# ProblemSet1_soln_19.pdf - Fluid Mechanics(ECH 141 Problem...

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Fluid Mechanics (ECH 141) Problem Set #1 1. Deen 1.1 Solution a) Of interest is the time t h for an object of mass m to fall a distance h under the influence of gravity g, when it is initially at rest (so the velocity at t=0 is not a paramter). We have 4 dimensional quantities (n=4) and 3 dimensions (M, L and T, so j=3), which suggests there is only one dimensionless group. Suppose we form it as Π = m a h b g c t h . Writing the dimensions of the quantities shows that Π = M a L b L T 2 c T . We conclude that a=0, since the units of mass cannot be canceled by any other dimensional quantity. The other equations are b + c = 0 2c + 1 = 0 so that c=1/2 and b=-1/2, and Π = g h 1/2 t h . Since there are no other dimensionless groups, Π = K or t h = K h g 1/2 . b) We can still omit the mass, since there are no dimensional quantities with units of mass that could possibly be used to include it. We have 4 dimensional quantities: t h , g, h and U. We have two primary dimensions: L and T. Then we expect 2 dimensionless groups. As one dimensionless group, use the result from part (a), Π 1 = g h 1/2 t h .
To form the second, let Π 2 = h b 2 g ( ) c 2 U so Π 2 = L b 2 L T 2 c 2 L T from which b 2 + c 2 + 1 = 0 2c 2 1 = 0 . Then c 2 = 1 2 and b 2 = 1 2 so Π 2 = U hg . Then we expect Π 1 = f Π 2 ( ) or t h = h g 1/2 f U hg . c) Newtonian mechanics states that the mass times acceleration of an object equals the force acting on it. In this case, the force is that due to gravity, so if z is a coordinate aligned with gravity, m d 2 z dt 2 = mg or dz dt = gt + C 1 . Since dz/dt is the velocity U at t=0, C 1 =U and dz dt = gt + U , or
z(t) = 1 2 gt 2 + Ut + C 2 . Since z(0)=0 (we are measuring fall distance from the position at t=0), the second constant is zero and z(t) = 1 2 gt 2 + Ut . Then to fall a distance h in time t h we have 1 2 gt h 2 + Ut h h = 0 . Solving the quadratic equation yields t h = U g + U 2 g 2 + 2h g . Then Π 1 = t h g h 1/2 = U gh + U 2 gh + 2 , or Π 1 = −Π 2 + Π 2 2 + 2 .