Dimensional analysis Drag force on sphere depends on: The velocity U The sphere diameter D The viscosity m of the fluid The density r of the fluid Other characteristics of secondary importance F = f ( U, D, m, r ) but dimensional analysis gives: Dimensional Analysis Jacob Y. Kazakia © 2002

Buckingham P Theorem Say we have : The parameters can be grouped in n - m dimensionless groups Where m is the number of independent dimensions needed to express the dimensions of all q ’s. We then have a relationship: Dimensional Analysis Jacob Y. Kazakia © 2002

Ex1. Journal bearing The load-carrying capacity W of a journal bearing depends on: The diameter D The length l The clearance c The angular speed w The viscosity m of the lubricant l D c We must now go through the following 6 steps: 1) The number of parameters n = 6 ( including W ) 2) Select as primary dimensions: M, L, t ( mass, length, time) Dimensional Analysis Jacob Y. Kazakia © 2002

Ex1 – cont. 3) Write the dimensions of all parameters in terms of the primary ones D  L l  L c  L  t –1  M L –1 t –1 W  M L t -2 All 3 primary dimensions are needed. r = 3 hence there must be n – r = 6 – 3 = 3 dimensionless groups. 4) Select repeating parameters: D, w, m 5) Find the groups: D a w b m g l  La t-b M g L-g t-g L = L 0 t 0 M 0 L: a - g + 1 = 0  a = -1 t : - b - g = 0  b = 0 M : g = 0  g = 0 P1 = D-1 l Dimensional Analysis Jacob Y. Kazakia © 2002

Ex1 – cont -1. We found P 1 = l / D we similarly get P 2 = c / D third group: D a w b m g W  La t-b M g L-g t-g M L t -2 = L 0 t 0 M 0 L: a - g + 1 = 0  a = -2 t : - b - g -2 = 0  b = -1 M : g +1 = 0  g = -1 or P3 = D-2 w-1 m-1 W P3 = W / (D2 w m ) 6) Write the relation among the P ‘s : Dimensional Analysis Jacob Y. Kazakia © 2002

Ex2. Extension of ex1 Suppose we had r in the list of the parameters in the previous example. We would then have an additional P ( nondimensional group) D a w b m g r  La t-b M g L-g t-g M L-3 = L 0 t 0 M 0 L: a - g - 3 = 0  a = 2 t : - b - g = 0  b = 1 M : g +1 = 0  g = -1 or P4 = D2 w r / m (Reynolds number) P4 = D2 w m-1 r Resulting in: Dimensional Analysis Jacob Y. Kazakia © 2002

Ex3. Belt in viscous fld. A continuous belt moving vertically through a bath of viscous liquid drags a layer of thickness h along with it. The volume flow rate Q of the liquid depends on m , r , g , h , and V where V is the belt speed. Predict the dependence of Q on the other variables. There are six parameters. The primary dimensions are : F, L, and t ( 3 of them) The repeating parameters will be: h, V, r We must find 6 – 3 = 3 groups. [Q] = L3 / t [m] = F t / L2 [r] = F t2 / L4 [g] = L / t2 [h] = L [V] = L / t h a V b r g Q  La Lb t-b F g t2g L-4g L3 t-1 = L 0 t 0 F 0 or by inspection we obtain: P1 = Q / ( V h2 ) Dimensional Analysis Jacob Y. Kazakia © 2002

Ex3. – cont. h a V b r g m  La Lb t-b F g t2g L-4g F t L-2 = L 0 t 0 F0 L: a + b - 4g - 2 = 0  a = -1 t : - b + 2g +1 = 0  b = -1 F : g +1 = 0  g = -1 P2 = m / ( hVr ) Or equivalently P2 = hVr / m h a V b r g g  by inspection we get: P3 = V2 / ( g h ) Result: Dimensional Analysis Jacob Y. Kazakia © 2002

Ex4. Choice of repeating parameters A fluid of density r and viscosity m flows with speed V under pressure p. Determine a relation between the four parameters [V] = L / t [r] = M / L3 [p] = M / (L t2 ) [m] = M / ( L t ) V a r b p g m  La t-a M b L-3b M g L-g t-2g M L-1 t -1 = L 0 t 0 F0 L: a - 3b - g - 1 = 0 t : - a - 2g -1 = 0 M : b + g +1 = 0 But this system has no solution. See next slide  Dimensional Analysis Jacob Y. Kazakia © 2002

Ex4. –cont. a - 3b - g - 1 = 0 - a - 2g -1 = 0 b + g +1 = 0 Inconsistent system. No solution is possible But if we chose out repeated variables differently: V, r, m then everything is O.K. V a r b m g p  La t-a M b L-3b M g L-g t-g M L-1 t -2 = L 0 t 0 F0 Produces: a = -2 , b = -1, g = 0 and hence p / r V2 it is now easy to see why the previous choice did not work. Dimensional Analysis Jacob Y. Kazakia © 2002

Model flow Prototype torpedo: D = 533 mm , Length = 6.7 ft Speed in water: 28 m/sec Model: Scale 1/5 in wind tunnel Max wind speed : 110 m/sec Temperature : 20 0C Force on model: 618 N Find : a) required wind tunnel pressure for dynamically similar test b) Expected drag force on prototype We assume the parameters: F, V, D, r, m and we get by the earlier method: To attain dynamically similar model test we must have equal Reynolds numbers in both flow situations. Dimensional Analysis Jacob Y. Kazakia © 2002

Model flow – cont. Water at 20 0C mp = 1 x 10-3 Pa sec Air at 20 0C mM = 1.8 x 10–5 Pa sec We must now find the pressure that will give this type of air density. IDEAL GAS LAW: (about 20 atmospheres) The force: Dimensional Analysis Jacob Y. Kazakia © 2002