1 MAE 21A FINAL EXAM OLUTION PROBLEM 1: Dimensional analysis of the foling of paper (2 points) (a) We wish to simplify the relation between the fol length l f an the other variables: The imensional matrix is written L M T Θ l f = F (EI, v, s, g, h). (1) [l f ] [EI] [v] [ s ] [g] [h] 1 2 1 2 1 1 1 1 2 1 2 We have N + 1 = 6, an the rank is easily foun to be r = 3. Therefore, we can construct 6 3 = 3 imensionless groups. Choosing s, g an h as variables with inepenent imensions, we easily fin: The relation (1) therefore simplifies to where the function G remains unknown. π = l f h, π 1 = EI s gh 3, π 2 = l f EI h = G s gh 3, v (gh) 1/2 (2) v. (3) (gh) 1/2 (b) The parameter π 2 is similar to a Froue number an compares the relative magnitue of inertial forces to gravitational forces. If v is so small that inertial forces are negligible compare to gravitational forces, this implies that π 2, in which case the imensionless fol length becomes solely a function of π 1 : l f EI h = H s gh 3 (5) (c) In the limit of large h, the epenence upon h isappears in equation (5) only if H(π 1 ) C π 1/3 1 where C is a imensionless constant. This implies (4) l f EI 1/3 h C s gh 3, (6) EI 1/3 l f C (7) s g PROBLEM 2: Win-riven flow insie a lake (3 points) (a) We write own the x an y components of the Navier-tokes momentum equation for an incompressible flow: ( u t + u u x + v u ) y x + µ u x 2 + 2 u y 2, (8) ( v t + u v x + v v ) y y g + µ v x 2 + 2 v y 2. (9)
2 For a parallel flow of the form u(x, y, t) = u(y)e x, these simplify to: p x = u µ2 y 2, (1) p = g. y (11) (b) We solve for the pressure by integrating equation (11) from y to h(x) (keeping x constant): p(x, h(x)) p(x, y) = g(h(x) y). (12) Noting that the pressure at the free surface is atmospheric: p(x, h(x)) = p atm, this gives: p(x, y) = g(h(x) y) + p atm (13) (c) At the bottom of the lake, the no-slip conition applies: u() =. At the free surface (y = h(x) h ), the viscous stress insie the water balances the stress applie by the win: τ xy (h ) = µ u y (h ) =, i.e. u y (h ) = µ (14) () To fin the velocity profile, we turn to equation (1). Using equation (13) for the pressure, it can be written This can be integrate twice: 2 u y 2 = g h µ x. (15) u(y) = g h 2µ x y2 + Ay + B. (16) We use the bounary conitions of part (b) to solve for the integration constants A an B. First, the no-slip conition u() = gives B =. The conition on the shear stress at the free surface gives: 2 u y 2 (h ) = g h µ x h + A = µ i.e. A = µ g h µ x h. (17) The velocity profile is therefore given by: u(y) = µ y + g µ h y 2 x 2 yh (18) Note that the slope of the free surface remains unknown. (e) The volumetric flow rate per unit wih in the z irection is All calculations one, Q = u(y) y = [ µ y + g ] h y 2 µ x 2 yh y. (19) Q = h2 2µ g h 3µ x h3 (2)
3 y u(y) x (f) At steay state, we must have Q = (zero flow rate across any vertical cross-section), which allows us to solve for the slope of the free surface: h 2 2µ g h 3µ x h3 =, i.e. h x = 3 2gh (21) ubstituting this into equation (18) yiels the final expression for the velocity profile: u(y) = h 4µ y ( 3 y ) h h 2 (22) PROBLEM 3: queeze flow (2 points) (a) If the top surface is moving own at a constant velocity U, the gap height ecreases linearly with time as h(t) = h U t (23) (b) We apply conservation of mass to a eforming cylinrical control volume that is circular with raius r in the plane of the plates an spans the height h(t). Conservation of mass is written V = n (w u), (24) V (t) where w is the surface velocity of the control volume an u is the flui velocity. The flui has constant ensity, therefore the left-han sie simplifies to V = (h(t)πr2 ) = πr 2 h = πr2 U. (25) V (t) In the right-han sie, we have that n (w u) = on the top an bottom bounaries (or there woul a mass flux through the walls), an w = on the circular bounary. Therefore, n (w u) = u r (r, t)2πrh(t). (26) Equating (25) an (26) an simplifying leas to u r (r, t) = U r 2h(t) (c) Differentiate with respect to time: u r t = U r 1 = U ( r 1 ) h 2 h(t) 2 h(t) 2 which simplifies to u r t = (27) (28) U 2 r 2h(t) 2 (29)
4 PROBLEM 4: Unsteay Couette flow (3 points) (a) The x-component of the momentum equation is written: ( u t + u u x + v u ) y x + µ u x 2 + 2 u y 2. (3) For uniirectional flow of the form u(x, t) = u(y, t)e x, this simplifies to The bounary conitions are no-slip at the top an bottom bounaries: u t = ν 2 u y 2 (31) u(, t) = an u(h, t) = U = at (32) (b) We assume the form u(y, t) = u (y) + tu 1 (y). ubstitute this into the momentum equation: t (u + tu 1 ) = ν 2 y 2 (u + tu 1 ), (33) For this relation to hol for all values of t, we must have u 1 = ν 2 u y 2 + tν 2 u 1 y 2. (34) 2 u 1 y 2 = an 2 u y 2 = 1 ν u 1 (35) (c) We first integrate the first equation for u 1 twice, yieling Now, substitute this into the secon equation for u, which becomes u 1 (y) = Ay + B. (36) 2 u y 2 = 1 (Ay + B). (37) ν Integrating twice, we get u (y) = 1 ν ) (A y3 6 + B y2 2 + Cy + D. (38) The total velocity fiel is therefore expresse as u(y, t) = 1 ν ) (A y3 6 + B y2 2 + Cy + D + t(ay + B). (39) To etermine the four integration constants, we apply the bounary conitions. At y =, we have u(, t) = = D + tb. (4) ν
5 For this to hol for all t, we must have D = B =. imilarly, at y = h: u(h, t) = at = 1 ) (A h3 ν 3 + Ch + tah, (41) which implies a = Ah an A h3 3 + Ch =, (42) A = a h an C = Ah2 6 Inserting these values into equation (39), we obtain the velocity fiel () The shear stress is given by At the bottom wall (y = ), this gives u(y, t) = a 6νh (y3 h 2 y) + at y h = ah 6. (43) τ xy (y, t) = µ u y = µa 6νh (3y2 h 2 ) + µ at h = a 6h (3y2 h 2 ) + µ at h. (45) (44) τ xy (, t) = ah 6 + µat h (46) which is zero at t = h 2 /6µ. At the top wall (y = h), we fin τ xy (h, t) = ah 3 + µat h (47) which is zero at t = h 2 /3µ. (e) Integrate the velocity profile: from which Now, we also have Therefore, u(y, t) y = terms inepenent of t + u(y, t) y = a y h at y y, (48) h y = ah 2. (49) τ xy (h, t) τ xy (, t) = ah 3 + ah 6 = ah 2. (5) u(y, t) y = τ xy (h, t) τ xy (, t) (51) This can be interprete as Newton s secon law (conservation of momentum) applie to a vertical slice of flui of unit wih x = 1, where the only forces applie on the flui are shear stresses at the top an bottom bounaries.