# 2 deen 13 use the whisk wire diameter viscosity and

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2. Deen 1.3 (Use the whisk wire diameter, viscosity and surface tension as recurring variables)
Π 2 = L a 2 M LT b 2 M T 2 c 2 L T . For the first dimensionless group we find that L a 1 b 1 + 3 = 0 M b 1 + c 1 = 0 T b 1 2c 1 = 0 so that b 1 = c 1 = 0 and a 1 = 3 yielding Π 1 = V D 3 . For the second group we find that L a 2 b 2 + 1 = 0 M b 2 + c 2 = 0 T b 2 2c 2 1 = 0 . Then c 2 = 1, b 2 = 1 and a 2 = 0 . So the second group is the Capillary number in Table 1.6, Π 2 = μ U γ 3. Deen 1.8 Hints: For part (a), find k dimensionless groups as ratios of the internal dimensions to the pipe diameter D, before beginning the usual procedure to find the P groups. One of the remaining groups should correspond to one of the well-known groups in Table 1.6—which one would you expect to be relevant? P Solution 3 1
pressure drop D P and the viscosity μ . Of the latter two, the pressure drop is chosen because it is the measurable quantity in the experiments, and is of primary interest in the design. It is not necessary to choose the viscosity as a non-recurring variable, but that choice is suggested by the statement in part (c) that the pressure drop becomes independent of viscosity at high Reynolds number. If the viscosity only appears in one dimensionless group, then that observation can be modeled by using equations that render one dimensionless group irrelevant when Re>>1. The recurring quantities are then D, r , and U. As suggested in the hint, k dimensionless groups can be found immediately as ratios of the internal dimensions to D, or Π i = d i D for 1 i k . The two remaining groups have the form Π k + 1 = D a 1 ρ b 1 U c 1 μ and Π k + 2 = D a 2 ρ b 2 U c 2 Δ P . Substituting dimensions, Π k + 1 = L a 1 M L 3 b 1 L T c 1 M LT and Π k + 2 = L a 2 M L 3 b 2 L T c 2 M LT 2 . Finding the coefficients shows that Π k + 1 = μ ρ UD = 1 Re and Π k + 2 = Δ P ρ U 2 .