MIT Department of Mechanical Engineering 2.25 Advanced Fluid Mechanics Problem 7.03 This problem is from “Advanced Fluid Mechanics Problems” by A.H. Shapiro and A.A. Sonin A metal ball falls at steady speed in a large tank containing a viscous liquid. The ball falls so slowly that it is known that the inertia forces may be ignored in the equation of motion compared with the viscous forces. (a) Perform a dimensional analysis of this problem, with the aim of relating the speed of fall V , to the diameter of the ball D, the mass density of the ball ρb , the mass density of the liquid ρl , and any other variables which play a role. Note that the ”eﬀective weight” of the ball is proportional to (ρb − ρl )g. (b) Suppose that an iron ball (sp. gr.=7.9, D=0.3 cm) falls through a certain viscous liquid (sp. gr. = 1.5) at a certain steady-state speed. What would be the diameter of an aluminum ball (sp. gr. = 2.7) which would fall through the same liquid at the same speed assuming inertial forces are negligible in both ﬂows? 1 Solution: (a) Non-dimensional Groups In steady state, the body force (weight, W ) must be balanced with buoyancy (FB ) and drag (FD ) forces. 3 4 D π = FD (7.03a) Wef f = (ρb − ρl )g 3 2 ρl [ML −3 µ ] −1 [ML V T −1 D −1 ] [LT ] 1 [L ] FD [ML1 T−2 ] Thus we have n = 5 variables k = 3 primary variables ⇒ j = 2 dimensionless group For our primary variables, we choose (1) a ﬂuid property: ρl , (2) a ﬂow parameter: V , and (3) a geometric parameter: D. Therefore, the ﬁrst dimensionless group is µ = f1 (ρl , V, D) or Π1 = K1 µρal V b Dc where K1 is a constant. Thus, M :0=1+a L : 0 = −1 − 3a + b + c T : 0 = −1 − b ⇒ a = b = c = −1 Π1 = K1 µ K1 = Re ρl V D (7.03b) Similarly, we can obtain the second non-dimensional parameter. Π2 = K2 FD ρal V b Dc ⇒ Π2 = K2 FD ρl V 2 D 2 When K2 = 2, this becomes the drag coeﬃcient CD , i.e., CD = where A is a characteristic cross-section area. 2 FD 1 2 2 ρV A (7.03c) (b) Example of Similarity In part (a), we obtained two non-dimensional variables. In highly viscous ﬂows or fast speed ﬂows, the drag force is a function of the Reynolds number. However, if the speed of the ball is very small (Re << 1), then the drag force is no longer a function of Reynolds number. When the non-dimensional parameters are consistent in two situations, the ﬂow ﬁelds are also similar. Let’s make the drag coeﬃcients are the same in the two cases. (ρi − ρl )g CD = ρV Di 3 4 3π 2 2 D2 i (ρa − ρl )g = 4 3π Da 3 2 ρV 2 Da2 where the subscripts i and a denote the iron and aluminum. ⇒ (7.9 − 1.5) × (0.3)3 (2.7 − 1.5) × Da3 = 2 Da2 (0.3) Therefore, the diameter of an aluminum ball which satisﬁes the similarity is Da = 1.6 cm (7.03d) D 3