flow occurs for pipes with very smooth walls; as the wall roughness increases, the value of n increases up to its maximum value of 2. If we gradually reduce the velocity from a high value, the points will not return along line BC. Instead, the points will lie along curve CA. We call point B the higher critical point, and A the lower critical point. However, velocity is not the only factor that determines whether the flow is laminar or turbulent. The criterion is Reynolds number.

Laminar and Turbulent Flow in Pipes

Reynolds Number Ratio of inertia forces to viscous forces is called Reynolds number. Where we can use any consistent system of units, because R is a dimensionless number.

Significance of Reynolds Number To investigate the development of Laminar and Turbulent flow Investigate Critical Reynold’s Number Develop a relationship between head loss (h. L) and velocity.

Critical Reynolds Number The upper critical Reynolds number, corresponding to point B of Fig. 8. 1, is really indeterminate and depends on the care taken to prevent any initial disturbance from effecting the flow. Its value is normally about 4000, but experimenters have maintained laminar flow in circular pipes up to values of R as high as 50, 000. However, in such cases this type of flow is inherently unstable, and the least disturbance will transform it instantly into turbulent flow. On the other hand, it is practically impossible for turbulent flow in a straight pipe to persist at values of R much below 2000, because any turbulence that occurs is damped out by viscous friction. This lower value is thus much more definite than the higher one, and is the real dividing point the two types of flow. So we define this lower value as the true critical Reynolds number.

Critical Reynolds Number It will be higher in a converging pipe and lower in a diverging pipe than in a straight pipe. Also, it will be less for flow in a curved pipe than in a straight one, and even for a straight uniform pipe it may be as low as 1000, where there is excessive roughness. However, for normal cases of flow in straight pipes of uniform diameter and usual roughness, we can take the critical value as Rcrit = 2000

Problem: Q: In a refinery oil (ν = 1. 8 x 10 -5 m 2/s) flows through a 100 -mm diameter pipe at 0. 50 L/s. Is the flow laminar or turbulent? Solution: Q = 0. 50 Liter/s = 0. 0005 m 3/s D = 100 mm = 0. 1 m Q = AV = (πD 2/4)V, V = (4 Q)/(πD 2) V = (4 x 0. 0005)/(3. 14 x 0. 1) = 0. 0637 m/s R = (DV)/ν = (0. 1 x 0. 0637)/(1. 8 x 10 -5) = 354 Since R < Rcrit=2000, the flow is laminar.

Hydraulic Radius For conduits having non-circular cross sections, we need to use some value other than the diameter for the linear dimension in the Reynolds number. The characteristic dimension we use is the hydraulic radius, defined as Rh = A/P Where A is the cross sectional area of the flowing fluid, and P is the wetted perimeter, that portion of the perimeter of the cross section where the fluid contacts the solid boundary, and therefore where friction resistance is exerted on the flowing fluid. For a circular pipe flowing full, Full-pipe flow: Rh = (π r 2)/(2πr) = r/2 = D/4

Friction Head Loss in Conduits This discussion applies to either laminar or turbulent flow and to any shape of cross section. Consider steady flow in a conduit of uniform cross section A, not necessarily circular as shown Fig. below. The pressures at sections 1 and 2 are p 1 and p 2, respectively. The distance between the sections is L.

Friction Head Loss in Conduits For equilibrium in steady flow, the summation of forces acting on any fluid element must be equal to zero (i. e. , ΣF=ma=0). Thus, in the direction of flow, p 1 A – p 2 A – γLAsinα – τ’ 0(PL) = 0 …………… (1) where we define τ’ 0, the average shear stress (average shear force per unit area) at the conduit wall. Nothing that sinα = (z 2 – z 1)/L and dividing each term in eq. (1) by γA gives, p 1 A/(γA) – p 2 A/(γA) – γLA(z 2 - z 1)/(γAL) = τ’ 0(PL)/(γA) p 1/γ – p 2/γ – z 2 + z 1 = τ’ 0(PL)/(γA) ……………. (2) From the left hand sketch of the Fig. , we can see that the head loss due to friction at the wetted perimeter is hf = (z 1 + p 1/γ) – (z 2 + p 2/γ) …………………. . (3)

Friction Head Loss in Conduits The eq. (3) equation indicates that hf depends only on the values of z and p on the centerline, and so it is the same regardless of the size of the cross-sectional area A. Substituting hf from eq. (3) and replacing A/P by Rh in eq. (2), we get, hf = τ’ 0 L/(Rhγ) ……………… (4) This equation is applicable to any shape of uniform cross section, regardless of whether the flow is laminar or turbulent.

