ASTR 320: Solutions to Problem Set 3

Transcription

1 ASTR 30: Solutions to Problem Set 3 Problem : The Venturi Meter The venturi meter is used to measure the flow speed in a pipe. An example is shown in Fig., where the venturi meter (indicated by the dashed lines) is placed in a pipe of diameter A. The venturi meter itself consists of a pipe of diameter A, as indicated. The pipe transports an incompressible fluid of density ρ with a flow velocity u (this is the quantity to be meaured). The flow velocity in the narrow pipe is u. The main and narrow pipes are connected via a U-shaped pipe that is filled with a fluid of density ρ > ρ. The quantity to be measured is h, the difference in the heights of the colums of the dense fluid. a) Assume that the venturi meter of Fig. is located in the Earth s gravitational field, with the direction of gravitational acceleration perpendicular to the flow direction. Express the pressure difference at points and in terms of u, u, and ρ. You can assume that the flow is laminar. SOLUTION: Using the fact that the Bernoulli constant is conserved along streamlines, we have that u + h + Φ = u + h + Φ Using that Φ = Φ, and that for an incompressible fluid the pressure potential h = P/ρ, the above reduces to u + P ρ = u + P ρ from which we infer that P P = ρ ( u u )

2 Figure : Illustration of a venturi meter. b) Derive an expression for u as function of h, the densities ρ and ρ, and the areas A and A. SOLUTION: The flow rate through the main pipe is ρ u A and has to be equal to the flow rate through the narrow pipe, which is ρ u A. Hence, we have that u = u (A /A ). In addition, we have that P P = ( ρ ρ) g h with g the gravitational acceleration. This follows from demanding that the pressure at the two points along the U-shaped pipe connected by the long dotted line is equal to each other. Equating this to the expression for the pressure difference derived under a), we have that Hence, we have that ( ρ ρ) g h = ρ ( u u ) = ρ u [ (A ) ] A ( ) [ (A ) ρ u = g h ρ ] A

3 Problem : Vorticity in a thin disk Consider an infinitessimally thin disk of fluid in rotation around the disk s (vertical) symmetry axis. It is given that the flow is symmetric around the same symmetry axis, and that the circulation around any curve C on the disk is zero, as long as the curve does not enclose to symmetry point of the disk, R = 0. Derive an expression for the velocity field u( x), i.e., what are the various components of u as function of location in the disk? SOLUTION: First we note that cylindrical coordinates are the coordinates of choice here. We pick the z-axis to be along the symmetry axis of the disk, and set z = 0 for the disk plane. The circulation corresponding to some curve C is given by Γ C = u d l = w ds C where S is the surface bound by C (which, for simplicity, we take to be in the disk plane). Since Γ C = 0 for any C that does not enclose R = 0, we must have that the vorticity w = 0 everywhere, except possibly at R = 0. In cylindrical coordinates, the vorticity is given by [ w = R R (R u θ) R S u R θ Due to the azimuthal symmetry of the flow, we have that u R / θ = 0 (i.e., there can t be any θ-dependence. Hence, using that the vorticity has to vanish everywhere except at R = 0, we have that R (R u θ) = 0 which implies that u θ /R. In addition, we have u z = 0 (to keep the disk infinitessimally thin), while u R is unconstrained. NOTE: u R is not necessarily zero. It can be non-zero and with a dependence on R; it is not allowed, however, to have a dependence on θ (violation of azimuthal symmetry). ] ê z Problem 3: Lift on an airfoil 3

4 Consider the wing of airplane, a cross section of which is depicted below. Let c be the chord of the wing. Because of the shape of the airfoil, air flows over the wing with a speed u + u, and under the wing with a speed u u. Here u is the speed of the airplane with respect to the air far from the wing. a) Give an expression for the circulation, Γ, around the airfoil in terms of c, u and u. In doing so, you may ignore the thickness of the wing. SOLUTION: The circulation is defined as Γ = u d l Since we may ignore the thickness of the wing, the contour consists only of the part over the wing, combined with that under the wing. Hence, using that the circularity is defined in the counter-clockwise direction, we have that where u u + u. Γ = (u + u ) c + (u u ) c = c u b) Use Bernoulli s theorem to calculate the lift (=force) per unit wing-length on the airfoil. You may assume that the plane is flying sub-sonic, so that the air may be treated as incompressible. Express your answer in terms of u, Γ, and the density of the air, ρ. SOLUTION: According to Bernoulli s theorem, the Bernoulli function B = u / + Φ + h is constant along the flow. If we assume that the flows that go over and under the wings have identical values for B (this *is* a questionable assumption!!!) and we ignore differences in the gravitational potential above and below the wing (this is a reasonable assumption, since we were told that we may ignore the thickness of the wing), then we have that u + h = u + h where u = u + u and u = u u = u u. In addition, we have that h = ε + P /ρ and h = ε + P /ρ. Since ε = ε(t), and we may 4

5 ignore differences in temperature and density above and below the wing (the latter derives from the fact that the air is incompressible), we thus have that ρu + P = ρ(u u) + P If we assume that u u then this translates to P P P = ρu u The pressure is the force per unit area. The area of the wing is the product of the chord, c, and the length of the wing L. Hence, the force per unit length (=the lift) can be written as Lift = c P = cρu u = ργu where in the last step I have once again used that u u, such that u u. 5