Homework #1 Solution Problem 1

Transcription

1 Homework #1 Solution Problem 1 In this problem, we are asked to consider the design of a straw by a fast-food company for a new milkshake. We need to perform a dimensionless analysis so that there is sufficient volumetric flow through the final product, given some reasonable amount of suction. The system is illustrated below: a.) The first step is to list all of the relevant parameters. In this case, the dependent variable is the volumetric flowrate at steady-state, which we will denote as Q. Other possible parameters include the straw diameter and length, as shown in the drawing above. In this problem, the person using the straw creates suction (i.e. a region of low pressure) at the top, which drives the flow. This induced pressure difference Δp between the top and bottom of the straw is one potential parameter. There are two things that will oppose the flow: gravity and viscous drag at the wall. This suggests that two other possibly relevant parameters include the gravitational constant g and the fluid viscosity, μ. Finally, the gravitational force experienced by the fluid depends on its density, ρ. All of these parameters are important, and there is no unique answer to this problem. From a practical standpoint, as an example, we may want to maximize the flowrate Q by varying the straw diameter d, since this is something we can easily specify in our final design. Note that there is no surface tension in our analysis; surface tension is

2 a property of interfaces between two different fluids. Thus, it may play a role during the start-up process when there is still air inside the straw (and hence an air-milkshake interface), but becomes irrelevant at steady-state, which is the case we are considering. b.) Summarizing the most important parameters: Parameter g d Dimensions L t 2 L Q L 3 t ρ m L 3 µ m Lt c.) According to the Buckingham Pi theorem, the total number of dimensionless groups we can form from these parameters equals the number of parameters (in this case 5) minus the number of fundamental dimensions (in this case 3), # of dimensionless groups = N D = 5 3 = 2 Let's find the two dimensionless groups. First, we must choose our recurring parameters, which are selected such that each fundamental dimension (here, mass, length, and time) is represented in the set. For instance, we can choose d, ρ, and µ. Note that we have not chosen the dependent variable Q. group: First, we non-dimensionalize the flowrate Q to obtain our first dimensionless Q = Q(d a ρ b μ c ) We substitute the dimensions for all the variables: Simplifying the right-hand side, m 0 L 0 t 0 = L3 t (L)a ( m L 3) b ( m Lt ) c

3 This equality suggests that: m 0 L 0 t 0 = m b+c L 3+a 3b c t 1 c b + c = a 3b c = 0 1 c = 0 The last equation tells us that c = 1. We can substitute this into the first equation to find that b = 1. Finally, we substitute b and c into the second equation to solve for a. Doing this, we find that Summarizing, Therefore, our first dimensionless group is a = 3b + c 3 = 3(1) 1 3 = 1 a = 1 b = 1 c = 1 Q = Qd 1 ρ 1 μ 1 = Qρ dμ If we substitute the dimensions of each parameter in this group, we can check our work and find that Q * is indeed dimensionless. This ratio tells us whether the flow we are studying will be inertia-dominated, or viscous-force dominated. We have just one dimensionless group to derive. The only non-recurring parameter that remains is the gravitational constant g. Therefore, the dimensionless group we have to derive is g = g(d a ρ b μ c ) m 0 L 0 t 0 = L t 2 (L)a ( m L 3) b ( m Lt ) c Once again, we simplify the right-hand side to obtain a system of equation we must solve for the exponents. Thus we have the system of equations: m 0 L 0 t 0 = m b+c L 1+a 3b c t 2 c b + c = 0

4 Solving, we find that 1 + a 3b c = 0 2 c = 0 a = 3 b = 2 c = 2 Therefore, the second dimensionless group is g = g(d 3 ρ 2 μ 2 ) = gd3 ρ 2 μ 2 We are done! This particular group tells us the relative importance of gravity versus viscous forces in our system (i.e. if this dimensionless group is very small, it suggests that viscous forces dominate over gravity). What if we had included the pressure difference Δp as one of our parameters? In this case we have a total of 6 parameters, and the Buckingham Pi theorem tells us that we can derive three dimensionless groups. In addition to the groups we already obtained, we can derive the final dimensionless group by non-dimensionalizing the pressure difference: Solving, we would find that Thus, Δp = Δp(d a ρ b μ c ) m 0 L 0 t 0 = m Lt 2 (L)a ( m L 3) b ( m Lt ) c = m 1+b+c L 1+a 3b c t 2 c a = 2 b = 1 c = 2 Δp = Δp(d 2 ρ 1 μ 2 ) = Δpd2 ρ μ 2 This particular group tells us about the relative importance of pressure forces versus viscous forces in our system.

5 d.) See part c.)

6 Problem 2 Here, we are considering the spread of a syrup drop along a wooden surface. After initially coming into contact with the surface, the liquid drop will spread until it reaches some equilibrium shape, illustrated below: θ c h D Here we have shown some potential choices for parameters to characterize the drop shape. D is the drop diameter, h is the height, and θ c is the contact angle. In this problem, there are two competing forces acting on the drop: 1) we have gravity, which is pulling down on the fluid so as to make it flat. However, this increases the surface area of the air-water interface, which is unfavorable. 2) Thus, we also have surface tension, which seeks to minimize the surface area of the air-syrup interface. At equilibrium, these two competing forces will balance to give some drop shape. The preceding discussion suggests that the surface tension of the air-syrup interface σ and the gravitational constant g are important to this problem. Since gravity is important in this problem, this suggests the density ρ of the fluid (which is related to the gravitational force it experiences) is important as well. It may seem reasonable to include the fluid viscosity as an important parameter, and this would be the case if we were looking at the process of a drop spreading on a surface. However, the viscosity only affects how quickly the drop spreads, but does not influence the final equilibrium shape. Since we are only interested in the final shape, viscosity is not relevant in this problem. Summarizing the most important parameters: Parameter h θ c g Dimensions L none L t 2

7 σ ρ Note that we have included two dependent parameters, θ c and h, since the drop shape cannot be characterized using just one parameter. Let s again apply the Buckingham Pi theorem. As in the previous problem, we have N = 5 parameters and D = 3 fundamental dimensions. Therefore, we can derive N D = 5 3 = 2 dimensionless groups. As our recurring parameters, let s choose the parameters that are not dependent, which are the density, surface tension, and gravitational constant. We nondimensionalize h, m t 2 m L 3 h = h(g a σ b ρ c ) m 0 L 0 t 0 = L ( L a t 2) ( m b t 2) ( m c L 3) = m b+c L 1+a 3c t 2a 2b Therefore, we have the equations b + c = a 3c = 0 2a 2b = 0 According to the last equation, b = a Plugging this into the first equation tells us that, a = c Therefore, we can solve the second equation, 1 + a 3a = 0 a = 1/2 Thus, b = 1/2 c = 1/2 We now have our dimensionless group, h = hg 1/2 σ 1/2 ρ 1/2

8 h = h2 gρ σ Since the contact angle does not have units, we don t need to non-dimensionalize it. Suppose we squared the dimensionless group h * (note that the square of a dimensionless quantity is still dimensionless, and therefore a valid dimensionless group). Doing this we find, (h ) 2 = h2 gρ σ Bo This is the Bond number, which we previously discussed in class. The Bond number tells us the relative importance of forces due to surface tension compared to body forces like gravity. Since these two forces are the most important in this problem, the Bond number has again emerged from our analysis.