Exam in Fluid Mechanics SG2214

Transcription

1 Exam in Fluid Mecanics G2214 Final exam for te course G / Examiner: Anders Dalkild Te point value of eac question is given in parentesis and you need more tan 20 points to pass te course including te points obtained from te omework problems. Copies of Cylindrical and perical Polar Coordinates, wic will be supplied, can be used for te exam as well as a book of basic mat formulas and a calculator. 1. iscous flow in a vertical cannel (10 p.) Consider te steady state flow of an incompressible, viscous fluid between two infinite, vertical parallel plates. Te pressure gradient is zero and te flow is driven by gravity. One of te plates is moving at constant speed in te vertical direction, wereas te oter is stationary. Determine te velocity of te moving plate suc tat te volumetric flux in te cannel is zero x U 0 g y Figure 1. Flow configuration for exercise 1.

2 2 2. Temperature field in a porous cannel Consider te incompressible flow between two porous, parallel infinite surfaces at distance apart. Te surfaces are at rest and te fluid flows troug te pores of te surfaces at constant velocity 0 from te lower to te upper surface. Te velocity parallel to te surfaces is zero. Te lower surface is at temperature T 0 and te upper at temperature T 0 + T 0. It is assumed tat te incoming fluid takes te temperature of te lower surface and te outgoing fluid as te temperature of te upper surface. y T 0 +ΔT 0 v 0 0 x T 0 a) (4 p.) erify tat te temperature field at steady state ( ) e ρcp 0 y k 1 T (y) = T 0 + T 0 e ρcp 0 k 1 satisfies te termal energy equation ρc p DT Dt = k 2 T + Φ were te velocity field is u = (0, 0 ). b) (4 p.) Calculate te total eat flux density vector (convective plus diffusive) ρuc p T k T and sow tat it is constant between te surfaces. c) (2 p.) Discuss te direction of te total eat flux.

3 3. Energy equation (5 p) Give a pysical interpretation to eac of te integrals in te equation D ρ (e + 12 ) Dt u iu i d = ρg i u i d + u i T ij n j d q i n i d, were is a fix control volume and its surface. q i = k T x i, T ij = pδ ij + τ ij, τ ij = 2µe ij + µ b δ ij e kk. e ij = e ij 1 3 δ ij u k ; e ij = 1 2 ( u i x j + u j x i ). (5 p) Apply te Gauss teorem f i f i n i d = d x i to eac of te surface integrals in te equation. ow tat p τ ij (u i T ij ) = u j + u i p u j + 2µe ij e ij + µ b x j x j x j x j ( ) 2 uk, and give a pysical interpretation to eac of te five terms. 3

4 4 4. Potential flow An irrotational vortex line is positioned at a distance above a orizontal wall. Obtain te complex potential of te flow by superposition of an artificial mirror image of te original vortex. (For a single vortex at z 0 te complex potential is given by F (z) = iγ 2π ln(z z 0)) Γ + (4 p.) Determine te velocity field in Cartesian coordinates. (4 p.) Let te vortex ( and its mirror) move parallel to te wall at constant velocity U to te left. Wat is te pressure difference p A p between te point A at te wall rigt below te vortex and a location infinitely far away from te wall? (2 p.) We may consider tis a simplistic model of a two-dimensional lifting airfoil moving at constant eigt above a wall. (Tis approximation is valid as long as te airfoil cord c << ) Te pressure difference p A p can be related to te angle of attack α of te airfoil. Derive tis relation wen te lift coefficient is C L = 2π sin α. By definition C L L 1 2 ρu c, 2 were c is te given cord of te airfoil and L is te lift force per unit widt. It is also a fact tat for a two-dimensional airfoil L = ργu.

5 5 5. Energy balance for te tokes problem Te budget for te mecanical energy E M in a fixed control volume wit surface is given by (integral form of te energy equation) d E M d + E M u j n j d = ρg i u i d + p u k d + u i T ij n j d dt were E M = 1 2 ρu iu i, T ij = pδ ij + τ ij, τ ij = 2µe ij. e ij = e ij 1 3 δ ij u k ; e ij = 1 2 ( u i x j + u j x i ), Φ = 2µe ij e ij Φd, Consider te tokes problem of an impulsively started, infinite flat plate of velocity U 0 in an incompressible fluid of kinematic viscosity ν = µ/ρ and density ρ. Te exact solution is given by u(y, t) = U 0 F (η); η = y δ(t) ; δ = 4νt; were F ( ) = 0. y F (η) = 1 2 π η 0 e η2 dη, U 0 x a) (4 p.) Use a control volume adjacent to te plate and extending into te undisturbed fluid far above te plate, and calculate te work rate per unit area A needed to move te plate (at any given time t).

6 6 b) (3 p.) ow tat a fraction 1/ 2 of te applied work rate is dissipated into eat in te fluid. c) (3 p.) Wat appens to te remaining part of te applied work rate? Te answer sould be motivated by identifying te terms in te mecanical energy equation associated to te quantities previously calculated in a) and b). (Hint: wat terms in te mecanical energy equation are zero?)