Friction Head Loss in Conduits For a smooth-walled conduit, where we can neglect wall roughness, we might assume that the average fluid shear stress τ’ 0 at the wall is some function of ρ, μ, V and some characteristic linear dimension, which we will here take as the hydraulic radius Rh. Thus τ’ 0 = f(ρ, μ, V, Rh) Using the pi theorem of dimensional analysis to better determine the form of this relationship, we choose ρ, Rh and V as primary variables, so that ∏ 1 = μ ρa 1 Rhb 1 Vc 1 ∏ 2 = τ’ 0 ρa 2 Rhb 2 Vc 2 With the dimensions of the variables being ML-1 T-1 for μ, ML-1 T-2 for τ’ 0, ML-3 for ρ, L for Rh, and LT-1 for V,

Friction Head Loss in Conduits the dimensions for ∏ 1 are ∏ 1 = μ ρa 1 Rhb 1 Vc 1 M 0 L 0 T 0 = (ML-1 T-1) (ML-3)a 1 (L)b 1 (LT-1)c 1 For M: 0 = 1 + a 1 For L: 0 = -1 – 3 a 1 + b 1 + c 1 For T: 0 = -1 – c 1 The solution of these simultaneous equations is a 1 = b 1 = c 1 = -1, from which ∏ 1 = μ ρ-1 Rh-1 V-1 ∏ 1 = μ /(ρ Rh V) = R-1 where (ρ Rh V)/μ is a Reynolds number with Rh as the characteristic length.

Friction Head Loss in Conduits the dimensions for ∏ 2 are ∏ 2 = τ’ 0 ρa 2 Rhb 2 Vc 2 M 0 L 0 T 0 = (ML-1 T-2) (ML-3)a 2 (L)b 2 (LT-1)c 2 For M: 0 = 1 + a 2 For L: 0 = -1 – 3 a 2 + b 2 + c 2 For T: 0 = -2 – c 2 The solution of these simultaneous equations is a 2 = -1, c 2 = -2, b 2 = 0, from which ∏ 2 = τ’ 0 ρ-1 V-2 ∏ 2 = τ’ 0 /(ρV 2) We can write ∏ 2 = Ф( ∏ 1 -1), which results in τ’ 0 = ρ V 2 Ф (R)

Friction Head Loss in Conduits Setting the dimensionless term Ф (R) = ½ Cf , this yields τ’ 0 = Cf ρ V 2/2 Where Cf = average friction-drag coefficient for total surface (dimensionless) Inserting this value of τ’ 0 and γ = ρg, in eq. (4), which is hf = τ’ 0 L/(Rhγ) , we get hf = Cf (L/Rh)(V 2/2 g) ……………… (5) which can apply to any shape of smooth-walled cross section. From this equation , we may easily obtain an expression for the slope of the energy line, S = hf / L = Cf /Rh (V 2/2 g) …………… (6) which we also know as the energy gradient.

Friction in Circular Conduits Head loss due to friction, hf = Cf (L/Rh)(V 2/2 g) Energy gradient, S = hf / L = Cf /Rh (V 2/2 g) For a circular pipe flowing full, Rh = D/4, and f = 4 Cf , where f is friction factor (also some times called the Darcy friction factor) is dimensionless and some function of Reynolds number. Substituting values of Rh and Cf into above equations, we obtain (for both smooth-walled and rough-walled conduits) the well known equation for pipe-friction head loss, Circular pipe flowing full (laminar or turbulent flow): hf = f (L/D) (V 2/2 g) ………………. (7) and hf /L = S = f /D (V 2/2 g) ……………. . (8)

Friction in Circular Conduits For a circular pipe flowing full, by substituting Rh = r 0/2, where r 0 is the radius of the pipe in the eq. (4), we get hf = τ’ 0 L/(Rhγ) = 2τ0 L/(r 0γ) where the local shear stress at the wall, τ0, is equal to the average shear stress τ’ 0 because of symmetry.

Friction in Circular Conduits The shear stress is zero at the center of the pipe and increases linearly with the radius to a maximum value τ0 at the wall as shown in Fig. 8. 3. This is true regardless of whether the flow is laminar or turbulent. From eq. (4) hf = τ0 L/(Rhγ), we have τ0= hf (Rhγ)/L, substituting eq. (7) and Rh = D/4 into this, we obtain τ0= f (L/D)(V 2/2 g)(D/4)(γ/L) τ0= (f /4) γ (V 2/2 g) or τ0= (f /4) ρ (V 2/2) where γ=ρg With this equation, we can compute τ0 for flow in a circular pipe for any experimentally determined value of f.

Friction in Circular Conduits For laminar flow under pressure in a circular pipe, We may use the pipe-friction equation (7) with this value of f as given by the above equation.

