Review of Undergraduate Material - The Basics of Fluid Dynamics | MAE 643, Study notes of Dynamics

Download Review of Undergraduate Material - The Basics of Fluid Dynamics | MAE 643 and more Study notes Dynamics in PDF only on Docsity! Review of undergraduate material August 27, 2007 2 The Basics of Fluid Dynamics Graduate students come with varied backgrounds. It is assumed here that all have had one undergraduate course in fluid mechanics, but the perspective in such courses, and what students have retained from them, is also quite diverse. This document, supplemented by Sections 4.1 and 4.2 of the Lec- ture Notes, is intended as a summary of methods and ideas assumed to be known, with a few fine points thrown in for good measure. Very little class time is devoted to this chapter, to be covered in independent study mode. Undegraduate textbooks are on reserve at the library. Bernoulli’s equation (if applicable), dimensional analysis (Ch.4) and control volume analysis are mathematically simple, and students in this field must be able to rely on them: no excuses. 0.1 Bernoulli’s equation Obtained from streamwise momentum: assume no viscous forces, incom- pressible steady state. Project the momentum equations on the streamline and integrate between two points A and B. The result is ρV 2A 2 + ρgzA + pA = ρV 2B 2 + ρgzB + pB (1) This equation, and its modified form including losses (below) is one of the most useful relations for undergraduate fluid mechanics. One of the reasons is that it is algebraic rather than differential. It also contains the hydrostatic equation as a particular case. But its limitations should be noticed, and a few details about its proper use will be emphasized. Let us consider (Fig. 0.1) the example of a water tank with an orifice B located a distance H below the free surface A. Assume the bulk of the water in the tank to be motionless (how good an approximation is this?), and take the free surface as reference level (zA = 0) at reference pressure (pA = 0). The jet issues into ambient air and is therefore at pB = 0 (neglecting change in pressure between A and B: is this accurate?). Then, Bernoulli’s equation gives VB = √ 2gH (2) known as Torricelli’s formula. 0.1. BERNOULLI’S EQUATION 5 Figure 3: A pipe flow with a manometer 10 2 10 3 10 4 10 5 10 6 10 7 10 −2 10 −1 Reynolds number D ar cy fr ic tio n fa ct or Figure 4: Sketch of the Moody diagram 6 tables: entrance, exit, bends, fittings, etc. Note that exit loss (Km = 1) is equivalent to the exit kinetic energy: use one or the other, not both, depend- ing on whether point B is the pipe exit or a point at rest in some discharge reservoir. Thus Bernoulli’s equation with losses gives a modified Torricelli relation as VB = √ √ √ √ 2gH 1 + f L D + ∑ Km (7) where the hidden dependence of f on VB requires iterative solution. For turbulent flow, the dependence is weak, but if we assume laminar pipe flow and ignore entrance effects, then f = 64 Re = 64 ν VBD (8) (See below for control volume analysis of Poiseuille flow). A summary of relevant ideas could start with the mind-map of Fig. 5: build your own, add ideas, express them in your own way. 0.2 Reynolds Transport Theorem Consider a generic property P (scalar of vector) with specific value per unit mass p: P = ∫ p ρ dV (9) In very few instances, the evolution of the property is governed by an equation of the type dtP = RHS (10) where the right-hand-side (r.h.s.) RHS = ∫ rhs ρ dV represents the various mechanisms by which the value of P can be modified. For example, in a bank account, there can be deposits and withdrawals and assorted fees, and the r.h.s. would reflect these operations and account for changes in daily balance. Any perplexity at apparent leakage is due to our incomplete accounting of the various mechanisms at play. This is the spirit of conservation laws in physics. Within classical physics (i.e. excluding the implications of E = mc2), mass is conserved through motion and chemical reactions, and we have dtm = dt ∫ ρ dV = 0 (11) 0.2. REYNOLDS TRANSPORT THEOREM 7 Figure 5: Some Bernoulli-related ideas: add your own! 10 Figure 7: Control volume and flux through the boundary combining the element area and the direction of its normal. Then, the dot product with velocity captures the relation between the directions and mag- nitudes of the flow and the surface in the vicinity of a given point: it measures how much volume leaves the system. That volume carries the properties of its contents with it. A differential form of the conservation laws for flow systems is easily obtained. We will assume that the control volume itself is fixed (i.e. the control surface is not moving), a restriction that can be removed at the cost of additional terms. Then the time-derivative can move in and out of the volume integral. Making use of the divergence theorem (Gauss), we obtain: ∫ [∂t(ρp) − ρ rhs + ∇ · (ρpU)] dV = 0 (17) Since this holds for any fixed volume, the integrand vanishes ∂t(ρ p) + ∇ · (ρ p U) = ρ rhs (18) 0.2.1 Application: a simple 2-D flow Consider the right-angled corner dividing a uniform steady flow as shown (Fig. 8). Assume uniform flow at the exit points. The flow is exposed to ambient pressure except