Problem: Q: Stream with a specific weight of 0. 32 lb/ft 3 is flowing with a velocity of 94 ft/s through a circular pipe with f = 0. 0171. What is the shear stress at the pipe wall? Solution: γ = 0. 32 lb/ft 3 V = 94 ft/s f = 0. 0171 g = 32. 2 ft/s 2 τ0= ? τ0= (f /4) γ (V 2/2 g) τ0= (0. 0171/4)(0. 32)(94 x 94)/(2 x 32. 2) τ0= 0. 187 lb/ft 2

Problem: Q: Stream with a specific weight of 38 N/m 3 is flowing with a velocity of 35 m/s through a circular pipe with f = 0. 0154. What is the shear stress at the pipe wall? Solution: γ = 38 N/m 3 V = 35 m/s f = 0. 0154 g = 9. 81 m/s 2 τ0= ? τ0= (f /4) γ (V 2/2 g) τ0= (0. 0154/4)(38)(35 x 35)/(2 x 9. 81) τ0= 9. 13 N/m 2

Problem: Q: Oil of viscosity 0. 00038 m 2/s flows in a 100 mm diameter pipe at a rate of 0. 64 L/s. Find the head loss per unit length. Solution: ν = 0. 00038 m 2/s D = 100 mm = 0. 1 m Q = 0. 64 L/s = 0. 00064 m 3/s g = 9. 81 m/s 2 hf /L = ? Q = AV = (πD 2/4)V, V = (4 Q)/(πD 2) V = (4 x 0. 00064)/(3. 14 x 0. 1) = 0. 0815 m/s R = (DV)/ν = (0. 1 x 0. 0815)/(0. 00038) = 21. 45

Problem: f = 64 / R = 64/21. 45 = 2. 983 hf /L = S = f /D (V 2/2 g) hf /L = (2. 983/0. 1)(0. 0815 x 0. 0815)/(2 x 9. 81) hf /L = 0. 010 m/m

Friction in Non-Circular Conduits Most closed conduits we use in engineering practice are of circular cross section; however we do occasionally use rectangular ducts and cross sections of other geometry. We can modify many of the equations for application to non circular sections by using the concept of hydraulic radius. For a circular pipe flowing full, that Rh = A/P = (π D 2/4)/(πD) = D/4 D = 4 Rh This provides us with an equivalent diameter, which we can substitute into eq. (7) to yield hf = f (L/4 Rh)(V 2/2 g)

Friction in Non-Circular Conduits and when substitute into equation of Reynolds number, we get R = (DVρ)/μ = (4 Rh. V)/ν This approach gives reasonably accurate results for turbulent flow, but the results are poor for laminar flow, because in such flow viscous action causes frictional phenomena throughout the body of the fluid, while in turbulent flow the frictional effect occurs largely in the region close to the wall; i. e. , it depends on the wetted perimeter.

Entrance Conditions in Laminar Flow In the case of a pipe leading from a reservoir, if the entrance is rounded so as to avoid any initial disturbance of the entering stream, all particles will start to flow with the same velocity, except for a very thin film in contact with the wall. Particles in contact with the wall have zero velocity and with the slight exception, the velocity is uniform across the diameter.

Entrance Conditions in Laminar Flow As the fluid progresses along the pipe, friction origination from the wall slows down the streamlines in the vicinity of the wall, but since Q is constant for successive sections, the velocity in the center must accelerate, until the final velocity profile is a parabola as shown in Fig. 8. 3. Theoretically, this requires an infinite distance, but both theory and observation have established that the maximum velocity in the center of the pipe wall reach 99% of its ultimate value in a distance Le = 0. 058 RD We call this distance the entrance length. For a critical value of R = 2000, the entrance length Le equals 116 pipe diameters. In other cases of laminar flow with Reynolds number less than 2000, the distance Le will be correspondingly less in accordance with the above equation.

Entrance Conditions in Laminar Flow Within the entrance length the flow is unestablished; that is the velocity profile is changing. In this region, we can visualize the flow as consisting of a central inviscid core in which there are no frictional effects, i. e. , the flow is uniform, and an outer, annular zone extending from the core to the pipe wall. This outer zone increases in thickness as it moves along the wall, and is known as the boundary layer. Viscosity in the boundary layer acts to transmit the effect of boundary shear inwardly into the flow. At section AB the boundary layer has grown until it occupies the entire cross section of the pipe. At this point, for laminar flow, the velocity profile is a perfect parabola. Beyond section AB, for the same straight pipe the velocity profile does not change, and the flow is known as (laminar) established flow or (laminar) fully developed flow.

Entrance Conditions in Laminar Flow The flow will continue as fully developed so long as no change occurs to the straight pipe surface. When a change occurs, such as at a bend or other pipe fitting, the velocity profile will deform and will require some more flow length to return to established flow. Usually such fittings are so far apart that fully developed flow is common; but when they are close enough it is possible that established flow never occurs.