at the points of contact with the wedge. The control surface should include the entrance and exit sections, con- nected by arbitrary surfaces across which there is no flow. Applying Bernoulli’s 0.2. REYNOLDS TRANSPORT THEOREM 11 Figure 8: Control volume and flow around a wedge. equations between 1 and 2 (and 1 and 3), one finds that V1 = V2 = V3, so that single V is adequate notation in this case. The mass flux integral gets contributions only from the parts of the control surface where there is normal velocity, and the total mass is constant for the control volume: ∂tm + ∮ ρ(U · dA) = 0 + ( ∫ A1 + ∫ A2 + ∫ A3 )ρ(U · dA) = 0 (19) Because the flux term is a dot product, it is independent of any choice of axes (as yet unspecified!); furthermore, the dot product is always negative at an entrance point, and positive at an exit point; finally, only the velocity component normal to the surface contributes to the dot product (Fig. 9). Thus, we have ρ(−A1V1 + A2V2 + A3V3) = 0 (20) and therefore A1 = A2 + A3 (21) Momentum balance requires a little more care. Ignore friction? What role does it play? A free body diagram reflects the forces applied by the 12 Figure 9: FBD for wedge flow; the force is applied to the fluid, its reaction to the wedge. wedge to the fluid in order to deflect the flow; and a choice of cartesian axes is also required (Fig. 9). For steady flow, the RTT reduces to ( ∫ A1 + ∫ A2 + ∫ A3 )ρU(U · dA) = F (22) which can be broken into x- and y-components: ( ∫ A1 + ∫ A2 + ∫ A3 )ρUx(U · dA) = Fx (23) ( ∫ A1 + ∫ A2 + ∫ A3 )ρUy(U · dA) = Fy (24) The important thing to notice is that the dot product retains all components, with the signs as for mass balance (i.e. independent of axes); whereas the vector for momentum is split into components for which the value and sign depend on the choice of axes. Putting the flux term in parentheses is a useful reminder. Therefore, we have ρV1(−A1V1) + ρV2 cos α(A2V2) + ρV3 sin α(A3V3) = Fx (25) 0.2. REYNOLDS TRANSPORT THEOREM 15 Figure 11: Control volume in fully developed pipe flow distance x (fully developed) or on angular position θ by symmetry. (Note that cylindrical coordinates do simplify the problem!). Steady flow is assumed. Then, consider a cylindrical element of radius r and length dx centered in the pipe (Fig. 11). One consequence of fully developed flow is that there can be no radial component of velocity. Indeed, assume an outward flow at any point: by symmetry, it must apply along the circumference, which would increase mass flow rate (i.e. velocity) outside and decrease it inside: if this were true, u would depend on x, which means the flow is evolving (see entrance flow, next chapter). Then, since the mass flow rates ∫ ρ(±u(r)2πr dr) cancel out through the bases of the cylindrical c.v., and is zero through the side surface, mass is conserved. Similarly, momentum balance gives ∫ x ρu(r)(−u(r)2πr dr)+ ∫ x+dx ρu(r)(+u(r)2πr dr) = −πr2 dp dx dx−2πr dxτ(r) (31) The viscous stress is given by Newton’s formula τ = µdru(r) (32) The momentum fluxes cancel out (fully developed, again), and we are left with −r2dxp = 2µrdru (33) Now take a derivative with respect to r (which amounts to taking two c.v.’s of slightly different rs, taking the difference and keeping the leading term: 16 −0.2 0 0.2 0.4 0.6 0.8 1 1.2 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 r/ R u . 4 µ / p‘ R2 Figure 12: Poiseuille profile in a circular pipe an annular c.v. of thickness dr). We get −dxp = −p ′ = µ 1 r dr(rdru), (34) which expresses the balance of pressure drop and viscous resistance. For fully developped flow, dxp = p ′ must be a constant (and negative if the motion is in the direction of increasing x). Then, integration is straight- forward, and gives − r2 4 p′ µ = u(r) + A ln r + B (35) The boundary conditions are that u should be finite at the centerline, and vanishes (no-slip) at r = R. Then the solution takes the form of the Poiseuille solution (Fig. 12): u(r) = p′R2 4µ (1 − r2 R2 ) (36) It is easy to verify that the average velocity is half the centerline velocity (this is different in the case of the plane channel, make sure you understand why). We can also calculate the wall stress as τw = − p′R 2 (37) 0.2. REYNOLDS TRANSPORT THEOREM 17 Figure 13: About fully-developed pipe flows 20 Figure 14: Figures relative to the problems as labeled. 0.4. FOOD FOR THOUGHT 21 below the surface. A manometer tube of diameter d is connected to the top of the siphon, as shown. Determine the manometer level ` relative to the free surface. (Neglect friction.) 11. A jet of liquid (density ρ) and area A strikes a block and splits into two jets, as on Fig. 14. Neglect friction. The upper jet of area a exits at an angle β while the lower jet is turned 90o downward. Neglecting fluid weight, derive the formula for the components of force applied ot the block. 12. A curved nozzle assembly that discharges to the atmosphere is shown on Fig. 14. The nozzle weight is W . Determine the relation between the force applied by the nozzle on the coupling to the inlet pipe and the problem parameters. 13. Consider the siphon pictured on Fig. 14. Determine the relation be- tween the discharge velocity and the problem parameters.